unit 3 lecturer notes of transportation problem of or by dr s v prakash

8/3/2019 Unit 3 Lecturer notes of Transportation problem of OR by Dr S V Prakash

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1

06IP/IM74 OPERATIONS RESEARCH

UNIT - 3: Transportation Problem

(By Dr. S.V Prakash, Asst. Prof. (Mech Dept.), MSRIT, Bangalore)

Introduction:

The objective of the transportation problem is to transport various quantities of a

single homogenous commodity, which are initially stored at various origins to various

destinations in such a way that the total transportation cost is minimum.

Definitions:

Basic Feasible solution: A feasible solution to a m-origin, n-destination problem is said to

be basic if the number of positive allocations are equal to (m+n-1).

Feasible Solution: A set of positive individual allocations which simultaneously removes

deficiencies is called a feasible solution.

Optimal Solution: A feasible solution (not basically basic) is said to be optimal if it

minimises the total transportation cost.

Mathematical Formulation of Transportation Problems

• Suppose there are ‘m’ ware houses (w1,w2,w3, _, _, wm),

where the commodity is stocked and ‘n’ markets where it is needed.

• Let the supply available in wear houses be a1, a2, a3, _,_,_ am and

• The demands at the markets (m1, m2, m3, _, _, mn) be b1, b2, b3, _, _ , _ bn.

• The unit cost of shipping from ware house i to a market j is Cij (C11,C12,_, _ Cn),

• Let X11, X12,X13,_, _, Xmn be the distances from warehouse to the markets

• we want to find an optimum shipping schedule which minimises the total cost of

transportation from all warehouses to all the markets



2

The total minimum transportation cost is

Z = i=1Σm

j=1Σn

Xij * Cij i.e. Z = X11C11 + X12C12 + _ _ _ + XmnCmn

Types of Transportation Problems

1. Minimisation Balanced Transportation Problems

2. Minimisation Unbalanced Transportation Problems

3. Maximisation Balanced Transportation Problems

4. Maximisation unbalanced Transportation Problems

5. All the above models with degeneracy.

Ware

houses

Markets

m1 m2 m3 - - mn

Supplies

W1

W2

W3

-

-

wm

C11X11 C12X12 C13X13 - - - C1nX1n

C21X21 C22X22 C23X23 - - - C1nX1n

C31X31 C32X32 C33X33 - - - C1nX1n

-

-

Cn1Xm1 Cn2Xm2 Cn3Xm3 - - - CmnXmn

a1

a2

a3

-

-

am

Demand b1 b2 b3 - - - - bn i=1Σm

ai = j=1Σn

b j



3

Methods of solving Transportation Problems

1. North- West Corner Rule method

2. Row-minima Method

3. Column minima method

4. Matrix Minima Method or least cost method

5. Vogel's Approximation method (VAM)

Methods for checking Optimality

1. Modified Distribution Method, UV or MODI method

PROBLEMS:

1. Solve the following transportation problem by

North-West corner rule, Row Minima, Column Minima,

Matrix Minima and VAM Method:

Solution:

This is a balanced transportation problem, since supply is equal to demand

Factories W1 W2 W3 W4 Supply

F1

F2

F3

6 4 1 5

8 9 2 7

4 3 6 2

14

16

05

Demand 6 10 15 4 35



4

1. North-West corner rule Method:


F1

F2

F3

6(6) 4(8) 1 5

8 9(2) 2(14) 7

4 3 6(1) 2(4)

14

16

05

Demand 06 10 15 04 35

The Total feasible transportation cost

= 6(6) + 4(8) + 9(2) + 2(14) + 6(1) + 2(4) = Rs. 128/-

2. Row Minima Method:


F1

F2

F3

6 4 1(14) 5

8(6) 9(5) 2(1) 7(4)

4 3(5) 6 2

14

16

05

Demand 06 10 15 04 35


= 1(14)+8(6)+9(5)+2(1)+7(4)+3(5)

= Rs.152/-



5

3. Column-Minima Method:


= 6(1)+4(10)+1(3)+2(12)+7(4)+4(5)

= Rs. 121/-

4. Matrix-Minima Method or Least Cost method:


F1

F2

F3

6 4 1(14) 5

8(6) 9(9) 2(1) 7

4 3(1) 6 2(4)

14

16

05

Demand 06 10 15 04 35


= 1(14)+8(6)+9(9)+2(1)+3(1)+2(4)

= 156/-


F1

F2

F3

6(1) 4(10) 1(3) 5

8 9 2(12) 7(4)

4(5) 3 6 2

14

16

05

Demand 06 10 15 04 35



6

5. VAM- Vogel’s approximation method:

Step-I: Against each row and column of the matrix, denote the difference between the two

least cost in that particular row and column.

