The pivot row is the third row corresponding tox4 and the pivot column is 1.
The third tableau follows naturally.
Third Tableau (Corresponds to PointC)
x1 x2 x3 x4 x5 rhs
f 0 0 0.077 0 0.615 −7.077
x2 0 1 −0.231 0 0.154 3.231
x4 0 0 [−0.154] 1 −0.231 −0.846
x1 1 0 0.077 0 −0.385 1.923
Fourth Tableau (Corresponds to PointD)
x1 x2 x3 x4 x5 rhs
f 0 0 0 0.5 0.5 −7.5
x2 0 1 0 −1.5 0.5 4.5
x3 0 0 1 −6.5 1.5 5.5
x1 1 0 0 0.5 −0.5 1.5
Primal optimality has been achieved while maintaining the dual feasibility. The solution isx1 =1.5,x2=4.5, and the function value is 7.5 (flipping the sign for the minimum problem).
In summary, the dual simplex method has the same steps as the primal simplex except that we decide the exiting variable first and then the entering variable. This method is attractive when all constraints are of the GE type. The method is useful when a GE type constraint is added to the current optimal tableau.
4.12 Sensitivity Analysis 167 try variations in one or more entities while others are held at their current values.
We now attempt to answer some questions posed in the preceding. The discussion follows the theoretical development ideas of the revised simplex method. This sec- tion further assumes that degeneracy is absent.
Change of the Objective Coefficients
We note thatcTBB−1bis the function value andrTN=cTBB−1N−cTN is the vector of reduced coefficients. Or, ifvis the optimal solution of the dual, then vT =cTBB−1. Thus, the dual optimal solutionv, which is obtained in the final simplex tableau, plays an important role in sensitivity analysis. As discussed earlier in detail,vi is the change in the objective function for unit change inbi, and is referred to as the shadow price. The results are summarized as follows.
vT=cTBB−1 (4.35)
f =vTb (4.36)
Thejth reduced coefficient is given by
rj =vTN.j−cj (4.37)
whereN.jis thejthcolumn ofN. If the objective coefficient corresponding to a basic variable is changed, it results in the change of the optimum value of the objective function. In addition, the reduced coefficients have to be calculated and checked if any of them are nonpositive. If a nonbasic coefficient is changed, there is no change in the function value – but we still need to check if any of therjis negative, in which case the current point is no more optimum.
Change of the Right-Hand Side of a Constraint
From the results presented in the revised simplex method, the change in the right- hand side results in the change in the basic solutionxB=B−1b. If this basic solution is feasible for the alteredb, then thisxBis optimal since change inbdoes not affect the reduced coefficients. No further steps are needed. The function value is given by cTBxB. If any of the basic variables become negative, we need to apply dual simplex steps in obtaining the optimum.
Change of Constraint Coefficients
We present the case of coefficient change by considering the change in a column ofA. Let us denote this column asaj. Ifxj is a nonbasic variable, then we need to calculate only the reduced coefficientrjand check if it is nonnegative.
rj =vTaj−cj (4.38)
Ifrjis negative, the simplex steps are performed to find the optimum. Ifxjis a basic variable, it implies that an oldajis replaced by a newaj. The product form of the inverse presented in Eqs.4.23and4.24can be used to updateB−1. The reduced coefficients are calculated using Eqs 4.34–4.36 to see if further steps of the simplex method are needed. The function value is easily calculated.
Adding a New Variable
Adding a new variable can be treated as adding a new column to the nonbasic set.
This is readily treated by using the discussion in the preceding section.
