0000000565-dmm presentation final

8/13/2019 0000000565-DMM Presentation Final

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Transshipment problem -I

By-

Udit Anand (211152)

Arjun Kapoor(211173)


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Here

P1,P2- Represent Manufacturing plants

W1,W2,W3- Represent Warehouses

The lines represent distances between

different plants and warehouses

The Case

This is a classical case of transshipment problem in which we have been given this

figure and we need to find out the minimum transportation cost in transporting

units from plants to warehouses with following assumptions.

(a) Inter plant and interwarehouse transfers are not allowed.

(b) The units may be transshipped.


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Since it is given that transportation costs are exactly proportional to the

distances , taking proportionality constant as unity .

Unit transportation cost=Distance b/w plants and warehouses given in

figure(in miles)

Objective- To minimize transportation costs as per given assumptions .

According to assumption (a),Inter-plant and Inter-warehouse transportation

cost are assumed to be to avoid allocations to those cells.

The total number of starting nodes (sources) as well as the total number of

ending nodes (destinations) of this transshipment problem is 2+3=5.

Converting figure into transportation table ,we get


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P1 P2 W1 W2 W3 Supply

P1 0 100 90 60 240

P2 0 120 140 110 160

W1 100 120 0 -

W2 90 140 0 -

W3 60 110 0 -

Demand - - 80 120 200

DESTINATION

SOURCE

Total Supply = 400 Total Demand = 400

Since Supply == Demand, it is a balanced problem

The value of buffer stock to be added to all the rows and columns is

400 units.


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P1 0 100 90 60 240+400

=640

P2 0 120 140 110 160+400

=560

W1 100 120 0 400

W2 90 140 0 400

W3 60 110 0 400

Demand 400 400 80+400

=480

120+40

0=520

200+4

00=60

0

DESTINATION

SOURCE

Applying VAM to find out initial basic feasible solution


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P1 P2 W1 W2 W3 Supply Penalties

P1 0 100 90 60 640 240

120

(60),(60),(60),(6

0),(60),(30),(40)

P2 0 120 140 110 560 160

80 80

(110),(10),(10),(

10),(10),(10),(10),(10)

W1 100 120 0 400 (100),(100)

W2 90 140 0 400 (90), (90),(90)

W3 60 110 0 400 (60), (60), (60),

(60)

Dema

nd

400 400 480 80

520 120

600 , 200

80

Penalt

ies

(60),(60),

(60),

(60),()

(110) (100),(1

00),(20),

(20),(20)

,(20),(20),(120)

(90),(90),(

90),(50),(

50),(50)

(60),(60),(6

0),(60),(50)

,(50),(50),(

50),(110)

400120 120

8080400

400

400

400


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P1 P2 W1 W2 W3

P1 400 ----- ----- 120 120

P2 ----- 400 80 ----- 80

W1 ----- ----- 400 ----- -----

W2 ----- ----- ----- 400 -----

W3 ----- ----- ----- ----- 400

Initial basic feasible solution

Test for degeneracym+n-1 = 5+5-1 = 9

Therefore, it is non-degenerate.

Solution= 0+ 0 + 9600 + 0 + 10800+0+ 7200 + 8800+0

=36400


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Test for Optimality

Applying MODI method

Cost of Occupied cells

1 2 3 4 5 ui

1 0 - - 90 60 60

2 - 0 120 - 110 110

3 - - 0 - - -10

4 - - - 0 - -30

5 - - - - 0 0

Vj -60 -110 10 30 0

Cost of unoccupied cells

1 2 3 4 5 ui

1 - 100 - - 60

2 - - 140 - 110

3 100 120 - -10

4 90 140 - -30

5 60 110 - 0

Vj -60 -110 10 30 0


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Net Evaluation table cij=cij-(ui+vj)- 30 - -

- - -

170 240 -

180 280 -

120 220 -

0

Since all values of the table are non-negative . IBFS is an optimal

solution.

Optimal Cost= 0+ 0 + 9600 + 0 + 3600 + 11200 + 0 + 12000+0

=36400

But since we have one 0 in the table , an alternate solutionexists.


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P1 P2 W1 W2 W3

P1 400 ----- ----- 120- 120+

P2 ----- 400 80 80-

W1 ----- ----- 400 ----- -----

W2 ----- ----- ----- 400 -----

W3 ----- ----- ----- ----- 400

Putting =80 in this solution ,we get

Applying Stepping Stone Method

Alternate Optimal solution


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P1 P2 W1 W2 W3

P1 400 ----- ----- 40 200

P2 ----- 400 80 80 -----

W1 ----- ----- 400 ----- -----

W2 ----- ----- ----- 400 -----

W3 ----- ----- ----- ----- 400

Test for degeneracy = m+n-1 = 5+5-1 = 9Therefore, it is non-degenerate.