Step-II: Select the maximum value noted as per step-I, in this row or column select the cell

which has the least cost

Step-III: Allocate the maximum possible quantity

Step-IV: After fulfilling the requirements of that particular row or column, Ignore that

particular row or column and recalculate the difference by the two lowest cost for each of

the remaining rows or columns, Again select the maximum of these differences and allocate

the maximum possible quantity in the cell with the lowest cost in that particular /

corresponding row or column.

Step-V: Repeat the procedure till the initial allocation is completed

5. VAM- Vogel’s approximation method:


F1

F2

F3

6 (4) 4(10) 1 5

8(1) 9 2(15) 7

4(1) 3 6 2(4)

14

16

05

Demand 06 10 15 04 35



7


= 6(4) + 4(10) + 8(1) + 2(15) + 4 (1) + 2(4)

= 114/-

II- Check for degeneracy:

If (m+n-1) is not equal to the number of allocated cells, then it is called degeneracy in

transportation problems,

Where m= number of rows, n= number of columns.

This will occur if the source and destination is satisfied simultaneously.

The degeneracy can be avoided by introducing a dummy allocation cell. To equate the

number of allocated cells equal to (m+n-1)

For the above problem (m+n-1) = (3+4-1) =6 = Number of allocations = 6

Hence there is no degeneracy.

III- Checking optimality using MODI method:

• For allocated cells Cij – (Ui +V j) = 0

• For unallocated cells Cij – (Uij) +Vj) ≥ 0


= 6(4)+4(10)+8(1)+2(15)+4(1)+2(4)

= 114/-

V1=0 V2=-2 V3 =-6 V4= -2

6(4) 4(10) 1 5

8(1) 9 2(15) 7

4(1) 3 6 2(4)

U1=6

U2=8

U3= 4



8

Problem.2

There are 3 Parties who supply the following quantity of coal P1= 14t, P2=12t, P3= 5t.

There are 3 consumers who require the coal as follows C1=6t, C2=10t, C3=15t. The cost

matrix in Rs. Per ton is as follows. Find the schedule of transportation policy which

minimises the cost:

Solution:

Factories W1 W2 W3 Supply

F1

F2

F3

6 8(5) 4(9)

4(6) 9 3(6)

1 2(5) 6

14

12

05

Demand 6 10 15 31



9

Therefore the total feasible transportation cost

= 8(5)+ 4(9) + 4(6) +3 (6) + 2(5)

= Rs. 128/-

II. Check for Degeneracy:

(m+n-1) = (3+3-1) = 5 = Number of allocations

Hence there is no degeneracy


•

For allocated cells Cij – (Ui +V j) = 0

• For unallocated cells Cij – (Uij) +V j) ≥ 0

Since all the marginal costs for the unallocated cells are positive, it gives an optimal

solution and the total minimum transportation cost = Rs. 128/-

V1=5 V2=8 V3=4

6 8(5) 4(9)

4(6) 9 3(6)

1 2(5) 6

U1= 0

U2=-1

U3=-6



10


F1

F2

F3

5 7 3 8

4 6 9 5

2 6 4 5

300

500

200

Demand 200 300 400 100 1000

Since the unit cost of production is Rs.4, 3 and 5

at the three factories, therefore Total cost = Production cost + cost


F1

F2

F3

9 11 7(300) 12

7(100) 9(300) 12 8(100)

7(100) 11 9(100) 10

300

500

200

Demand 200 300 400 100 1000

Therefore the total feasible solution = 7(300) + 7(100) +9(300) + 8(100) +7(100) + 9(100)

= Rs. 7900/-



11

Therefore the total feasible transportation solution= 7(300) + 7(100) + 9(300) + 8(100) + 7(100) + 9(100)

= Rs.7900/-





• For allocated cells Cij – (Ui +V j) = 0

• For unallocated cells Cij – (Uij) +Vj) ≥ 0

Since all the marginal costs for the unallocated cells are positive, it gives an optimal

solution and the total minimum cost = Rs. 7900/-


F1

F2

F3

9 11 7(300) 12

7(100) 9(300) 12 8(100)

7(100) 11 9(100) 10

300

500

200

Demand 200 300 400 100 1000

V1 V2 V3 V4

9 11 7(300) 12

7(100) 9(300) 12 8(100)

7(100) 11 9(100) 10

U1= -2

U2=0

U3=0



12

Problem-4

A company has three plants supplying the same product to the five distribution centers.

