ISyE 4231: Engineering Optimization Georgia Tech, School of ISyE Fall 2004 Instructor: Zhaosong Lu Class # 11 SM: Multiple Optimal Solutions, Unbounded Solution, Degeneracy

Announcement Quiz #2 solution is posted Exam 1 will be on this Thursday

Objectives Quick review: simplex methods for max/min LPs; Understand how the SM detects multiple optimal solutions for max/min LPs (Chapter 4.7) Understand how the SM detects an unbounded solution (Chapter 4.8) Understand the concept of degeneracy in LP (Ch. 4.11)

Review of SM Let us summarize the steps of the SM for solving Max LP when used in a tabular format: Step 1: Convert the LP to standard, then canonical form. Set up the tableau. Step 2: Obtain a BFS (if possible) from the standard form. If all constraints were < and all RHS values were nonnegative, this is easy: the slacks are all BV’s, while the original variables are all NBV’s. Step 3: Determine if the current BFS is optimal. If all of the coefficients in Row 0 (except for z) are  0 , we are optimal (why?); otherwise, our basis is not optimal. If we are optimal, remember to return to the original variables! Step 4: If the current BFS is not optimal, determine the entering variable. Choose a variable whose coefficient in Row 0 is < 0. Next, determine the leaving variable. Look down the column of the entering variable; for all coefficients > 0, determine the ratio of RHS/coefficient. The minimum such ratio determines the leaving variable. Step 5: Use ERO’s to update the tableau and obtain the new BFS, and go to Step 3.

Review of SM (cont.) The steps 3 and 4 of Simplex method for max LPs need to be modified as follows in order to solve Min LPs. The other steps are same. Step 3 Determine if the current BFS is optimal. If all of the coefficients in Row 0 (except for z) are < 0, we are optimal (why?); otherwise, our basis is not optimal. If we are optimal, remember to original variables! Step 4: If the current BFS is not optimal, determine the entering variable. Choose a variable whose coefficient in Row 0 is > 0. Next, determine the leaving variable. Look down the column of the entering variable; for all coefficients > 0, determine the ratio of RHS/coefficient. The minimum such ratio determines the leaving variable.

SM and Multiple Optimal Solutions As we discussed, when the SM is used for a minimization problem, a sufficient condition for optimality is that all Row 0 coefficients be < 0. Further, if all of the Row 0 coefficients corresponding to NBV’s are < 0, the solution is optimal and unique! To see this, note that the only way to move away from the current solution is to increase a NBV; however, because all of the NBV Row 0 coefficients are < 0, this means that the OF will strictly increase, so the only optimal solution is the current point.

SM and Multiple Optimal Sol’ns (cont.) Now if the SM tableau for a minimization problem meets the following conditions: All Row 0 coefficients (except for z) < 0 One NBV Row 0 coefficient = 0 then we have the possibility that multiple optimal solutions exist. If, in addition, the current tableau has all RHS > 0 (except Row 0), then we are guaranteed to have multiple optimal solutions!

SM and Multiple Optimal Sol’ns (cont.) Example: Notice that the current tableau is optimal, since all Row 0 coefficients are < 0. However, suppose we bring x2 into the basis. We can then pivot on the circled coefficient, so x1 leaves the basis. We then end up with the following tableau: It is clear that this tableau is also optimal! z x1 x2 x3 s1 S2 s3 RHS 1 -10 -280 -2 2 -8 24 -4 8 1.25 -0.5 1.5 z x1 x2 x3 s1 s2 s3 RHS 1 -10 -280 1.6 1.2 -5.6 27.2 -1.6 11.2 0.8 -0.4

SM and Multiple Optimal Sol’ns (cont.) The first tableau corresponds to the BFS v := (x1, x2, x3, s1, s2, s3) = (2, 0, 8, 24, 0, 0) The second tableau corresponds to the BFS w := (x1, x2, x3, s1, s2, s3) = (0, 1.6, 11.2, 27.2, 0, 0) Both have the same objective function value (z = -280) and are optimal. It follows from the convexity of the LP feasible region that is optimal for all l between 0 and 1.

SM and Multiple Optimal Sol’ns (cont.) Note that three conditions guarantee that multiple optimal solutions exist: The current tableau is optimal (i.e. all Row 0 coefficients (except for z) are < 0) One Row 0 coefficient for a NBV is = 0 All RHS (except Row 0) are positive We can tighten this somewhat: #3 can read “All RHS corresponding to positive coefficients in the entering variable column are positive” If a RHS corresponding to a positive coefficient in the entering variable column is 0, we can’t guarantee multiple optimal solutions!

SM and Multiple Optimal Sol’ns (cont.) Note that for max LPs, three conditions guarantee that multiple optimal solutions exist: The current tableau is optimal (i.e. all Row 0 coefficients (except for z) are  0 One Row 0 coefficient for a NBV is = 0 All RHS (except Row 0) are positive We can tighten this somewhat: #3 can read “All RHS corresponding to positive coefficients in the entering variable column are positive” If a RHS corresponding to a positive coefficient in the entering variable column is 0, we can’t guarantee multiple optimal solutions!

