mp-qmb 2

Question 1 Solve the following transportation problem for the optimum cost:

D1 D2 D3 D4 Supply

01 6 5 1 3 100

02 4 8 7 2 125

03 6 3 9 5 75

Demand 70 90 80 60 Solution The given transportation problem s of minimization type which is a balanced one. Voggles Approximation Method

D1 D2 D3 D4 ai

Penalty

01

6

5

1

3

100

2

2

1

1

02

4

8

7

2

125

2

2

4

-

03

6

3

9

5

75

2

2

3

3

bj

70

90

80

60

300

300

2 2 6 1 2 2 - 1 2 2 - - 0 2 - -

80

65

75

15 05

60

Total Transportation Cost = 30+75+260+120+225+80 = 790 RIM = m + n – 1 = 3 + 4 – 1 = 6

D1 D2 D3 D4 ai ui

01

6 -

5

1

3 +

-1 100

0

02

4 +

8

5

7

8

2 -

125

-2

03

6

2

3

9

10

5

3 75

-2

bj

70

90

80

60

300

300

vj

6

5

1

4

Since the cell evaluations for 01 - D4 is negative, optimum solution is not attained. In cost = 1 x 5 = 5 Therefore, new cost = 790 – 5 = 785

D1 D2 D3 D4 ai ui

01

6

1

5

1

3

100

0

02

4

8

4

7

7

2

125

-1

03

6

3

3

9

10

5

4 75

-2

bj

70

90

80

60

300

300

vj

5

5

1

3

65

75

80 15 05

60

80 15

70 55

75

05

Since all the cell evaluations are positive, optimum solution is attained. Therefore, the optimum transportation cost is Rs 785.

Question 2 Solve the following transportation problem for the optimum cost:

D1 D2 D3 D4 Supply

01 3 1 3 5 120

02 2 6 1 3 95

03 5 1 4 8 85

Demand 50 60 90 100 Solution The given transportation problem is of minimization type which is a balanced one i.e. total capacity = total demand. Phase I Vogal’s Approximation Method

D1 D2 D3 D4 ai

Penalty

01

3

1

3

5

120 95 0

0

0

2

2

02

2

6

1

3

95 5 0

1

1

1

1

03

5

1

4

8

85 25 0

3

1

3

bj

50 25 0

60 0

90 0

100 5 0

300

300

1 0 1 2 1 2 2 1 - 2 1 - 2

90

25 60

95 25

5

Total Transportation Cost = (3*25) + (5*25)+ (1*60) + (1*90) + (5*95) + 3*5)=75+125+60+90+475+15 = 840 RIM = m + n – 1 = 3 + 4 – 1 = 6

D1 D2 D3 D4 ai ui

01

3 -

1 2

3 0

5 +

120

0

02

2

6

9

1 +

3 -

95

-2

03

5 +

1

4 -

-1

8

1 85

2

Bj

50

60

90

100

300

300

Vj

3

-1

3

5

Since the cell evaluation for 03 – D3 is negative, optimum solution is not attained. In cost = 1 x 25 =25 Therefore, new cost = 840 – 25 = 815

D1 D2 D3 D4 ai ui

01

3

1 1

3 + 0

5 -

120

0

02

2 1

6

8

1 -

3 +

95

-2

03

5

1

1

4

8

2 85

1

Bj

50

60

90

100

300

300

Vj

3

0

3

5

25 60

90

95 25

5

25

65

50

30

60

70

Since all the cell evaluations are positive, optimum solution is attained. Therefore, the optimum transportation cost is Rs 785.

Question 3. Solve the following transportation problem for the optimum cost:

D1 D2 D3 D4 Supply ai

01 4 5 3 6 50

02 3 6 7 3 70

03 1 4 1 2 80 Demand

bj 50 40 90 20 200 SOLUTION: The given transportation problem is a balanced one which is of minimization type. Phase 1: To get Initial Basic Feasible Solution using VAM

D1 D2 D3 D4 ai

Penalty

01

4

5

3

6

50

1

1

1

1

02

3

6

7

3

70

0

0

0

3

03

1

4

1

2

80

0

-

-

-

bj

50

40

90

20

200

200

2 1 2 1 1 1 4 3 1 1 - 3 1 1 - -

10

50

40

20

80

Total transportation cost / Initial Basic Feasible solution = (3*50) + (5*40) + (3*10) + (1*80) + (3*20) = 150 + 200 + 30 + 80 + 60 = Rs. 520/- RIM condition = m + n – 1, Where m = number of rows and n = number of columns Here, number of allocations = 5 Since m + n – 1 ≠ number of allocations, degeneracy occurs. Phase 2: MODI method

D1 D2 D3 D4 ai ui

01

4

1

5

3

6

3 50

0

02

3

6

1

7

4

3 70

0

03

1

4

1

1

2

1 80

-2

bj

50

40

90

20

200 200

vj

3

5

3

3

Since all the cell evaluations are positive, optimum solution is attained. Therefore, IBFS = Optimum cost = Rs. 520/- Thus, the total minimum transportation cost is Rs. 520/-

10 40

E

20 50

80

Question 4. The given TP is of minimization type which is balanced one. Therefore, we can solve the problem by VAM. Phase I VAM I II III Capacity Penalty

A 10 7 40

8 5

45 5 1/1/1

B 15 12 9 15

15 3/3/3

C 7 25

8 15

12 40 15

1/4

Demand 25 55 40

20 100

Penalty 3 1/1/5 1/1/1

TOTAL TRANSPORTATION COST= 8X5 + 9X15 + 7X25 + 7X40 + 8X15 =750 RIM CONDITION m+n-1 3+3-1 =5 No. of allocations=5 RIM condition=No. of allocation

I II III Capacity A 10 7 8 45 B 15 12 9 15 C 7 8 12 40 Demand 25 55 20

Phase II: Modi Method

I II III ai ui A 10

4

7 40

8 5

45 0

B 15 8

12 4

9 15

15 1

C 7 25

8 15

12 3

40 1

Bj 25 55 20 100 vj 6 7 8

Since all the cell evaluations are non-negative, the optimum solution is Rs. 750.

Question 5. National Oil Company(NOC) has three refineries and four depots. Transportation costs per ton,capacities and requirements are as follows: Determine optimal allocation of output.

D1 D2 D3 D4 Capacity (tons)

R1 5 7 13 10 700

R2 8 6 14 13 400

R3 12 10 9 11 800

Requirements (tons)

300 600 700 400

D1 D2 D3 D4 CAPACITY

R1 5 7 13 10 700

R2 8 6 14 13 400

R3 12 10 9 11 800

DEMAND 300 600 700 400 1900

2000

Step 1: Unbalanced The problem is not balanced i.e. Condition 1 is not satisfied. Therefore a Dummy is created. Step 2: Minimisation

PHASE I: Initial Basic Feasible Solution: 1500+2000+1400+2400+6300+1100+0 = 14700 PHASE II: Test for Optimality:

1. No. of occupied cells= m+n-1 1. = 4+4-1 2. = 7

2. All cells are at independent positions.

D1 D2 D3 D4 CAPACITY PENALTY R1 5 7 13 10 700 2/2/2/3/3

R2 8 6 14 13 400 2/2/2/7

R3 12 10 9 11 800 1/1/1/1/1

DUMMY 0 0 0 0 100 0

DEMAND 300 600 700 400 2000 PENALTY 5/3/3 6/1/1/1/3 9/4/- 10/1/1/1/1

300 200

400

200

700 100

100

U

V1=5 V2=7 V3=8 V4=10

U1=0 5 7 13 10

U2=-1 8 6 14 13

U3=1 12 10 9 11

U4=-10 0 0 0 0

All CE > = 0 Therefore there is an optimum solution. Optimum Cost = Rs 14,700/-

V

300

4

6 2

3

7

2

4400

200 200

700 100

100

5

5

Question6. A company has three plants located at different places but producing an identical product the cost of production, distribution cost for each plant to the three different warehouses, the sale price at each warehouse and the individual capacities for both the plant and warehouse are given as follows: ' Plant F1 F2 F3 Raw material 15 18 14 Other expenses 10 9 12 Distribution Sale price Warehouse Cost to warehouse Capacity 1 3 9 5 34 80 2 1 7 4 32 110 3 5 8 3 31 150 Capacity of Plant 150 100 130

1) Establish a suitable table giving net profit / loss for a unit produced at different plants and Distributed at different warehouse

2) Introduce a suitable dummy warehouse / plant so as to match capacities of plants and Warehouses

3) Find a distribution pattern so as to maximize profit

Solution

PROFIT MATRIX

Plant 1 2 3 Dummy Capacity

F1 6 6 1 0 150

F2 -2 -2 -4 0 100

F3 3 2 2 0 130

Demand 80 110 150 40 380

Convert maximization problem into minimization problem by subtracting all the elements of the matrix form the highest value i.e. 6

REGRET MATRIX

Plant 1 2 3 Dummy Capacity Penalty F1 040 0110 5 6 150 0/5/- F2 840 8 1020 640 100 2/2/2/4 F3 3 4 4130 6 130 1/1/1/- Demand 80 110 150 40 380 Penalty 3/3/5/- 4/-/-/- 1/1/6 0/0/0 STEP 1 – Balanced STEP 2 – Minimization PHASE 1 – Initial Basic Feasible Solution Production Schedule 40*0 - 0 40*8 - 320 110*0 - 0 20*10 - 200 130*4 - 520 40*6 - 240

Total 1280 PHASE 2 – Test for optimality No. of occupied cells = m + n – 1 = 3+ 4 – 1 = 6 All cells are at independent positions

V1=0 V2=0 V3=2 V4= -2 U1=0 0 -

0

5-(2) 3

6-(-2)

8

U2= 8 8

8-(8)

0

10

6

U3= 2 3-(2)

1

4-(2)

2

12

6-(0)

6

Since all cells are non negative there is optimum solution i.e. optimum transportation cost is Rs. 1280

40

130

110

40 20 40

Question 7. A company manufactures a certain product at two factories and distributes it to three warehouses. Each factory runs both a regular shift and when necessary, an overtime shift for any remaining production requirements. Production cost differ between factories, and the selling price of the product varies at the different warehouse locations. The pertinent information is given in Table 1 Table 1 Factory Weekly Production

Capability(Units) Unit Production Cost

Regular Overtime Regular Overtime 1 100 40 Rs. 17 Rs. 24 2 150 30 Rs. 18 Rs. 23 Warehouse Weekly Requirements Units Unit selling Price 1 80 Rs. 35 2 120 Rs. 37 3 70 Rs. 34 The objective is to select that combination of regular and overtime production which when allocated to shipping in an optimal fashion, will maximise profit. Questions: 1 Set up the transportation table for an initial solution 2. Obtain an optimum solution (Z = 3720) Solution Since the objective of the problem is to maximise profits, we form the profit matrix. To calculate the profit/unit: Add the regular/overtime production cost to the transportation cost of the respective factory and subtract the total from the unit selling price of the respective warehouses. Profit Matrix: Warehouse

Factory 1 2 3 ai R1 12 13 11 100 O1 5 6 4 40 R2 13 17 8 150 O2 8 12 3 30 bj 80 120 70 270/320 Since the give TP is unbalanced, we first balance it by creating a dummy row.

Warehouse

Factory 1 2 3 Dummy ai R1 12 13 11 0 100 O1 5 6 4 0 40 R2 13 17 8 0 150 O2 8 12 3 0 30 bj 80 120 70 50 320/320 Since the Tp is a maximisation one, we convert it to minimisation type by forming a regret matrix and then solve it by VAM to get Initial Feasible Solution. Phase I: To get IFS Regret Matrix Warehouse Penalty Factory 1 2 3 Dummy Ai I II III IV V R1 5

30 4 6

60 17

100 30 1 1 1 12

O1 12 11 13

17 50

40 1 1 1 5 5

R2 4 0 120

9 17 150 30 4 5

O2 9 20

5

14 17 20

30 10 4 5 5 8 8

Bj 80 50 70

120

70 50 320

I 1 4 3 0

II 1 3 0

III 4 7 0

IV 4 0

V 3 0

Total regret is 12*30=360 11*70=770 0*40=0 13*30=390 17*120=2040 8*20=160 0*10=0 =Rs 3720

Rim Conditions: No. of allocation = m+ n – 1 = 7 Therefore we can solve it by Modi method. Phase II:Modi Method Warehouse Factory 1 2 3 Dummy ai Ui F1/ R 5

30 4

36

7017

4100 0

F1/OT 12 3

11 6

13 3

17 40

40 4

F2/R 4 30

0 120

9 4

17 5

150 -1

F2OT 9 20

5 0

14 4

17 10

30 4

Bj 80 120 70 50 320 Vj 5 1 6 13 Optimality Test: Since all cell evaluations are positive/non negative, optimum solution is obtained. However since the cell evaluation for O2-2 is 0, an alternate optimum solution exists. Total Profit: From To Quantity Profit/Unit Total F1/R W1 30 12 300 F1/R W3 70 11 770 F1/O D 40 0 0 F2/R W1 30 13 390 F2/R W2 120 17 2040 F2/O W1 20 8 160 F2/O D 10 0 0 Total Profit: Rs. 3720

Question 8.

The given Problem is of minimization type and a balanced one thus we move on further with the Vogels Approximation Method.

VOGELS APPROXIMATION METHOD D1 D2 D3 D4 CAPACITY PENALTYO1 6 (5) 5 (15) 1 (80) 7 100 2,2,1,1 O2 4 (65) 8 7 2 (60) 125 2,2,4 O3 6 3 (75) 9 5 75 2,2,3,3 DEMAND 70 90 80 60 300 PENALTY 2,2,2,0 2,2,2,2 6 1,1

PHASE 1 IBF SOLUTION = 6*5 + 15*5 + 80*1 + 65*4 + 60*2 + 75*3 = 30+75+80+260+120+225 Ans =Rs.790 Phase II- Number of occupied cells M+ N -1=1 6=4+3-1 6=6

Since Phase II is satisfied we move on to ui,vj method

U/V D1 D2 D3 D4 U

D1 D2 D3 D4 CAPACITY 01 6 5 1 3 100 02 4 8 7 2 125 03 6 3 9 5 75 DEMAND 70 90 80 60 300

O1 6 (5) - -- -5 - (15)--- 1 -- (80)-- --3 - + |-1|

0

O2 4 -- (65) --+

8----------- 7----------- 2 -- (60) - -2

O3 6 |2| 3 (75) 9 |10| 5 |3| -2

V 6 5 1 4

Cell evaluation for all non allocation cells cost – (u +v) Cell evaluation for allocation cells cost = u +v Minimum from negative i.e. 5 or 60 Thus minimum is 5 Therefore, reduction in cost = 5 *1 = Rs.5 New Cost = 790-5 = Rs.785 Since it is not yet optimal we create a new table again. D1 D2 D3 D4 U O1 6 |1| 5 (15) 1 (80) 3 (5) 0

O2 4 (70) 8 |4| 7 |7| 2 (55) -1

O3 6 |3| 3 (75) 9 |10| 5 |4| -2

V 5 5 1 3

Since all cell evaluations are positive optimum solution is attained Therefore Transportation cost = Rs. 785

Question9: Destination Origin D1 D2 D3 D4 Supply

O1 4 2 1 3 120

O2 1 6 3 2 95

O3 3 5 4 7 85

Demand 70 90 80 60 300 300

Solution The given transportation problem is of minimization type which is a balanced one. I) VAM Destination Origin

D1 D2 D3 D4 Supply

(ai)

I

II

III

IV

O1 4 2 1 3

120 1 2 2 3

O2 1 6 3 2

95 1 1 1 2

O3 3 5 4 7

85 1 1 3 7

Demand (bj)

70 90 80 60 300 300

I 2 3 2 1

II 2 - 2 1 III - - 2 1 IV - - - 1

Total Transportation Cost = (2 X 90) + (3 X 30) + (1 X 70) + (2 X 25) + (TTC) (80 X 4) + (7 X 5) = 180+90+70+50+320+35

90 30

2570

580

TTC = 745

Rim Condition = m + n - 1 where: = 3 + 4 – 1 m = no. of rows

= 7 –1 n = no. of columns = 6 II) Destination Origin

D1 D2 D3 D4 Supply

(ai) ui

O1

4 2

2 1 1

3 120 0

O2

1 -O

6 5

3 4

2 + O 95 -1

O3 3 + O -3

5 -1

4 7 - O 85 4

Demand (bj)

70 90 80 60 300 300

vj 2 2 0 3

Since CELL EVALUATION O3D1 is negative, therefore optimum solution is not attained. O = Minimum {70, 5} O = 5 Decrease in cost = (5 X 3) = 15 New Cost = 745 – 15 = 730 New Cost = 730

90 30

2570

580

III) Destination Origin

D1 D2 D3 D4 Supply

(ai) ui

O1

4 2

2 1 + O -2

3 - O 120 0

O2

1 -O

6 5

3 1

2 + O

95 -1

O3 3 + O

5 2

4 - O

7 3

85 1

Demand (bj)

70 90 80 60 300 300

vj 2 2 3 3

Since CELL EVALUATION O1D3 is negative, therefore optimum solution is not attained. O = Minimum {30, 65, 80} O = 30 Decrease in cost = (30 X 2) = 60 New Cost = 730 – 60 = 670 New Cost= 670

90 30

3065

805

IV) Destination Origin

D1 D2 D3 D4 Supply

(ai) ui

O1

4 4

2 1

3 2

120 0

O2

1

6 3

3 1

2

95 1

O3 3

5 0

4

7 3

85 3

Demand (bj)

70 90 80 60 300 300

vj 0 2 1 1

Since all the CELL EVALUATIONS are positive, optimum solution is obtained.

90

6035

5035

30

Optimum Transportation Cost = Rs. 670/-

Question10. Solve the following cost minimization transportation problem. 1 2 3 4 SUPPLY A 7 3 8 6 60 B 4 2 5 10 100 C 2 6 5 1 40 DEMAND 20 50 50 60 180\200 Solution As the problem is unbalanced, we balance it and then solve the as the problem as minimization type. Phase I: We solve the problem by using VAM Method.

1 2 3 4 Dc SUPPLY

Penalty

A 7 3 (20) 8 6 (20) 0 (20) 60 3 3 3 4 X XB 4 (20) 2 (30) 5 (50) 10 0 100 2 2 2 2 2 2C 2 6 5 1 (40) 0 40 1 X X X X X

DEMAND 20 50 50 60 20 200 2 1 0 5 0

3 1 3 4 0 3 1 3 X 0 3 1 3 X X 4 2 5 X X 4 2 X X X

Thus, total transportation cost: 3 x 20 = 60 6 x 20 = 120 0 x 20 = 0 4 x 20 = 80 2 x 30 = 60 5 x 50 = 250 1 x 40 = 40 Rs 160 Rim Conditions: m + n – 1 = 3 + 5 -1 = 7 Number of allocations = 7 Thus, m + n – 1 = Number of allocations, rim conditions are satisfied. Phase II:

1 2 3 4 Dc SUPPLY Ui

A 7 2 3 (20) 8 2 6 (20) 0 (20) 60 0

B 4 (20) 2 (30) 5 (50) 10 5 0 1 100 -1

C 2 2 6 8 5 4 1 (40) 0 5 40 -5

DEMAND 20 50 50 60 20 200\200

Vj 5 3 6 6 0

Note: Numbers in blue = Allocated cells Numbers in red = Cell evaluations Optimality Test: As all the cell evaluations are non negative, the optimum solution is attained. Ans: The total minimum transportation cost is Rs 610

Question11. Solve the following transportation problem for maximum profit. In the following table profit is given in Rs. 100/- Solution: The given transportation problem is of maximization type. Therefore we convert it into minimization type and solve using VAM method. A B C D SUPPLY PENALTY

X

200

X

200 0 /0 /0 /0

Y

180

100

400

500 8/ 8 /2 /3

Z

20

100

300 6 /6 /3 / 8

DEMAND 180 320 100 400 1000 PENALTY 2 /6 /6 /0 11 /4 /4 /4 1 /1 /1 /1 5 /2 /0 /0

X → Β = 7 x 200 = 1,400 Y → Β = 18 x 100 = 1,800 Y → D = 7 x 400 = 2,800 Z → A = 11 x 180 = 1,980 Z → B = 20 x 22 = 440 Z → C = 14 x 100 = 1,400 TTR REGRET SOLN 9,820 RIM CONDITIONS: No. of allocations = 6 m + n – 1 = 6 Therefore no. of allocations = m + n – 1 Therefore it is a GENERATE Solution and we can check the optimality using MODI method.