Due to peculiarities inherent in the set of cost of manufacturing, the cost/ unit will vary

from plant to plant. Which is given below. There are restrictions in the monthly capacityof each plant, each distribution center has a specific sales requirement, capacity

requirement and the cost of transportation is given below.

The cost of manufacturing a product at the different plants is Fixed cost is Rs 7x105,

4x 105

and 5x 105.

Whereas the variable cost per unit is Rs 13/-, 15/- and 14/- respectively.

Determine the quantity to be dispatched from each plant to different distribution centers,

satisfying the requirements at minimum cost.

Solution:

Factories W1 W2 W3 W4 W5 Supply

F1

F2

F3

5 3 3 6 4

4 5 6 3 7

2 3 5 2 3

200

125

175

Demand 60 80 85 105 70 400 500

Factories W1 W2 W3 W4 W5 D Supply

F1

F2

F3

18 16 16 19 17 0

19 20 21 18 22 0

16 17 19 16 17 0

200

125

175

Demand 60 80 85 105 70 100 500



13

Therefore the total feasible transportation cost =

= 16(55) + 16(85) + 17(60) + 20(25) + 0(100) +16(60) + 16(105) +17(10) = Rs6570/-




Factories W1 W2 W3 W4 W5 D Supply

F1

F2

F3

18 16(55) 16(85) 19 17(60) 0

19 20(25) 21 18 22 0(100)

16 17 19 6(105) 17(10) 0

200

125

175

Demand 60 80 85 105 70 100 500



14



15



16



17

Therefore total feasible solution

= 1(15) +4(10)+4(35)+2(5)+0+3(22)+2(20)+4(3)= Rs. 323/-

Therefore total feasible solution

= 1(15) +4(10)+4(35)+2(5)+0+3(22)+2(20)+4(3)= Rs. 323/-

II. Check for degeneracy:



III. Check for optimality:

V1=3 V2=1 V3=2 V4=4 V5=4 V6=2

4 1(15) 3 4(10) 4(35) 0(-2)

2 3 2 2(5) 3 0(30)

3(22) 5 2(20) 4(3) 4 0(-2)

U1=0

U2=-2

U3=0



18

V1=3 V2=1 V3=2 V4=4 V5=4 V6=2

4 1(15) 3 4(10-x) 4(35) 0(x)

2 3 2 2(5+x) 3 0(30-x)

3(22) 5 2(20) 4(3) 4 0(-2)

U1=0

U2=-2

U3=0

10-x=0 30-x=0 Min. value of x=10

V1=1 V2=1 V3=0 V4=2 V5=4 V6=0

4 1(15) 3 4 4(35) 0(10)

2 3 2 2(15) 3(-1) 0(20)

3(22) 5 2(20) 4(3) 4 0(-2)

U1=0

U2=0

U3=2

20-X = 0: 3-x=0: therefore x=3

V1=3 V2=1 V3=2 V4=3 V5=4 V6=0

4 1(15) 3 4 4(35) 0(10)

2 3 2 2(18) 3(-1) 0(17)

3(22) 5 2(20) 4 4 0(3)

U1=0

U2=-1

U3=0



19

35-x=0 and 17-x = 0, therefore x= 17

V1=3 V2=1 V3=2 V4=3 V5=4 V6=0

4 1(15) 3 4 4(18) 0(27)

2 3 2 2(18) 3(17) 0

3(22) 5 2(20) 4 4 0(3)

U1=0

U2=-1

U3=0

27-x=0 : 22-X= 0: 17-X=0: therefore X=17

V1=3 V2=1 V3=2 V4=3 V5=4 V6=0

4 1(15) 3 4 4(35) 0(10)

2(17) 3 2 2(18) 3 0

3(5) 5 2(20) 4 4 0(20)

U1=0

U2=-1

U3=0



20

Factories A B C Supply

M1

M2

M3

139140 137

209207 210

254 255 255

160

120

140

Demand 90 210 120 420



21


M1

M2

M3

116115(70) 118(90)

4(90) 48 45(30)

1 0(140) 0

160

120

140

Demand 90 210 120 420


M1

M2

M3

116115 118

46 48 45

1 0 0

160

120

140

Demand 90 210 120 420



22

V1=4 V2=42 V3=45

116 115(70+x) 18(90-x)

4(90) 48 45(30)

1 0(140-x) 0(x)

