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## PowerPoint Slideshow about ' Chapter 8 Sensitivity Analysis' - shelley-greer

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Chapter 8Sensitivity Analysis

- Bottom line:
- How does the optimal solution change as some of the elements of the model change?
- For obvious reasons we shall focus on Linear Programming Models.

Ingredients of LP Models

- Linear objective function
- A system of linear constraints
- RHS values
- Coefficient matrix (LHS)
- Signs (=, <=, >=)
- How does the optimal solution change as these elements change?

Parametric Changes

- Changes in one or more of the coefficients of the objective function (cj)

8.4.3 Changes in the elements of the cost vector, c.

- Suppose that the value of ck changes for some k. How will this affect the optimal solution to the LP problem?
- We therefore can distinguish between two cases:

(1) xk is not in the old basis

(2) xk is in the old basis

Observations

- r’j = 0 for basic variables xj.
- (ek)j = 0 for all nonbasic variables xj.
- if opt=max all the old reduced costs are non-negative
- if opt=min all the old reduced costs are non-positive.

8.4.3 Example

Suppose that the reduced costs in the final simplex tableau are as follows:

r = (0,0,0,2 3 4)

with IB=(2,3,1), namely with x2,x3 and x1 comprising the basis.

What would happen if we change the value of c4 ?

First we observe that x4 is not in the basis (why?) and that the opt=max

The recipe for this case, namely (8.20) is that the old optimal solution remains optimal as long as r4 ≥ , or in our case, 2 ≥ .

- Note that we do not need to know the current (old) value of c4 to reach this conclusion.
- Next, suppose that consider changes in c1, recalling that x1 is in the basis.

Preliminary Analysis

- We see that in order to analyse this case we have to know the entries in the row of the final tableau which is represented by x1 in the basis (tp.).
- What is the value of p?
- Since IB=(2,3,1), this is row p=3.
- Suppose that this row is as follows:
- t3. = (0,0,1,3,-4,0)

We can display this in a "tableau" form as follows:

If we add to the old c1, we would have instead

correction

So we now have to restore the canonical form

of the x1 column.

end result ....

- To ensure that the current basis remains optimal we have to make sure that all the reduced costs are non-negative (opt=max). Hence,
- 2+3 ≥ 0 and 3-4 ≥ 0
- Thus,

3/4 ≥ ≥ -2/3

in words, ....

- the old optimal solution will remain optimal if we keep the increase in c1 in the interval [-2/3, 3/4]. If is too small it will be better to enter x4 into the basis, if is too large it will better to put x5 into the basis.

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