transpotation algorithm

8/14/2019 Transpotation Algorithm

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Lecture 15

Transportation Algorithm

October 14, 2009


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Lecture 15

Outline

Recap the last lecture

Selection of the initial basic feasible solution

Northwest-corner method

Computing reduced costs of nonbasic variables

Thorugh the use of shadow prices Basis change

Operations Research Methods 1


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Lecture 15

Sun-Ray Transportation ModelGrain from Silos to Mills - Balanced problem

Mill 1 Mill 2 Mill 3 Mill 4 Supply

Silo 1 10 2 20 11 15

Silo 2 12 7 9 20 25

Silo 3 4 14 16 18 10

Demand 5 15 15 15

Northwest-corner method

Mill 1 Mill 2 Mill 3 Mill 4

Silo1 10 x11= 5 2 x12= 10 20 11

Silo 2 12 7 x22= 5 9 x23= 15 20 x24= 5

Silo 3 4 14 16 18 x34= 10

Working with the simplex method would require 12 variables, of which 6

are basic variables. We resort to a more compact representation:

- the use of the preceding table.



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Lecture 15

Simplex Method

Having the initial table (with initial basic feasible solution), we perform thetypical simplex iteration

Step 1Reduced Cost Computation

Compute the reduced costs of the nonbasic variables

Step 2Optimality CheckLooking at the reduced cost values, we check the optimality

In the transportation minimization problem: optimality requires non-

negative reduced cost

Step 3Basis ChangeIf not optimal, we perform change of basis and update the table

In the transportation minimization problem: solution is not optimal

when reduced cost is positive for some nonbasic variable



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Lecture 15

Reduced Cost Computation

We do not have Basis Inverse, so we have to rely on the dual problemandthe fact that the reduced costs of the basic variables are zero

We use shadow prices - hence, we need to look at the dual of the

transportation problem

minimizem

i=1

n

j=1

cijxij

subject ton

j=1

xij =bi for i= 1, . . . , m (ui)

m

i=1

xij =

dj for

j= 1

, . . . , n (vj)

xij 0 for all i, j



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Lecture 15

The dual of the general transportation problem

maximizem

i=1

biui+n

j=1

djvj

subject to ui+ vj cij for all (i, j) pairs (xij)

Reduced cost cij of variable xij is given by

cij =ui+ vj cij

where cij is the original cost of the variable xij as given in the objectiveof the transportatation problem



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Lecture 15

Step 1: Computing the reduced costs of nonbasic

variables

1. Determine the shadow prices of the problem corresponding to the current

basic feasible solution, using the fact that the reduced costs of the basic

variables are 0, i.e.

cij = ui+ vj cij = 0 for all basic variablesxij

We have m + n unknowns ui, vj, and m + n 1equations

One degree of freedom: set u1= 0 (or other than u1 variable)

2: Using the shadow prices determined in item 1, compute the reducedcosts of the nonbasic variables

cij =ui+ vj cij for all nonbasic variables xij



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Lecture 15

Back to Sun-Ray Case: Reduced cost


Silo1 10 x11= 5 2 x12= 10 20 11

Silo 2 12 7 x22= 5 9 x23= 15 20 x24= 5

Silo 3 4 14 16 18 x34= 10

1. Determining the shadow prices fromcij = 0for currently basic variables(with u1= 0)

x11 u1+ v1= 10 & u1= 0 v1= 10

x12 u1+ v2= 2 u1= 0 v2= 2

x22 u2+ v2= 7 v2= 2 u2= 5x23 u2+ v3= 9 u2= 5 v3= 4

x24 u2+ v4= 20 u2= 5 v4= 15

x34 u3+ v4= 18 v4= 15 u3= 3



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Lecture 15

2. Using the shadow prices, compute the reduced costs cij for nonbasic

variables

cij =ui+ vj cij


Silo1 10 x11= 5 2 x12= 10 20 11

Silo 2 12 7 x22= 5 9 x23= 15 20 x24= 5

Silo 3 4 14 16 18 x34= 10

x13 c13= u1+ v3 c13= 0 + 4 20 = 16

x14 c14= u1+ v4 c14= 0 + 15 11 =4

x21 c21= u2+ v1 c21= 5 + 10 12 =3

x31 c31= u3+ v1 c31= 3 + 10 4 =9

x32 c32= u3+ v2 c32= 3 + 2 14 = 9

x33 c33= u3+ v3 c33= 3 + 4 16 = 9



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Lecture 15

Step 2: Optimality Check

x13 c13= 16

x14 c14=4

x21 c21=3

x31 c31=9

x32 c32= 9

x33 c33= 9

We can choose any ofx14, x21, and x31.

We may choose the one with the largest reduced cost - x31.



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Lecture 15

Simplex Method

Having the initial table (with initial basic feasible solution), we perform the

typical simplex iteration

Step 1Reduced Cost Computation (DONE)

Compute the reduced costs of the nonbasic variables

Step 2Optimality Check (DONE)

Looking at the reduced cost values, we check the optimality

In the transportation minimization problem: optimality requires non-

negative reduced cost

Step 3Basis Change (We are HERE, x31 enters the basis)

If not optimal, we perform change of basis and update the table



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Lecture 15

Step 3: SunRay Basis Change


Silo1 10 x11= 5 2 x12= 10 20 11

Silo 2 12 7 x22= 5 9 x23= 15 20 x24= 5

Silo 3 4 14 16 18 x34= 10



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Lecture 15

Figure 1: Graph representation of the current basic faesible solution



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Lecture 15

Figure 2: x31 entering the current basis. Flow push along a cycleformed by arcs of the old solution and the new arc (3,1) of thevariable entering the basis



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Lecture 15

Figure 3: Table representation of the cycle and the flow push.

The maximum flow that can be send along the cycle cannot exceed the

amount of the flow on the backward traversed arcs (corresponds to removal

of the existing flow on these arcs):

5 0, 5 0, 10 0 = = 5



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Lecture 15

Figure 4: The resulting table after sending 5 units of flow along thecycle. There are two variables that can leave the basis: those whoseflow dropped to 0. Thus, either x11 or x22 may leave the basis.

Suppose we choose x11 to leave.



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Lecture 15

Figure 5: The resulting basic feasible solution after x11 left the basis.

Now we have completed a simplex iteration.

We go to the next iteration:We repeat steps 1,2,3 for this basic solution.


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