The following tableau format of the problem shows that the starting basic solution (x4, x5, x6) is both nonoptimal (because of nonbasic x3) and infeasible (because of basic x4).
Basic x1 x2 x3 x4 x5 x6 Solution
z 0 0 -2 0 0 0 0
x4 1 -2 2 1 0 0 -8
x5 -1 1 1 0 1 0 4
x6 2 -1 4 0 0 1 10
We can solve the problem without the use of any artificial variables or artificial constraints, first securing feasibility using the dual simplex and then seeking optimality using the primal simplex.
The dual simplex selects x4 as the leaving variable. The entering variable can be any nonbasic variable with a negative constraint coefficient in the x4-row (recall that if no negative constraint coefficient exists, the problem has no feasible solution). In the present example, x2 has a negative co- efficient in the x4-row and is selected as the entering variable. The next tableau is thus computed as
Basic x1 x2 x3 x4 x5 x6 Solution
z 0 0 -2 0 0 0 0
x2 -12 1 -1 -12 0 0 4
x5 -12 0 2 12 1 0 0
x6 3
2 0 3 -12 0 1 14
The new solution is now feasible but nonoptimal, and we can use the primal simplex to de- termine the optimal solution. In general, had we not restored feasibility in the preceding tableau, we would repeat the procedure as necessary until feasibility is satisfied or until there is evidence that the problem has no feasible solution.
remarks. The essence of the generalized simplex method in Example 4.4-2 is that the simplex algorithm is not rigid. The literature abounds with variations of the simplex method (e.g., the primal–dual method, the criss-cross method, and the multiplex method) that give the impression that each procedure is fundamentally different, when, in fact, they all seek corner-point (basic) solutions, with a slant toward automated computations and, perhaps, computational efficiency.
4.5 Post-oPtimal analysis
In Section 3.6, we dealt with the sensitivity of the optimum solution by determining
the ranges for the different LP parameters that would keep the optimum basic vari-
ables unchanged. In this section, we deal with making changes in the parameters of the
model and finding the new optimum solution. Take, for example, a case in the poultry
industry, where an LP model is commonly used to determine the optimal feed mix per
broiler (see Example 2.2-2). The weekly consumption per broiler varies from .26 lb
(120 g) for a 1-week-old bird to 2.1 lb (950 g) for an 8-week-old bird. Additionally, the
cost of the ingredients in the mix may change periodically. These changes require peri-
odic re-calculation of the optimum solution. Post-optimal analysis determines the new
solution in an efficient way. The new computations are rooted in the use duality and the
primal–dual relationships given in Section 4.2.
The following table lists the cases that can arise in post-optimal analysis and the actions needed to obtain the new solution (assuming one exists):
Condition after parameters change Recommended action Current solution remains optimal and feasible. No further action is necessary.
Current solution becomes infeasible. Use dual simplex to recover feasibility.
Current solution becomes nonoptimal. Use primal simplex to recover optimality.
Current solution becomes both nonoptimal and infeasible.
Use the generalized simplex method to recover optimality and feasibility.
The first three cases are investigated in this section. The fourth case, being a combina- tion of cases 2 and 3, is treated in Problem 4-47.
The TOYCO model of Example 4.3-2 will be used to explain the different pro- cedures. Recall that the problem deals with the assembly of three types of toys: trains, trucks, and cars. Three operations are involved in the assembly. The model and its dual are repeated here for convenience.
TOYCO primal TOYCO dual
Maximize z =3x1 +2x2 +5x3 subject to
Minimize z =430y1 +460y2+ 420y3 subject to
x1+ 2x2 + x3 …430 1Operation 12 3x1 +2x3 …460 1Operation 22 x1+ 4x2 …420 1Operation 32
x1, x2, x3Ú 0
y1 +3y2 + y3Ú 3 2y1 +4y3Ú 2 y1 +2y2 Ú 5
y1, y2, y3Ú 0
Optimal solution: Optimal solution:
x1 =0, x2 =100, x3 =230, z = $1350 y1 =1, y2 =2, y3= 0, w=$1350
The associated optimum tableau for the primal is given as
Basic x1 x2 x3 x4 x5 x6 Solution
Z 4 0 0 1 2 0 1,350
X2 -14 1 0 1
2 -14 0 100
X3 32 0 1 0 12 0 230
X6 2 0 0 -2 1 1 20
4.5.1 Changes affecting feasibility
The feasibility of the current optimum solution is affected only if the right-hand
side of the constraints is changed, or a new constraint is added to the model. In both
cases, infeasibility occurs when one or more of the current basic variables become
negative.
