4.5 Post-Optimal Analysis 189 problem completely anew, the constraint is added to the current optimum tableau as follows (x7 is a slack variable):
Basic x1 x2 x3 x4 x5 x6 x7 Solution
z 4 0 0 1 2 0 0 1350
x2 -14 1 0 12 -14 0 0 100
x3 32 0 1 0 12 0 0 230
x6 2 0 0 -2 1 1 0 20
x7 3 3 1 0 0 0 1 500
This means that that x7 = 500 is not consistent with the values of x1, x2 and x3 in the rest of the tableau. To effect consistency, the x7-row must be “conditioned” by performing the following row operations:
New x7@row= Old x7@row - [3 * 1x2@row2 + 1 * 1x3@row2]
These operations are the same as the ones used in the M-method (Section 3.4.1) to zero out the coefficients of the artificial variables in the objective function and are exactly equivalent to using the substitutions
x2= 100 - 1-14 x1 + 12 x4 - 14 x52 x3= 230 - 132 x1 + 12 x52 The new (consistent) tableau is thus given as
Basic x1 x2 x3 x4 x5 x6 x7 Solution
z 4 0 0 1 2 0 0 1350
x2 -14 1 0 1
2 -14 0 0 100
x3 32 0 1 0 12 0 0 230
x6 2 0 0 -2 1 1 0 20
x7 94 0 0 -32 1
4 0 1 -30
Application of the dual simplex method will produce the new optimum solution x1= 0, x2 = 90, x3 = 230, and z = $1330 (verify!). The solution shows that the addition of the non- redundant constraint of operation 4 is not recommended because, as expected, it lowers the revenues from $1350 to $1330.
4.5.2 Changes affecting optimality
This section considers making changes in the objective coefficients and the addition of a new economic activity (variable).
Changes in the objective function coefficients. These changes affect only the opti-
mality of the solution and require recomputing the z-row coefficients (reduced costs)
according to the following procedure:
If the new z-row satisfies the optimality condition, the solution remains unchanged (the optimum objective value may change, however). If it is not, the primal simplex is used to recover optimality.
example 4.5-3
Situation 1. In the TOYCO model, suppose that the company is instituting a revised pricing policy to meet the competition. The new unit revenues are $2, $3, and $4 for train, truck, and car toys, respectively.
The new objective function is
Maximize z = 2x1 + 3x2 + 4x3
Thus,
1New objective coefficients of basic x2, x3, and x62 = 13, 4, 02 Using Method 2, Section 4.2.3, the new dual variables are computed as
1y1, y2, y32 = 13, 4, 02£
12 -14 0 0 12 0 -2 1 1
≥ =
1
32, 54, 02
The z-row coefficients are determined as the difference between the left- and right-hand sides of the dual constraints (Formula 2, Section 4.2.4). It is not necessary to recompute the objective-row coefficients of the basic variables (x2, x3, and x6) because they are always zero regardless of any changes made in the objective coefficients (verify!).
1Reduced cost of x12 = y1 + 3y2 + y3 - 2 = 32 + 3
1
542
+ 0 - 2 = 1341Reduced cost of x42 = y1 - 0 = 32
1Reduced cost of x52 = y2 - 0 = 54
Note that the right-hand side of the first dual constraint is 2, the new coefficient in the modified objective function.
The computations show that the current solution, x1= 0 train, x2 = 100 trucks, and x3 = 230 cars, remains optimal. The corresponding new revenue is computed as 2 * 0+ 3 * 100 +4*230 = $1220. The new pricing policy is not recommended because it lowers revenue.
Situation 2. Suppose now that the TOYCO objective function is changed to Maximize z = 6x1 + 3x2 + 4x3
Will the optimum solution change?
We have
1y1, y2, y32 = 13, 4, 02£
12 -14 0 0 12 0 -2 1 1
≥ =
1
32, 54, 02
1Reduced cost of x12 = y1 + 3y2 + y3 - 6 = 32 + 3
1
542
+ 0 - 6 = -341Reduced cost of x42 = y1 - 0 = 32
1Reduced cost of x52 = y2 - 0 = 54
The new reduced cost of x1 shows that the current solution is not optimum.
4.5 Post-Optimal Analysis 191
To determine the new solution, the z-row is changed as highlighted in the following tableau:
Basic x1 x2 x3 x4 x5 x6 Solution
z -34 0 0 32 54 0 1220
x2 -14 1 0 1
2 -14 0 100
x3 3
2 0 1 0 1
2 0 230
x6 2 0 0 -2 1 1 20
The highlighted elements are the new reduced costs and the new objective value. All the re- maining elements are the same as in the original optimal tableau. The new optimum solution is then determined by letting x1 enter and x6 leave, which yields the solution x1 = 10, x2 = 102.5, x3 = 215, and z = $1227.50 (verify!). Although the new solution recommends the production of all three toys, the optimum revenue is less than that when only two toys are manufactured.
addition of a new activity. A new activity signifies adding a new variable to the model.
