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Computational Methods for Management and Economics Carla Gomes. Module 8b The transportation simplex method. The transportation and assignment problems. Special types of linear programming problems.
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Module 8b
The transportation simplex method
Insert table 8.13
Note:all entries not shown are 0.
Insert table – 8:14
Almost the entire simplex tableau can be eliminated!
Transportation simplex methodDoes the transportation simplex method
do all the general simplex method operations?
NO!
Basically by solving a simple procedure that allows us to find the solution to the set of equations for the basic variables that have a coefficient 0 in row 0, i.e., cij – ui – vj=0.
Assume m = 10 n = 100 and see the difference!
Insert table 8.15
(*) Because there are m+n equations a BF solution in this case only has m+n1
basic variables – the other variable can be obtained as a function of the m+n1 variables
From the rows and columns still under consideration select the next basic variable (allocation) according to some criterion
Make that allocation the largest possible (the smallest of the remaining supply in its row vs. remaining demand in its column)
Eliminate the row or column that zeroed the corresponding supply/demand (in case of tie, randomly pick between the row vs. column; this is a case of a degenerate basic solution, i.e, a basic variable with value 0).
If only one row or column remains under consideration, then the procedure is complete by selecting every remaining variable associated with that row or column to be basic with the only feasible allocation. Otherwise return to step 1.
There are several methods. Here are three:
Northwest Corner Method
Minimum Cost Method
Vogel’s Method
To find the bfs by the NWC method:
Begin in the upper left (northwest) corner of the transportation tableau and set x11 as large as possible (here the limitations for setting x11 to a larger number, will be the demand of demand point 1 and the supply of supply point 1. Your x11 value can not be greater than minimum of this 2 values).
The Northwest Corner Method dos not utilize shipping costs. It can yield an initial bfs easily but the total shipping cost may be very high. The minimum cost method uses shipping costs in order come up with a bfs that has a lower cost. To begin the minimum cost method, first we find the decision variable with the smallest shipping cost (Xij). Then assign Xijits largest possible value, which is the minimum of si and dj
After that, as in the Northwest Corner Method we should cross out row i and column j and reduce the supply or demand of the noncrossedout row or column by the value of Xij. Then we will choose the cell with the minimum cost of shipping from the cells that do not lie in a crossedout row or column and we will repeat the procedure.
Begin with computing each row and column a penalty. The penalty will be equal to the difference between the two smallest shipping costs in the row or column.
Identify the row or column with the largest penalty. Find the first basic variable which has the smallest shipping cost in that row or column. Assign the highest possible value to that variable, and crossout the row or column as in the previous methods.
Compute new penalties and use the same procedure.
For each row and column remaining under consideration determine its ui – the largest cij still in that row, and its vj  the largest cij still in that column.
For each variable xij not previous selected in these rows and columns calculate ij = cij  ui  vj.
Select the variable xij with the largest (in absolute terms) negative ij (ties my be broken randomly).
21 = 14 19 – M =  5 – M
31 = 19 M– M = 19 – 2 M
41 = M M– M = – M
12 = 16 22 –19= – 25
22 = 14 19 – 19 = – 24
32 = 19 M– 19 =  M
42 = 0 M– 19= – M  19
13 = 13 22 –M= –9  M
23 = 13 19 – M = – 6  M
33 = 20 M– M = 20 – 2 M
43 = M M– M = – M
13 = 22  22 –23= –23
23 = 19  19 – 23 = – 23
33 = 23  M– 23 = – M
43 = 0  M– 23 = – M 23