We do not need to apply MODI method as this is an alternate solution ofan optimal table.

Optimal Solution =36400

Alternate optimal solution


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P1

P2

W2

W3

W1

120

120

80

80

Diagrammatic representation of optimal solution(IBFS)


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Since it is given that transportation costs are exactly proportional to the

distances , taking proportionality constant as unity .

Unit transportation cost=Distance b/w plants and warehouses given in

figure(in miles)

Objective- To minimize transportation costs as per given assumptions .

According to assumption (b) that the units may be

Transhipped i.e. Inter-plant and Inter-warehouse transportation cost are not but their actual allocations to those cells.

The total number of starting nodes (sources) as well as the total number of

ending nodes (destinations) of this transshipment problem is 2+3=5.

Converting figure into transportation table ,we get


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P1 0 80 100 90 60 240

P2 80 0 120 140 110 160

W1 100 120 0 60 80 -

W2 90 140 60 0 90 -

W3 60 110 80 30 0 -

Demand - - 80 120 200

Total Supply = 400

Total Demand = 400

Since Supply == Demand, it is a balancedproblem.

SOURCE

DESTINATION


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P1 0 80 100 90 60 240+400

=640

P2 80 0 120 140 110 160+400

=560

W1 100 120 0 60 80 400

W2 90 140 60 0 30 400

W3 60 110 80 30 0 400

Demand 400 400 80+400

=480

120+40

0=520

200+4

00=60

0

DESTINATION

SOURCE

P1 P2 W1 W2 W3 S l P lti


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P1 P2 W1 W2 W3 Supply Penalties

P1 0 80 100 90 60 640 240

40

(60),(60),(30),(3

0),(30),(30),(10)

P2 80 0 120 140 110 560 160

80 80

(80),(30),(10),(1

0),(10),(10),(20),

(20),(20)

W1 100 120 0 60 80 400 (60),(60),(60)

W2 90 140 60 0 30 400 (30),

(30),(30),(30)

W3 60 110 80 30 0 400 (30), (30), (30),

(30),(30)

Dema

nd

400 400 480 80 520 120

80

600 , 200

80

Penalt

ies

(60),(60) (80) (60),(60)

,(60),(20

),(20),(20),(20),(

(30),(30),(

30),(30),(

60),(50),(50),(140)

(30),(30),(3

0),(30),(60)

,(50)

400

400

400

400

400

20040

8080


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400 ----- ----- 40 200

----- 400 80 80 ------

----- ----- 400 ----- -----

----- ----- ----- 400 -----

----- ----- ----- ----- 400

Initial basic feasible solution

Test for degeneracy = m+n-1 = 5+5-1 = 9Therefore, it is non-degenerate.

Solution= 0+3600 + 12000 + 0 + 9600 + 11200 + 0 + 0+0

=36400


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Test for Optimality

Applying MODI method

Cost of Occupied cells

1 2 3 4 5 ui

1 0 - - 90 60 60

2 - 0 120 140 - 110

3 - - 0 - - -10

4 - - - 0 - -30

5 - - - - 0 0

Vj -60 -110 10 30 0

Cost of unoccupied cells

1 2 3 4 5 ui

1 - 80 100 - - 60

2 80 - - - 110 110

3 100 120 - 60 80 -10

4 90 140 60 - 30 -30

5 60 110 80 30 - 0

Vj -60 -110 10 30 0


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P1 P2 W1 W2 W3

P1 400 ----- ----- 40+ 200-

P2 ----- 400 80 80-

W1 ----- ----- 400 ----- -----

W2 ----- ----- ----- 400 -----

W3 ----- ----- ----- ----- 400




Al i l l i


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P1 P2 W1 W2 W3

P1 400 ----- ----- 120 120

P2 ----- 400 80 ---- 80

W1 ----- ----- 400 ----- -----

W2 ----- ----- ----- 400 -----

W3 ----- ----- ----- ----- 400

Test for degeneracy = m+n-1 = 5+5-1 = 9





l l l


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P1 P2 W1 W2 W3

P1 400 ----- ----- 40- 200+

P2 ----- 400 80 80 ----

W1 ----- ----- 400 ----- -----

W2 ----- ----- ----- 400 -----

W3 ----- ----- ----- 400-




Alt t ti l l ti


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P1 P2 W1 W2 W3

P1 400 ----- ----- 0 240

P2 ----- 400 80 ---- 80

W1 ----- ----- 400 ----- -----

W2 ----- ----- ----- 400 -----

W3 ----- ----- ----- ----- 360

Test for degeneracy = m+n-1 = 5+5-1 = 9






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P1

P2

W3

W2

W1

200

40

80

80


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Any Queries ???

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