7 17

11

18 15

13 7 19 0

22 14 5

MODI METHOD: A B C D SS Ui

X

200

X

200 0

Y θ+

100

θ−

400

500 11

Z

180

θ−

20

100

θ+

300 15

DD 180 320 100 400 1000 Vj - 4 7 - 1 - 4

Since, the cell evaluation of D2 is -6, it implies the solution is not optimum. Hence we form a loop. θ = min {20, 400} = 20 Therefore the urgent reduces by = 20 x 6 = 120 Therefore Next Regret = 9,820 – 120 = 9,700

13 7 19 0

7 17

11

18 15

22 14 5

17 20 4

10 5

-6 18

To check if the solution is optimum we use MODI Method again: A B C D SS Ui

X

200

X

200 0

Y

120

380

500 11

Z

180

100

20

300 15

DD 180 320 100 400 1000 Vj - 4 7 - 1 - 4

Since all the cell evaluations are non negative, optimum solution is obtained.

FROM TO QUANTITY PROFIT/UNIT TTR PROFITX B 200 18 3,600Y B 120 7 840Y D 380 18 6,840Z A 180 14 2,520Z C 100 11 1,100Z D 20 20 400

Rs. 15,300 Therefore maximum profit is Rs. 15,300/- To check: Optimum profit = ( max. profit/unit x total quantity) – optimum regret = (25 x 1000) – 9,700 = 25,000 – 9,700 = 15,300

13 7 19 0

7 17

11

18 15

22 14 5

17 20 4

10 5

0

Question 12: Given below is the transportation problem with transportation costs and initial feasible solution.

D1 D2 D3 D4 SUPPLY

O1 5 10 4 (100)

5 100

O2 6 (200)

8 7 2 (50)

250

O3 4 (50) 2 (100)

5 (50) 7 200

DEMAND

250 100 150 50

State with reasons, whether:

1) The given solution is feasible and degenerate 2) The solution is optimal

3) Can there be more than one optimal solution in this problem?

4) How will you test the optimality test of the solution is the cell cost changes? (take

example, O3 – D3 changes from Rs.5/- to Rs. 2/- ) SOLUTION

D1 D2 D3 D4 SUPPLY

O1 5 10 4 (100)

5 100

O2 6 (200)

8 7 2 (50)

250

O3 4 (50) 2 (100)

5 (50) 7 200

DEMAND

250 100 150 50 550 550

TOTAL TRANSPORATION COST O1 – D3 4 X 100 400 O2 – D1 6 X 200 1200 O2 – D4 2 X 50 100 O3 – D1 4 X 50 200 O3 – D2 2 X 100 200 O3 – D3 5 X 50 250 TOTAL Rs. 2350/- RIM CONDITION: - m + n – 1 = no. of allocations, Where, M = no. of origins N = no. of destinations Therefore, m + n – 1 = 3 + 4 – 1 = 6 No. of allocations = 6 Therefore the rim conditions have been satisfied as m +n – 1 = no. of allocations = 6. Therefore, the solution is non- degenerate. PHASE II (MODI METHOD)

D1 D2 D3 D4 ai ui

O1 5 10 4 (100) 5 100 0

O2 6 (200) 8 7 2 (50) 250 3

O3 4 (50) 2 (100) 5 (50) 7 200 1

bj

250 100 150 50 550 550

Vj 3 1 4 -1

Therefore, Cell Evaluation (CE) = cost of the cell – (u + v) for all non- allocated cells.

D1 D2 D3 D4 ai ui

O1 5 2 10 9 4 (100) 5 6 100 0

O2 6 (200) 8 4 7 0 2 (50) 250 3

O3 4 (50) 2 (100) 5 (50) 7 7 200 1

bj

250 100 150 50 550 550

Vj 3 1 4 -1

Therefore, since all the Cell Evaluations are non-negative, optimum solution is attained. ANSWER: The Total minimum transportation cost is Rs. 2350/- However, since one Cell Evaluation (CE) of O2 – D3 is ZERO, an alternate optimum solution exists

D1 D2 D3 D4 ai ui

O1 5 2 10 9 4 (100) 5 6 100 0

O2 6 (200) - 0

8 4 7 0 + 0

2 (50) 250 3

O3 4 (50) + 0

2 (100) 5 (50) - 0

7 7 200 1

bj

250 100 150 50 550 550

Vj 3 1 4 -1

θ = minimum of the negative ‘theta’ values, therefore θ = min { 50, 200 } θ = 50

Therefore, reduction in cost = 50 x 0 Reduction in cost = Rs. 0. Therefore, new cost = 2350 – 0 New Cost = Rs. 2350/-

D1 D2 D3 D4 ai ui

O1 5 2 10 9 4 (100) 5 6 100 0

O2 6 150 8 4 7 50 2 (50) 250 3

O3 4 100 2 (100) 5 7 7 200 1

bj

250 100 150 50 550 550

Vj 3 1 4 -1

ANSWERS:

1) The initial solution was feasible because, all the Cell Evaluations were non-negative. The solution was also non-degenerate because the Rim Conditions ie. m + n – 1 = no. of allocations = 6 has been satisfied.

2) The initial feasible solution was the optimal solution for the problem as all the

Cell Evaluations ( CE) were non-negative.

3) Yes, there can be more than one optimal solution for this problem as one of the Cell Evaluation ( O2 – D3 ) was ZERO. Therefore an Alternate Optimum Solution exists. The working of which has been shown.

4) If the cost of O3 – D3 changes from Rs. 5/- to Rs. 2/- then the problem will be

solved in the following manner. (next page)

Question13: Firm PQR has the following schedule for transporting inventory in the network. W1 W2 W3 W4 SUPPLY F1 6 2 3 8 100 F2 12 13 6 13 100 F3 8 3 8 4 150 F4 5 4 1 6 250 DEMAND 180 160 160 100 Find an initial feasible solution by using VAM Answer the following questions:

1. Is the above solution feasible?

2. Is it a degenerate solution?

3. Find an Optimum Solution to the above problem?

4. Find the associated costs to this solution?

5. Is there any alternate solution to the above problem?

6. What is the opportunity cost to the route F1 to W3?

7. If the management wants to transport to the route F2 to W2 will that increase the

cost? What will the rate of increase in the cost?

8. A transporter is willing to offer discount on route F1 to W4. By what rate he must

drop the rate?

9. If he wants a minimum of 10 units quantity should the offer be accepted?

10. The production capacity of F1 is reduced by 2 units and to compensate this F3

capacity is raised. What impact will this have on the transportation cost?

Solution:

The above transportation problem is balanced and is of minimization type:

PHASE I: to get IBFS using VAM

W1 W2 W3 W4 SUPPLY penalty F1 6

2

100 3 8

100 1/1/4/4

F2 12

13 6 100

13 100 6

F3 8

3 50

8 4 100

150 1/1/5

F4 5 180

4 10

1 60

6 250 3/3/3/1

DEMAND 180 160 160 100 600

Penalty 1/1/1 1/1/1 2/2 2/2 Total transportation cost:

UNITS C.P.U TOTAL

100 2 200

100 6 600

50 3 150

100 4 400

180 5 900

10 4 40

60 1 60

TOTAL 2350/-

Ans 1) the above solution is feasible as the quantities transported to the various

destinations are positive & also the Solution so obtained is non negative.

Ans 2) the solution is non-degenerate as m + n – 1 = no of allocations.

4 + 3 – 1 = 7

7 = 7

Therefore solution is non-degenerate.

Ans 3)

PHASE II: To test for optimum solution using MODI method.

W1 W2 W3 W4 Ai Ui F1 6

3

2 100

3 4

8

5

100 0

F2 12

2

13

4

6 100

13

3

100 7

F3 8

4

3 50

8

8

4 100

150 1

F4 5 180

4 10

1 60

6

1

250 2

bj 180 160 160 100 600

Vj

3 2 -1 3

Since, all Cell Evaluation (CE) are positive, optimum solution is attained.

Optimum transportation cost = Rs. 2350/-

Ans 4) associated cost to the above IBF solution is Rs. 2350/-

Ans 5) there is no alternate optimum solution to the above problem as none of the CE are

zero i.e. CE>0

Ans 6) the opportunity cost from F1 to W3 is Rs. 4/ unit.

Ans 7) if the management wants to transport to the route F2 to W2, it will increase the

cost and the rate at which it will increase is RS 4/- unit as the CE is Rs. 4

Ans 8) if the management wants to transport to the route F2 to W4, the rate that it should

drop should be less than 10.

I.e. cost =9 OR <9 (u + v = 10)

Ans 9) if the transporter wants to transport minimum 10 units on F2 to W4, the

transportation cost would be: (if costs are Rs 9)


3

2 100

3

4

8

5

100 0

F2 12

2

13

4

6 100

9

-1

100 7

F3 8

4

3

50

8

8

4

100

150 1

F4 5

180

4

10

1

60

6

1

250 2

bj 180 160 160 100 600

Vj

3 2 -1 3

= 10 min (100, 10, 100)

Reduction in cost = 2350 – (1 * 10)

= Rs. 2340/-


2

2 100

3

3

8

5

100 0

F2 12

2

13

5

6 90

9 10

100 6

F3 8

3

3 60

8

7

4 90

150 1

F4 5 180

4

1

1 70

6

2

250 1

bj 180 160 160 100 600

Vj

4 2 0 3

Since, all Cell Evaluation (CE), optimum solution is attained.

When the transportation cost is reduced and minimum of 10 units are to be transported on

the route F2 to W4, reduction in costs are Rs. 10/-

Optimum transportation cost = Rs. 2340/-

Ans 10) if production capacity at F1 is reduced to 98 and increased in F3 to 153, then

transportation cost.

Phase I: solve by VAM

W1 W2 W3 W4 SUPPLY penalty F1 6

2

983 8

98 1/1/4/4/4

F2 12

13 6 100

13 100 6

F3 8

3 52

8 4 100

152 1/1/1/5

F4 5 180

4 10

1 60

6 250 3/3/1/1

DEMAND 180 160 160 100 600

Penalty 1/1/1//1 1/1/1/1 2/2/0 2/2/2 Transportation cost

UNITS C.P.U TOTAL

98 2 196

100 6 600

52 3 156

100 4 400

180 5 900

10 4 40

60 1 60

TOTAL 2352/-

The transportation cost is increased by Rs 2/-.

The above solution is non-degenerate as Rim conditions are satisfied (m+n-1= no of

allocations).

Phase II: since there is no change in the cost per unit & therefore CE >= 0

Thus optimum transportation cost in this case is Rs.2352/-

Question14: Solve the following T.P for optimum cost

D1 D2 D3 D4 SUPPLY/CAPACITY

O1 4 5 1 2 120 O2 1 3 4 5 85 O3 3 1 6 3 95

DEMAND 70 80 50 100 SOLUTION: The given T.P is of minimisation type and a balanced one. Therefore we solve it by using VAM PHASE 1- VAM , To get initial feasible solution D1 D2 D3 D4 Ss penalty

O1 4 5 x

1 50

2 70

120 1/2/2 x

O2 1 3 4 x

5 15

85 2/2/4 70 x

O3 3 x

1 80

6 x

3 15

95 1/2/0

Dd 70 80 50 100 300\300 penalty 2/2/2 2/2/x 3/x 1/1/1

TOTAL TRANSPORTATION COST O1→D3 = 1*50 = 50 O1→D4 = 2*70 = 140 O2→D1 = 1*70 = 70 O2→D4 = 5*15 = 75 O3→D2 = 1*80 = 80 O3→D4 = 3*15 = 45 Rs. 460 Rim conditions m + n – 1 = no. of allocations = 6 therefore it is a non-degenerate solution.

Therefore we now solve it by modi method. PHASE 2→ MODI METHOD

D1 D2 D3 D4 Ss Ui O1 4 5 1 2 120 0

6 5 50 70 O2 1 3 4 5 85 3

70 0 0 15 O3 3 1 6 3 95 1

4 80 4 15

Dd

70

80

50

100

300\300

Vj -2 0 1 2 Since all the cell evaluations are non negative, optimum solution is attained. OPTIMUM TRANSPORTATION COST = Rs. 460/-

Question 15. D1 D2 D3 D4 Supply Capacity

O1 4 5 3 6 50 O2 3 6 7 3 70 O3 1 4 1 2 80

Demand 50 40 90 20 200 Solution: The given Transportation Problem is of minimization type which is balanced one (as Total capacity = Total demand) Phase 1: To get the initial feasible solution, solving by Vogel’s Approximation Method or VAM. D1 D2 D3 D4 ai Penalty

O1 4

5

40

3

10

6 50

1/1/1/-

O2 3

50

6 7 3

20

70

0/0/0/0/-

O3 1

4 1

80

2 80

0/-

bj 50 40 90 20 200 Penalty 2/1/1/1/- 1/1/1/1/- 2/4/- 1/3/3/-

Calculation of Total Transportation Cost:

Route Cost (Rs.) Quantity (units) Total (Rs.) O1-D2 5 40 200 O1-D3 3 10 30 O2-D1 3 50 150 O2-D4 3 20 60 O3-D3 1 80 80

Total Transportation Cost 520 Rim Conditions: No. of allocations = 5 m + n – 1 = 3 + 4 – 1 = 6 Therefore, m + n – 1 = No. of allocations

Degeneracy occurs and the solution that is obtained is called degenerate solution. To remove this degeneracy, we introduce epsilon (E) so that m + n – 1 = No. of allocations. The position of Epsilon is chosen in such a way that it does not form a closed loop. Phase 2: Modi Method D1 D2 D3 D4 ai Ui

O1 4

1

5

40

3

10

6

3

50

0

O2 3

50

6

1

7

4

3

20

70

0

O3 1

E

4

1

1

80

2

1

80

-2

bj 50 40 90 20 200 Vj 3 5 3 3

Since all the cell evaluations are positive, optimum solution is attained. Optimum Transportation Cost = Rs. 520

Question 16.

D1

D2

D3

D4

SUPPLY

O1

1

3

4 2 50

O2

5 3

6 1 70

O3

8 7 1

2

180

TOTAL DEMAND

50

70

80

100

300

Find the optimum solution and also the transportation cost :: SOLUTION: The above Transportation problem is of the minimization type and is a balanced one. Phase1: Solving by Vogel’s Approximation Method or VAM.

D1

D2

D3

D4

SUPPLY

PENALTY

O1

1 50

3

4 2 50

1

O2

5 3 70

6 1 70

2/2/2

O3

8 7 1 80

2 100

180

1/1/5

TOTAL DEMAND

50

70

80

100

300

PENALTY

4

0/4/4

3/5

1/1

Therefore Total Transportation cost is = 1 X 50 + 3 X 70 + 1X 80 + 2 X 100 = 50 + 210 +80 +200 = Rs. 540/- Phase 2: Test for optimality using MODI method.

Since, RIM conditions are not satisfied i.e. m + n –1 not equal no. of allocations( i.e. 4 in this case), degeneracy occurs. Therefore, EPSILON IS USED (E)

D1

D2

D3

D4

ai

ui

O1

1 50

3 – 4 + Q - 1

4 – 1 3

2 – Q E

50

0

O2

5 – 0 5

3 – Q 70

6 – 0 6

1 + Q E

70

-1

O3

8 – 1 7

7 – 4 3

1 80

2 100

180

0

TOTAL DEMAND

bj

50

70

80

100

300

vj

1 4

1

2

Q= min (E, 70) Q= E Reduction in cost = 540 –(1 X E) = 540 – 0 = Rs. 540.

Since, Cell evaluations are positive, there is optimal solution. Therefore, optimum solution or optimum transportation cost = Rs. 540/-

D1

D2

D3

D4

ai

ui

O1

1 50

3 E

4 – 0 4

2 - 1 1

50 0

O2

5 – 1 4

3 70

6 – 0 6

1 E

70

0

O3

8 – 2 6

7 – 4 3

1 80

2 100

180 1

bj

50

70

80

100

300

vj

1 3

0

1

2

Question17. Cement manufacturing company wishes to transport its three factories P,Q and R to five distribution depots situated at A,B,C,D and E. The quantities at the factories per week, requirements at the depots per week and respective transportation costs in Rs. Per tonne are given in the table below: Factories DEPOTS Tonnes

Available A B C D E P Q R

4 2 3

1 3 5

3 2 2

4 2 4

4 3 4

60 35 40

Tonnes Required

22 45 20 18 30 135

Determine the least cost distribution programme for the company. If the transportation cost from R to D is changed to Rs.2/- per tonne, does this affect the optimal allocation? If so, determine the revised schedule. SOLUTION: The above Transportation problem is of the minimization type and is a balanced one. Phase1: Solving by Vogel’s Approximation Method or VAM.