14 = 17  22 –M= –5M
24 = 15  19 – M = –4M
34 = M  M– M = – M
44 = 0  M– M = – 2M
An example for Russell’s MethodStep 2: For each variable xij not previous selected in these rows and columns calculate ij = cij  ui  vj.
21 = 14 19 – 19=  24
31 = 19 M– 19 = – M
12 = 16 22 –19= – 25
22 = 14 19 – 19 = – 24
32 = 19 M– 19 =  M
13 = 13 22 –20= – 29
23 = 13 19 – 20= –26
33 = 20 M– 20 = – M
14 = 22  22 –23= –23
24 = 19  19 – 23 = – 23
34 = 23  M– 23 = – M
15 = 17  22 –M= –5M
25 = 15  19 – M = –4M
35 = M  M– M = – M
An example for Russell’s MethodStep 2: For each variable xij not previous selected in these rows and columns calculate ij = cij  ui  vj.
Note: v1 and v3 changed
An example for Russell’s MethodStep 2: For each variable xij not previous selected in these rows and columns calculate ij = cij  ui  vj.
11 = 16 22 –19 =  25
21 = 14 19 – 19=  24
31 = 19 M– 19 = – M
12 = 16 22 –19= – 25
22 = 14 19 – 19 = – 24
32 = 19 M– 19 =  M
13 = 13 22 –20= – 29
23 = 13 19 – 20= –26
33 = 20 M– 20 = – M
14 = 22  22 –23= –23
24 = 19  19 – 23 = – 23
34 = 23  M– 23 = – M
(*) In particular ij = cij  ui – vj estimate the relative values of cij  ui – vj
Optimality test – A BF solution is optimal iff
cij  ui – vj 0 for every (i,j) such that xij is nonbasic .
Note: the cij  ui – vj correspond to the coefficients in row 0.
So what do we have to do?
Optimality test – A BF solution is optimal iff
cij  ui – vj 0 for every (i,j) such that xij is nonbasic .
Note: the cij  ui – vj correspond to the coefficients in row 0.
x31: 19 = u3 + v1; set u3=0 v1 = 19;
x32: 19 = u3 + v2; v2 = 19;
x34: 19 = u3 + v4; v4 = 23;
x21: 14 = u2 + v1; v1 = 19; u2 = 5;
x23: 13 = u2 + v3; u2 = 5; v3 = 18;
x13: 13 = u1 + v3; v3 = 18; u1 = 5;
x15: 17 = u1 + v5; u1 = 5; v5 = 22;
x45: 0 = u4 + v5; v5 = 22; u4 = 22;
Now we can compute the cij  ui – vj for the nonbasic variables.
The one with the largest cij  ui – vj in absolute value
x11: 16 +5  19 = 2;
x12: 16 + 5 – 19 = 2;
x14: 22 + 5 – 23 = 4;
x22: 14 + 5 –19 = 0; v1 = 19;
x24: 19 + 5 –23 = 1; + v3;
x25: 15 + 5 –22 = 2;
x33: 20 –0 –18 = 2;
x41: M + 22 – 19 = M – 3;
x42: 0 + 22 – 19 = 3;
x43: M + 22 – 18 = M + 4;
x43: 0 + 22 – 23 = 1;
Is it optimal?
Which nonbasic variable will enter the basis?
Select as the entering variable the variable with the largest (in absolute value) negative cij  ui – vj value.
Select as the leaving variable. Increasing the entering variable from zero sets off a chain reaction of compensating changes in other basic variables. The first variable to be decreased to zero then becomes the leaving basic variable.
Find the chain involving the entering variable and some of the basic variables.
Counting the cells in the chain, label them alternating as (+) and (–) cells, starting with the entering variable with label (+);
3. Find the () cells whose variable assumes the smallest value. Call this value . The variable corresponding to this () cell will leave the basis.
4. To perform the pivot, decrease the value of each () cell by and increase the value of each + cell by . The variables that are not in the loop remain unchanged. The pivot is now complete.
( If =0, the entering variable will equal 0, and a  variable that has a current value of 0 will leave the basis. In this case a degenerate bfs existed before and will result after the pivot. If more than one  cell in the loop equals , you may arbitrarily choose one of these odd cells to leave the basis; again a degenerate bfs will result)
Ent.
Variable
+

+

+

Example of the chain
What is the value of the entering variable?
New bfs after x14 is pivoted into basis. Since There is no loop involving the cells (1,1), (1,4), (2,1), (2,2), (3,3) and (3, 4) the new solution is a bfs.
After the pivot the new bfs is x11=15, x14=20, x21=30, X22=20, X33=30 and X34=10.
In the pivoting procedure:
Since each row has as many +20s as –20s, the new solution will satisfy each supply and demand constraint.
By choosing the smallest () variable (X23) to leave the basis, we ensured that all variables will remain nonnegative.