The problem of finding the minimum-cost distribution of a given commodity from a group of supply centers ( sources ) i=1,…,m to a group of receiving centers ( destinations ) j=1,…,n
Each source has a certain supply (s i )
Each destination has a certain demand (d j )
The cost of shipping from a source to a destination is directly proportional to the number of units shipped
6.
Simple Network Representation 1 2 m 1 2 n Sources Destinations … … Supply s 1 Supply s 2 Supply s m Demand d 1 Demand d 2 Demand d n x ij Costs c ij
Since any transportation problem can be formulated as an LP, we can use the simplex method to find an optimal solution
Because of the special structure of a transportation LP, the iterations of the simplex method have a very special form
The transportation simplex method is nothing but the original simplex method, but it streamlines the iterations given this special form
16.
Transportation Simplex Method Initialization (Find initial CPF solution) Is the current CPF solution optimal? Move to a better adjacent CPF solution Stop No Yes
17.
The Transportation Simplex Tableau v j Z = d n … d 2 d 1 Demand s m c mn … c m2 c m1 m … … … … … … s 2 c 2n … c 22 c 21 2 s 1 c 1n … c 12 c 11 1 n … 2 1 Source u i Supply Destination
Holiday shipments of iPods to distribution centers
Production at 3 facilities,
A, supply 200k
B, supply 350k
C, supply 150k
Distribute to 4 centers,
N, demand 100k
S, demand 140k
E, demand 300k
W, demand 250k
Total demand vs. total supply
19.
Prototype Problem W 90 0 0 0 0 Dummy 250 10 15 17 v j Z = 300 140 100 Demand 150 23 20 9 C 350 19 13 14 B 200 22 13 16 A E S N Source u i Supply Destination
In the regular simplex method, we needed to check the row-0 coefficients of each nonbasic variable to check optimality and we have an optimal solution if all are 0
There is an efficient way to find these row-0 coefficients for a given BFS to a transportation problem:
Given the basic variables, calculate values of dual variables
u i associated with each source
v j associated with each destination
using c ij – u i – v j = 0 for x ij basic, or u i + v j = c ij
(let u i = 0 for row i with the largest number of basic variables)
Row-0 coefficients can be found from c ’ ij =c ij -u i -v j for x ij nonbasic
37.
Optimality Test (1) 90 140 100 60 140 210 50 W 90 0 0 0 0 Dummy 250 10 15 17 v j 300 140 100 Demand 150 23 20 9 C 350 19 13 14 B 200 22 13 16 A E S N Source u i Supply Destination
Select the variable with the largest negative c ’ ij
Find the leaving basic variable
Determine the chain reaction that would result from increasing the value of the entering variable from zero
The leaving variable will be the first variable to reach zero because of this chain reaction
42.
Initial Solution Obtained by the Northwest Corner Rule 100 150 10 300 40 100 90 W - 4 2 - 1 -15 90 0 0 0 0 Dummy 15 250 10 15 17 2 19 13 16 v j 300 140 100 Demand 9 12 - 2 -5 150 23 20 9 C - 2 0 350 19 13 14 B 3 0 200 22 13 16 A E S N Source u i Supply Destination
43.
Iteration 1 W -15 90 0 0 0 0 Dummy 15 250 10 15 17 19 13 16 v j 300 140 100 Demand -5 150 23 20 9 C 0 350 19 13 14 B 0 200 22 13 16 A E S N Source u i Supply Destination 100 150 10 300 40 100 90 ? + - + -
44.
End of Iteration 1 100 150 100 210 40 100 90 W 90 0 0 0 0 Dummy 250 10 15 17 v j 300 140 100 Demand 150 23 20 9 C 350 19 13 14 B 200 22 13 16 A E S N Source u i Supply Destination
45.
Optimality Test 100 150 100 210 40 100 90 W 4 6 3 -19 90 0 0 0 0 Dummy 15 250 10 15 17 2 19 13 16 v j 300 140 100 Demand 9 12 - 2 -5 150 23 20 9 C - 2 0 350 19 13 14 B 3 0 200 22 13 16 A E S N Source u i Supply Destination
46.
Iteration 2 100 150 100 210 40 100 90 ? + - + - W -19 90 0 0 0 0 Dummy 15 250 10 15 17 19 13 16 v j 300 140 100 Demand -5 150 23 20 9 C 0 350 19 13 14 B 0 200 22 13 16 A E S N Source u i Supply Destination
47.
End of Iteration 2 60 150 100 210 140 90 40 W 90 0 0 0 0 Dummy 250 10 15 17 v j 300 140 100 Demand 150 23 20 9 C 350 19 13 14 B 200 22 13 16 A E S N Source u i Supply Destination
48.
Optimality Test 60 150 100 210 140 90 40 Z = 10330 W 4 8 5 -19 90 0 0 0 0 Dummy 15 250 10 15 17 0 19 11 14 v j 300 140 100 Demand 9 14 0 -5 150 23 20 9 C 2 0 350 19 13 14 B 1 2 200 22 13 16 A E S N Source u i Supply Destination
49.
Optimal Solution A B C N S W Sources Destinations Supply = 200 Supply = 350 Supply = 150 E 60 140 40 100 150 210 Demand = 100 Demand = 140 Demand = 250 Demand = 300 (shortage of 90) Cost Z = 10330
Make the assignments one at a time in positions that have zero elements.
Begin with rows or columns that have only one zero. Cross out both the row and the column involved after each assignment is made.
Move on to the rows and columns that are not yet crossed out to select the next assignment, with preference given to any such row or column that has only one zero that is not crossed out.
Continue until every row and every column has exactly one assignment and so has been crossed out.
57.
Hungarian Method 4 4 2 Goyle 5 4 1 Draco 3 2 5 Harry Lupin McG Snape Student Mentor Goyle Draco Harry Lupin McG Snape Student Mentor Goyle Draco Harry Lupin McG Snape Student Mentor Goyle Draco Harry Lupin McG Snape Student Mentor
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