4.5 Post-Optimal Analysis 187
Changes in the right-hand side. This change requires recomputing the right-hand side of the tableau using Formula 1 in Section 4.2.4:
a
New right@hand side of
tableau in iteration i
b = aInverse in
iteration i
b * aNew right@hand side of constraints
bRecall that the right-hand side of the tableau gives the values of the basic variables.
example 4.5-1
Situation 1. Suppose that TOYCO is increasing the daily capacity of operations 1, 2, and 3 to 600, 640, and 590 minutes, respectively. How would this change affect the total revenue?
With these increases, the only change that will take place in the optimum tableau is the right- hand side of the constraints (and the optimum objective value). Thus, the new basic solution is computed as follows:
£ x2
x3
x6
≥ = £
12 -14 0 0 12 0 -2 1 1
≥£ 600 640 590
≥ = £ 140 320 30
≥
Thus, the current basic variables, x2, x3, and x6, remain feasible at the new values 140, 320, and 30 units, respectively. The associated optimum revenue is $1880.
Situation 2. Although the new solution is appealing from the standpoint of increased revenue, TOYCO recognizes that its implementation may take time. Another proposal shifts the slack capacity of operation 3 1x6 = 20 minutes2 to the capacity of operation 1. How would this change impact the optimum solution?
The capacity mix of the three operations changes to 450, 460, and 400 minutes, respectively.
The resulting solution is
£ x2
x3
x6
≥ = £
12 -14 0 0 12 0 -2 1 1
≥£ 450 460 400
≥ = £ 110 230 -40
≥
The resulting solution is infeasible because x6 = -40, which requires applying the dual sim- plex method to recover feasibility. First, we modify the right-hand side of the tableau as shown by the shaded column. Notice that the associated value of z= 3*0+2* 110+5 *230= $1370.
Basic x1 x2 x3 x4 x5 x6 Solution
z 4 0 0 1 2 0 1370
x2 -14 1 0 12 -14 0 110
x3 32 0 1 0 12 0 230
x6 2 0 0 -2 1 1 -40
Using the dual simplex, x6 leaves and x4 enters, which yields the following optimal feasible tab- leau (in general, the dual simplex may take more than one iteration to recover feasibility).
Basic x1 x2 x3 x4 x5 x6 Solution
z 5 0 0 0 52 12 1350
x2 1
4 1 0 0 0 1
4 100
x3 32 0 1 0 12 0 230
x4 -1 0 0 1 -12 -12 20
The optimum solution (in terms of x1, x2, and x3) remains the same as in the original model.
This means that the proposed shift in capacity allocation is not advantageous because it simply shifts the surplus capacity from operation 3 to a surplus capacity in operation 1. The conclusion then is that operation 2 is the bottleneck, and it may be advantageous to shift the surplus to operation 2 instead (see Problem 4-42).
addition of a new constraint. The addition of a new constraint can never improve the current optimum objective value. If the new constraint is redundant, it will have no effect on the current solution. Otherwise, the current solution does not satisfy the new constraint, and a new solution is determined by the dual simplex method.
example 4.5-2
Situation 1. Suppose that TOYCO is changing the design of its toys and that the change will require the addition of a fourth assembly operation. The daily capacity of the new operation is 500 minutes and the times per unit for the three products on this operation are 3, 1, and 1 minutes, respectively.
The new constraint for operation 4 is
3x1 + x2 + x3 … 500
This constraint is redundant because it is satisfied by the current optimum solution x1= 0, x2 = 100, and x3= 230. Hence, the current optimum solution remains unchanged.
Situation 2. Suppose, instead, that TOYCO unit times on the fourth operation are changed to 3, 3, and 1 minutes, respectively. All the remaining data of the model remain the same.
The new constraint for operation 4 is
3x1 + 3x2 + x3… 500
This constraint is not satisfied by the current optimum solution; namely, for x1 = 0, x2 = 100, and x3 = 230,
x7= 500 - 13 * 0 + 3 * 100 + 1 * 2302 = -30
This means that the new constraint is not redundant. Off hand, this is not good news be- cause it indicates that the additional constraint will worsen the optimum objective value (remember the intuitive argument that additional constraints can never improve the opti- mum objective value). Nonetheless, to obtain the new solution without having to solve the
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.