chap 7 transportation problems

8/7/2019 CHAP 7 TRANSPORTATION PROBLEMS

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TRANSPORTATION PROBLEMS

Transportation problem is a case of LPP in which the objective is to transport

various amounts of a single homogeneous commodity to various destinations

for a minimum transport cost.

1. Feasible Solution

For a set of non negative values xij that satisfies the constraints are called the

feasible solution of the transportation problem.

2. Basic Feasible Solution

A feasible solution that contains not more than m+n-1 non negative allocation

is called BF solution to the TP.

3. Optimal Solution A feasible solution is said to be optimal if it minimizes the total transportation

cost.

4. Degenerate & Non- degenerate Basic Feasible Solution

A BF solution to the m × n TP that contains exactly m+n-1 allocation in

independent position is called non degenerate BF solution.

A feasible solution that contains less than m+n-1 non negative allocation is

said to be degenerating.

Methods to find initial basic feasible solution

1. North West Corner Rule (NWCR)

Ques : Find the IBFS whose cost matrix is given by

D1 D2 D3 D4 Capacity

6 4 1 5 14

8 9 2 7 16

4 3 6 2 15

6 10 15 14 Demand



Soln:

Number of allocation = m+n-1 = 4+3-1 = 6

IBFS Cost = (6 ×××× 6) + (8 ×××× 4) + (2 ×××× 9) + (14 ×××× 2) + (1 ×××× 6) +

(14 ×××× 2)

= 148

2. Matrix Minima or Low Cost Entry Method

Ques:

1 2 3 4 Supply

A 21 16 15 3 11

B 17 18 14 23 13

C 32 27 18 41 19

Demand 6 10 12 15

1 2 3 4 Cap

1 6

6

4

8

1

×××× 5

××××

14

2 8

××××

92

214

7

××××

16

3 4

××××

3

××××

6

1

2

14

15

Dem 6 10 15 14 45

45



Soln:

Number of allocation = m+n-1= 4+3-1 = 6

IBFS Cost = (11 ×××× 3) + (1 ×××× 17) + (12 ×××× 14) + (5 ×××× 32)+ (10 ×××× 27) + (4 ×××× 41)

= 812

3. Vogel’s Approximation Method

Ques:

1 2 3 4 Supply

A 21 16 15 3 11

B 17 18 14 23 13

C 32 27 18 41 19

Demand 6 10 12 15

21

××××

16

××××

15

×××× 3

11

11

17

1

18

××××

14

12

23

××××

13

32

5

27

10

18

××××

41

4

19

6 10 12 15 43

43



Soln:

12 |

3 | 3 | 3 | 4 |

9 | 9 | 9 | 9| 9 | 9 | 18 |

Number of allocation = m+n-1 = 4+3-1 = 6

IBFS Cost = (11 ×××× 3) + (6 ×××× 17) + (3 ×××× 18) + (4 ×××× 23)

+ (7 ×××× 27) + (12 ×××× 18)

= 686

Problems:

1. Find the IBFS whose cost matrix is given below by NWCR, MMM, VAM

21

××××

16

××××

15

×××× 3

11

11

17

6

18

3

14

××××

23

4

13

32

××××

27

7

18

12

41

××××

19

6

××××

10 12 15 43

43

4 2 1 20

15 9 4 18

15 9 4

9 4

27 18

18



E F G H I J Supply

A 14 19 32 9 21 0 200

B 15 10 18 7 11 0 225

C 20 12 13 18 16 0 175

D 11 32 14 14 18 0 350

Demand 130 110 140 260 180 130

Soln:

a. NWCR:

IBFS Cost = (130 ×××× 14) + (70 ×××× 19) + (40 ×××× 10) + (140 ×××× 18)

+ (45 ×××× 7) + (175 ×××× 18) + (40 ×××× 14) + (180 ×××× 18)

+ (0 ×××× 130)

= 13335

14

130

19

70

32

××××9

××××

21

××××

0

××××200

15

××××

10

40

18

140

7

45

11

××××

0

××××

225

20

×××× 12

××××

13

××××

18

175

16

××××

0

××××

175

11

××××32

××××

14

××××

14

40

18

180

0

130

350

130 110 140 260 180 130



b. VAM:

IBFS Cost = 10315

14

××××

19

××××

32

×××× 9

35

21

35

0

130 200 9 5 5

15××××

10××××

18××××

7225

11××××

0××××

225 7 3 3 3

20

×××× 12

110

13

65

18

××××

16

××××

0

××××

175 12 1 1 1 1

11

130 32

××××

14

75

14

××××

18

145

0

××××

350 11 3 3 3 3 3

130 110 140 260 180 130 950

3 2 1 2 5 0

3 2 1 2 5 0

3 2 1 2

4 2 1 7

9 20 1 4

11 32 14 14



c. MMM

c.