Intuitively, the addition of a new activity is desirable only if it is profitable. This condition can be checked by using Formula 2, Section 4.2.4, to compute the reduced cost of the new variable. The new activity is not profitable if it satisfies the optimality condition.
example 4.5-4
TOYCO recognizes that toy trains are not currently in production because they are not profitable.
The company wants to replace toy trains with a new product, a toy fire engine, to be assembled on the existing facilities. TOYCO estimates the revenue per toy fire engine to be $4 and the assembly times per unit to be 1 minute on each of operations 1 and 2, and 2 minutes on operation 3.
Let x7 represent the new fire engine product. Given that 1y1, y2, y32 = 11, 2, 02 are the op- timal dual values, we get
1Reduced cost of x72 = 1y1 + 1y2+ 2y3 - 4 = 1* 1 + 1 * 2 + 2 * 0 - 4= -1 The result shows that it is profitable to include x7 in the optimal basic solution. To obtain the new optimum, we first compute its column constraint using Formula 1, Section 4.2.4, as
x7@constraint colum = £
12 -14 0 0 12 0 -2 1 1
≥ £ 1 1 2
≥ = £
14 12
1
≥
Thus, the current simplex tableau must be modified as follows2
Basic x1 x2 x3 x7 x4 x5 x6 Solution
z 4 0 0 -1 1 2 0 1350
x2 -14 1 0 14 12 -14 0 100
x3 32 0 1 12 0 12 0 230
x6 2 0 0 1 -2 1 1 20
2As a side observation, variable x1 can be eliminated from the tableau altogether, thus reducing the size of the tableau and hence the associated amount of computations.
The new optimum is determined by letting x7 enter the basic solution, in which case x6 must leave. The new solution is x1 = 0, x2 = 0, x3= 125, x7 = 210, and z = $1465 1verify!2, which improves the revenues by $115.
remarks. The heart of the post-optimal computations is the inverse matrix of the opti- mal tableau; meaning that for the mathematics to work correctly, post-optimal sensitivity analysis cannot include changes in the data of the original problem that affect the inverse matrix (recall from Sections 4.2.2 and 4.2.3 that the inverse is computed from the basic
matrix composed of constraint columns of the original problem). So, even though the post-optimal analysis in this chapter is more encompassing than the presentation in Sections 3.6.2 and 3.6.3 in that it allows simultaneous changes in both the objective function and the constraints, it still has the shortcoming of not allowing changes in the constraint columns of basic variables. And herein lies a typical problem where mathematics is not sufficiently responsive to practical needs; meaning that, in a practical sense, we cannot use the technical excuse that the changes cannot be made “because the associated variable is basic”! Instead, the changes have to be tested in a different manner and, as stated in Chapter 3, a viable alternative in this case calls for solving the proposed LP scenario totally anew.
bibliograPhy
Bazaraa, M., J. Jarvis, and H. Sherali, Linear Programming and Network Flows, 4th ed., Wiley, New York, 2009.
Bradley, S., A. Hax, and T. Magnanti, Applied Mathematical Programming, Addison-Wesley, Reading, MA, 1977.
Diwckar, U., Introduction to Applied Optimization, Kluwer Academic Publishers, Boston, 2003.
Nering, E., and A. Tucker, Linear Programming and Related Problems, Academic Press, Boston, 1992.
Vanderbei, R., Linear Programming: Foundation and Extensions, 3rd ed., Springer, New York, 2008.
Problems
Section Assigned Problems Section Assigned Problems
4.1 4-1 to 4-6 4.3.2 4-31 to 4-34
4.2.1 4-7 to 4-7 4.4.1 4-35 to 4-39
4.2.2 4-8 to 4-9 4.4.2 4-40 to 4-41
4.2.3 4-10 to 4-18 4.5.1 4-42 to 4-49
4.2.4 4-19 to 4-27 4.5.2 4-50 to 4-56
4.3.1 4-28 to 4-30
4-1. In Example 4.1-1, derive the associated dual problem if the sense of optimization in the primal problem is changed to minimization.
*4-2. In Example 4.1-2, derive the associated dual problem given that the primal problem is augmented with a third constraint, 3x1 + x2 = 4.
4-3. In Example 4.1-3, show that even if the sense of optimization in the primal is changed to minimization, an unrestricted primal variable always corresponds to an equality dual constraint.
