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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 P2 W1 W2 W3 Supply
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 P2 W1 W2 W3 Supply
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 P2 W1 W2 W3 Supply
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
Putting =80 in this solution ,we get
Applying Stepping Stone Method
Alternate Optimal solution
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
Therefore, 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
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-
Putting =40 in this solution ,we get
Applying Stepping Stone Method
Alternate Optimal solution
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
Therefore, 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
W3
W2
W1
200
40
80
80
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Any Queries ???