operations management mba sem ii

Operations ManagementMBA Sem II

Module IV

Transportation

…the physical distribution of goods and services from several supply centers to several demand centres.

Transportation

The structure of transportation problem involves a large no. of shipping routes from several supply origins to several demand centres.

Objective : To determine the number of units of an item that should be shipped from an origin to a destination in order to satisfy the required quantity of goods or services at each destination centre.

Method

1. Formulate the problem and arrange the data in matrix form

2. Obtain an initial basic feasible solution by

- North West Corner Method

- Least Cost method

- Vogel’s Approximation Method

The solution must satisfy all the supply and demand constraints

The number of positive allocations must be m+n-1, where m is the no. of rows

and n is the no. of columns

3. Test the initial solution for optimality (MODI)

4. Update the solution

Initial basic solution by North west corner method

Q

A company has three production facilities S1,S2, S3 with production capacity of 7,9 and 18 units (in 100’s) per week of a product, respectively. These units are to be shipped to four warehouses D1, D2, D3 and D4 with requirement of 5,6,7 and 14 units (in 100’s) per week, respectively.

The transportation costs(in rupees) per unit between factories to warehouses are

given in the table in the next slide.

Minimize the total transportation cost.

D1 D2 D3 D4 CAPACITY

S1 19 30 50 10 7

S2 70 30 40 60 9

S3 40 8 70 20 18

Demand 5 8 7 14 34

• The no. of positive allocations (occupied cells) = m + n -1 = 6

D1 D2 D3 D4 CAPACITY

S1 19 5 30 2 50 10 7

S2 70 30 6 40 3 60 9

S3 40 8 70 4 20 14 18

Demand 5 8 7 14 34

Total cost = 5 *19 + 2 * 30 + 6*30 + 3 * 40 +

4 * 70 + 14* 20 = Rs. 1,015

Initial basic solution by VAM

Choose largest value 22 in columnD2

Choose cell with lowest cost

Satisfy it

D1 D2 D3 D4 capacity Row differences

S1 19 30 50 10 7 9

S2 70 30 40 60 9 10

S3 40 8 70 20 18 12

Demand 5 8 7 14 34Column differences

21 22 10 10

ghj

D1 D2 D3 D4 cap Row diff

S1 19 30 50 10 7 9

S2 70 30 40 60 9 20

S3 40 8 70 20 18 20

Demand 5 8 7 14 34Column differences

21 22 10 10

8

5

D1 D2 D3 D4 capacity

Row diff

S1 19 30 50 10 7 40

S2 70 30 40 60 9 20

S3 40 8 70 20 18 50

Demand 5 8 7 14 34

Column differences

10 10

8

5

10

D1 D2 D3 D4 capacity

Row diff

S1 19 30 50 10 7 40

S2 70 30 40 60 9 20

S3 40 8 70 20 18

Demand 5 8 7 14 34Column differences

10 50

5

8 10

2

D1 D2 D3 D4 capacity

Row diff

S1 19 30 50 10 7 40

S2 70 30 40 60 9 20

S3 40 8 70 20 18

Demand 5 8 7 14 34

Column differences

10 50

5

2

27

8 10

Total cost : 5*19 + 2*10 + 7 *40 + 2*60

+ 8 *8 + 10 *20 = Rs 779

Q

consider the following transportation problem involving 3 sources and four destinations. The cell entries represent the cost of transportation per unit. Obtain the initial basic feasible soln using

1. NWCM2. VAMFind the optimal soln (after NWCM) using

UV method

Dest

source

1 2 3 4 supply

1 3 1 7 4 300

2 2 6 5 9 400

3 8 3 3 2 500

Demand 250 350 400 200 1200

• By VAM

Dest

source

1 2 3 4 supply

1 3 1 7 4 300

2 2 6 5 9 400

3 8 3 3 2 500

Demand 250 350 400 200 1200

300

250 150

50 250 200

m+n-1 = 6

Hence this is a feasible soln

Total Cost = Rs. 2850

By NWCM

Dest

source

1 2 3 4 supply

1 3 1 7 4 300

2 2 6 5 9 400

3 8 3 3 2 500

Demand 250 350 400 200 1200

250 50

300 100

300 200

• Total cost = Rs. 4400

Optimal soln

Row 1,2,3 are assigned values U1,U2,U3

AND

Col 1, col2, col3 and col 4 are assigned variables V1,V2,V3,V4

• FOR BASIC cells Ui + Vj = cij

• Take U1 = 0

V1=3 V2=1 V3=0 V4=-1

1 2 3 4 supply

U1=0 1 3 1 7 4 300

U2=5 2 2 6 5 9 400

U3=3 3 8 3 3 2 500

250 350 400 200 1200

25050

300100

200300

If all pij<=0; optimality is reached

Compute Pij (penalties) for the non basic cells by using the formula :

Pij = Ui + Vj - cij

V1=3 V2=1 V3=0 V4=-1

1 2 3 4 supply

U1=0 1 3 1 7 4 300

U2=5 2 2 6 5 9 400

U3=3 3 8 3 3 2 500

250 350 400 200 1200

250 50

300 100

300 200

6

-ve -ve

-ve

-ve 1

If all Pij are <= 0; then optimality is reached;

Cell (2,1) with penalty 6, has the most positive penalty. So, there is scope for improving soln.