Factories

DEPOTS

Tonnes

Available

PENALTY

A B C D E

P 4

45 1

3

4

15 4

60 15

2/1/1/0/0

Q

17 2

3

2

18 2

3

35 17

1/0/0/1/-

R

5 3

5

202

4

15 4

40 20 15

1/1/1/1/1

Tonnes Required

22 5

45

20 18 30 15

135

PENALTY 1/1/1/1/ 1

2/-/-/-/-

0/1/1/-/-

2/2/-/-/-

1/1/1/1/0

Therefore Total Transportation cost is = 45*1+15*4+17*2+18*2+5*3+20*2+15*4 = 45+60+34+36+15+40+60 =290

(1) Z= Rs. 290

Phase 2: Test for optimality using MODI method. Rim Conditions: No. of allocations = 7 m + n – 1 = 5 + 3 – 1 = 7 Therefore, m + n – 1 = No. of allocations A B C D E ai Ui

P

4 -3 1

1 45

3 -2 1

4 -3

1

4

15

60

0

Q

2

17

3 -0

3

2 -1

1

2

18

3 -3 0

35

-1

R

3

5

5 -1

4

2

20

4 -3

1

4

15

40

0

bj 22 45 20 18 30 135

Vj 3 1 2 3 4

Since all the cell evaluations are positive, optimum solution is attained. Optimum Transportation Cost = Rs. 290 If the cost from R to D is changed to Rs. 2/- per tonne, then the corresponding CE becomes (-1). As CE is –ve, therefore optimum solution does not remain optimum, so form a loop. A B C D E ai Ui

P

4 -3 1

1 45

3 -2 1

4 -3

1

4

15

60

0

Q

2

17

3 -0

3

2 -1

1

2

18

3 -3 0

35

-1

R

3

5

5 -1

4

2

20

2 -3

-1

4

15

40

0

bj 22 45 20 18 30 135

Vj 3 1 2 3 4

Reduction in cost = 290-1(5) =285 A B C D E ai Ui

P

4 -2 2

1 45

3 -2 1

4 -2

2

4

15

60

0

Q

2

22

3 -1

2

2 -2

0

2

13

3 -4 -1

35

0

R

2 -2

0

5 -1

4

2

20

2

5

4

15

40

0

bj 22 45 20 18 30 135

Vj 2 1 2 2 4

Reduction in cost = 285 – 1(13) =272 A B C D E ai Ui

P

4 -3 1

1 45

3 -2 1

4 -4

0

4

15

60

0

Q

2

22

3 -0

3

2 -1

1

2 -1

1

3

13

35

-1

R

3 -3

0

5 -1

4

2

20

2 18

4

2

40

0

bj 22 45 20 18 30 135

Vj 3 1 2 2 4

Since all the cell evaluations are positive, optimum solution is attained. Optimum Transportation Cost = Rs. 272

Question 18. D1 D2 D3 D4 Supply

capacity O1 1 2 6 1 105 O2 4 1 2 5 110 O3 3 3 4 2 85 Demand 70 80 100 40 300 Solution: D1 D2 D3 D4 Dummy

column Supply capacity

O1 1 2 6 1 0 105 O2 4 1 2 5 0 110 O3 3 3 4 2 0 85 Demand 70 80 100 40 10 300 The above transportation problem is of minimization type which is a balanced one. D1 D2 D3 D4 Dummy Supply

capacity I II III IV V

O1 1 70

2 6 1 35

0 105 35 1 - 1 1 1

O2 4 1 10

2 100

5 0 110 10 1 1 1 4 -

O3 3 3 70

4 2 5

0 10

85 75 5

2 1 1 1 1

demand 70 80 70 100 40 5 10 300 I 2 1 2 1 0

II 2 1 2 1

III 1 2 1 IV 1 2 1

V 1 2 1 Total transportation cost is 1*70 =70 1*35 =35 1*10=10 2*100=200 3*70=210 2*5=10 0*10=0 = 535

No. of allocation = m+ n – 1 = 7 Therefore we can solve it by modi method. D1 D2 D3 D4 Dummy Supply

capacity Ui

O1 1 70

2 6 1 35

0 5000 0

O2 4 1 10

2 100

5 0 4000 -1

O3 3 3 70

4 2 5

0 10

7000 1

demand 3000 2500 3500 4000 3000 16000 Vj 1 2 3 1 -1 Cost evaluation- since all the cell evaluations are positive, optimum solution is attained Therefore the optimum transportation cost is Rs. 535

Question 19 D1 D2 D3 D4 Supply

capacity O1 15 24 11 12 5000 O2 25 20 14 16 4000 O3 12 12 22 13 7000 demand 3000 2500 3500 4000 16000 Solution: D1 D2 D3 D4 Dummy


O1 15 24 11 12 0 5000 O2 25 20 14 16 0 4000 O3 12 12 22 13 0 7000 demand 3000 2500 3500 4000 3000 16000 The above transportation problem is of minimization type which is a balanced one. D1 D2 D3 D4 Dummy Supply


O1 15 24 11 2500

12 2500

0 5000 2500 11 1 1 1 1

O2 25 20 14 1000

16 0 3000 4000 1000 14 2 2 2 2

O3 12 3000

12 2500

22 13 1500

0 7000 4500 1500

12 0 1 9

demand 3000 2500 3500 4000 2500

3000 16000

I 3 8 3 1 0

II 3 8 3 1

III 3 3 1 IV 3 1

V 3 4 Total transportation cost is 3000*12=36000 2500*12=30000 2500*11=27500 1000*14=14000 2500*12=30000 1500*13=19500

=15700 No. of allocation = m+ n – 1 = 7 Therefore we can solve it by modi method. D1 D2 D3 D4 Dummy


Ui

O1 15 4 24 13 11 2500 12 2500 0 3 5000 0 O2 25 11 20 6 14 1000 16 1 0 3000 4000 3 O3 12 3000 12 2500 22 10 13 1500 0 3 7000 1 demand 3000 2500 3500 4000 3000 16000 Vj 11 11 12 -3 Cost evaluation- since all the cell evaluation are positive, optimum solution is attained Therefore the optimum transportation cost is Rs. 157000

Question 20. D1 D2 D3 Supply

capacity O1 7 10 5 90 O2 12 9 4 50 O3 7 3 11 80 O4 9 5 7 60 Demand 120 100 110 330\300 Solution: Since the above transportation problem is of minimization type but an unbalanced one, we first balance it by adding a dummy row. D1 D2 D3 Supply

capacity O1 7 10 5 90 O2 12 9 4 50 O3 7 3 11 80 O4 9 5 7 60 Dummy 0 0 0 50 Demand 120 100 110 330 Phase I: To get Initial Feasible Solution Solve by VAM D1 D2 D3 Supply


O1 7 30

10 5 60

90 30 2 2 2 2 2

O2 12 9 4 50

50 5 5

O3 7 3 80

11 80 4 4 4

O4 9 40

5 20

7 80 40 2 2 2 2 2

Dummy 0 50

0 0 50

Demand 120 70

100 20

110 60

330

I 7 3 4

II 0 2 1

III 0 2 2

IV 0 5 2

V 0 2 Total transportation cost is 7*30=210 5*60=300 4*50=200 3*80=240 9*40=360 5*20=100 0*50=0 =Rs 1410 Optimality Test: Rim Conditions: No. of allocation = m+ n – 1 = 7 Therefore we can solve it by Modi method. Phase II:Modi Method D1 D2 D3 A i Ui O1 7

30 10

7 5

6090 0

O2 12 6

9 7

4 50

50 -1

O3 7 +θ

0

3 -θ

80

11 6

80 0

O4 9 -θ

40

5 +θ

20

7 0

60 2

Dummy 0 50

0 4

0 2

50 -7

B j 120 100 110 330 Vj

7 3 5

Since, Cell Evaluation O3-D1 is 0, there is alternate optimum solution. θ = min {40, 80 } = 40 Reduction in Cost = 1410 – (0*40) = 1410

D1 D2 D3 A i Ui O1 7

30 10

7 5

6090 0

O2 12 6

9 7

4 50

50 -1

O3 7 40

3 40

11 6

80 0

O4 9 0

5 60

7 0

60 2

Dummy 0 50

0 2

0 2

50 -7

B j 120 100 110 330 Vj

7 3 5

Cost evaluation- since all the cell evaluation are positive, optimum solution is attained Therefore the optimum transportation cost is Rs. 1410

Question 21. A company has three factories at locations, A, Band C, which supplies three warehouses located at D,E & F. Monthly factory capacities are 10, 80 & 15 units respectively. Monthly warehouse Requirements are 75, 20 & 50 units respectively. Unit shipping costs in Rs. are given below: D E F Supply

capacity A 7 10 5 90 B 12 9 4 50 C 7 3 11 80 9 5 7 60 Demand 120 100 110 330\300 The penalty costs for not satisfying the demand at warehouses D, E & F are Rs.5/-. Rs. 3/- & Rs. 2/- per unit respectively. Determine the optimal distribution for the company using any of the known algorithms. (Z=Rs.595) Solution: Step 1: Balance Step 2: Minimize Phase I: To get Initial Feasible Solution Solve by VAM D E F Supply

capacity I II III IV

A 5

1 10

7

10 4 x

B 6 60

4 10

6

10

80 2 2 2 2

C 3 15

2

5 15 1 1 1 x

PENALTY (Dummy)

5

3

2 40

40 1 1 x

Demand 75

20

50

145

I 2 1 3

II 2 1 3

III 3 2 1 IV 6 4 6

Initial basic feasible solution: 1*10=10 6*60=360 4*10=40 6*10=60 3*15=45 2*40=80 =Rs 595 Optimality Test: Rim Conditions: No. Of allocations = m+ n – 1 = 5 Therefore we can solve it by Modi method. Phase II: Modi Method

v u

Vi=3 V2=1 V3=3

Ui=0 5-3 2

1 10

7-3 4

U2=3

6 60

4 10

6 10

U3=0

3 15

2-1 1

5-3 2

U4=-1

5-3 2

3-1 2

2 40

Since, all cell evaluations are positive. Thus this is the optimum solution. Thus, optimum cost is Rs.595/- (answer)

Question 22. A company has factories at F1, F2, F3 which supply warehouses W1, W2, W3. Weekly factory requirements are 200, 160 & 90 units respectively. Unit shipping costs in rupees are as follows

W1 W2 W3

F1 16 20 12

F2 14 8 18

F3 26 24 16 Determine the optimal distribution cost for this company to minimize shipping cost. Note- Since demand was not stated in the sum, I have assumed it to be 150.

W1 W2 W3 Supply

F1 16 20 12 200

F2 14 8 18 160

F3 26 24 16 90

Demand 150 150 150

Solution The given transportation problem s of minimization type which is a balanced one. Vogel’s Approximation Method

W1 W2 W3 ai

F1

16

20

12

200

4

4

4

F2

14

8

18

160

6

4

4

F3

26

24

16

90

8

10

-

bj

150

150

150

450

450

2 12 4 2 - 4 2 4

Total Transportation Cost = 2240+140+1200+720+1440 = 5740 RIM = m + n – 1 = 3 + 3 – 1 = 5

60

10

140

150

90

W1 W2 W3 ai ui

V1

16

20 10

12

200

0

V2

14

8

18

8160

-2

V3

26

6

24 10

16 90

4

bj

150

150

150

450

450

vj

16

10

12

Since the cell evaluations are positive, optimum solution is attained. Therefore, the optimum transportation cost is Rs 5740.

10

60 140

150

90

Question23 The following table shows all the necessary information on the availability of supply of each warehouse, the requirement of each market and the unit transportation cost in rupees from each warehouse to each market: P Q R S Supply

capacity A 6 3 5 4 22 B 5 9 2 7 15 C 5 7 8 6 8 Requirement 7 12 17 9 45 The shipping clerk has worked out the following schedule from experience: 12 units from A to Q, 1 unit from A to R, 9 units from A to S, 15 units from B to R, & units from C to P and 1 unit from C to R.

1) Check and see if the clerk has optimal schedule 2) Find the optimal schedule and minimum total transportation cost 3) If the clerk is approached by a carrier of route C to Q who is willing to reduce the

rate in hope of getting some business, by how much the rate should be reduced before the clerk will offer him business.

Solution

1) Shipping Clerk’s Matrix

P Q R S Supply capacity

A 6 3 12 5 1 4 9 22 B 5 9 2 15 7 15 C 5 7 7 8 1 6 8 Requirement 7 12 17 9 45 Total transportation cost as per the Shipping Clerk A-Q 3*12 = 36 A-R 5*1 = 5 A-S 4*9 = 36 B-R 2*15 = 30 C-P 5*7 = 35 C-R 8*1 = 8 Total 150

Check for Optimality P Q R S Supply

capacity Ui

A 6 4 3 12 5 1 4 9 22 0 B 5 6 9 9 2 15 7 6 15 -3 C 5 7 7 1 8 1 6 -1 8 3 Requirement 7 12 17 9 45 Vj 2 3 5 4 Since all the cell evaluation are not positive optimum solution is not attained. 2) Therefore we solve it by Modi method Modi Method P Q R S Supply

capacity Ui

A 6 4 3 12 5 +Q 1 4 -Q 9 22 0 B 5 6 9 9 2 15 7 6 15 -3 C 5 7 7 1 8 -Q 1 6 +Q -1 8 3 Requirement 7 12 17 9 45 Vj 2 3 5 4 Min Q = (1, 9) Q = 1 Reduction in Cost = 1*1 = 1 New Cost = Old Cost – Reduction in Cost 150 – 1 = 149 Check for Optimality P Q R S Supply

capacity Ui

A 6 3 3 12 5 2 4 8 22 0 B 5 5 9 9 2 15 7 6 15 -3 C 5 7 7 2 8 1 6 1 8 2 Requirement 7 12 17 9 45 Vj 3 3 5 4 Since all the cell evaluations are positive, optimum solution is attained. Therefore the optimum transportation cost is Rs.149. 3) The carrier should reduce the rate by Rs 3 to get some business from the clerk. Carrier’s rate should be Rs

Question 24. A company manufacturing air coolers has two plants located at Mumbai and Kolkata with a weekly capacity of 200 units and 100 units respectively. The company supplies air coolers to their 4 showrooms at Ranchi, Delhi, Lucknow and Kanpur which have a demand of 75,100,100 & 50 units respectively. The cost of transportation per unit is shown in the following table: RANCHI DELHI LUCKNOW KANPUR MUMBAI 90 90 100 100 KOLKATA 50 70 130 85 The given transportation problem is of minimization type but is unbalanced since the demand is not equal to the supply so we add a dummy row. Showrooms Plants Ranchi Delhi Lucknow Kanpur Supply Mumbai 90 90 100 100 200 Kolkata 50 70 130 85 100 Dummy 0 0 0 0 25 Demand 75 100 100 50 325 The transportation problem is of minimization type and is balanced (demand = supply) Phase 1 To get initial basic feasible solution VAM/Penalty Method Plants Ranchi Delhi Lucknow Kanpur Supply Penalty Mumbai 90 90-75 100-75 100-50 200 0/0/10/10 Kolkata 50-75 70-25 130 85 100 20/20/15/15Dummy 0 0 0-25 0 25 0/- Demand 75 100 100 50 325 Penalty 50 70 100 85 40 20 30 15 - 20 30 15 20 - 15 Total Transportation cost Kolkata –Ranchi 50 *75 = 3750 Mumbai-Delhi 90 *75 = 6750 Kolkata-Delhi 70 *25 = 1750 Mumbai-Lucknow 100*75 = 7500 Mumbai-Kanpur-100*50 = 5000

Dummy-Lucknow0 *25 = 0___ Rs 24, 750 RIM CONDITION If number of allocations is equal to m+n-1. we can use Modi Method .m+n-1 =6 Phase 2- Modi Method/Test for optimality 1.No. of occupied cells = m+n-1 2 All cells are at independent positions. ui 90- (70) 90 100 + 100 - U1=0 - 20 75 75 50 50 70 130 – (80) 85-(80) U2=-20 75 25 50 5 0- (-30) 0- (-10) 0 - 0-(0) + U3=-100 30 10 25 0 V1= 70 V2=90 V3=100 V4=100 Vj Since all evaluations are positive, optimum solution is attained. Therefore the optimum transportation cost is Rs 24,750. ALTERNATE OPTIMUM SOLUTION If any of the cell evaluation is zero, alternate optimum solution exists. Cell evaluation for the box U3-V4 is zero, alternate optimum solution exists. To find alternate optimum solution θ = min (25,50) = 25 ui 90- (70) 90 100 100 U1=0 - 20 75 100 25 50 70 130 – (80) 85-(80) U2=-20 75 25 50 5 0- (-30) 0- (-10) 0 –(0) 0 U3=-100 30 10 0 25 V1= 70 V2=90 V3=100 V4=100 Vj Reduction in cost is O*25 = 0 Therefore the transportation cost is Rs 24,750

Question 25.

A company has four manufacturing plants and five warehouses. Each plant manufactures

the same product, which is sold at different prices at each warehouse area. The cost of

manufacturing and cost of raw materials are different in each plant due to various factors.

The capacities of the plant are also different. The data is given in the following table: -

Plant 1 Plant 2 Plant 3 Plant 4 Manufacturing cost in Rs./unit

90 90 100 100

Raw material cost in Rs./unit

50 70 130 85

Capacity per unit time

100 200 120 80

The company has 5 warehouses. The sales prices, transportation costs and

demands are given in the table below: -

Warehouses Transportation costs in Rs./unit 1 2 3 4 Sales

price in Rs./unit

Demand

A 4 7 4 3 30 80 B 8 9 7 8 32 120 C 2 7 6 10 28 150 D 10 7 5 8 34 70 E 2 5 8 9 30 90

(i) Formulate this into a transportation problem to maximize profit

(ii) Find the solution using VAM method

(iii) Test for optimality and find the optimal solution

SOLUTION The total cost matrix derived is a follows: -

A B C D E Supply

I 144 148 142 150 142 100

II 167 169 167 167 165 200

III 234 237 236 235 238 120

IV 188 193 195 193 194 80

Demand 80 120 150 70 90 510/500

Therefore, the profit matrix derived is equal to SP – Cost

The Profit Matrix is as follows: -

A B C D E Supply

I -114 -116 -114 -116 -112 100

II -137 -137 -139 -133 -135 200

III -204 -205 -208 -201 -208 120

IV -158 -161 -167 -159 -164 80

Demand 80 120 150 70 90 510/500

The above is an unbalanced, maximization type of problem.

A B C D E Supply

I -114 -116 -114 -116 -112 100

II -137 -137 -139 -133 -135 200

III -204 -205 -208 -201 -208 120

IV -158 -161 -167 -159 -164 80

Dummy

(D1)

0 0 0 0 0 10

Demand 80 120 150 70 90 510

Regret Matrix A B C D E Supply

I 94 92 94 92 96 100

II 71 71 69 75 73 200

III 4 3 0 7 0 120

IV 50 47 41 49 44 80

Dummy

(D1)

208 208 208 208 208 10

Demand 80 120 150 70 90 510

Thus, the matrix is now balanced and a minimization type of problem.

VAM A B C D E Supply Penalty

I 94 92 94 92 96 100 0/0/0/0/0

II 71 71 69 75 73 200 0/2/2/12/4/X

III 4 3 0 7 0 120 0/0/X

IV 50 47 41 49 44 80 3/3/6/X

Dummy

(D1)

208 208 208 208 208 10 0/0/0/0/0

Demand 80 120 150 70 90 510

Penalty 46/X 44/44/24/24/21/X 41/41/28/28/28/X 42/42/26/26/17/X 44/44/29/X

RIM Conditions are Fulfilled: 9 = 10 – 1 = order of the matrix

Initial Basic Feasibility Solution for Regret Matrix: - Cell Number Total

I – B 92 * 100 9200

II – B 71 * 10 710

II – C 69 * 120 8280

II – D 75 * 70 5250

III – A 4 * 80 320

III – E 0 * 40 0

IV – C 41 * 30 1230

IV – E 44 * 50 2200

D1 - B 208 * 10 2080

Total Regret = Rs. 29270

Checking for optimum regret: -

VA = 97 VB = 92 VC = 90 VD = 92 VE = 93

UI = 0 94

(-97) = -3

92

40

94

(-90) = 4

92

60

96

(- 93) = 3

UII = -21 71

(-76) = -5

71

80

69

120

75

(-75) = 0

73

(- 72) = 1

UIII = -93 4

80

3

(- -1) = 4

0

(- -3) = 3

7

(-3) = 4

0

40

UIV = -49 50

(-48) = 2

47

(-43) = 4

41

30

49

(- 47) = 2

44

50

UD1 = 116 208

(- 213) = -5

208

(-208) = 0

208

( - 206) = 2

208

10

208

(-209) = -1

Since UI – VA, UII – VA, UD1 – VE, UD1 – VA all have a negative CE, the solution is

not optimal.

θ = minimum (50, 120, 80)

θ = 50

Reduction in Regret = 29270 – (50 * 5) = Rs. 29020


VA = 97 VB = 92 VC = 90 VD = 92 VE = 93

UI = 0 94

(-97) = -3

92

40

94

(-90) = 4

92

60

96

(- 93) = 3

UII = -21 71

50

71

80

69

70

75

(-71) = 4

73

(- 72) = 1

UIII = -93 4

30

3

(- -1) = 4

0

(- -3) = 3

7

(- -1) = 8

0

90

UIV = -49 50

(-48) = 2

47

(-43) = 4

41

80

49

(- 43) = 6

44

(- 44) = 0

UD1 = -115 208

(- -18) = 226

208

(- -23) = 231

208

(- -25) = 233

208

10

208

(- -22) = 230

Since CE of UI – VA is < 0, the solution is not optimum.