IBFS Cost = (35 ×××× 9) + (35 ×××× 21) + (130 ×××× 0) + (225 ×××× 7)

+ (110 ×××× 12) + (65 ×××× 13) + (130 ×××× 11) + (75 ×××× 14)

+ (145 ×××× 18)

= 9880

14

××××

19

××××

32

×××× 9

35

21

35

0

130 200

15

××××

10

××××

18

××××

7

225

11

××××

0

××××

225

20

×××× 12

110

13

65

18

××××

16

××××

0

××××

175

11

130 32

××××

14

75

14

××××

18

145

0

××××

350

130 110 140 260 180 130



Test for optimality

1. Number of allocation = m+n-1

2. These allocations must be in independent positions

Non independent

Non independent

Independent

Prob: Test the optimality of the given problem by Modi method or UVmethod

19 30 50 10 7

70 30 40 60 9

40 8 70 20 18

5 8 7 14



Soln:

IBFS Cost = (19 ×××× 5) + (10 ×××× 2) + (40 ×××× 7) + (2 ×××× 60)

+ (8 ×××× 8) + (65 ×××× 13) + (20 ×××× 10)

= 779

Test by Modi Method

19

5

30

+50

+

10

2u1 = 10

90

+

30

−−−−

40

7

60

2 -0u2 = 60

40

+

8

8888

70

+

20

10 +0u3 = 20

v1 =9 v2= -12 v3= -20 v4= 0

For allocated cells we have cij = ui + v j

C14 = U1 + V4 U1 = 10 10 = U1 + 0

19

5

30

××××

50

×××× 10

2

7 9 9 40 40

70

××××

30

××××

40

7

60

2

9 10 20 20 20

40

×××× 8

8

70

××××

20

10

18 12 20 50

54 8 7 14 3454 8 7 14

21 22 10 10

21 10 10

10 10

10 50



C24 = U2 + V4 U2 = 10 60 = U2 + 0

C34 = U3 + V4 U3 = 0 20 = U3 + 0

C11 = U1 + V1 V1 = 9 19 = V1 + 10

C23 = U2 + V3 V3 = -20

40 = V3 + 60

C32 = U3 + V2 V2 = -12 8 = V1 + 20

For empty cells

dij

= cij – ( u

j + v

j )

d12 = 30 – (10 – 12 )d12 = 32

d13 = 50 – (10 – 20 )d13 = 60

d21 = 90 – (9 + 60 ) d22 = 30 – (60 – 12 )d21 = 21 d22 = -18

d31 = 40 – (9 + 20 ) d33 = 70 – (20 – 20 )d31 = 11 d33 = 70 Ans: 743



Unbalanced Transportation Problem

Prob 1: A company has 4 factories A,B,C,D which supply to

warehouses E,F,G,H,I. Monthly capacities are 200,225,175,350 units.

Monthly ware house requirements are 130,110,140,260 & 180 units

respectively. The shipping cost in Rs is given below. Find the minimumtransportation cost.

Soln: Since the capacity is not equal to demand. This is an unbalanced TP.

Balance it by adding a dummy column having a demand of 950 – 820 =

130 units.

The cost values in the dummy column are taken as zeroes.