Adding and Dropping Inequality Constraints
Adding a new constraint or dropping an existing constraint poses some interesting challenges. We consider only inequality constraints that are brought to LE form. In the development of this treatment, we make use of the following matrix identities for updating the inverse when the dimension is increased by 1 or the dimension is collapsed by 1. Let [Bd ce] and [Pr qs] be two partitioned matrices withcandqcolumn vectors,dandrrow vectors, andeandsscalars, such that
B= B c d e
and B−1= P q r s
(4.39)
IfB−1is known andc,d, andeare added,B−1can be written using the relations
s= 1
e−dB−1c q= −sB−1c
r= −sdB−1 P=B−1+qr
s (4.40)
The above relations correspond to increase of dimension by 1. These relations pro- vide an alternative way of finding the inverse of a matrix. If the dimension is col- lapsed, thenP,q,r, andsare known andB−1 is to be determined. From the last relation in Eq.(4.40), we get
B−1 =P−qr
s (4.41)
Addition of a constraintbrings in a new row with an additional variable. The column corresponding to the new variable will have a 1 in that row and the rest of the elements in the column are zeroes. By rearranging the index set we can adjust the new basis matrixBto the form given in Eq. (4.39)so that c=0,e=1, and
4.12 Sensitivity Analysis 169 dcorresponding to the variables in the basis. The updated inverse is constructed using Eq.(4.40)noting thats=1,q=0,r= −dB−1,P=B−1. Since the objective coefficient for the new basic variable is zero, there is no change in the reduced coeffi- cients. Thus, optimality condition is satisfied. Once this is done we need to check for updatedxB=B−1b, in which only the basic variable of the last row variable needs to be checked. If this is nonnegative, the basis is optimal, otherwise dual simplex iteration needs to be applied.
Dropping a constrainthas to be treated with some care. If the variable corre- sponding to the constraint isin the basis, then it is not active. For LE type constraint, the original column corresponding to the basic variable has 1 at the constraint row and zeroes in other locations. Thus,c=0ande=1. From Eq.(4.40), we observe thatq=0and it follows from Eq.(4.41)thatB−1 =P. There is no change in the partitioned inverse. Dropping of the row corresponding to the constraint and the column corresponding to the basic variable can be dropped. There is no change in the optimum point, the function value, and the reduced coefficients. This case is shown inFig. 4.7a. If the variable corresponding to the constraint isnonbasicimply- ing that the constraint is active, it leads to a two-step process. First, this variable has to be brought into basis with a move to an adjacent point, which is currently infeasible. This is done by performing the ratio test in the following way. Ifxjis the variable to be brought into the basis, we perform the ratio testfor only negative aij
andicorresponding to the leastbi/(−aij) is chosen as the pivot row. This permits the optimum point to become infeasible with respect to the deleted constraint and keep feasibility with respect to the others. If there is no negative coefficient appear- ing in the column, the removal of this constraint leads to unbounded state. Once the pivot is identified, the operation is same as the interchange of columnjwith column iof the basis. Updating of the inverse is done using the technique suggested in the revised simplex method. Now that the constraint to be dropped is in the basis, the corresponding row and column can be deleted.
Example 4.4
Consider again the example
maximize 2x1 + x2
subject to 2x1 − x2≤ 8 [1]
x1 +2x2≤14 [2]
−x1 + x2≤ 4 [3]
x1≥0,x2≥0
We know the optimum solution to be xB = [x1, x2, x5], xNB = [x3, x4].
If using the revised simplex method, we also know the current B−1 matrix.
x2 x2
x1 x6 = 0
x5 = 0 x4 = 0
x3 = 0
(a)
Inactive Constraint Dropped (x5 in basis)
New Optimum Active Constraint Dropped x4 = 0
(b)
x3 = 0 x6 = 0 x4 = 0
x5 = 0
x1
Figure 4.7. Dropping a constraint.
For the purposes of this example, we simply note that, at the optimum, B=
2 −1 0
1 2 0
−1 1 1
,b=
8 14 4
,cB=
2 1 0
,N=
1 0 0 1 0 0
cN=[0, 0]T, andf=16.
Assume that we change the constraint limits tobnew=[10, 12, 4]T. Then, the new basic solution isxnewB =B−1bnew=[6.4, 2.8, 7.6]T. SincexB≥0, we can determine the new objective to befnew=cTBxnewB =2(6.4)+1(2.8)=15.6.
Now assume that we change the objective ascnewB =[2+, 1, 0]Tand wish to determine the limits on, wherein the optimal basis does not change and where sensitivity may be used. We havevT=cTBB−1=[0.6+0.4, 0.8+0.2, 0] andfopt=vT b. The limits onare obtained from requiring thatr=vTN
−cN ≥0 :≥ −1.5 and≥ – 4. Thus, ≥–1.5. With this, sensitivity can be used to predict the solution to the above problem with the objective f = (2+)x1+x2 : fopt=16+6.