U1=73

U2=0

U3=-42

90-X=0 and 140-X=0, therefore X=90

V1=4 V2=45 V3=45

116 115(160) 18

4(90) 48 45(30)

1 0(50) 0(90)

U1=70

U2=0

U3=-45



23


M1

M2

M3

139 140 137

209 207 210

254 255 255

90

210

120

Demand 160 120 140 420



24


M1

M2

M3

116 115 118

46 48 45

1 0 0

90

210

120

Demand 160 120 140 420


M1

M2

M3

116(90) 115 118

46(70) 48 45(140)

1 0(120) 0

90

210

120

Demand 160 120 140 420



25


M1

M2

M3

116(90) 115 118

46(70) 48 45(140)

1(θ) 0(120) 0

90

210

120

Demand 160 120 140 420

V1=0 V2=-1 V3= -1

116(90) 115 118

46(70) 48 45(140)

1(θ) 0(120) 0

U1=116

U2=46

U3=1



26

Factories A B C D Supply

I

II

III

D

21 26 20 21

22 24 20 19

18 20 19 20

00 00 00 00

450

300

150

020

Demand 200 300 150 270 920

A B C D Supply

4(150) 0(300) 6 5

5(50) 2 6(130) 7(120)

8 6 7 6(150)

0 0 0(20) 0

450

300

150

020

200 300 150 270 920

Factories A B C D Supply

I

II

III

22 26 20 21

21 24 20 19

18 20 19 20

450

300

150

Demand 200 300 150 270 920 900



27

V1= 5 V2=1 V3= 6 V4=7

4(150) 0(300) 6 5(-1)

5(50) 2 6(130) 7(120)

8 6 7 6(150)

0 0 0(20) 0

U1=-1

U2=0

U3=-1

U4=-6

V1= 5 V2=1 V3= 6 V4=7

4(150-X) 0(300) 6 5(X)

5(50+X) 2 6(130) 7(120-X)

8 6 7 6(150)

0 0 0(20) 0

U1=-1

U2=0

U3=-1

U4=-6

150-X= 0, 120-X= 0, therefore X=120



28

V1= 5 V2=1 V3= 6 V4=7

4(150-X) 0(300) 6 5(X)

5(50+X) 2 6(130) 7(120-X)

8 6 7 6(150)

0 0 0(20) 0

U1=-1

U2=0

U3=-1

U4=-6

V1= 4 V2=0 V3=5 V4=5

4(30) 0(300) 6 5(120)

5(170) 2 6(130) 7

8 6 7 6(150)

0 0 0(20) 0

U1=0

U2=1

U3=1

U4=-5



29



30



31

III. CHECK FOR OPTIMALITY



32

Factories D E F G Supply

A

B

C

OT.A

OT.B

11 13 17 14

16 18 14 10

21 24 13 10

15 17 21 18

21 23 19 15

250

350

400

50

75

Demand 200 225 275 300 1125

Factories D E F G D1 Supply

A

B

C

OT.A

OT.B

11(200) 13(50) 17 14 0(θ)

16 18(175) 14 10(175) 0

21 24 13(275) 10(125) 0

15 17 21 18 0(50)

21 23 19 15 0(75)

250

350

400

50

75



33

Factories D E F G D1 Supply

A

B

C

OT.A

OT.B

11(200) 13(50) 17 14 0

16 18(175) 14 10(175) 0

21 24 13(275) 10(125) 0

15 17 21 18 0(50)

21 23 19 15 0(75)

250

350

400

50

75

Demand 200 225 275 300 125 1125



34

V1=16 V2=18 V3=13 V4=10 V5=0

11(200) 13(50+θ) 17 14 0

16 18(175-θ) 14 10(175) 0(θ)

21 24 13(275) 10(125) 0

15 17(-1) 21 18 0(50)

21 23 19 15 0(75)

U1=-5

U2=5

U3=5

U4=0

U5=0



35

11(200) 13(50+θ) 17 14 0

16 18(175-θ-x) 14 10(175) 0(θ+x)

21 24 13(275) 10(125) 0

15 17(x) 21 18 0(50-x)

21 23 19 15 0(75)



36



37



38



Reference Books:1. Taha H A, Operation Research - An Introduction, Prentice Hall of India, 7th edition, 2003

2. Ravindran, Phillips and Solberg, Operations Research : Principles and Practice, John

Wiely & Sons, 2nd Edition

3. D.S.Hira, Operation Research, S.Chand & Company Ltd., New Delhi, 2004

4.

unit 3 lecturer notes of transportation problem of or by dr s v prakash

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