θ = minimum (40, 50)

θ = 40

Reduction in regret = 29020 – (40*3) = Rs. 28900


VA = 94 VB = 94 VC = 92 VD = 92 VE = 90

UI = 0 94

40

92

(- 94) = -2

94

(-92) = 2

92

60

96

(- 90) = 6

UII = -23 71

10

71

120

69

70

75

(-69) = 6

73

(- 67) = 5

UIII = -90 4

30

3

(- 4) = -1

0

(- 2) = -2

7

(- 2) = 5

0

90

UIV = -51 50

(-43) = 7

47

(-43) = 4

41

80

49

(- 41) = 8

44

(- 39) = 5

UD1 = 116 208

(- 210) = -2

208

(- 210) = -2

208

(- 208) = 0

208

10

208

(- 206) = 2

Since all CE are not > or equal to 0, therefore, optimum solution is not obtained


θ = 10

Reduction in regret = 28900 – (10 * 2) = 28880


VA = 94 VB = 94 VC = 92 VD = 92 VE = 90

UI = 0 94

30

92

(- 94) = -2

94

(-92) = 2

92

70

96

(- 90) = 6

UII = -23 71

10

71

120

69

70

75

(-69) = 6

73

(- 67) = 5

UIII = -90 4

30

3

(- 4) = -1

0

(- 2) = -2

7

(- 2) = 5

0

90

UIV = -51 50

(-43) = 7

47

(-43) = 4

41

80

49

(- 41) = 8

44

(- 39) = 5

UD1 = 114 208

10

208

(- 208) = 0

208

(- 206) = 2

208

(-206) = 2

208

(- 204) = 4



θ = 30



VA = 94 VB = 94 VC = 92 VD = 92 VE = 92

UI = 0 94

30

92

(- 94) = -2

94

(-92) = 2

92

70

96

(- 92) = 4

UII = -23 71

40

71

120

69

40

75

(-69) = 6

73

(- 69) = 4

UIII = -92 4

(-2) = 2

3

(- 2) = 1

0

30

7

(- 0) = 7

0

90

UIV = -51 50

(-43) = 7

47

(-43) = 4

41

80

49

(- 41) = 8

44

(- 41) = 3

UD1 = 114 208

10

208

(- 208) = 0

208

(- 206) = 2

208

(-206) = 2

208

(- 206) = 2


θ = minimum (30, 120)

θ = 30



VA = 92 VB = 92 VC = 90 VD = 92 VE = 90

UI = 0 94

(-92) = 2

92

30

94

(-90) = 4

92

70

96

(- 90) = 6

UII = -21 71

70

71

120

69

40

75

(-71) = 4

73

(- 69) = 4

UIII = -90 4

(- 2) = 2

3

(- 2) = 1

0

30

7

(- 2) = 5

0

90

UIV = -49 50

(-43) = 7

47

(-43) = 4

41

80

49

(- 43) = 6

44

(- 41) = 3

UD1 = 116 208

10

208

(- 208) = 0

208

(- 206) = 2

208

(-208) = 0

208

(- 206) = 2

All CE are > or equal to 0, therefore optimum regret is achieved.

Regret solution = Rs. 28760

Therefore, Profit = (maximum profit per unit * total quantity) – optimum regret

= (-112 * 510) – 28760

= -57120 – 28760

= Rs. –85880

Thus, the firm makes a loss of Rs. 85880.

Question 26. Suppose that England,france and spain produce all the wheat,barley and oats needed in the world. The world demand for wheat requires 125 million acres of land devoted to wheat production. Similarly, 60 milion for Barley and 75 milion acres of land for Oats are required. The total amount of land available for these purpose in England, France and Spain is 70 milion, 110 milion and 80 milion acres respectively. The number of hours needed in England France and Spain to produce an acre of wheat is 18 hrs, 13 hrs and 16 hrs respectively. The number of hours needed for barley are 15, 12 and 12 hrs. The number of hours for oat are 12,10 and 16 hrs respectively. The labour cost for wheat is $3, $2.4 and and $3.3 respectively.Labour cost for barley is $2.7, $3 and $2.8. Labour cost in producing oats is $2.3 , $2.5 & $2.1 respectively. The problem is to allocate land use in each country so as to meet world’s food requirement and minimize the total labour cost

1) Formulate the problem 2) Solve it for optimum cost

Ans) Labour Cost in $ (Multiplying By 10) Wheat Barely Oats Capacity Penalty England 540 405 276 (70) 70 129/129/x France 312 (110) 360 250 110 62/x Spain 528 (15) 336 (60) 336 (5) 80 0/0/0 Demand 125 60 75 Penalty 216/12 24/69 29/60 Step 1 Balanced Step 2 Minimization Phase 1 276 * 70 + 312 * 110 + 528 * 15 + 336 * 60 + 336 * 5 = Rs 83400 Phase 2 Test for Optimality 1) No. of occupied cells = m+n-1 = 3+3-1 = 5 2) All cells are at independent positions

u v V1=468 V2=276 V3=276

U1=0 540 – 468 = 72 405 – 276 = 129 276 (70)

U2=-156 312 (110) 360 – 120 = 240 250 – 120 = 130

U3=60 528 (15) 336 (60) 336 (5)

All Cell Evaluations ≥ 0, there is optimum solution. Therefore, optimum cost= 83400/10= Rs 8340

Question 27. A wholesale distributor has three houses W1,W2 & W3 whose stocks are distributed to meet the demands of four market region M1,M2,M3 & M4. The weekly supplies available and the require demands and the unit cost of transportation are displayed below Warehouses Market Supplies(uts)W1 12 6 10 8 440 W2 20 18 4 14 300 W3 20 14 16 12 160 DEMAND(Uts) 140 240 340 180 Presently the wholesaler is following the schedule of sending supplies from warehouses W1, 240 uts to M2, 20 uts from M3 and 180 uts M4. From W2 all the 300 uts to M3. W3 supplies 20 uts to M3 and 140 uts to M1

1) Compute the total transportation cost of the present schedule 2) Is it possible to get a better schedule that will reduce present total cost?If so,

find the best solution and the saving resulting there from. 3) Suppose the transportation cost from W3 to W2 becomes 9 per unit,would this

change the optimal solution? If so, derive the new solution and the associated cost.

Ans) M1 M2 M3 M4 Supply W1 12 6 (240) 10 (20) 8 (180) 440 W2 20 18 4 (300) 14 300 W3 20 (140) 14 16 (20) 12 160 Demand 140 240 340 180 900 Phase 1 Calculation of Initial Basic Feasible Solution IBF Solution= 6*240+10*20+8*180+4*300+20*140+16*20 ` = Rs. 7400 Phase 2 Test for Optimality 1) No. of occupied cells = m+n-1 = 3+4-1 = 6 2) All cells are at independent positions

u v V1=14 V2=6 V3=10 V4=8

U1=0 12 – 14 = -2 6 (240) 10 (20) 8 (180)

U2=-6 20 – 8 = 12 18 – 0 = 18 4 (300) 14 – 12 = 12

U3=6 20 (140) 14 – 12 = 2 16 (20) 12 – 14 = -2

Ø= min {20,180} Ø= 20

Reduction in cost = 7400 -2*20 = 7400-40 = RS 7360 u v V1=16 V2=6 V3=10 V4=8

U1=0 12 – 16 = -4 6 (240) 10 (40) 8 (160)

U2=-6 20 – 10 = 10 18 – 0 = 18 4 (300) 14 – 12 = 12

U3=4 20 (140) 14 – 10 = 4 16 – 14 = 2 12 (20)

Ø= min {140,160} Ø= 140

Reduction in cost = RS 7360 – 4*140 = Rs 6800 u v V1=16 V2=6 V3=10 V4=8

U1=0 12 (140) 6 (240) 10 (40) 8 (20)

U2=-6 20 – 6= 14 18 – 0 = 18 4 (300) 14 – 12 = 12

U3=4 20-6=14 14 – 10 = 4 16 – 14 = 2 12 (160)

All C.E. ≥ 0, optimumsolution is attained Optimum Cost is Rs.6800 3) If the cost of route W2 to M2 is changed to Rs 9 per ut the change in optimum solution would be u v V1=16 V2=6 V3=10 V4=8

U1=0 12 (140) 6 (240) 10 (40) 8 (20)

U2=-6 20 – 6= 14 18 – 0 = 18 4 (300) 14 – 12 = 12

U3=4 20-16=4 9 – 10 = -1 16 – 14 = 2 12 (160)

Ø= min {240,160} Ø= 160

Reduction in cost = 6800 -1*160 = Rs. 6640

Question 28. Solve the following problem using transportation algorithm. Use VAM for finding the initial feasible solution. The cell entries in the table are unit costs.

FROM TO I II III IV V SUPPLY

O1 80 69 103 64 61 12 O2 47 100 72 65 40 16 O3 16 103 87 36 94 20 O4 86 15 57 19 25 8 O5 27 20 72 94 19 8

DEMAND 16 14 18 6 10 SOLUTION:

FROM TO I II III IV V SUPPLY Penalty

O1 80

69 10312 64 6 1 12 3/3/3/8/34/34

O2 47 100 726 65 4010 16 6

7/25/25/32/28/28

O3 16 16 103 87 364 94 20 4

20/51/-/-/-/-

O4 86 156 57 192 25 8 6

4/4/4/10/42/42

O5 27 208 72 94 19 8

1/1/1/1/52/-/-

DEMAND 16 14 6 18 6 2 10

Penalty 11/-/-/-/-/-

5/5/5/5/5/54 15/15/15/15/15/15

17/17/45/-/-/- 6/6/6/6/-/-

STEP 1: Balanced STEP 2: Minimisation

PHASE I: Initial Basic Feasible Solution O1-III --- 103 x 12= 1236 O2-III --- 72 x 6= 432 O2-V --- 40 x 10= 400 O3-I --- 16 x 16= 256 O3-IV--- 36 x 4= 144 O4-II --- 15 x 6= 90 O4-IV --- 19 x 2= 38 O5-II --- 20 x 8= 160 Rs. 2756 Thus, total production cost is Rs.2756. PHASE II: Test for Optimality

1. Number of occupied cells =m+n-1 = 5+5-1 =9 It is a degenerate solution since the number of allocations is 8 & not 9.

2. All cells are at independent positions. V1=56 V2=72 V3=103 V4=76 V5=71 U1=0 80-(56)

24

69-(72)

-3

103 - 64-(76)

+

-12

61-(71)

-10 U2=-31 47-(25)

22

100 –(41)

59

72

+ 65 –(45)

20

40 -

U3=-40 16

103-(32)

71

87-(63)

24

36 94-(31)

63

U4=-57 86-(-1)

87

15 +

57 (-46)

11

19 - 25-(14)

11

U5=-52 27-(4)

23

20

-

72-(51)

21

94-(24)

70

19

+ =Minimum{12,10,2,8} =2 Reduction in cost=2756-12x2 =Rs.2732

E

16

10

2

4

6

12

8

6

V1=44 V2=72 V3=103 V4=64 V5=71 U1=0 80-(44)

36

69-(72)

-3

103 - 64

61-(71)

+

-10 U2=-31 47-(13)

34

100 –(41)

59

72

+ 65 –(33)

32

40 -

U3=-28 16

103-(44)

59

87-(75)

12

36 94-(43)

51

U4=-57 86-(-13)

99

15

57 (-46)

11

19 –(7)

12

25-(14)

11

U5=-52 27-(-8)

35

20

72-(51)

21

94-(12)

82

19

=Minimum{10,8} =8 Reduction in cost=2732-10x8 =Rs.2652

All cell evaluations are greater than or equal to 0 Therefore, there is optimum solution. Optimum cost =Rs.2652

V1=44 V2=62 V3=103 V4=64 V5=61 U1=0 80-(44)

36

69-(62)

7

103 64

61

U2=-31 47-(13)

34

100 –(31)

69

72

65 –(33)

32

40-(30)

10

U3=-28 16

103-(34)

69

87-(75)

12

36 94-(33)

61

U4=-47 86-(-3)

89

15

57 (-56)

1

19 –(17)

2

25-(14)

11

U5=-42 27-(2)

25

20

72-(61)

11

94-(22)

72

19

2

16

8

4

8

10

6

8

2

2

16 4

16

2

6

8

2 8

Question 29. ABC Manufacturing company wishes to develop a monthly production schedule for the next three months. Depending upon the sales commitments, the company can either keep the production constant, allowing fluctuations in inventory, or inventories can be maintained at a constant level, with fluctuating production. Fluctuating production necessitates in working overtime, the cost of which is estimated to be double the normal production cost of Rs. 12per unit. Fluctuating inventories result in inventory carrying cost of Rs. 2 per unit if the company fails to fulfill its sales commitment, it incurs a shortage cost of Rs.4 per month. The production capacities for nest three months are shown below:

MONTH PRODUCTION CAPACITY SALES REGULAR OVERTIME

1 50 30 60 2 50 0 120 3 60 50 40

Determine the optimal production schedule.

SOLUTION

FROM TO SALES Production 1 2 3 Dummy Supply

R1 12 14 16 0 50 R2 16 12 14 0 50 R3 20 16 12 0 60 O1 24 26 28 0 30 O3 32 28 24 0 50

Demand 60 120 40 20 240

FROM TO SALES Production 1 2 3 Dummy Supply Penalty

R1 12 50 14 16 0 50 12/2/- R2 16 12 50 14 0 50 12/2/- R3 20 16 20 12 40 0 60 12/4/4/4/4 O1 24 10 26 20 28 0 30 24/2/2/2/2/2O3 32 28 30 24 0 20 50 24/4/4/4/4

Demand 60 120 40 20 240 Penalty 4/4/4/4/4/8 2/2/4/10/10/2 2/2/2/12 0

STEP 1 – Balanced STEP 2 – Minimisation

PHASE 1 – Initial Basic Feasible Solution Production Schedule 12*50 - 600 12*50 - 600 16*20 - 320 12*40 - 480 24*10 - 240 26*20 - 520 28*30 - 840 0*20 - 0 Total 3600 PHASE 2 – Test for optimality No. of occupied cells = m + n – 1 = 5 + 4 – 1 = 8 All cells are at independent positions

V1=12 V2=14 V3=10 V4= -14 U1=0 12 -

14-(14) +

0

16-(10) 6

0-(-14)

14

U2= -2 16-(10)

6

12

14-(8)

6

0 –(-16)

16

U3= 12 20-(14)

6

16

12

0-(-12)

12

U4= 12 24 +

26 -

28 (-22)

6

0-(-2)

2

U5= 14 32-(26)

16

28

24-(24)

0

0

If any of the cell evaluation is 0 alternate optimum solution exists. To find alternate optimum solution = min (50,20) = 20

20

30

10

50

20

50

40

20

Reduction in cost = 3600 – (20*0) 3600

V1=12 V2=14 V3=10 V4= -20 U1=0 12

14

16-(10) 6

0

20

U2= -2 16-(10)

6

12

14-(8)

6

0

22

U3= 2 20-(14)

6

16 +

12 -

0

18

U4= 12 24

26

0

28 (-22)

6

0

8

U5= 20 32

0

28

-

24

+

-8

0

= min (30,40) = 30 Reduction in cost = 3600 – (8*30) 3360

V1=12 V2=14 V3=10 V4= -14 U1=0 12

14

6 6

0

14

U2= -2 16-(10)

6

12

14-(8)

6

0

16

U3= 2 20-(14)

6

16

12

0

12

U4= 12 24

26

0

28 (-22)

6

0

2

U5= 14 32

6

28

24

0

= min (20,30) = 20 Reduction in cost = 3360 – (0*20) 3360

30

30

30

20

50

40

20

20

30

30

50

50

10

20

20

30

Question 30. A company has three factories that supply to four marketing areas. The transportation cost of shipping from each factory to different marketing areas is given below. The availability at each factory and requirements at different markets is also given.

Table 4.16 Factory Marketing area Supply

M1 M2 M3 M4 F1 19 30 50 10 1600

F2 70 30 40 40 1200

F3 40 8 70 20 1700

Demand 1000 1500 800 1200 4500 Find initial feasible solution using VAM Is the solution obtained optimum? If the solution is not optimal carry out improvements for optimality using MODI method. Solution The given transportation problem is of minimization type and is a balanced one. Phase 1: To get initial basic feasible solution using VAM method.

Factory Marketing area Supply Penalty M1 M2 M3 M4

F1 19 30 50 10 1600 9/9/40/40 1000 600

F2 70 30 40 40 1200 10/10/0/0 800 400

F3 40 8 70 20 1700 12/20/50/ 1500 200

Demand 1000 1500 800 1200 4500

Penalty 21/21 22 10/10/10/10 10/10/10/30

Total transportation cost: Units Cost per unit Total 1000 19 19000 1500 8 12000 800 40 32000 600 10 6000 400 40 16000 200 20 4000 Total transportation cost: Rs 89000 Rim condition: No. of allocations = 6 m+n-1 = 3 + 4 – 1 = 6 Since, no of allocations = m + n – 1, hence can solve by MODI method. Phase 2: MODI method

Factory Marketing area Supply ui M1 M2 M3 M4

F1 19 30 50 10 1600 0 1000 32 40 600

F2 70 30 40 40 1200 30 21 2 800 400

F3 40 8 70 20 1700 10 11 1500 50 200

Demand 1000 1500 800 1200 4500 vj 19 -2 10 10

1. ui + vj = cost of allocated cells. 2. Cell evaluation: CE = Cost – (u + v), for all non allocated cells.

E.g. for F2 – M1 cell evaluation: 70 – (30+19) = 21. Since all cell evaluations are positive, the solution obtained is an optimum solution. Therefore, optimum transportation cost = Rs 89000.

Question 31. There are three canning factories around a state which need baskets of strawberries. Three orchards supply these to the factories. Their costs for supplying these baskets are as follows:

Table 4.18

Orchard Price in Rs (per basket)Annual

capacity A 19 300 B 20 700 C 21 1350

The cost of transportation (per basket in Rs) from each orchard to each factory is given below:

Table 4.19 To Factory

From X Y Z A 2 4 1 B 5 3 6 C 3 2 7

The annual requirements of three factories is 300, 600, 1200 baskets respectively. How many baskets should be purchased from each orchard by each factory to minimize the total cost? Solution: The above problem is of minimization type. However, the problem is unbalanced since supply is greater than demand. We take care of this imbalance by introducing a 4th dummy factory having a requirement of 250 baskets. Phase 1: to get initial basic feasible solution Orchard Factory Supply Penalty

X Y Z Dummy A 21 23 20 0 300 20/1/ 300 B 25 23 26 0 700 23/2/2 700 C 24 23 28 0 1350 23/1/1/1 300 600 200 250

Demand 300 600 1200 250 2350

Penalty 3/3/1 0/0/0 6/6/2 0/

Total transportation cost: Units Cost per unit Total 300 24 7200 600 23 13800 300 20 6000 700 26 18200 200 28 5600 250 0 0 Total transportation cost: Rs 50800 Rim condition: No. of allocations = 6 m+n-1 = 3 + 4 – 1 = 6 Since, no of allocations = m + n – 1, hence can solve by MODI method. Phase 2: MODI method Orchard Factory Supply ui

X Y Z Dummy A 21 23 20 0 300 0 5 8 300 8 B 25 23 26 0 700 6 3 2 700 2 C 24 23 28 0 1350 8 300 600 200 250

Demand 300 600 1200 250 2350 vj 16 15 20 -8


Since all cell evaluations are positive, the solution obtained is an optimum solution. Therefore, optimum transportation cost = Rs 50800

Question 32 A city bus service has two bus depots where the buses are parked at night. Early morning the buses have to reach to three different starting points. The distance in km between the depot and the starting points is given below. Find the optimum routing of buses from depot to the starting point so as to minimize the total distance traveled by empty buses. STARTING POINTS Bus Depot A B C Availability X 2 8 4 25 Y 3 7 3 10 Requirements 15 8 12 35 Solution: The given transportation problem is of minimization type which is a balanced one as Total Requirement is equal to Total Availability. Phase I: To get Initial Basic Feasible Solution (IBF) using Vogle’s Approximation Method/ Penalty Method. Bus Depot A B C Availability Penalty X 2 (15) 8 (8) 4 (2) 25

2/4/4

Y 3 7 3 (10) 10

0/4/-

Requirement 15 8 12 35 35

Penalty 1 -

1 1

1 1

IBF Solution = (15*2) + (8*8) + (4*2) + (3*10) = 132 kms. Phase II: Test for Optimality

1. No. of cells = m+n-1 = 4 2. All cells are at independent positions.

The above conditions are fulfilled. Therefore, we solve and find the optimum solution using Modified Distribution (MODI) Method or UV Method. V U

V1=2 V2=8 V3=4

U1 = 0

2 (15)

8 (8)

-θ

4 (2)

+θ

U2 = -1

3-(1) 2

7-(7)

+θ 0

3 (10)

-θ

θ = 8 Reduction in Cost = 132 – (0*8) = 132 – 0 = 132 kms. Ans. The total minimum and optimum distance traveled by the buses is 132 kms.