Construct a loop using one basic cell and other non basic cells.

Assign + and – signs alternatively to the basic cells.

Choose the minimum among the –vely assigned basic cells = 250

Add to all +vely cells.

Subtract from all –vely assigned cells.

V1=3 V2=1 V3=0 V4=-1

1 2 3 4 supply

U1=0 1 -

3

+ 1 7 4 300

U2=5 2 2 + -

6

5 9 400

U3=3 3 8 3 3 2 500

250 350 400 200 1200

250

6

-ve

50

300

1

-ve

100

300

-ve

-ve

200

V1=-3 V2=1 V3=0 V4=-1

1 2 3 4 supply

U1=0 1 3 1 7 4 300

U2=5 2 2 -

6

5

+

9 400

U3=3 3 8 + 3 X 3 - 2 500

250 350 400 200 1200

-ve

300

-ve -ve

250 50 100

-ve

-ve 1300 200

V1=-2 V2=1 V3=1 V4=0

1 2 3 4 supply

U1=0 1 3 1 7 4 300

U2=4 2 2 6 5 9 400

U3=2 3 8 3 3 2 500

250 350 400 200 1200

-ve300

-ve -ve

250150

-ve

-ve 50

-ve

250 200

All penalties <=0;

Hence optimum solution is reached

Total = 2850

Q

Consider the transportation problem as shown in the table. Find the initial basic solution using

1. NWCM

2. VAM

APPLY UV method to find the optimal solution.

1 2 3 4 5 supply

1 10 2 16 14 10 300

2 6 18 12 13 16 500

3 8 4 14 12 10 825

4 14 22 20 8 18 375

Demand

350 400 250 150 400

The given problem is unbalanced because the total demand(1550) < total supply (2000)

Convert to a balanced transportation problem.

1 2 3 4 5 6 Supply

1 10 2 16 14 10 0 300

2 6 18 12 13 16 0 500

3 8 4 14 12 10 0 825

4 14 22 20 8 18 0 375

Demand

350 400 250 150 400 450 2000

Initial basic soln using NWCM

1 2 3 4 5 6 supply

1 10 2 16 14 10 0 300

2 6 18 12 13 16 0 500

3 8 4 14 12 10 0 825

4 14 22 20 8 18 0 375

Demand

350 400 250 150 400 450

300

50 400 50

200 150 400 75

375

Total cost = Rs. 19,700

Initial basic solution using VAM

1 2 3 4 5 6 Supply

1 10 2 16 14 10 0 300

2 6 18 12 13 16 0 500

3 8 4 14 12 10 0 825

4 14 22 20 8 18 0 375

Demand 350 400 250 150 400 450 2000

300

350 75 75

100175 150 400

375

Total cost = Rs. 12250

Asssign Ui & Vj

V1=6 V2=2 V3=12 V4=10

V5=8

V6=0

1 2 3 4 5 6 Supply

U1=0 1 10 2 16 14 10 0 300

U2=0 2 6 18 12+ 13 16 0 - 500

U3=2 3 8 4 14 - 12 10 0 + 825

U4=0 4 14 22 20 8 18 0 375

D 350 400 250 150 400 450 2000

300

350 75 75

100 175 150 400

2 375

- ve -ve-ve -ve

-ve -ve -ve

-ve

-ve -ve -ve -ve

0

2

V1=6 V2=2 V3=12 V4=10 V5=8 V6=-2

1 2 3 4 5 6 S

U1=0 1 10 2 16 14 10 0 300

U2=0 2 6 18 12 13 16 0 500

U3=2 3 8 4 14 -

12

10 + 0 825

U4=2 4 14 22 20 8

+

18 0

-

375

D

350 400 250 150 400 450 2000

-ve -ve -ve -ve -ve

-ve -ve -ve -ve

-ve -ve -ve -ve

300

350 150

0100 100 150 400 75

4 375

V1=6 V2=2 V3=12 V4=6

V5=8 V6=-2

1 2 3 4 5 6 S

U1=0 1 10 2 16 14 10 0 300

U2=0 2 6 18 12 13 16 0 500

U3=2 3 8 4 14 -

12

10 0 825

U4=2 4 14 22 20 8 18 0

-

375

D

350 400 250 150 400 450 2000

-ve -ve -ve -ve -ve

-ve -ve -ve -ve

-ve

-ve -ve -ve -ve

300

350 150

0 100 100 400 225

150 225

• Total cost = Rs. 11,500