14 19 32 9 21 0

15 10 18 7 11 0

26 12 13 18 16 0

11 22 14 14 18 0

E F G H I Cap

A 14 19 32 9 21 200

B 15 10 18 7 11 225

C 26 12 13 18 16 175

D 11 22 14 14 18 350Dem 130 110 140 260 180 950

820



14

××××

19

××××

32

×××× 9

200

21

××××

0

×××× 200

15

××××

10

××××

18

××××

7

45

11

80

0

××××

225

26

×××× 12

45

13

××××

18

××××

16

××××

0

130

175

11

130 22

65

14

140

14

15

18

××××

0

××××

350

130 110 140 260 180 130

14

××××

19

××××

32

×××× 9

200

21

××××

0

×××× 200

15

××××

10

××××

18

××××

7

45

11

80

0

××××

225

26

×××× 12

45

13

××××

18

××××

16

××××

0

130

175

11

130 22

65

14

140

14

15

18

××××

0

××××

350

14

××××

19

××××

32

×××× 9

200

21

××××

0

××××15

××××

10

××××

18

××××

7

60

11

165

0

××××

26

×××× 1245

13

××××

18

××××

16

××××

0130

11

130 22

65

14

40

14

××××

18

15

0

××××



Test for optimality

14

+19

+32

+

9

200

21

+

u1 = -5

5

+

30

−−−−

40

7

60

2 -0

11

165

u2 = -7

26

+

8

8888

70

+

20

10 +0

16

+u3 = -10

11

130

22

65

14

140

14

0

18

100 u4 = 0

v1 =11 v2= 22 v3= 14 v4= 14 v5= 18 v6= 10

(45 + 0) Q -Q (130-0)

(65 – 0) –Q Q

130 – Q = 0And, 65 – 0 =0

Q = 65 & 130

Qmin = 65

Q 65 - Q

15 – Q 65 + Q

Qmin = 15



Min Tc = 8935

Maximisation TP

Prob 1: A firm manufacturing a certain product has 3 plants A, B & C.

The firm has to supply to 3 customers C1, C2 & C3. The profits expected

per unit transported from the 3 plants to the 3 customers are given

below. The requirements of the customers are 100, 70, & 120 units

respectively & the capacities of the plants A, B & C are 80,130 & 100

units respectively. Determine the optimum no. of units to be transplantedfrom each of the plants to each of the customers. Does this problem have

more than one optimum solution? If yes find another solution for the

problem

C1 C2 C3 Capacity

A 4 1 3 80

B 2 3 2 130

C 3 5 2 100

Demand 100 70 120 130

290

14

+19

+32

+

9

200

21

+

0

+

u1 = -4

15

+

10

+

18

+

7

60

11

165

0

+

u2 = -6

26+ 1210101010 1350505050 18+ 1615 0+ u3 = -1

11

130

22

+

14

90

14

+

18

+

0

130 u4 = 0

v1 =11 v2= 13 v3= 14 v4= 13 v5= 17 v6= 0



Soln: Since the capacity is not equal to demand. This is an unbalanced

TP. Balance it by adding a dummy column having a demand of 310 –

290 = 20 units.

C1 C2 C3 C4 Capacity

A 4 1 3 0 80

B 2 3 2 0 130

C 3 5 2 0 100

Demand 100 70 120 20

Since the given matrix is a profit matrix it has to be maximized. The

given matrix is converted to solve for maximization.

(1) Select highest element in the matrix and subtract all other

elements from it.

For the revised matrix apply regular procedure

IBFS Cost = (70 ×××× 4) + (10 ×××× 3) + (110 ×××× 2) + (20 ×××× 0)

+ (30 ×××× 3) + (70 ×××× 5)

= 970

1

70

4

××××

10

×××× 5

××××

80 1 1 1 3

3

××××

2

××××

3110

520

130 1 0 1 2

2

30 0

70

3

××××

5

××××

100 2 1

100 70 120 20

1 2 1 0

1 1 0

2 1 0

1 0



Test for optimality

Since, empty cell evaluations are all non negative the solution is optimal.

Hence, there is zero evaluation for 2 empty cells which suggest thisproblem has more than one optimal solution.

10 – Q Q

110+ Q 20 – Q

Q =10

1

70 4

+2

10

5

0u1 = - 1

3+ 2+ 3110 520 u2 = - 1

2

30303030 0

70707070

3

+

5

0u3 = 0

v1 =2 v2= 0 v3= 3 v4= 6

1

70 4

+2

10

5

0u1 = - 1

3

+

2

+

3

120

5

10u2 = - 1

2

30303030 0

70707070

3

- 5

-u3 = 0

v1 =2 v2= 0 v3= 4 v4= 6



70 + Q 10 – Q

10 + Q

120 –Q

30 –Q Q Q =10

Degeneracy in TP

Degeneracy occurs in a TP whenever the number of occupied cells is

less than m + n – 1. The degeneracy can develop in 2 ways

(1) Basic feasible solution may degenerate from the initial stage

onwards.

