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Decision Making Model

Transshipment Problem

Case Presentation

Submitted to: Prof. Hitesh Arora

Presented by :

Pankaj Bansal221084Piyush Jain221089

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Case Description

There are two Suppliers (S1, S2) three plants (P1, P2, P3)and two warehouses(W1, W2).

Warehouse receive the raw material from the two suppliers

and ship them to the three plants.

Supply value of Suppliers S1- 1000 units S2 - 1500 units

Demand of plants

P1800 units P21200 units P31000

units

Objective : To calculate the optimal shipping plan from

different sources(suppliers) to different destinations

(plants) through the transient nodes(warehouses).

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Transportation cost per unit b/w different sources

and destinations combination can be

represented as :

P1 P2 P3 W1 W2 TotalSupply

S1 17 16 1000

S2 11 13 1500

W1 15 8 20 0 13

W2 18 10 9 14 0

Total

Demand

800 1200 1000

Here, Total Supply = 2500 units and Total Demand = 3000 units

Total supply is not equal to total demand, therefore its an

Unbalanced Transportation Problem

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Balanced Transportation Problem

P1 P2 P3 W1 W2Total

Supply

S1 17 16 1000

S2 11 13 1500

Sd 0 0 500

W1 15 8 20 0 13 3000

W2 18 10 9 14 0 3000

Total

Demand

800 1200 1000 3000 3000

Cost Table

BACK 1

BACK 2

BACK 3

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P1 P2 P3 W1 W2 Supply Penalty

S1 17 16 1000 1

S2 11 13 1500 2

Sd 0 0 500 0

W1 15 8 20 0 13 3000 8

W2 18 10 9 14 0 3000 9

Demand 800 1200 1000 3000 3000

Penalty 3 2 11 0 0

1000 2000

1

USING VAM METHOD

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S1 17 16 1000 1

S2 11 13 1500 2

Sd 0 0 500 0

W1 15 8 20 0 13 3000 8

W2 18 10

1000

9

14

0 2000 10

Demand 800 1200 1000 3000 3000

Penalty 3 2 11 0 0

2000

1000

2

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S1 17 16 1000 1

S2 11 13 1500 2

Sd 0 0 500 0

W1

15

8 20 0 13

3000

8

W2 18 10

1000

9

14 2000

0 3000 10

Demand 800 1200 1000 3000 1000

Penalty 11 0 0

800 2200

3

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S1 17 16 1000 1

S2 11 13 1500 2

Sd

0 0 500

0

W1 800

15 8 20 0 13 2200

8

W2 18 10

1000

9 14

2000

0 3000 10

Demand 800 1200 1000 3000 1000

Penalty 11 0 13

1200 1000

4

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S1 17 16 1000 1

S2 11 13 1500 2

Sd

0 0

500

0

W1 800

15

1200

8 20 0 13 1000 13

W2 18 10

1000

9 14

2000

0 2000 10

Demand 800 1200 1000 3000 1000

Penalty 11 0 13

500

500

5

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S1 17 16 1000 1

S2 11 13 1500 2

Sd

0

500

0

500

0

W1 800

15

1200

8 20 0 13 1000 13

W2 18 10

1000

9 14

2000

0 3000 10

Demand 800 1200 1000 3000 500

Penalty 11 11 0

1000

2000

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S1 17 16 1000 1

S2

11 13 1500 2

Sd

0

500

0 500 0

W1 80015 12008 20 0 13 3000 13

W2 18 10

1000

9 14

2000

0 3000 10

Demand 800 1200 1000 2000 1000

500

Penalty 11 6 3

1500

500

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S1

17 16

1000

1S2

1500

11 13 1500 2

Sd

0

500

0 500 0

W1 80015

12008 20 0 13 3000 13

W2 18 10

1000

9 14

2000

0 3000 10

Demand 800 1200 1000 500 1000

500

Penalty 11 17 16

500

500

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S1

500

17 16

1000

500 16

S2

1500

11 13 1500 2

Sd

0

500

0 500 0

W1 800

15

1200

8 20 0 13 3000 13

W2 18 10

1000

9 14

2000

0 3000 10

Demand 800 1200 1000 3000 1000

500

Penalty 11 17 16

500

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S1

500

17

500

16

1000

16

S2

1500

11 13 1500 2

Sd

0

500

0 500 0

W1 800

15

1200

8 20 0 13 3000 13

W2 18 10

1000

9 14

2000

0 3000 10

Demand 800 1200 1000 3000 3000

Penalty 11 17 16

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Initial Basic Feasible Solution

(IBFS)

P1 P2 P3 W1 W2

S1 - - - 500 500

S2 - - - 1500 -

Sd - - - - 500

W1 800 1200 - 1000 -

W2 - - 1000 - 2000

Here, Number of Basic Variables = 9

Also M+N-1 = 9

Hence, Its a Non-Degenerate Solution .