Problems 193 4-4. Write the dual for each of the following primal problems:
(a) Maximize z= 66x1- 22x2
subject to
-x1 + x2 … -2 2x1 + 3x2 … 5
x1, x2 Ú 0 (b) Minimize z= 6x1+ 3x2
subject to
6x1 - 3x2 + x3 Ú 25 3x1 + 4x2 + x3 Ú 55
x1, x2, x3 Ú 0 (c) Maximize z= x1 + x2
subject to
2x1 + x2 = 5 3x1 - x2 = 6 x1, x2 unrestricted
*4-5. Consider Example 4.1-1. The application of the simplex method to the primal requires the use of an artificial variable in the second constraint of the standard primal to secure a starting basic solution. Show that the presence of an artificial primal in equation form variable does not affect the definition of the dual because it leads to a redundant dual constraint.
4-6. True or False?
(a) The dual of the dual problem yields the original primal.
(b) If the primal constraint is originally in equation form, the corresponding dual variable is necessarily unrestricted.
(c) If the primal constraint is of the type …, the corresponding dual variable will be nonnegative (nonpositive) if the primal objective is maximization (minimization).
(d) If the primal constraint is of the type Ú, the corresponding dual variable will be nonnegative (nonpositive) if the primal objective is minimization (maximization).
(e) An unrestricted primal variable will result in an equality dual constraint.
4-7. Consider the following matrices:
a = £
1 4
2 5
3 6
≥, p1= a10
20b, p2 = £ 10 20 30
≥ V1 = 111, 222, V2 = 1-2, -4, -62
In each of the following cases, indicate whether the given matrix operation is legitimate, and, if so, calculate the result.
*(a) aV1 (b) ap1
*
(c) ap2 (d) V1a
*(e) V2a (f) p1p 2 (g) V1p1
4-8. Consider the optimal tableau of Example 3.3-1.
*(a) Identify the optimal inverse matrix.
(b) Show that the right-hand side equals the inverse multiplied by the original right-hand side vector of the original constraints.
4-9. Repeat Problem 4-8 for the last tableau of Example 3.4-2.
4-10. Find the optimal value of the objective function for the following problem by inspecting only its dual. (Do not solve the dual by the simplex method.)
Minimize z= 10x1 + 4x2 + 5x3
subject to
5x1- 7x2 + 3x3 Ú 20 x1, x2, x3 Ú 0
4-11. Solve the dual of the following problem, and then find its optimal solution from the solution of the dual. Does the solution of the dual offer computational advantages over solving the primal directly?
Minimize z = 50x1 + 60x2 + 30x3 subject to
5x1 + 5x2 + 3x3 Ú 50 x1 + x2 - x3 Ú 20 7x1 + 6x2 - 9x3 Ú 30 5x1 + 5x2 + 5x3 Ú 35 2x1 + 4x2 - 15x3 Ú 10 12x1 + 10x2 Ú 90 x2 - 10x3 Ú 20 x1, x2, x3 Ú 0
*4-12. Consider the following LP:
Maximize z = 5x1 + 2x2 + 3x3
subject to
x1 + 5x2+ 2x3 = 15 x1 - 5x2- 6x3 … 20
x1, x2, x3 Ú 0
Given that the artificial variable x4 and the slack variable x5 form the starting basic variables and that M was set equal to 100 when solving the problem, the optimal tableau is given as:
Basic x1 x2 x3 x4 x5 Solution
z 0 23 7 105 0 75
x1 1 5 2 1 0 15
x5 0 -10 -8 -1 1 5
Write the associated dual problem, and determine its optimal solution in two ways.
4-13. Consider the following LP:
Minimize z= 4x1+ x2
subject to
3x1 + x2 = 30 4x1 + 3x2 Ú 60 x1 + 2x2 … 40
x1, x2 Ú 0
The starting solution consists of artificial x4 and x5 for the first and second constraints and slack x6 for the third constraint. Using M = 100 for the artificial variables, the optimal tableau is given as
Basic x1 x2 x3 x4 x5 x6 Solution
z 0 0 0 -98.6 -100 -.2 34
x1 1 0 0 .4 0 -.2 4
x2 0 1 0 .2 0 .6 18
x3 0 0 1 1 -1 1 10
Write the associated dual problem, and determine its optimal solution in two ways.