(2) They may become degenerate at any inter mediate stage.

Resolution of degeneracy

To reduce degeneracy allocate an extremely small quantity of goods

(usually denoted by or ε ) to one or more of the empty cells so that the

numbers of occupied cells become equal to m + n – 1. The small

quantity is introduced on the least cost independent cell.

1

80

4

+

2

0

5

0

u1 = - 1

3

+

2

+

3

110

5

20u2 = 0

2

20202020 0

70707070

3

10101010

5

0u3 = 0

v1 =2 v2= 0 v3= 3 v4= 5



Test for optimality

Prob 1: The following table given the cost of transporting a certain item

from A, B, C & D to destination E, F, G & H. The demand at E, F, G &

H are 10, 20, 30, & 40 units respectively. The capacities of A, B, C & D

are 10, 30, 20 & 60 units respectivelya. Determine the optimum transportation schedule and minimum total

cost.

b. If the cost from C to G is increased to 12 per unit does the optimal

schedule change if yes find the new optimal solution? Does it have

an alternate solution?

Soln: Rs 730

Prob 2: A company has factories A, B, C which supplies to ware houses

at D, E & F. Monthly capacities are 150,200 & 250 units respectively on

regular production. If are time is utilized factories A & B can produce 50

& 100 units respectively. The expected profits per unit for different

combination of factories to warehouses are given below. If overtime is

utilized, the expected profit/unit decreases by 5% & 4% respectively per

factories A & B. Determine the optimum distribution plan for thecompany & the max profit.

E F G H

A 8 10 12 17 0 10B 15 13 18 11 0 30

C 14 20 6 10 0 20

D 13 19 7 5 0 60

10 20 30 40 20



D E F

A 11 13 17 150

B 16 18 14 200

C 21 20 13 250

300 200 200

D E F

A 11 13 17 150

B 16 18 14 200

C 21 20 13 250A0 6 8 12 50

B0 12 14 10 100

Considering over time we take 2 fictitious factories having capacities 50

& 100 units respectively.

The expected profits per unit for these factories are estimated by

decreasing Rs.5 & 4 from factories A & B without over time.

This is an unbalanced problem & hence to be balanced.

D E F Gf

A 11 13 17 0 150

B 16 18 14 0 200

C 21 20 13 0 250

A0 6 8 12 0 50

B0 12 14 10 0 100

300 200 200 50



No. of allocations = m+n-1 = 8, but m + n – 1 =7

This is a case of degeneracy.

Since the empty cell evaluations are non negative the solution is optimal.

IBFS Cost = (150 ×××× 17) + (50 ×××× 16) + (150 ×××× 18)

+ (250 ×××× 21) + (50 ×××× 12) + (50 ×××× 14)

+ (50 ×××× 0)

= 12600Max Profit = 12600

D E F Gf

A 10

××××

8

××××

4

150 21

××××

150

B 550

3150

7

××××

21

××××

200

C 0

250 1

××××

8

××××

21

××××

250

A0 15

×××× 13

××××

9

50

21

××××

50

B0 9

××××

750

11

××××

2150

100

300 200 200 50



D E F Gf

A 10

+

8

+

4

50 21

+

u1 = -3

B 550

3150

7

21+

u2 = 0

C 0

250 1

+

8

+

21

+

u3 = -5

A0 15

+ 13

+

9

50

21

+

u4 = 2

B0

90 750 11+ 2150

u5 = 4

v1 =5 v2= 3 v3= 7 v4= 17

13

+

11

1

16

+ 23

+

u1 = -3

11

1

19

4

26

+

16

-

u2 = 8

12

+ 11

2

4

5

9

3

u3 = 0

7

8 15

0

9

+

14

+

u4 = 4

v1 = 3 v2= 11 v3= 4 v4= 9

chap 7 transportation problems

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