Example 4.5
Consider again Example 4.1.
(i) To what extent can the strength of cable 1 be reduced without changing the optimum basis?
At optimum, we know that cable 1 constraint is inactive. That is, force in cable 1 is less than its strength. Cable 1 strength may be reduced as long asxB≥ 0. Thus,
xB=B−1b=
⎡
⎢⎣
1 0 0 4 3 0 4 1 1
⎤
⎥⎦
−1⎡
⎢⎣ 200 1280 960−
⎤
⎥⎦≥
⎡
⎢⎣ 0 0 0
⎤
⎥⎦
4.12 Sensitivity Analysis 171 gives≤80. At=80, cable 1 constraint becomes 4W1 +W2 ≤880 and is activeat optimum. The slack corresponding to cable 1 is a basic variable with zero value, referred to as a degenerate solution (the reader is encouraged to plot the constraint set for this problem).
(ii) Assume the cable 1 constraint changes to 4W1 +W2≤850. Knowing the optimum to the problem with 4W1+W2 ≤960, efficiently determine the new optimum.
We may interpret the solution to the original problem as “near-optimal” to the current perturbed problem. Thus, the task is one of driving the current opti- mal solution to a feasible solution. From the basis (revised simplex or tableau) at optimumcorresponding to the originalb, we have: basic variables [x1,x2,x5], nonbasic variables [x3,x4], and
B=
⎡
⎢⎣
0 1 0 4 3 0 4 1 1
⎤
⎥⎦, b=
⎡
⎢⎣ 200 1280 850
⎤
⎥⎦, cB=
⎡
⎢⎣ 1 1 0
⎤
⎥⎦, cN= 0 0
, N=
⎡
⎢⎣ 1 0 0 1 0 0
⎤
⎥⎦
Using Eqs.(4.34)–(4.36), the new values of variables and reduced cost coeffi- cients are:
xB=
⎡
⎢⎣ 170 200
−30
⎤
⎥⎦, rN= .25 .25
The reduced cost coefficients remain positive, indicating “optimality”, as expected since they are independent of the right-hand side b. On the other hand, the point is “infeasible” since basic variablex5 = −30 (<0). Thus, the dual simplex method is efficient here. A tableau-based dual simplex procedure was presented in Section 4.8, and will also be used in this example, although a matrix-based revised dual simplex is computationally more efficient. The start- ing step is the final primal simplex tableau corresponding to the original con- straint 4W1+W2 ≤960. It is left as an exercise to the reader to show that this tableau is:
0 0 2 −1 1 80 (x5basic)
0 1 1 0 0 200 (x2basic)
1 0 −0.75 0.25 0 170 (x1basic)
0 0 0.25 0.25 0 370 (rNrow,f=370)
We now change 80 to−30 to obtain a starting tableau for the dual simplex method:
0 0 2 −1 1 −30 (x5basic)
0 1 1 0 0 200 (x2basic)
1 0 −0.75 0.25 0 170 (x1basic)
0 0 0.25 0.25 0 370 (rNrow,f=370)
Thus,x5has toleavethe basis. The onlyaijwith a negative coefficient is col- umn 4 in row 1. Thus, the pivot is as indicated in the preceding tableau, implying thatx4entersthe basis. This pivot operation is achieved by premultiplying the tableau byT(see Chapter 1) where
T=
⎡
⎢⎢
⎢⎣
−1 0 0 0
0 1 0 0
.25 0 1 0
.25 0 0 1
⎤
⎥⎥
⎥⎦
to obtain
0 0 −2 1 −1 30 (x4basic)
0 1 1 0 0 200 (x2basic)
1 0 −0.25 0 0.25 162.5 (x1basic)
0 0 0.75 0 0.25 362.5 (rNrow,f=362.5)
Thus, a single pivot operation yields the new optimum,x1=162.5,x2=200, f=162.5. Cable 2 constraint is slack. It was not necessary to resolve the entire problem.