4-14. Consider the following LP:
Maximize z = 2x1 + 4x2 + 4x3- 3x4
subject to
x1 + x2+ x3 = 4 x1 + 4x2 + x4= 8
x1, x2, x3, x4 Ú 0
Using x3 and x4 as starting variables, the optimal tableau is given as
Basic x1 x2 x3 x4 Solution
z 2 0 0 3 16
x3 .75 0 1 -.25 2
x2 .25 1 0 .25 2
Write the associated dual problem, and determine its optimal solution in two ways.
Problems 195
*4-15. Consider the following LP:
Maximize z = x1 + 5x2 + 3x3
subject to
x1 + 2x2+ x3 = 3 2x1 - x2 = 4
x1, x2, x3 Ú 0
The starting solution consists of x3 in the first constraint and an artificial x4 in the second constraint with M = 100. The optimal tableau is given as
Basic x1 x2 x3 x4 Solution
z 0 2 0 99 5
x3 1 2.5 1 -.5 1
x1 0 -.5 0 .5 2
Write the associated dual problem, and determine its optimal solution in two ways.
4-16. Consider the following set of inequalities:
2x1 + 3x2… 12 -3x1 + 2x2 … -4 3x1 - 5x2 … 2 x1 unrestricted
x2Ú 0
A feasible solution can be found by augmenting the trivial objective function, maximize z= x1 + x2, and then solving the problem. Another way is to solve the dual, from which a solution for the set of inequalities can be found. Apply the two methods.
4-17. Estimate a range for the optimal objective value for the following LPs:
*(a) Minimize z = 5x1+ 2x2 subject to
x1 - x2 Ú 3 2x1 + 3x2 Ú 5
x1, x2 Ú 0 (b) Maximize z = x1 + 5x2 + 3x3
subject to
x1 + 2x2 + x3 = 30 2x1 - x2 = 40
x1, x2, x3 Ú 0
(c) Maximize z= 2x1 + x2 subject to
x1 - x2… 2
2x1 … 8
x1, x2 Ú 0 (d) Maximize z= 3x1 + 2x2
subject to
2x1 + x2 … 3 3x1 + 4x2 … 12
x1, x2 Ú 0
4-18. In Problem 4-17(a), let y1 and y2 be the dual variables. Determine whether the following pairs of primal–dual solutions are optimal:
*(a) (x1 = 3, x2 = 1; y1= 4, y2= 1) (b) (x1 = 4, x2 = 1; y1= 1, y2= 0) (c) (x1 = 3, x2 = 0; y1= 5, y2= 0)
4-19. Generate the first simplex iteration of Example 4.2-1 (you may use TORA’s Iterations 1M@method with M = 100 for convenience), then use Formulas 1 and 2 to verify all the elements of the resulting tableau.
4-20. Consider the following LP model:
Maximize z = 4x1 + 14x2 subject to
2x1 + 7x2 + x3 = 21 7x1 + 2x2 + x4= 21
x1, x2, x3, x4Ú 0
Check the optimality and feasibility of each of the following basic solutions:
(a) Basic variables = 1x2, x42, Inverse = a
17 0 -27 1b (b) Basic variables = 1x2, x32, Inverse = a0 12
1 -72b (c) Basic variables = 1x2, x12, Inverse = a
457 -452
-452 7 45b (d) Basic variables = 1x1, x42, Inverse = a 12 0
-72 1b 4-21. Consider the following LP model:
Maximize z = 3x1 + 2x2 + 5x3
*
Problems 197
subject to
x1 + 2x2+ x3 + x4 = 30
3x1 + 2x3 + x5 = 60
x1 + 4x2 + x6 = 20
x2, x2, x3, x4, x5, x6 Ú 0
Check the optimality and feasibility of the following basic solutions:
(a) Basic variables = 1x4, x3, x62, Inverse = °
1 -12 0
0 12 0
0 0 1
¢
(b) Basic variables = 1x2, x3, x12, Inverse = °
14 -18 1 3 8
2 -14 -34
-1 12 12
¢
(c) Basic variables = 1x2, x3, x62, Inverse = °
12 -14 0 0 12 0 -2 1 1
¢
*4-22. Consider the following LP model:
Minimize z= 2x1+ x2 subject to
3x1 + x2 - x3 = 3 4x1 + 3x2 - x4 = 6 x1 + 2x2 + x5 = 3
x1, x2, x3, x4, x5 Ú 0
Compute the entire simplex tableau associated with the following basic solution, and check it for optimality and feasibility.
Basic variables = 1x1, x2, x52, Inverse = °
35 -15 0 -45 3
5 0
1 -1 1
¢ 4-23. Consider the following LP model:
Maximize z= 5x1 + 12x2 + 4x3
subject to
x1 + 2x2 + x3 + x4 = 5 2x1 - x2 + 3x3 = 1
x1, x2, x3, x4 Ú 0