Question 33 A cigarette mfg company has factories in 3 different cities. Hyderabad, Bangalore and belgaum. It sells its product in 3 differesnt markets.the cost ofd raw materials, labour & transportation are different and the prices at which the packets are sold in different markets is also not uniform. The profits therefore vary form palce of manufacture and markets. They are as follows. (profits are in Rs. 10 per packet )

markets Availability M1 M2 M3

Hyderabad 29 28 30 2000 Bangalore 25 27 23 2000 Belgaum 35 37 38 2000

1500 3000 1500 Formulate the problem and find initial feasible solution using VAM. SOLUTION The given problem is a maximization type. We convert it into minimization type by forming a regret matrix. REGRET MATRIX

markets Availability M1 M2 M3

Hyderabad 9 10 8 2000 Bangalore 13 11 15 2000 Belgaum 3 1 0 2000 Demand 1500 3000 1500 6000

Now, the sum is of minimization tyoe and Is balanced, thus we can solve it by VAM Initial feasible solution

markets Availability Penalty M1 M2 M3

Hyderabad 9 10 8 0 1 1 1 - 500 1500

Bangalore 13 11 15 0 2 2 2 2 1000 1000

Belgaum 3 1 0 0 1 - - - 2000

Demand 0 0 0

Penalty

6 9 8

6 1 8 6 1 - - 1 -

TOTAL REGRET COST (9*500) + (8*1500) + (13*1000) + (11*1000) + (1*2000) = 42500 RIM CONDITION CHECK Number of allocations = 5 Degree of matrix = M+N-1 = 5 Since, number of allocations = degree of matrix Feasible solution is attained

FROM TO QUANTITY PROFITTOTAL PROFIT

Hyderabad M1 500 29 14500Hyderabad M3 1500 30 45000Bangalore M1 1000 25 25000Bangalore M1 1000 27 27000Belgaum M2 2000 37 74000

Total 185500 THUS, total maximum profit = 185000

Question 34. Given below is a transportation problem with transportation cost and initial feasible solution. Destination origin

D1 D2 D3 D4 Supply

O1

5 10 4100

5 100

O2

6 200

8 7 2 50

O3

4 50

2100

550

7 200

Demand 250 100 150 50 550 State with reason, whether:

(1) The given feasible solution is degenerate. (2) The solution is optimal. (3) Can there be more than one optimal solution in this problem. (4) How will you test the optimality of the solution when one of the cell cost change?

Solution

(1) The number of occupied cells is 6 And m + n -1 = 4 + 3 – 1 = 6 Since number of occupied cells = m + n – 1 = 6, the rim requirements are satisfied. Therefore, the solution obtained is a non-degenerate solution.

(2) To check whether the solution is optimal solution or not, the modi method is adopted.

Modi method

destination origin

D1 D2 D3 D4 Supply Ui

O1

5 2

109

4100

56

100 0

O2

6 200

85

70

250

250 3

O3

4 50

2100

550

70

200 1

250

Demand

250 100 150 50 550

Vj

3 1 4 -1

Optimality test Since all the cell evaluations are positive/ non- negative optimum solution is obtained. (3) yes, there can be more than one optimal solution in the above case as the cell

evaluations for the cells 02 d3 and o3 d4 is 0 each. Therefore, alternate optimum solutions exits. Now the total transportation cost = 100*4 + 200*6 + 50*2 + 50*4 + 100*2 + 50*5 = 2350 Alternate optimum solution 1

Destination origin


O1

5 2

109

4100

56

100 0

O2

- 6 200

85

+ 70

250

250 3

O3

4 + 50

2100

550 -

70

200 1

Demand

250 100 150 50 550

Vj

3 1 4 -1

∂ = min (50,200) = 50 Therefore reduction in cost = 50*0 = 0 Therefore new cost = 2350 – 0 = 235

Alternate optimum solution 2

Destination Origin


O1

5 2

109

4100

56

100 0

O2

+ 6 200

85

70

- 250

250 3

O3

4 - 50

2100

550

70 +

200 1

Demand

250 100 150 50 550

Vj

3 1 4 -1

∂ = min (50, 50) = 50 Therefore reduction in cost = 50*0 = 0 Therefore new cost = 2350 – 0 = 2350

(4) When one of the cell costs change, there are two possibilities to be considered i.e. A) the change of cost is for occupied cell

Or b) The change of cost is for an unoccupied cell In the first case we have to compute the initial basic feasible solution by the vam method and calculate new ui and vj and cell evaluations to check the optimality of the solution. In the second case we just have to calculate new cell evaluation for the changed cell and find out whether the solution is optimal.

Question 35. Mr. Pinto is the production supervisor at chips and chips electronics company in Mumbai. On arriving at work on one fine morning hr finds the foloowing pallet information

Dept Pallets

available Dept Pallets reqd.

A 28 G 14 B 27 H 12 C 21 I 23 J 17

The time to move a pallet from 1 dept to another is as follows From A A B B B B A A C C C C To G H G H I J I J G H I J min 13 25 18 23 14 9 12 21 23 15 12 13 Find the distribution plan using it as a transportation problem so as to minimize the total time required for distribution plan

SOLUTION The Matrix showing time required for distribution from each source to each destination along with demand and supply for pallets is as follows To G H I J Supply From A 13 25 12 21 18B 18 23 14 9 27C 23 15 12 13 21 Demand 14 12 23 17 66 The above problem is of minimization type and a balanced one. Hence, solving it with VAM method

To G H I J Supply Penalty From A 13 25 12 21 1/1/1/1 14 4 18 B 18 23 14 9 5/5/4/4 10 17 27 C 23 15 12 13 1/1/11/ 12 9 21 Demand 14 12 23 17 66 Penalty 5/5/5/10 8/ 0/0/0/2 4/4/ Total Transportation Time required is 14*13 + 15*12 + 12*4 + 14*10 + 12*9 + 17*9 =811 minutes = 13.51 hrs = 13hrs 30 min RIM condition check M+n-1 = 6 = number of allocations Therefore, by Modi Method To G H I J Supply ui From A 13 25 12 21 14 10 4 14 18 0B 18 23 14 9 3 6 10 17 27 2C 23 15 12 13 10 12 9 6 21 0Demand 14 12 23 17 66 vj 13 15 12 7 Since, all cell valuations are non negetive the solution so obtained is the optimum feasible solution. Answer: The distribution plan should be as follows A to G, A to I, B to I, C to H, C to I, B to I.

Question 36. “YOURS OWN” garment manufacturing firm of Mumbai wishes to develop a monthly production schedule for the next three months. Depending on sales commitments, the company can either keep the production constant, and allowing the fluctuations in inventory or maintained inventories at a constant level, with fluctuating production. The fluctuating production necessitates, working overtime, the cost of which is estimated to be double the normal production cost of Rs. 10 per unit. Fluctuating inventories result in inventory carrying cost of Rs. 4 per unit. If the company fails to fulfill its sales commitment, it incurs a shortage cost of Rs. 5 per unit per month. The production capacities for the next three months are shown in the following table:

Production Capacity Months 1 2 3

Regular 50 50 60 Overtime 30 00 50 Sales 60 120 40 Formulate it as a Transportation Problem to obtain an optional production schedule. Solution: The following Transportation problem is an unbalanced problem and of minimization type. We therefore formulate the COST MATRIX by adding a dummy column ‘D’. Production

Months Months Dummy

4 Suppy

1 2 3 R1 10 14 18 0 50 O1 20 24 28 0 30 R2 M 10 15 0 50 O2 M 20 25 0 00 R3 M M 10 0 60 O3 M M 20 0 50

Demand 60 120 40 20 240 Explanation: We add inventory carrying cost to production cost in the succeeding months (i.e. R1:10, 10+4, 10+4+4 and O1:20, 20+4, 20+4+4). In addition to this there is shortage cost that occurs in month 2, because production in months 1 & 2 together fall short of commitment of sales in month 2 if the commitment of sales in month 1 is carried out. This shortage cost is Rs. 5 per unit per month which is added to the cost in the month 3 (i.e. R2-3 and O2-3). Finally, production in a given month cannot be used in the preceding month; therefore we get a cost infinity which is denoted by “M” for the earlier months.

Question 37. Find the initial feasible solution for the following transportation using VAM . DESTINATION ORIGIN D1 D2 D3 D4 D5 SUPPLY O1 2 11 10 3 7 4 O2 1 4 7 2 1 8 O3 3 9 4 8 12 9 DEMAND 3 3 4 5 6 21 SOLUTION DESTINATION PENALTY ORIGIN D1 D2 D3 D4 D5 SUPPLY O1 2 11 10 3 7 4 1 1 1 1 - O2 1 4 7 2 1 8 1 1 - - - O3 3 9 4 8 12 9 1 1 1 5 5 DD 3 3 4 5 6 21 PENALTY 1 5 3 1 6 1 5 3 1 - 1 2 6 5 - 1 2 - 5 - 3 9 - 8 - STEP 1): Balanced 2):Minimisation PHASE 1:Initial Basic Feasible Solution = 3*4 + 1*6 + 4*2 + 3*3 + 9*1+ 4*4+ 8*1 =12+6+8+9+9+16+8 =Rs 68 Initial Basic Feasible Solution is Rs 68

321 4

4

16

Question 38. There are 4 bus depots where the buses and parked for night. These empty buses should reach the starting points early in the morning to start the bus service on various routes. The cost of per unit transportation for empty buses from the depots to starting points is given below. Find the optimum movement of empty buses from the depots to starting point so as to minimize the total transportation cost.

Starting Point Bus Depot

1 2 3 4 5 6 Supply

A 10 12 11 14 15 12 30 B 12 13 12 11 14 13 50 C 14 12 15 19 16 12 75 D 12 11 17 13 14 16 20 Demand 20 40 30 10 50 25 175

SOLUTION

Starting Point

Bus Depot

1 2 3 4 5 6 Supply Penalty

A 10 20 12 11 10 14 15 12 30 10 1/1/1/1/-

B 12 13 12 20 11 10 14 20 13 50 40

20 1/1/1/1/1/0

C 14 12 20 15 19 16 30 12 25 75 55

25 0/0/0/0/0/0

D 13 11 20 17 13 14 16 20 2/2/-/-/- Demand 20 40 20 30 20 10 50 30 25 175 Penalty 2/-/-/-

/- 1/1/0/0/1/1

1/1/1/1/3 2/2/3/-/-

0/0/1/1/2/2

0/0/0/0/1/1

STEP 1 – Balanced STEP 2 – Minimisation

PHASE 1 – Initial Basic Feasible Solution 20*10 + 10*11 + 20*12 +10*11+20*14 + 20*12 + 30*16 + 25*12 + 20*11 =Rs.2180/- PHASE 2 – Test for optimality No. of occupied cells = m + n – 1 = 6+4-1 = 9 All cells are at independent positions

U / V V1=10 V2=9 V3=11 V4= 10 V5= 13 V6= 9 U1=0 10

12 -(9)

3

11

14 –(10)

4

15 –(13)

2

12 –(9)

3

U2= 1 12 -(11)

1

13 –(10)

3

12

11

1

14

13 –(10)

3

U3= 3 14 -(13)

1

12

+

15 –(14)

1

19 –(13)

6

16 -

12

U4= 2 13 –(12)

1

11 -

17 -(13))

4

13 –(12)

1

14 –(15)

+

-1

16 -(11)

5

= 20 Therefore Reduction In cost = 20 Therefore the new cost = 2180 – 20 Therefore New Cost =Rs.2160

20

20

20

10

20 10 20

30 25

U / V V1=10 V2=9 V3=11 V4= 10 V5= 13 V6= 9 U1=0 10

12 –(9)

3

11

14 –(10)

4

15 –(13)

2

12 –(9)

3

U2= 1 12 –(11)

1

13 –(10)

3

12

11

14

13 –(10)

3

U3= 3 12 –(13)

1

12

15 –(14)

1

19 –(13)

6

16

12

U4= 1 13 –(11)

1

11 –(10)

1

17 –(12)

5

13 –(11)

2

14

16 –(10)

6

All Cell Evaluations are greater than or equal to 0 Hence there is optimum solution. The Optimum Cost is Rs.2160/-.

20

40

10

20 10 20

10

20

25

Question 39. Find the optimal solution to the following transportation problem. The figures give the transportation cost per unit of the product from origin to destination

Destination Origin 1 2 3 4 Supply

O1 13 17 6 8 30 O2 2 7 10 41 40 O3 12 18 2 22 53

Demand 22 35 25 41 123 Use VAM and MODI method. Solution : The given transportation problem is of minimization type and is a balanced one. PHASE I – Initial Basic Feasible Solution VAM Method

Origin Destination Supply (ai) Penalty D1 D2 D3 D4 1 13 17 6 8 30 2/- 30 2 2 7 10 41 40 5/5/5/8 5 35 3 12 18 2 22 53 10/10/10/10 17 25 11

Demand 22 35 25 41 123 (bj)

Penalty 10/10/10/10 10/4/4/- 4/8/8/8 14/19/-/- Basic Feasibility Solution: O1-D4 = 30*8 = 240 O2-D1 = 2*5 = 10 O2-D2 = 7*35 = 245 O3-D1 = 12*17 = 204 O3-D3 = 25*2 = 50 O3-D4 = 22*11 = 242 991 Number of occupied cells = m+n-1 , i.e. 3+4-1 = 6 Therefore, all cells are at independent position.

Phase 2: MODI method

Origin Destination Supply ui 1 2 3 4

O1 13 17 6 8 30 0 15 14 18 30

O2 2

7 10 41 13 4

5 35 18 29

O3 12 18 2 22 14

17

1 25 11

Demand 22 35 25 41 113 vj -2 3 -12 8


Since all cell evaluations are non-negative, the solution obtained is an optimum solution. Therefore, optimum solution is Rs. 991/-

Question 40. Find the optimum feasible solution to the following degenerate transportation problem.

Retail outlets Distribution

center 1 2 3 4 Supply

1 10 7 3 6 3 2 1 6 8 3 5 3 7 4 5 3 7

Demand 3 2 6 4 15 The figures show the per unit transportation cost of the product from the distribution center to different retail outlets. Solution The given transportation problem is of minimization type and is a balanced one. Phase 1: To get initial basic feasible solution using northwest corner rule. Distribution

center Retail outlets Supply 1 2 3 4 1 3 3 10 7 3 6 2 3 2 5 1 6 8 3 3 2 3 2 7 7 4 5 3

Demand 3 2 6 4 15

Total transportation cost: Units Cost per unit Total 3 3 9 3 1 3 2 3 6 2 4 8 3 5 15 2 3 6 Total transportation cost: Rs.47/-

Rim condition:

1. No. of allocations = 6 m+n-1 = 3 + 4 – 1 = 6

2. All cells are at independent conditions Phase 2: MODI method

Origin Retail outlets Supply ui 1 2 3 4 1

10

7 3

6 3 0

11 5 3 5 2

1

6

8 3 5 2

3 2 3 2 3

7 4 5 3 7 2

1 2 3 2

Demand 3 2 6 4 15 vj -1 2 3 1


Since all cell evaluations are non-negative, no decrease in cost can be obtained Therefore, optimum transportation cost = Rs.47/-

Question 41. The following data gives transportation cost per unit of product from origin to destination as well as the supply and demand of the product at each end.

Destination Origin 1 2 3 4 Supply

O1 7 1 35 20 15 O2 8 9 4 21 13

Demand 9 6 7 6 28

i) Obtain initial feasible solution. You may use northwest corner rule. ii) Test this solution for optimality using MODI method.

Solution The given transportation problem is of minimization type and is a balanced one. Phase 1: To get initial basic feasible solution using northwest corner rule.

Origin Destination Supply 1 2 3 4

O1 7 1 35 20 15 9 6

O2 8 9 4 21 13 7 6

Demand 9 6 7 6 28

Total transportation cost: Units Cost per unit Total 9 7 63 6 1 6 7 4 28 6 21 126 Total transportation cost: Rs.223/- Rim condition: No. of allocations = 4 m+n-1 = 2 + 4 – 1 = 5 Since, no of allocations is not equal to m + n – 1, the rim condition is not satisfied. Therefore ‘DEGENERACY’ occurs. To remove this degeneracy we introduce Epsilon

(E) so that the number of allocations = m+n-1. The position of E is chosen in such a way that it is the least cost and it does not form a closed loop. Phase 2: MODI method

Origin Destination Supply ui 1 2 3 4

O1 7 1 35 20 15 0 9 6 32 0

O2 8

9 4 21 13 1

E 7 7 6

Demand 9 6 7 6 28 vj 7 1 3 20


E.g. for O2 – 2 cell evaluation: 9 – (1+1) = 7. Since all cell evaluations are non-negative, the solution obtained is an optimum solution. Therefore, optimum transportation cost = Rs.223/-

Question 42. Three plants produce identical spare parts to be used in further industrial production in plants 1, 2 and 3. The production in these plants is 100, 50, 50 units respectively. The demand for these in three industries A, B, C is 50, 80, 70 units respectively. The transportation costs in Rs. per unit are as follows: From 1 to A = 100 1 to B = 25 1 to C = 150 From 2 to A = 150 3 to A = 125 2 to B = 100 3 to B = 60 2 to C = 175 3 to C = 130 Find how many parts should be sent from each plant to respective industry so as to minimize the total cost of transport?

Solution:

A B C SUPPLY 1 100 25 150 100 2 150 100 175 50 3 125 60 130 50

DEMAND 50 80 70 200/200 The given transportation problem is a minimization type which is a balanced one. ( as total supply = total demand = 200 PHASE I – Initial Basic Feasible Solution VAM Method

Plant Industries Supply (ai) Penalty A B C 1 100 25 150 100 75/50 20 80 2 150 100 175 50 50/25/25/25 30 20 3 125 60 130 50 65/5/5 50

Demand 50 80 70 200 (bj)

Penalty 25/25/25 35 20/20/45

Total transportation Cost = (20*100) + (80*25) + (30*150) + (175*20) + (130*50) = 2000 + 2000 + 4500 + 3500 + 6500 = Rs. 18500/- RIM condition m + n - 1 3 + 3 – 1 = 5 (equal to no. of allocations) Hence MODI method can be used. PHASE II – MODI method

Plant Industries Supply (ai) ui A B C 1 100 25 150 100 0 20 80 25 2 150 100 175 50 50 30 25 20 3 125 60 130 50 5 20 30 50

Demand 50 80 70 200 (bj) vj 100 25 125

1) u + v = cost for all ALLOCATION CELLS 2) Cell Evaluation CE = Cost – (u + v) for all NON-ALLOCATION CELLS

Since the cell evaluations are positive (non-negative), optimum solution is attained.