BACK

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MODI Method

Ui

- - - 17 16 16

- - - 11 - 10

- - - - 0 0

15 8 - 0 - -1

- - 9 - 0 0

Vj 16 9 9 1 0

Cost for Occupied

CellsUi

- - 16

- 13 10

0 - 0

- - 20 - 13 -1

18 10 - 14 - 0

Vj 16 9 9 1 0

Cost for Unoccupied Cells

Ui + Vj= Cij IBFS COST TABLE

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p1 p2 p3 w1 W2

S1 - -

S2 - 3

Sd -1 -

W1 - - 12 - 14

W2 2 1 - 13 -

MODI Method

Net Evaluation

Cij = Cij(Ui + Vj )

As there is a negative element ( i.e. all Cij > 0condition is not satisfied) innet evaluation table, thus it is NOT AN OPTIMAL SOLUTION.

Need to introduce at the negative element position.

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P1 P2 P3 W1 W2

S1 - - - 500- 500+

S2 - - - 1500 -

Sd - - - 500-

W1 800 1200 - 1000 -

W2 - - 1000 - 2000

Minimal possible value of = 500

Two basic cells will become zero.

Introduction of Theta ()

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Basic Feasible Solution (NBFS) -

1

P1 P2 P3 W1 W2

S1 - - - - 1000

S2 - - - 1500 -

Sd - - - 500 -

W1 800 1200 - 1000 -

W2 - - 1000 - 2000


Also M+N-1 = 9

Hence, Its a Degenerate Solution .

Degeneracy is resolved by introduction of Epsilon ()

BACK

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Introduction of Epsilon()

P1 P2 P3 W1 W2

S1 - - - - 1000

S2 - - - 1500 -

Sd - - - 500 -

W1 800 1200 - 1000

W2 - - 1000 - 2000


Also M+N-1 = 9

Hence, Its a Non-Degenerate Solution

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MODI Method

Ui

- - - - 16 16

- - - 11 - 24

- - - 0 - 13

15 8 - 0 13 13

- - 9 - 0 0

Vj 2 -5 9 -13 0

Ui

17 - 16

- 13 24

- 0 13

- - 20 - - 13

18 10 - 14 - 0

Vj 2 -5 9 -13 0

Cost for occupied

cells

Cost for unoccupied cells

Ui + Vj= Cij BFS COST TABLE

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MODI Method

Net Evaluation

Cij = Cij(Ui + Vj )

p1 p2 p3 w1 W2

S1 14 -

S2 - -11

Sd - -13

W1 - - -2 - -

W2 3 15 - 21 -

As there are negative elements ( i.e. all Cij > 0condition is not satisfied) innet evaluation table, thus it is NOT AN OPTIMAL SOLUTION.

Need to introduce at the most negative element position.

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P1 P2 P3 W1 W2

S1 - - - - 1000

S2 - - - 1500 -

Sd - - - 500-

W1 800 1200 - 1000 + -

W2 - - 1000 - 2000

Minimal possible value of =

One basic cell will become zero.

Introduction of Theta ()

B i F ibl S l ti (BFS)

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Basic Feasible Solution (BFS)

- 2P1 P2 P3 W1 W2

S1 - - - - 1000

S2 - - - 1500 -

Sd - - - 500 -

W1 800 1200 - 1000 + -

W2 - - 1000 - 2000


Also M+N-1 = 9

Hence, Its a Non-Degenerate Solution .

BACK

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MODI Method

Ui

- - - - 16 16

- - - 11 - 11

- - - 0 0 0

15 8 - 0 - 0

- - 9 - 0 0

Vj 15 8 9 0 0

Ui

17 - 16

- 13 11

- - 0

- - 20 - 13 0

18 10 - 14 - 0

Vj 15 8 9 0 0

Cost for occupied

cells

Cost for unoccupied cells

Ui + Vj= Cij BFS COST TABLE

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MODI Method

Net Evaluation

Cij = Cij(Ui + Vj )

p1 p2 p3 W1 W2

S1 1 -

S2 - 2

Sd - -

W1 - - 11 - 13

W2 3 2 - 14 -

As there is no negative costs ( i.e. all Cij > 0) in net evaluationtable,

thus it is an OPTIMAL SOLUTION.

T t l

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P1 P2 P3 W1 W2Total

Supply

S1 17 16 1000

S2 11 13 1500

Sd 0 0 500

W1 15 8 20 0 13 3000

W2 18 10 9 14 0 3000

Total

Demand

800 1200 1000 3000 3000

P1 P2 P3 W1 W2

S1 - - - - 1000

S2 - - - 1500 -

Sd - - - 500 -

W1 800 1200 -1000 +

-

W2 - - 1000 - 2000

Cost Table

Optimal Table

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Final Optimal Cost

The optimal cost for transshipment fromsources to destinations through transient

nodes is :

Z* = 16*1000 + 11*1500 + 0*(500-) + 0* +

15*800 + 8*1200 + 0*(1000 + ) +

9*1000 + 0*2000

= Rs. 63100 ( is considered as negligible)

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Diagrammatic Representation

S1

S2

P1

P2

P3

1500

1000

800

1200

1000

W

W 2

Sd500

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THANK YOU

221084-89 dmm

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