Multiple Optimal Sol’ns: 4.7.5 Show that the following LP has multiple solutions:

Multiple Optimal Sol’ns: 4.7.5 (cont.) Converting to standard form, then canonical form, we have the initial tableau: We pivot on the circled coefficient to obtain the following tableau: This tableau is optimal; further, it meets the three conditions given (bringing x2 into the basis), so multiple optimal solutions exist. z x1 x2 s1 s2 RHS 1 -2 6 2 13 z x1 x2 s1 s2 RHS 1 2 12 6 -1 -2

SM and Unboundedness The SM can detect an unbounded solution. Let us illustrate this with a sample tableau: In this case, either x3 or s2 can enter the basis. If we bring x3 into the basis, we can increase it without bound, and as a result, we can increase z without bound. To see this, let x3 = t while keeping all other NBVs at zero. Then we have: Thus we can send and while still remaining feasible. z x1 x2 x3 x4 s1 s2 RHS 1 2 -9 12 -14 100 -6 6 20 -1 5

SM and Unboundedness (cont.) Notice that the SM tableau detects an unbounded solution for max LPs under the following two conditions: A variable is eligible to enter the basis, i.e. its Row 0 coefficient is < 0 All of the constraint coefficients in that variable’s column are < 0 (so the ratio test has no ratios to check) Notice that the SM tableau detects an unbounded solution for min LPs under the following two conditions: A variable is eligible to enter the basis, i.e. its Row 0 coefficient is > 0 The second condition is as same as the one above

SM and Unboundedness: 4.8.3 Prove that the following LP has an unbounded solution:

SM and Unboundedness: 4.8.3 (cont.) After converting to standard form, we have the following initial tableau: Notice that we can bring either x1 or x2 into the basis. If we bring x2 in, we see that we can increase it without bound, while decreasing z without bound => unbounded solution! From the tableau, we see this since the Row 0 coefficient of x2 is = 2 > 0, while the constraint coefficients of x2 are -1 and 0 (both < 0). z x1 x2 s1 s2 RHS 1 3 2 -1 4

Degeneracy An LP is said to be degenerate if one of its BFS’s has a basic variable = 0 Degeneracy can cause significant problems for the Simplex Method. To see this, consider the following LP:

Degeneracy (continued) Initial tableau: Notice that our initial solution is z=0, (x1, x2, s1, s2) = (0,0,6,0) Here our minimum ratio is 0. Let’s see what happens with the next tableau… We bring x1 into the basis… z x1 x2 s1 s2 RHS 1 -5 -2 6 -1 Ratio Test: 6/1 = 6 0/1 = 0*

Degeneracy (continued) Our next tableau: Notice that our solution hasn’t changed—we still have z = 0, (x1, x2, s1, s2) = (0, 0, 6, 0) Only our basic variables have changed (now our BV’s are x1 and s1) In addition, our objective function remained unchanged z x1 x2 s1 s2 RHS 1 -7 5 2 -1 6

Degeneracy and Cycling In a nondegenerate pivot, our objective function will strictly improve—as a result, we can never visit the previous basis again. Under degeneracy, we can find ourselves returning to an earlier basis. This is called CYCLING. It does happen occasionally in practice, but there are strategies to prevent it from happening. (Notice that in the first tableau, if I let x2 be my entering variable, my minimum ratio is 6, not 0, so I get strict improvement in my objective function.)

Degeneracy and Cycling (cont.) Assume that slack and excess variables are numbered as xn+1, xn+2, … Anti-cycling technique (Bland rule) for Max LP Choose as the entering variable the variable with a negative coefficient in row 0 that has the smallest subscript; If there is a tie in the ratio test, then break the tie by choosing the winner of the ratio test so that the leaving variable has the smallest subscript

Why Does Degeneracy Occur? Degeneracy occurs because an excess number of constraints are binding at a vertex. Here is the feasible region of our previous problem: F Note that at the point (0,0), the following constraints are binding: x1 > 0 x2 > 0 x1 – x2 < 0 Only two of these need to be binding to uniquely determine the point… x1 - x2 < 0 x1 > 0 x1 + x2 < 6 (0,0) x2 > 0

Summary For a SM tableau to indicate multiple optimal solutions, for max LPs the following conditions are sufficient: The tableau is optimal: all Row 0 coeff. (except z)  0 At least one Row 0 coeff. corresponding to a NBV is = 0 In that column, look at all positive constraint coefficients; if the corresponding RHS coef.’s are > 0, we are guaranteed to have multiple optimal solutions For a SM tableau to indicate an unbounded solution for max LPs the following conditions are sufficient: At least one NBV has Row 0 coefficient <0 In a NBV column with Row 0 coef. > 0, all of the constraint coefficients in that column are < 0 Degeneracy: it might cause cycle for simplex method

Quick review for exam #1 Modeling part includes: Geometry of LP Diet, Blending, working schedule, financial planning, production process, multi-period (inventory, financial investments, working schedule) problems Geometry of LP Extreme point of LP How to find all extreme points for 2-variable LP Graphical method to solve 2-variable LP Classification of LP (using graphical method) Standard form LP, BFS, relationship between BFS and extreme points Simplex method for LP If LP is not in standard form, need to convert to standard form.