Optimum Transportation Cost = Rs. 18500/-

Question 43. A manufacturer has 5 production units and five wholesale depots. The demand at these depots is as follows: W1 W2 W3 W4 W5 80 60 20 210 80 The production units have the following production capacity. P1 P2 P3 P4 150 30 120 130 The transportation cost per unit of product in Rs. from the production units to the depot is as follows: The blank indicates non availability of that route. Find the optimum allocation of units from producing point to wholesale depot.

SOLUTION

STEP 1 – Balanced STEP 2 – Minimization Phase 1 : Initial Basic Feasible Solution 7*30- 210 8*70- 560 4*50- 200 0*30- 0 1*120- 120 0*20- 0 4*60- 240 3*50- 150

W1 W2 W3 W4 W5 P1 7 10 8 8 4 P2 3 16 10 9 0 P3 8 - 5 1 18 P4 3 4 0 - 6

From To Wholesale Depots Production W1 W2 W3 W4 W5 Supply Penalty P1 7 30 10 8 8 70 4 50 150 3/3/3/3/3/3P2 3 16 10 9 0 30 30 3/3/3/-/-/- P3 8 m 5 1 120 8 120 4/7/-/-/-/- P4 3 50 4 60 0 20 m 6 130 3/1/1/1/1/1Dummy 0 0 0 0 20 0 20 0/0/0/0/-/- Demand 80 60 20 210 80 450 Penalty 3/3/3/4/4/4 4/4/4/4/6/6 5/-/-/-

/-/- 1/1/1/8/m-8/-

4/4/4/2/2

0*20- 0 Total = 1480 PHASE 2 – Test for optimality (Modi Method) No. of occupied cells = m + n – 1 9 = 5 + 5 – 1 9 = 9 Therefore RIM condition satisfied All cells are at independent positions. V1=7 V2=8 V3=4 V4=8 V5= 4 U1=0 7

10-(8)

2

8-(4) 4

8

70

4

U2= -4 3-(3)

0

16-(4)

12

10-(0)

10

9-(4)

5

0

U3= -7 8-(0)

8

m-(1)

m-1

5-(-3)

8

1

8-(-3)

11

U4= -4 3

4

0

m-(4)

m-4

6-(0)

6

U5= -8 0-(-1)

1

0-(0)

0

0-(-4)

4

0 0-(-4)

4

All cell evaluations are positive therefore there is optimum solution.

30

60 50

50

30

120

20

20

Question 44. A cement company has three factories which manufacture cement which is then transported to four distinct centres. The quantity of monthly production of each factory, the demand of each distribution centre and the associated transportation cost per quintal are given below:

Factories Distribution Centre Monthly

Production (In Quintals) W X Y Z

A 10 8 5 4 7000

B 7 9 15 8 8000

C 6 10 14 8 10000

Monthly Demand

(In Quintals) 6000 6000 8000 5000

1.) Suggest the optimal transportation schedule.

2.) Is there any other transportation schedule which is equally attractive? If so, write that

Solution

The given Transportation Problem is o Minimization Type which is a Balanced one. Phase 1: To get the initial feasible solution by Voggles Approximation Method

W X Y Z ai

A

10

8

5

4

7000

1

-

-

B

7

9

15

8

8000

1

1

1

C

6

10

14

8

10000

2

2

2

bj

6000

6000

8000

5000

25000

25000

1 1 9 4 1 1 1 0 - 1 1 0

Total Transportation Cost = 5*7000 + 9*6000 + 15*1000 + 8*1000 + 6*6000 + 8*4000 =35000 + 54000 + 15000 + 8000 + 36000 + 32000 =1,80,000 RIM Condition No of allocations = m+n-1 = 6 Therefore, Rim Condition is satisfied.

7000

6000

1000

6000

4000

1000

Phase 2: Modi Method

W X Y Z ai ui

A

10 14

8 9

5

4

6 7000

0

B

7 1

9

15 - θ

8 +θ 8000

10

C

6

10 1

14 +θ

-14

8 -θ

10000

10

bj

6000

6000

8000

5000

25000 25000

vj

-4

-1

5

-2

Since the cell evaluations for C - Y is negative, optimum solution is not attained. θ = Minimum (1000,4000) θ = 1000 Therefore, In cost = 1000 x 1 = 1000 Therefore, New Cost = 1,80,000 – 1,000 = 1,79,00

1000

6000

6000

4000

1000

7000

W X Y Z ai ui

A

10 13

8 8

5

4

5 7000

0

B

7 1

9

15

1

8

8000

9

C

6

10 1

14

8

10000

9

bj

6000

6000

8000

5000

25000 25000

vj

-3

0

5

-1

Since all the cell evaluations are positive, optimum solution is attained. Ans. 1.) Therefore, the Optimum Transportation Cost is Rs. 1,79,000. Ans. 2.) No. There is no other Transportation schedule which is equally attractive. The above solution is unique since all delta ij values are negative, and none equals zero.

7000

6000 2000

6000 3000 1000

Question 45. The following data gives transportation cost per unit of product from origin to destination as well as the supply and demand of the product at each end.

1- obtain initial feasible solution. You may use north-west corner rule. 2- Test the solution for optimality using MODI method.

Solution: The given transportation problem is of minimization type which is a balanced one i.e. total capacity = total demand Phase I: To get initial feasible solution (basic feasible solution)

DESTINATIONS SUPPLY ORIGINS 1 2 3 4 O1 7

1 35 20 15

O2 8 9 4 21 13

DEMAND 9 6 7 6 28

DESTINATIONS SUPPLY ORIGINS 1 2 3 4 O1 7

1 35 20 15/9/-

O2 8 9 4 21 13/6/-

DEMAND 9/- 6/- 7/- 6/- 28

9 6

7 6

TOTAL TRANSPORTATION COST

FROM TO UNITS COST PER UNIT

TOTAL COSTS

O1 1 9 7 63 O1 2 6 1 06 O2 3 7 14 28 O2 4 6 21 126

TOTAL 223 Phase II: Test for optimality M= no of origins N= no of rows m+n = 4 (no of occupied cells) 4+2 = 4

Degeneracy exists

Thus we introduce an epsilon.

Therefore all cell evaluation >= 0 Optimum solution is attained Therefore optimum solution fot transportation cost is Rs. 223/-.

DESTINATIONS SUPPLY Ui ORIGINS 1 2 3 4 O1 7

1 35 20 15/9 0

O2 8 9

21 13 1

DEMAND 9 6 7 6 28 Vj 7 1 3 20

9 6

7 6E

7

32 0

Question 46. A company has four factories situated in different locations in the country and four sales agencies located in four other locations in the country. The cost of production (Rs per unit), the sales price (Rs per unit), shipping cost (Rs per unit), in the cells of the matrix, monthly capacities and monthly requirements are given below

Factory

Sales agency

Monthly Capacity

Cost of

Production

1 2

3

4

A 7

5

6

4

10

10

B 3

5

4

2

15

15

C

4

6

4

5

20

16

D

8

7

6

5

15

15

Monthly Requirement

8 12 18 22

Sales price 20 22 25 18

Find the monthly production and distribution schedule, which will maximize profits (CA, May 1996)

SOLUTION: Using the information given below, we may derive the profit matrix indicating the profit per unit obtainable when produced and sold in various combinations of factories and sales agencies. PROFIT MATRIX:

Factory

Sales agency

Supply

1 2 3 4

A

3

7

9

4

10

B

2

2

6

1

15

C 0 0 5 -3 20

D -3 0 4 -2 15

Initial basic feasible solution – VAM

Factory

Sales Agency

Supply

ui

1

2

3

4

A

6

10 2

0

5

10

0

B

7

7

3

15 8

15

4

C

2 9

9

+

18 4

-

12

20

4

D

6 12

2

- 9

+ 5

7 11

15

7

-1 0 -1

2 -1 1

-3 -4

2

Demand 8 12 18 22 60 vj

5 2 0 4

Opportunity loss matrix : Optimal solution

Factory

Sales Agency

Supply

ui

1 2 3 4

A

6

10 2

0

5

10

0

B

7

7

3

15 8

15

4

C

8 9

9

12 4

12

20

6

D 12

2 9

6 5

7 11

15

7

Demand 8 12 18 22 60

vj 3 2 -2 4

The optimal solution is expressed as follows: From: Factory To: Sales Agency Units Profit A 2 10 70 B 4 15 15 C 1 8 0 3 12 60 D 2 2 0 3 6 24 4 7 (14) TOTAL 155

-3

-2

-1

0 -1 -1

-1 -2

-2

LINEAR PROGRAMMING PROBLEM

INTRODUCTION In a decision‐making embroilment, model formulation is important because it represents the essence of business decision problem. The term formulation is used to mean the process of converting the verbal description and numerical data into mathematical expressions which represents the relevant relationship among decision factors, objectives and restrictions on the use of resources. Linear Programming (LP) is a particular type of technique used for economic allocation of 'scarce' or 'limited' resources, such as labour, material, machine, time, warehouse space, capital, energy, etc. to several competing activities, such as products, services, jobs, new equipment, projects, etc. on the basis of a given criterion of optimally. The phrase scarce resources means resources that are not in unlimited in availability during the planning period. The criterion of optimality, generally is either performance, return on investment, profit, cost, utilily, time, distance, etc.

George B Dantzing while working with US Air Force during World War II, developed this technique, primarily for solving military logistics problems. But now, it is being used extensively in all functional areas of management, hospitals, airlines, agriculture, military operations, oil refining, education, energy planning, pollution control, transportation planning and scheduling, research and development, etc. Even though these applications are diverse, all I.P models consist of certain common properties and assumptions. Before applying linear programming to a real‐life decision problem, the decision‐maker must be aware of all these properties and assumptions, which are discussed later in this chapter.

Before discussing in detail the basic concepts and applications of linear programming, let us be clear about the two words, linear and programming. The word linear refers to linear relationship among variables in a model. Thus, a given change in one variable will always cause a resulting proportional change in another variable. For example, doubling the investment on a certain project will exactly double the rate of return. The word programming refers to modelling and solving a problem mathematically that involves the economic allocation of limited resources by choosing a particular course of action or strategy among various alternative strategies to achieve the desired objective.

• STRUCTURE OF LINEAR PROGRAMMING General Structure of LP Model The general structure of LP model consists of three components. Decision variables (activities): We need to evaluate various alternatives (courses of action) for arriving at the optimal value of objective function. Obviously, if there are no alternatives to select from, we would not need LP. The evaluation of various alternatives is guided by the nature of objective function and availability of resources. For this, we pursue certain activities usually denoted by x1, x2…xn. The value of these activities represent the extent to which each of these is performed. For example, in a product‐mix manufacturing, the management may use LP to decide how many units of each of the product to manufacture by using its limited resources such as personnel, machinery, money, material, etc.

These activities are also known as decision variables because they arc under the decision‐maker's control. These decision variables, usually interrelated in terms of consumption of limited resources, require simultaneous solutions. All decision variables are continuous, controllable and non‐negative. That is, x1>0, x2>0, ....xn>0. The objective function: The objective function of each L.P problem is a mathematical representation of the objective in terms of a measurable quantity such as profit, cost, revenue, distance, etc. In its general form, it is represented as: Optimise (Maximise or Minimise) Z = c1x1 + c2X2. … cnxn where Z is the mcasure‐of‐performance variable, which is a function of x1, x2 ..., xn. Quantities c1, c2…cn are parameters that represent the contribution of a unit of the respective variable x1, x2 ..., xn to the measure‐of‐performance Z. The optimal value of the given objective function is obtained by the graphical method or simplex method. The constraints: There are always certain limitations (or constraints) on the use of resources, e.g. labour, machine, raw material, space, money, etc. that limit the degree to which objective can be achieved. Such constraints must be expressed as linear equalities or inequalities in terms of decision variables. The solution of an L.P model must satisfy these constraints. The linear programming method is a technique for choosing the best alternative from a set of feasible alternatives, in situations in which the objective function as well as the constraints can be expressed as linear mathematical functions. In order to apply linear programming, there are certain requirements to me met. (a) There should be an objective which should be clearly identifiable and measurable in

quantitative terms. It could be, for example, maximisation of sales, of profit, minimisation of cost, and so on.

(b) The activities to be included should be distinctly identifiable and measurable in quantitative terms, for instance, the products included in a production planning problem.

(c) The resources of the system which arc to be allocated for the attainment of the goal should also be identifiable and measurable quantitatively. They must be in limited supply. The technique would involve allocation of these resources in a manner that would trade off the returns on the investment of the resources for the attainment of the objective.

(d) The relationships representing the objective as also the resource limitation considerations, represented by the objective function and the constraint equations or inequalities, respectively must be linear in nature.

(e) There should be a series of feasible alternative courses of action available to the decision makers, which are determined by the resource constraints.

When these stated conditions are satisfied in a given situation, the problem can be expressed in algebraic form, called the Linear Programming Problem (LPP) and then solved for optimal decision. We shall first illustrate the formulation of linear programming problems and then consider the method of their solution.

ADVANTAGES OF LINEAR PROGRAMMING

Following are certain advantages of linear programming. 1. Linear programming helps in attaining the optimum use of productive resources. It also

indicates how a decision‐maker can employ his productive factors effectively by selecting and distributing (allocating) these resources.

2. Linear programming techniques improve the quality of decisions. The decision‐making approach of the user of this technique becomes more objective and less subjective.

3. Linear programming techniques provide possible and practical solutions since there might be other constraints operating outside the problem which must be taken into account. Just because we can produce so many units docs not mean that they can be sold. Thus, necessary modification of its mathematical solution is required for the sake of convenience to the decision‐maker.

4. Highlighting of bottlenecks in the production processes is the most significant advantage of this technique. For example, when a bottleneck occurs, some machines cannot meet demand while other remains idle for some of the time.

5. Linear programming also helps in re‐evaluation of a basic plan for changing conditions. If conditions change when the plan is partly carried out, they can be determined so as to adjust the remainder of the plan for best results.

LIMITATIONS OF LINEAR PROGRAMMING

In spite of having many advantages and wide areas of applications, there arc some limitations associated with this technique. These are given below. Linear programming treats all relationships among decision variables as linear. However, generally, neither the objective functions nor the constraints in real‐life situations concerning business and industrial problems are linearly related to the variables.

1. While solving an LP model, there is no guarantee that we will get integer valued solutions. For example, in finding out how many men and machines would be required lo perform a particular job, a non‐integer valued solution will be meaningless. Rounding off the solution to the nearest integer will not yield an optimal solution. In such cases, integer programming is used to ensure integer value to the decision variables.

2. Linear programming model does not take into consideration the effect of time and uncertainty. Thus, the LP model should be defined in such a way that any change due to internal as well as external factors can be incorporated.

3. Sometimes large‐scale problems can be solved with linear programming techniques even when assistance of computer is available. For it, the main problem can be fragmented into several small problems and solving each one separately.

4. Parameters appearing in the model are assumed to be constant but in real‐life situations, they are frequently neither known nor constant.

It deals with only single objective, whereas in real‐life situations we may come across conflicting multi‐objective problems. In such cases, instead of the LP model, a goal programming model is used to get satisfactory values of these objectives.

APPLICATION AREAS OF LINEAR PROGRAMMING

Linear programming is the most widely used technique of decision‐making in business and Industry and in various other fields. In this section, we will discuss a few of the broad application areas of linear programming.

Agricultural Applications

• These applications fall into categories of farm economics and farm management. The former deals with agricultural economy of a nation or region, while the latter is concerned with the problems of the individual farm.

• The study of farm economics deals with inter‐regional competition and optimum allocation of crop production. Efficient production patterns can be specified by a linear programming model under regional land resources and national demand constraints.

• Linear programming can be applied in agricultural planning, e.g. allocation of limited resources such as acreage, labour, water supply and working capital, etc. in a way so as to maximise net revenue.

Military Applications

• Military applications include the problem of selecting an air weapon system against enemy so as to keep them pinned down and at the same time minimising the amount of aviation gasoline used. A variation of the transportation problem that maximises the total tonnage of bombs dropped on a set of targets and the problem of community defence against disaster, the solution of which yields the number of defence units that should be used in a given attack in order to provide the required level of protection at the lowest possible cost.

Production Management

• Product mix A company can produce several different products, each of which requires the use of limited production resources. In such cases, it is essential to determine the quantity of each product to be produced knowing its marginal contribution and amount of available resource used by it. The objective is to maximise the total contribution, subject to all constraints.

• Production planning This deals with the determination of minimum cost production plan over planning period of an item with a fluctuating demand, considering the initial number of units in inventory, production capacity, constraints on production, manpower and all relevant cost factors. The objective is to minimise total operation costs.

• Assembly‐line balancing This problem is likely to arise when an item can be made by assembling different components. The process of assembling requires some specified sequcnce(s). The objective is to minimise the total elapse time.

• Blending problems These problems arise when a product can be made from a variety of available raw materials, each of which has a particular composition and price. The objective here is to determine the minimum cost blend, subject to availability of the raw materials, and minimum and maximum constraints on certain product constituents.

• Trim loss When an item is made to a standard size (e.g. glass, paper sheet), the problem that arises is to determine which combination of requirements should be produced from standard materials in order to minimise the trim loss.

Financial Management • Portfolio selection This deals with the selection of specific investment activity

among several other activities. The objective is to find the allocation which maximises the total expected return or minimises risk under certain limitations.

• Profit planning This deals with the maximisation of the profit margin from investment in plant facilities and equipment, cash in hand and inventory.

Marketing Management

• Media selection Linear programming technique helps in determining the advertising media mix so as to maximise the effective exposure, subject to limitation of budget, specified exposure rates to different market segments, specified minimum and maximum number of advertisements in various media.

• Travelling salesman problem The problem of salesman is to find the shortest route from a given city, visiting each of the specified cities and then returning to the original point of departure, provided no city shall be visited twice during the tour. Such type of problems can be solved with the help of the modified assignment technique.

• Physical distribution Linear programming determines the most economic and efficient manner of locating manufacturing plants and distribution centres for physical distribution.

Personnel Management

• Staffing problem Linear programming is used to allocate optimum manpower to a particular job so as to minimise the total overtime cost or total manpower.

• Determination of equitable salaries Linear programming technique has been used in determining equitable salaries and sales incentives.

• Job evaluation and selection Selection of suitable person for a specified job and evaluation of job in organisations has been done with the help of linear programming technique.

Other applications of linear programming lie in the area of administration, education, fleet utilisation, awarding contracts, hospital administration and capital budgeting, etc.

METHODS OF SOLUTION: Solving an LP problem involves:

• Selection of appropriate method of solution and • Then obtain a solution to the problem with the help of selected method • Test whether this solution is optimal.

The problem can be solved by using: (1) Graphical method: This method can be used if there are only t• decision variables in the

LPP. (2) Simplex method: This method is useful in solving LP problems with two or more than two decision variables.

The Graphical method of solution:

This method can be used in case where LPP has only two decision variables. But there is no restriction on the number of constraints. The method uses the familiar graphical presentation with two axes. The method becomes unwieldy when there are three variables since we then need a three dimensional graph. The method cannot be used if the number of decision variables is more than three. In such a case we have to use a non graphical method to obtain a solution. The graphical method of solution to L.P. problem uses all the equations in a given problem, namely the equation expressing objective unction the constraints imposed in achieving the objective. These constraints can be of (i) greater than (ii) less than or (iii) strict equality type. There is also a non‐negativity restriction on the values of the decision variables. It implies that the solution of the problem lies in the first quadrant of the graph. All these relations are linear.

SPECIAL CASES IN LPP 1. Infeasibility 2. Unboundedness 3. Redundancy 3. Alternate optima (Alternate optimum solution)

INFEASIBILITY It is a case where there is no solution, which satisfies all the constraints at the same time. This may occur if the problem is not correctly formulated. Graphically, infeasibility is a case where there is no region, which satisfies all constraints simultaneously. UNBOUNDEDNESS

A LPP can fail to have an optimum solution if the objective can be made infinitely large without violating any of the constraints. If we come across unboundedness in solving real problems, then the problem is not correctly formulated. Since, no real situation permits any management to have infinite production of goods and infinite profits, unbounded solution results if in a maximization problem all constraints are of greater than or equal to type. In such a situation there is no upper limit on feasible region. Similarly, an unbounded solution occurs in a minimization problem if all constraints are of less than or equal to type.

REDUNDANCY

A constraint, which does not affect the feasible region, is called a redundant constraint. Such a constraint is not necessary for the solution of the problem. It can therefore be omitted while formulating the problem. This will save the computation time. In many LP problems, redundant constraints are not recognized as being redundant until the problem is solved. However, when computers are used to solve LPP, redundant constraints do not cause any difficulty.

ALTERNATIVE OPTIMA

[The slope of a line ax + by + c = 0 is defined as –b ]

The solution to a LPP shall always be unique if the slope of the objective function line is different from the slope of all of the constraint lines. Incase, the slope of objective function line is same as the slope of one of its constraint line, then multiple optimum solution might exist.

GRAPHICAL SOLUTION Extreme point enumeration approach Convex Polyhedron TYPES OF SOLUTION (a) Solution. Values of decision variables xj (j = 1, 2, 3, ….n) which satisfy the constraints of the

general L. P. P., is called the solution to that L. P. P. (b) Feasible solution. Any solution that also satisfies the nonnegative restrictions of the general

L. P. P. is called a feasible solution. (c) Basic Solution. For a set of m simultaneous equations in n unknowns (n> m). a

solution obtained by setting (n ‐ m) of the variables equal to zero and solving the remaining m equations in m unknowns is called a basic solution. Zero variables (n ‐ m) are called non‐basic variables and remaining m are called basic variables and constitute a basic solution.

(d) Basic Feasible Solution. A feasible solution to a general L.P.P. which is also basic solution is called a basic feasible solution.

(e) Optimum Feasible Solution. Any basic feasible solution which optimizes (maximizes or minimizes) the objective function of a general L.P.P. is known as an optimum feasible solution to that L.P.P.

(f) Degenerate Solution. A basic solution to the system of equations is called degenerate if one or more of the basic variables become equal to zero.

a

THE SIMPLEX METHOD OF SOLUTION:

The simplex method uses a simplex algorithm; which is an iterative, procedure for finding, in a systematic manner the optimal solution to a linear programming problem. The procedure is based on the observation that if a feasible solution to a linear programming exists; it is located at a corner point of the feasible region determined by the constraints of the problem. The simplex method, selects the optimal solution from among the set of feasible solution to the problem. The algorithm is very efficient as it considers only those feasible solutions, which are provided by the corner points. Thus, we need to consider a minimum number of feasible solutions to obtain an optimal one. The method is quite simple and the first step requires the determination of basic feasible solution. Then, with the help of a limited number of steps the optimum solution can be determined. Terminology of Simplex Method: Algorithm: A formalised systematic procedure for solving problem. Simplex Tableau: A table used to keep track of the calculations made b iteration of the simplex procedure and to provide basis for tableau revision.

Basis: The set of basic variables which are not restricted to zero in the basic solution and are listed in solution column.

The basic variables: The variables with non‐zero positive values make up the basis are called basic variables and the remaining variables are called non‐basic variables.

Iteration: A sequence of steps taken in moving from one basic. The solution to another basic feasible solution.

Cj row: A row in the simplex tableau which contains the co‐efficients variables in the objective function.

Zj row: A row in the simplex tableau whose elements represent the decrease (increase) of the value of the objective function if one unit of the jth variable is brought into the solution.

Cj ‐ Zj or j row: A row whose elements represent net per unit contribution of the jth variable in the objective function, if the variable is brought into the new basic solution. Positive value of j therefore indicates gain and negative value indicates loss in the total value Z obtained of the objective function. Key or pivot column: The column with the largest positive j and it indicates which variable will enter the next solution in a maximization case.

Key or pivot row: The row with the smallest positive value of the, replacement ratio 0 of the constraint rows. The replacement ratio is obtained by dividing elements in the solution column by the corresponding elements in the key column. The key row indicates the variable that will leave the basis to make room for new entering variable.

Key (pivot) element: The element at the intersection of key row and key column. In addition to these terms in a simplex tableau we have the follow terms which are necessary to make a linear programming problem fit to be solved by simplex method. Slack variable: A variable used to convert a less than or equal constraint () into equality

constraint. It is added to the left hand side of the constraint. Surplus variable: A variable used to convert a greater than or equal to (?) constraint into

equality constraint. It is subtracted from the left hand side of the constraint. Artificial variable: It is a variable added to greater than or type () constraint. This is in

addition to surplus variables used.

SOME TECHNICAL ISSUES: In the earlier chapter, we considered some special problems encountered in solving LPP using graphical method. Here we discuss, how the presence of these problems ‐ namely, infeasibility, unboundedness, multiple solutions, degeneracy, is indicated in a simplex tableau. Infeasibility: A solution is called feasible if it satisfies all the constraints and the non‐

negativity conditions. Sometimes it is possible that the constraints may be inconsistent so that there is no feasible solution to the problem. Such a situation is called infeasibility. In a graphical solution, the infeasibility is evident when there is no feasible region in which all the constraints can be satisfied simultaneously. However, problem involving more than two variables cannot be easily graphed and it may not be immediately known that the problem is infeasible, when the model is constructed. The simplex method provides information as to where the infeasibility lies. If the simplex algorithm terminates with one or more artificial variables at a positive value, then there is no feasible solution to the original problem.

Unboundedness: It occurs when there are no constraints on the solution. So that one or more of the decision variables can be increased indefinitely without violating any of the restrictions. Graphically the objective function line can be moved in the desired direction over the feasible region, without any limits. How do we recognize unboundedness in a simplex method? We know the replacement ratio determines the leaving variable in a simplex tableau. Now if there are no non‐negative ratios (i.e. ratios are negative) or they are equal to i.e. of the type say 60/0, then we have unbounded solution.

Alternative Optima: (Multiple optimum solution)

A solution to a linear programming problem may or may not be unique. This is indicated in a graphical solution by the slope of the line of the objective function which may coincide with the slope of one of the constraints.

In case of simplex method, Whenever a non basic variable (i.e. a variable which is not in the solution ) has a zero value in the j (i.e. cj j – zj j) row of an optimal tableau then bringing that variable into the solution will produce a solution which is also optimal.( Alternative Optimal solution )

Degeneracy: It occurs when one or more of the basis variables assume zero value. In

conditions of degeneracy, the solution would contain a smaller number of non – zero variables than the number of constraints i.e. if there are 3 constraints the number of non‐zero variables in the solution is less than 3.

Some obvious examples of degeneracy occur if: • One or more basic variable have a zero value in the optimal solution. • There is a tie in the replacement ratios for determining the leaving variable. The next

tableau gives the degenerate solution. • When algorithm pivots in a degenerate row, the objective function value in the next

tableau does not change i.e. there is the problem of cycling. ‐ The system moves along the same route and the cycle would be repeated forever. There are sophisticated rules

to handle the problem of cycling; however, they are outside the scope of this book. It is also observed that real life problems rarely cycle.

Explain the summary procedure for the maximization case of the simplex method.

Step 1 Formulate the problem

Translate the technical specifications of the problems into inequalities, and make a precise statement of the objective function.

Convert the inequalities into equalities by the addition of nonnegative slack variables. These inequalities should be symmetric or balanced so that each slack variable appears in each equation with a proper coefficient.

Modify the objective function to include the slack variables. Step 2 Design an initial program (A basic feasible solution)

Calculate the net evaluation row: To get a number in the net evaluation row under a column, multiply the entries in that column by the corresponding numbers in the objective column, and add the products. Then subtract this sum from the number listed in the objective row at the top of the column. Enter the result in the net evaluation row under the column.

Test : Examine the entries in the net evaluation row for the given simplex tableau. If all the zero or negative, the optimal solution has been obtained. Otherwise, the presence of any positive entry in the net‐evaluation row indicates that a better program can be obtained.

Step 3 Revise the program

Find the key column. The column under which falls the largest positive net‐evaluation‐ row entry is the key column.

Find the key row and the key number. Divide the entries in the “quantity” column by the corresponding nonnegative entries of the key column to form replacement ratios, and compare these ratios. The row in which falls the smallest replacement ratio is the key row. The number which lies at the intersection of the key row and the key column is the key number.

Transform the key row. Divide all the numbers in the key row (starting with and to right of the “quantity” column) by the key number. The resulting numbers form the corresponding row of the next tableau.

Transform the non‐key rows. Subtract from the old row number of a given key row (in each column ) the product of the corresponding key‐row number and the corresponding fixed ratio formed by dividing the old row number in the key column by the key number. The result will give the corresponding new row number. Make this transformation for all the non‐key rows.

Enter the results of (3) and (4) above in a tableau representing the revised program. Step 4 Obtain the optimal program

Repeat steps 3 & 4 until a program has been derived.

[ Linear‐programming problems involving the minimization of an objective function usually contain structural of the “greater than equal to” type. They can also be solved by the simplex method. The simplex procedure for solving a linear‐programming problem in which the objective is to minimize rather than maximize a given function, although basically the same as above, requires sufficient modifications to deserve the listing of a separate summary. ]

Summary procedure for the simplex method (minimization case)

Step 1 Formulate the problem

Translate the technical specification of the problem into inequalities, and make a precise statement of the objectivity function.

Convert the equalities into inequalities by the subtraction of nonnegative slack variables. Then modify these equations by the addition of nonnegative artificial slack variables. These equations should be asymmetric or balanced so that each slack and artificial slack variable appears in each equation with a proper coefficient.

Modify the objective functions to include all the slack artificial slack variables. Step 2 Design an initial program (a basic feasible solution).

Design the first program so that only the artificial slack variables are included in the solution. Place the program in a simplex tableau. In the objective row, above each column variable, place the corresponding coefficient of that variable from step 1.c.In particular, place a zero above each column containing an artificial slack variable.

Step 3 Test and revise the program.

Calculate the net evaluation row. Toto get a number in the net evaluation row under a column, multiplying the entries in that column by the corresponding number in the objective column, and add the products. Then subtract the sum from the number listed in the objective row above the column. Enter the result in the net‐evaluation row under the column.

Test. Examine the entries in the net‐evaluation row for the given simplex tableau. If all the entries are zero or positive, the optimum solution has been obtained. Otherwise, the presence of any negative entry in the net‐evaluation row indicates that a better program can be obtained.

Revise the program. Find the key column. The under which falls the largest negative net‐evaluation‐entry is the key column.

Find the key row and the key number. Divide the entries in the “Quantity” column by the corresponding nonnegative entries in the key column to form replacement ratios, and compare these ratios. The row in which the smallest replacement ratio falls is the key row. The number which lies at the intersection of the key row and the key column is the key number.

Transform the key row. Divide all the numbers in the key row (staring with and to the right to the of the “Quantity” column by the key number. The resulting numbers form the corresponding row of the next tableau.

Transform the non‐key rows. Subtract from the old row the number of a given non‐key row (in each column) the product of the corresponding fixed ratio formed by dividing the old row number in the key column by the number. The result will give the corresponding new row number. Make the transformation for all rows.

Enter the results of (3) and (4) above in a tableau representing the revised program. Step 4 Obtain the optimal program.

Repeat steps 3and 4 until an optimal program has been derived.

We repeat the following comments comparing the maximization and minimization problems as solved by the simplex method.

The procedure for calculating the net‐evaluation row is the same in both cases. However, whereas the largest positive value is chosen to identify the incoming product in a maximization problem, the most negative value is chosen in a minimization problem. The rest of the mechanics, namely, the transformation of the key and the non‐key rows, is exactly the same. The decision rule identifying the optimal solution is the absence of any positive value in the non‐evaluation row in the maximization problem, and the absence of any negative value in the minimization problem.

DUAL SIMPLEX In ordinary simplex method we start with feasible but non‐optimal solution while in Dual Simplex Method, we start with infeasible but optimal solution. Successive iterations will maintain optimality to remove infeasibility of the solution. The following steps are followed to arrive at optimal feasible solution.

(1) Write down objective function as maximization and all constraints as ≤ or =. (2) Construct first dual simplex table from the given problem in usual manner. (3) The leaving variable is the basic variable having the most negative value (Break ties, if

any, arbitrarily). If all the basic variables are non‐negative, the process ends and the feasible (optimal) solution is reached.

(4) To determine the entering variable take ratios of the coefficients of non‐basic variables in the objective function to the corresponding coefficients in the row associated with the leaving variable. Ignore the ratios with positive or zero denominators. The entering variable is the one with the smallest absolute value of the ratio. (Break ties, if any, arbitrarily). If all the denominators are zero or positive, the problem has no feasible solution.

After selecting the entering and leaving variables, row operations are applied as usual to obtain the next table.

Application of this Dual Simplex Method is useful in Sensitivity Analysis. For example, suppose a new constraint is added to the problem after the optimal solution is reached. If this constraint is not satisfied by the optimal solution, the problem remains optimal but it becomes infeasible. The Dual Simplex Method is then used to clear the infeasibility in the problem.

Example : Minimize Z = 2x1 + x2 subject to 3x1 + x2 ≥ 3, 4xx + 3x2 ≥ 6,

x1 + 2x2 ≤ 3, x1, x2 ≥ 0.

CPM AND PERT CRITICAL PATH ANALYSIS

• Looping: Normally in the network, the arrow points from left to right. This convention is to be strictly adhered, as this would avoid the illogical looping, as shown wrongly below.

• Dangling: The situation represented by the following diagram is also at fault, since the activity represented by the dangling arrow 9‐11 is undertaken with no result.

To overcome problems arising due to dangling arrows, we must make sure that

All events except the first and the last must have at least one activity entering and one activity leaving them.

All activities must start to finish within event.

• The critical path determination After having computed various time estimates, we are now interested in finding the critical Path of the network. A network will consist of a number of parts. A path is a continuous series of activities through the network that leads from the initial event (or node) of the network to its terminal event. For finding the critical Path, we list out all possible paths through a network along with their duration. In the network under consideration, various paths have been listed as follows: Path length in days 1‐2‐3‐5‐6 36 1‐2‐4‐5‐6 52 1‐2‐3‐4‐5‐6 50

1 2 3 4

6 7

8

9

Critical Path: a path in a project network is called critical if it is the longest Path. The activities lying on the critical Path are called the critical activities. In the above example, the Path 1‐2‐4‐5‐6 with the longest duration of 52 days is the critical Path and the activities 1‐2,2‐4, 4‐5 and 5‐6 are the critical activities. Calculation of floats It may be observed that for every critical activities in a network, the earliest start and latest start times are the same. This is so since the critical activities cannot be scheduled later than the earliest scheduled time without delaying the total project duration, they do not have any flexibility in scheduling. However, noncritical activities do have some flexibility. That is these activities can be delayed for sometime without affecting the project duration. This flexibility is termed as slack in case of an event and as floats in case of an activity. Some people do not make any distinction between a slack and a float. Slack time for an event

• The slack time or slack of an event in a network is the difference between the latest event time and the earliest event time.

• Mathematically it may be calculated using the formula Li – Ei where Li is the latest allowable occurrence time and Ei is the earliest allowable occurrence time of an event i.

Total float of an activity

• The total activity float is equal to the difference between the earliest and latest allowable start or finish times for the activity in question. Thus, for an activity (i‐j), the total float is given by:

• TFij = LST – EST or TFij = LFT – EFT • In other words, it is the difference between the maximum time available for the activity

and the actual time it takes to complete. Thus, total float indicates the amount of time by which the actual completion of an activity can exceed its earliest expected completion time without causing any delay in the project duration.

Free float

• It is defined as that portion of the total float within which an activity can be manipulated without affecting the float of the succeeding activities. It can be determined by subtracting the head event slack from the total float of an activity.

• i.e. FFij = TFij – (slack of event j) • The free float indicates the value by which an activity in question can be delayed beyond

the earliest starting point without affecting the earliest start, and therefore the total float of the activities following it.

Independent float

• It is defined as that portion of the total float within which an activity can be delayed for start without affecting float of the preceding activities. It is computed by subtracting the tail event slack from the free float.

• i.e. IFij = FFij – (slack of event i)

• The independent float is always either equal to a less than the free float of an activity. If a negative value is obtained, the independent float is taken to be 0.

Interfering float

• Utilisation of the float of an activity can affect the float of subsequent activities in the network. Thus, interfering float can be defined as that part of the total float which causes a reduction in the float of the successor activities. In other words, it can be defined as the difference between the latest finish time of the activity under consideration and the earliest start time of the following activity, or 0, whichever is larger. Thus, interfering float refers to that portion of the activity float which cannot be consumed without affecting adversely the float of the subsequent activity or activities.

Distinctions between PERT and CPM The PERT and CPM models are similar in terms of their basic structure, rationale and mode of analysis. However, there are certain distinctions between pert and CPM networks which are ennumerated below.

1. CPM is activity oriented that is CPM network is built on the basis of activities. Also results of various calculations are considered in terms of activities of the project. On the other hand, PERT is event oriented.

2. CPM is a deterministic model that is it does not take into account the uncertainties involved in the estimation of time for execution of a job or an activity. It completely ignores the probabilistic element of the problem. Pert, however is the probabilistic model. It uses three estimates of the activity time; optimistic, pessimistic and most likely; with a view to take into account time uncertainty. Thus, the expected duration of each activity is probabilistic and expected duration indicates that there is 50% probability of getting the job done within that time.

3. CPM places dual emphasis on time and cost and evaluates the trade‐off between project cost and project time. By deploying additional resources, it allows the critical path project manager to manipulate project duration within certain limits so that project duration can be shortened at an optimal cost. On the other hand, pert is primarily concerned with time. It helps the manager to schedule and coordinate various activities so that the project can be completed on schedule time.

4. CPM is commonly used for those projects which are repetitive in nature and where one has prior experience of handling similar projects. What is generally used for those projects with time required to complete various activities are not known before hand. Thus, pert is widely used for planning and scheduling research and development projects.

PROJECT CRASHING In some cases, there are compelling reasons to complete a project earlier than the originally estimated time duration of the critical path computed on the basis of normal activity times, by employing extra resources. An example would be introduction of a new project. The motives in hastening the project might be to ensure that the competitors do not steal a march. Increase or decrease in the total duration of the completion time for project is closely associated with cost considerations. In such cases when the total time duration is reduced, the project cost increases, but in some exceptional cases project cost is reduced as well. Production cost occurs in the cases of those projects which make use of a certain type of resources for example a machine and whose time is more valuable than the operator’s time. Some definitions:

• Activity cost: it is defined as the cost of performing and completing a particular activity or task.

• Crash cost, Cc: this is the direct cost that is anticipated in completing an activity within

the crash time.

• Crash time, Ct: This is the minimum time required to complete an activity.

• Normal cost Nc: this is the lowest possible direct cost required to complete an activity.

• Normal time Nt: this is the minimum time required to complete an activity at normal cost.

• Activity costs slope: the costs slope indicates the additional cost incurred per unit of

time saved in reducing the duration of an activity.

Let OA represent the normal duration of completing a job and OC the normal cost involved to complete the job. Assume that the management wish to reduce the time of completing the job to OB from normal time OA. Therefore under such a situation the cost of the project increases and it goes upto say OD (Crash Cost). This only amounts to saving that by reducing the time period by BA the cost has increased by the amount CD. The rate of increase in the cost of activity per unit decrease in time is known as cost slope and is described as follows.

Activity cost slope = OBOAOCOD

ABCD

−−

=

= CrashtimeNormaltime

NormalCosttCrash−

−cos

Crash cost

Normalcost

Crash Time

Normal Time

COST

DURATION FOR THE JOB

D E

C

F

ABO

Optimum duration: the total project cost is the sum of the direct and indirect costs. In case the direct cost varies with the project duration time, the total cost would have the shape as indicated in the above figure.

At Point A, the cost will be minimum. The time corresponding to this point Point A is called the optimum duration and the cost as optimum cost for the project.

COST

A

O

TOTAL PROJECT COST

DIRECT COST

INDIRECT COST

Crash Normal Optimal

TIME

TRANSPORTATION & ASSIGNMENT

What is an Assignment Problem?

• The assignment problem can be stated as a problem where different jobs are to be assigned to different machines on the basis of the cost of doing these jobs. The objective is to minimize the total cost of doing all the jobs on different machines

• The peculiarity of the assignment problem is only one job can be assigned to one machine i.e., it should be a one‐to‐one assignment

• The cost data is given as a matrix where rows correspond to jobs and columns to machines and there are as many rows as the number of columns i.e. the number of jobs and number of Machines should be equal

• This can be compared to demand equals supply condition in a balanced transportation problem. In the optimal solution there should be only one assignment in each row and columns of the given assignment table. one can observe various situations where assignment problem can exist e.g., assignment of workers to jobs like assigning clerks to different counters in a bank or salesman to different areas for sales, different contracts to bidders.

• Assignment becomes a problem because each job requires different skills and the capacity or efficiency of each person with respect to these jobs can be different. This gives rise to cost differences. If each person is able to do all jobs equally efficiently then all costs will be the same and each job can be assigned to any person.

• When assignment is a problem it becomes a typical optimization problem it can therefore be compared to a transportation problem. The cost elements are given and is a square matrix and requirement at each destination is one and availability at each origin is also one.

• In addition we have number of origins which equals the number of destinations hence the total demand equals total supply . There is only one assignment in each row and each column .However If we compare this to a transportation problem we find that a general transportation problem does not have the above mentioned limitations. These limitations are peculiar to assignment problem only.

2) What is a Balanced and Unbalanced Assignment Problem? A balanced assignment problem is one where the number of rows = the number of columns

(comparable to a balanced transportation problem where total demand =total supply)

Balanced assignment problem: no of rows = no of columns

Unbalanced assignment is one when the number of rows not equal to the number of columns and vice versa. e.g. The number of machines may be more than the number of jobs or the number of jobs may be more than the number of machines. In such a situation we introduce dummy row/column(s) in the matrix. These rows or columns have a zero cost element. Thus we can balance the problem and then use Hungarian method to find optimal assignment.

Unbalanced assignment problem: no of rows not equal to no of columns

3) What is a Prohibited Assignment Problem? A usual assignment problem presumes that all jobs can be performed by all individuals there can be a free or unrestricted assignment of jobs and individuals. A prohibited assignment problem occurs when a machine may not be in, a position to perform a particular job as there be some technical difficulties in using a certain machine for a certain job. In such cases the assignment is constrained by given facts. To solve this type problem of restriction on job assignment we will have to assign a very high cost M This ensures that restricted or impractical combination does not enter the optimal assignment plan which aims at minimization of total cost. 4) What are the methods to solve an Assignment Problem (Hungarian Method)? There are different methods of solving an assignment problem:

DIFFERENT METHODS OF ASSIGNMENT PROBLEM

Hungarian method

Simplex Method

Transportation Problem Complete

Enumeration

1)Complete Enumeration Method: This method can be used in case of assignment. problems of small size. In such cases a complete enumeration and evaluation of all combinations of persons and jobs is possible. One can select the optimal combination. We may also come across more than one optimal combination The number of combinations increases manifold as the size of the problem increases as the total number of possible combinations depends on the number of say, jobs and machines. Hence the use of enumeration method is not feasible in real world cases. 2) Simplex Method: The assignment problem can be formulated as a linear programming problem and hence can be solved by using simplex method.However solving the problem using simplex method can be a tedious job. 3)Transportation Method: The assignment problem is comparable to a transportation problem hence transportation method of solution can be used to find optimum allocation. Howver the major problem is that allocation degenerate as the allocation is on basis one to one per person per person per job Hence we need a method specially designed to solve assignment problems. 4 )Hungarian Assignment Method (HAM): This method is based on the concept of opportunity cost and is more efficient in solving assignment problems. Method in case of a minimization problem. As we are using the concept “opportunity this means that the cost of any opportunity that is lost while taking a particular decision or action is taken into account while making assignment. Given below are the steps involved to solve an assignment problem by using Hungarian method.

Step 1:

Determine the opportunity cost table

Step 2:

Determine the possibility of an optimal assignment

Step 3

Modify the second reduced cost table

Step 4:

Make the optimum assignment

Step 1: Determine the opportunity cost tableI

• Locate the smallest cost in each row and subtract it from each cost figure in that row. This would result in at least one zero in each row. The new table is called reduced cost table.

• Locate the lowest cost in each column of the reduced cost table subtract this figure from each cost figure in that column. This would result in at least one zero in each row and each column, in the second reduced cost table.

Step 2: Determine the possibility of an optimal assignment:

• To make an optimal assignment in a say 3 x 3 table. We should be in a position to locate 3 zero’s in the table. Such that 3 jobs are assigned to 3 persons and the total opportunity cost is zero .A very convenient way to determine such an optimal assignment is as follows:

• Draw minimum number of straight lines vertical and horizontal, to cover all the zero elements in the second reduced cost table. One cannot draw a diagonal straight line. The aim is that the number of lines (N) to cover all the zero elements should be minimum. If the number of lines is equal to the number of rows (or columns) (n) i.e N=n it is possible to find optimal assignment .

• Example :for a 3 x 3 assignment table we need 3 straight lines which cover all the zero elements in the second reduced cost table. If the number of lines is less than the number of rows (columns) N < n optimum assignment cannot be made. we then move to the next step.

Step 3: Modify the second reduced cost table:

• Select the smallest number in the table which is not covered by the lines. Subtract this number from all uncovered numbers aswell as from itself.

• Add this number to the element which is at the intersection of any vertical and horizontal lines.

• Draw minimum number of lines to cover all the zeros in the revised opportunity cost table.

• If the number of straight lines at least equals number of rows (columns) an optimum assignment is possible.

Step 4: Make the optimum assignment: If the assignment table is small in size it is easy to make assignment after step 3. However, in case of large tables it is necessary to make the assignments systematically. So that the total cost is minimum. To decide optimum allocation.

• Select a row or column in which there is only one zero element and encircle it Assign the

job corresponding to the zero element i.e. assign the job to the circle with zero element. Mark a X in the cells of all other zeros lying in the column (row) of the encircled zero. So that these zeros cannot be considered for next assignment.

• Again select a row with one zero element from the remaining rows or columns. Make the next assignment continue in this manner for all the rows.

• Repeat the process till all the assignments are made i.e. no unmarked zero is left. • now we will have one encircled zero in each row and’ each column of the cost matrix.

The assignment made in this manner is Optimal. • Calculate the total cost of assignment from the original given cost table.

Maximization method In order to solve a maximization type problem we find the regret values instead of opportunity cost. the problem can be solved in two ways

• The first method is by putting a negative sign before the values in the assignment matrix and then solves the sum as a minimization case using Hungarian methods as shown above.

• Second method is to locate the largest value in the given matrix and subtract each element in the matrix from this value. Then one can solve this problem as a minimization case using the new modified matrix.

Hence there are mainly four methods to solve assignment problem but the most efficient and most widely used method is the Hungarian method Q5 ) Note on Traveling Salesmen problem. Traveling salesman problem is a routine problem. It can be considered as a typical assignment problem with certain restrictions. Consider a salesman who is assigned the job of visiting n different cities. He knows (is given) the distances between all pairs of cities. He is asked to visit each of the cities only once. The trip should be continuous and he should come back to the city from where he started using the shortest route. It does not matter, from which city he starts. These restrictions imply.

(1) No assignment should be made along the diagonal. (2) No city should be included on the route more than once

This type of problem is quite simple but there is no general algorithm available for its solution. The problem is usually solved by enumeration method, where the number of enumerations is very large. For example for a salesman who is instructed to visit five cities we shall have to consider more than 100 possible routes. The method is therefore impractical for large size problems and it also implies approximations in finding route with minimum distance. The peculiar nature of the problem and the various restrictions imposed on resulting solution indicate that the method of solution to a traveling salesman problem should include: (1) Assigning an infinitely large element M in the diagonal of the distance matrix. (2) Solve the problem using Hungarian Method as it gives shortcut route but,

(3) Test the solution for feasibility whether it satisfies the condition of a continues route without visiting a city more than once. If the route is not feasible, make adjustments with minimum increase in the total distance traveled by the salesman. This is how one can solve traveling sales man problem 6) What is a Transportation Problem?

• A transportation problem is concerned with transportation methods or selecting routes in a product distribution network among the manufacturing plants and distribution warehouses situated in different regions or local outlets.

• In applying the transportation method, management is searching for a distribution route, which can lead to minimization of transportation cost or maximization of profit.

• The problem involved belongs to a family of specially structured LPP called network flow problems.

7) What is a Balanced and Unbalanced Transportation Problem? Balanced transportation problem Transportation problems that have the supply and demand equal is a balanced transportation problem. In other words requirements for the rows must equal the requirements for the columns. Unbalanced transportation problem An Unbalanced transportation problem is that in which the supply and demand are unequal. There are 2 possibilities that make the problem unbalanced which are

(i) Aggregate supply exceeds the aggregate demand or (ii) Aggregate demand exceeds the aggregate supply.

Such problems are called unbalanced problems. It is necessary to balance them before they are solved.

Aggregate supply exceeds the

aggregate demand

Aggregate demand exceeds the

aggregate supply

2 Possibilities That Make The Problem Unbalanced

Balancing the transportation problem

• Where total Supply exceeds total demand. In such a case the excess supply is, assumed to go to inventory and costs nothing for shipping(transporting). This type of problem is balanced by creating a fictitious destination. This serves the same purpose as the slack variable in the simplex/method A column of slack variables is added to the transportation tableau which represents a dummy destination with a requirement equal to the amount of excess supply and the transportation cost equal to zero. This problem can now be solved using the usual transportation methods.

• When aggregate demand exceeds aggregate supply in a transportation problem When aggregate demand exceeds aggregate supply in a transportation problem a dummy row is added to restore the balance. This row has an availability equal to the excess demand and each cell of this row has a zero transportation cost per unit. Once the problem is balanced it can be solved by the procedures normally used to solve a transportation problem. 8) What is a Prohibited Transportation Problem? Sometimes in a given transportation problem some route(s) may not be available. This could be due to a variety of reasons like‐

(i) Strikes in certain region (ii) Unfavorable weather conditions on a particular route (iii) Entry restriction. In such situations there is a restriction on the routes available for transportation. To

overcome this difficulty we assign a very large cost M or infinity to such routes. When a large cost is added to these routes they are automatically eliminated form the solution. The problem then can be solved using usual methods.

REASONS FOR UNAVAILABILITY OF ROUTES

Strikes in certain

region

Unfavorable weather

conditions on a

particular route

Entry restriction.

9) What is Degeneracy in a Transportation Problem? The initial basic feasible solution to a transportation problem should have a total number of occupied cell (stone squares) which is equal to the total number of rim requirements minus one i.e. m + n — 1. When this rule is not met the solution is degenerate. Degeneracy may occur‐ If the number of occupied cells is more than m + n — 1.

This type of degeneracy arises only in developing the initial solution. It is caused by an improper assignment of frequencies or an error in formulating the problem. In such cases one must modify the initial solution in order to get a solution which satisfies the rule m + n—i. The problem becomes degenerate at the‐

(i) Initial stage When in the initial solution the number of occupied cells is less than m + n — 1 (rim requirements minus 1) i.e. the number of stone squares in insufficient (ii) When two or more cells are vacated simultaneously Degeneracy may appear subsequently when two or more cells are vacated simultaneously in the process of transferring the units, along the closed loop to obtain an optimal solution.

When transportation problem becomes degenerate When transportation problem becomes degenerate it cannot be tested for optimality because it is impossible to compute u and, v values with MODI method. To overcome the problem of insufficient number of occupied cell we proceed by assigning an infinitesimally small amount (close to zero) to one or more (if needed) empty cell and treat that cell as occupied cell. This amount is represented by the Greek letter E (epsilon). It is an insignificant value and does not affect the total cost. But it is appreciable enough to be considered a basic variable. When the initial basic solution is degenerate, we assign c to an independent empty cell. An independent cell is one from which a closed loop cannot be traced. It is preferably assigned to a cell which has minimum per unit cost. After introducing e we solve the problem using usual methods of solution. 10) Steps to solve a Transportation Problem. A transportation problem can be solved in 2 phases PHASE I Step 1:

Step 1:

Check whether the given T.P. is balanced or not

Step 2:

Develop initial feasible solution by any of the five methods

Check whether the given T.P. is balanced or not. If it is unbalanced then balance it by adding a row or a column. Step 2: Develop initial feasible solution by any of the five methods: a) North West Corner Rule (NWCR) or South West Corner Rule (SWCR) b) Row Minima Method (RMM) c) Column Minima Method (CMM) d) Matrix Minima Method (MMM) e) Vogel’s Approximation Method (VAM) We discuss here the two commonly used methods to make initial assignments (1) Northwest corner rule (2) Vogel’s Approximation Method (VAM) (1) Northwest Corner Rule: Start with the northwest corner of the transportation tableau and consider the cell in the first column and first row. We have values a1 and b1 at the end on the first row and column i.e. the availability at row one is a1 and requirement of column 1 is b1. (i) If al > b1 assign quantity b1 in the cell, i.e. x1 b1. Then proceed horizontally to the next column in the first row until a1 is exhausted i.e. assign the remaining number a1 ‐ b1 in the next column. (ii) If al < b1 then put Xl al and then proceed vertically down to the next row until b1 is satisfied. i.e. assign b1 – a1 in the next row. (iii) If a1 = b1 then put XII = a1 and proceed diagonally to the next cell or square determined by next row and next column. In this way move horizontally until a supply source is exhausted, and vertically down until destination demand is completed and diagonally when a1 = b1, until the south‐east corner of the table is reached. (2) Vogel’s Approximation Method (VAM): The north‐west corner rule for initial allocation considers only the requirements and availability of the goods. It does not take into account shipping costs given in the tableau. It is therefore, not a very sound method as it ignores the important factor, namely cost whiçh we seek to minimize. The VAM, on the other hand considers the cost in each cell while making the allocations we explain below this method. (i) Consider each row of the cost matrix individually and find the difference between two least cost cells in it. Then repeat this exercise for each column. Identify the row or column with the Largest difference (select any one in case of a tie). (ii) Now consider the cell with minimum cost in that column (or row) and assign the maximum units possible to that cell.

(iii) Delete the row/column that is satisfied. (iv) Again find out the differences and proceed in the same manner as stated in earlier paragraph and continue until all units have been assigned. PHASE II – TEST FOR OPTIMALITY Before we enter phase II, the following two conditions should be fulfilled in that order. (i) Obtaining a basic feasible solution implies finding a minimum number of ij values. This minimum number is m + n ‐ 1. Where m is the number of origins n is the number of destinations. Thus initial assignment should occupy m + n ‐ I cells i.e. requirements of demand and supply cells minus — 1. (ii) These ij should be at independent positions These requirements are called RIM requirements. Test for optimality (or improvement): After obtaining the initial feasible solution, the next step is to test whether it is optimal or not. We explain here the Modified Distribution (MODI) Method for testing the optimality. If the solution is non‐optimal as found from MODI method then we improve the solution by exchanging non‐basic variable for a basic variable. In other words we rearrange the allocation by transferring units from an occupied cell to an empty cell that has the largest net cost change or improvement index, and then shift the units from other related cells so that all the rim (supply, demand) requirements are satisfied. This is done by tracing a closed path or closed loop.

Step 1:

Add a column u to the RHS of the transportation tableau and a row v at the bottom of the tableau.

Step 2:

Assign, arbitrarily, any value to u or v generally u = 0.

Step 3

Having determined u1 and v calculate ij = (u1 + v1) — for every

unoccupied cell. Step 4:

If the solution is not optimal select the cell with largest positive

improvement index.

Step 5:

Test the solution again for optimality and improve fit if necessary

Step I: Add a column u to the RHS of the transportation tableau and a row v at the bottom of the tableau. Step 2: Assign, arbitrarily, any value to u or v generally u = 0. This method of assigning values to u1 and v1 is workable only if the initial solution is non‐degenerate i.e., for a table there are exactly m + n ‐1 occupied cells. Step 3: Having determined u1 and v calculate ij = (u1 + v1) — for every unoccupied cell. This represents the net cost change or improvement index of these cells (1) If all the empty cells have negative net cost change ij, the solution is optimal and unique (2) If an empty cell has a zero Xij and all other empty cells have negative Xij the solution is optimal but not unique. (3) If the solution has positive Ai for one or more empty cells the solution is not optimal. Step 4: If the solution is not optimal select the cell with largest positive improvement index. Then trace a closed loop and transfer the units along the route. Tracing loop (closed path): 1) Choose the unused square to be evaluated. (2) Beginning with the selected unused square trace a loop via used squares back to the original unused squares. Only one loop exists for any unused square in a given solution. (3) Assign (+) and (—) signs alternately at each square of the loop beginning with a plus sign at the unused square. Assign these sign in clockwise or anticlockwise direction. These signs indicate addition or subtraction of units to a square. (4) Determine the per unit net change in cost as a result of the changes made in tracing the loop. Compare the addition to the decrease in cost. It will give the improvement index. (It is equivalent to j in a LPP). (5) Determine the improvement index for each unused square. (6) In a minimization case. If all the indices are greater than or equal to zero, the solution is optimal. If not optimal, we should find a better solution. We may also note the following points: (i) An even number of at least four cells participate in a closed loop. An occupied cell can be considered only once.

(ii) If there exists a basic feasible solution with m + n — 1 positive variables, then there would be one and only one closed loop for each cell. (iii) All cells that receive a plus or minus sign except the starting empty cell, must be the occupied cells. (iv) Closed loops may or may not be square or rectangular in shape. They may have peculiar configurations and a loop may cross over itself. Step 5: Test the solution again for optimality and improve fit if necessary. Repeat the process until an optimum solution is obtained. 11) SWhat are the differences between assignment problem and transportation problem? The differences between AP and TP are the following:

1. TP has supply and demand constraints while AP does not have the same. 2. The optimal test for TP is when all cell evaluation \s are greater than or equal to zero

whereas in AP the number of lines must be equal to the size of matrix. 3. A TP sum is balanced when demand is equal to supply and an AP sum is balanced when

number of rows are equal to the number of columns. 4. for AP. We use Hungarian method and for transportation we use MODI method 5. In AP. We have to assign different jobs to different entities while in transportation we

have to find optimum transportation cost.

mp-qmb 2

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