L09_Some Exceptional Cases in LPPs

L09_Some Exceptional Cases in LPPs - Problem 5 Maximize...

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Unformatted text preview: Problem 5 Maximize Problem Set 3.4B Pages 101-102 z = 2 x1 + 2 x2 + 4 x3 2 x1 + x2 + x3 ≤ 2 3 x1 + 4 x2 + 2 x3 ≥ 8 x1 , x2 , x3 ≥ 0 Subject to the constraints We shall solve this problem by two phase method. Phase I: Minimize r = R2 =2 Subject to the constraints 2 x1 + x2 + x3 + s1 3 x1 + 4 x2 + 2 x3 x1 , x2 , x3 , s1 , s2 , R2 ≥ 0 − s2 + R2 = 8 Here s1 is a slack variable, s2 is a surplus variable and R2 artificial variable. asic r x1 x2 3 4 r10 0 s1 R2 r s1 x2 02 03 10 0 5/4 0 3/4 1 4 0 0 1 x3 2 0 1 2 0 1/2 1/2 s1 s2 0 -1 00 1 0 0 1 0 -1 0 1/4 R2 0 -1 0 1 -1 -1/4 1/4 Sol. 8 0 2 8 0 0 2 0 -1/4 Phase I has ended. We now start Phase II. asic z x1 x2 -1/2 0 z 1 -2 -2 s1 x2 z x3 x2 0 5/4 0 3/4 17 0 5/2 0 -1/2 0 1 0 0 1 x3 -3 -4 1/2 1/2 0 1 0 s1 s2 0 -1/2 0 0 1 1/4 R2 Sol. 4 0 0 2 4 0 2 0 -1/4 6 2 1 1/2 -1 -1/2 This is the optimal tableau. Opt.Sol.: x1 = 0, x2 = 2, x3 = 0. Opt.Value: z = 4 We saw in Phase I, in the starting iteration, there was a tie for the leaving variable: namely s1 and R2. We rightly chose R2 to leave the basis. Now we shall see what happens if s1 was allowed to leave the basis. asic r x1 x2 3 4 r 10 0 s1 R2 r x2 R2 02 03 1 -5 02 0 -5 1 4 0 1 0 x3 2 0 1 2 -2 1 -2 s1 s2 0 -1 00 1 0 -4 1 -4 0 -1 -1 0 -1 R2 0 -1 0 1 0 0 1 Sol. 8 0 2 8 0 2 0 Phase I has ended. But we find the artificial variable R2 is still in the basis. Since it is at 0 level, we now start Phase II having the starting BFS as x2 and R2. (That is we have to keep the R2 column in Phase II also.) asic z x1 x2 20 z 1 -2 -2 x2 R2 z x2 x3 02 0 -5 17 0 -1/2 0 5/2 1 0 0 1 0 x3 -2 -4 1 -2 0 0 1 s1 s2 20 0 0 1 -4 0 -1 R2 0 0 0 1 -1 1/2 -1/2 Sol. 4 0 2 0 4 2 0 61 -1 -1/2 2 1/2 We now make R2 leave the basis. Remember that even though its coefficient in the pivot column is negative, it does not spoil the feasibility as RHS value is zero. As R2 has left the basis, we now forget the R2 column. The remaining tableau is optimal. Hence the optimal solution is: x1 = 0, x2 = 2, x3 = 0. And the optimum value: z = 4 (as before.) Now we shall look at four special cases that arise in the application of Simplex method. • Degeneracy • Alternative optimum solutions • Unbounded solutions • Infeasible solutions Degeneracy If in a basic feasible solution (BFS) one of the basic variables is present at 0 level (i.e. is zero) then we say we have a degenerate BFS. This usually happens when there is a tie for the leaving variable (as we saw in the previous example). There is nothing alarming in having a degenerate BFS except that in some cases it may lead to Cycling : a phenomenon in which the a leaving variable again enters the basis and again leaves repeatedly without optimality being reached. The Optimization people claim that this is a rare phenomenon and normally optimality is reached in a finite number of steps. Degeneracy can be permanent as in the previous example or can be temporary as the following example shows. Problem 2 Problem Set 3.5A Page 105 Maximize Subject to z = 3 x1 + 2 x2 4 x1 − x2 ≤ 8 4 x1 +3 x2 ≤12 4 x1 + x2 ≤ 8 x1 , x2 ≥0 We solve it by Simplex method. Basic z z 1 s1 s2 s3 z x1 s2 s3 0 0 0 1 0 0 0 x1 -3 4 4 4 0 1 0 0 x2 -2 -1 3 1 -11/4 -1/4 4 2 s1 0 1 0 0 3/4 1/4 -1 -1 s2 0 0 1 0 0 0 1 0 s3 0 0 0 1 0 0 0 1 Sol. 0 8 12 8 6 2 4 0 Note that the BFS obtained is degenerate as s3 = 0. (We also observe that there was a tie for the leaving variable.) Basic z z 1 x1 s2 x2 z x1 s1 x2 0 0 0 1 0 0 0 x1 0 1 0 0 0 1 0 0 x2 0 0 0 1 0 0 0 1 s1 -5/8 1/8 1 s2 0 0 1 s3 11/8 1/8 -2 1/2 1/8 3/8 -2 -3/2 Sol. 6 2 4 0 17/2 3/2 4 2 -1/2 0 0 5/8 0 1 0 -1/8 1 1/2 This is the optimal tableau. Note that the optimal BFS obtained is NOT degenerate. Alternative Optimal Solutions When the objective function is parallel to a binding constraint (i.e. a constraint that is satisfied as an equation), then the objective function will assume the same optimal value at more than one corner point. We say the problem has alternative optimal solutions. In this case there also will be an infinity of (non-basic) feasible solutions where the objective function has the same optimum value. Criterion for the existence of alternative optimal solutions from the optimal tableau: If ,in the optimal tableau, the coefficient of a non-basic variable is zero, then that can enter the basis and the objective function will not change (why?). Thus we shall have alternative optimal solutions as the following example illustrates. Maximize z = x1 + x2 x1 + x2 ≤ 6 x1 + 2 x2 ≤10 x1 , x2 ≥ 0 Subject to the constraints We shall solve this problem by the Simplex method. Basic z z 1 s1 s2 z x1 s2 0 0 1 0 0 x1 -1 1 1 0 1 0 x2 -1 1 2 0 1 1 s1 0 1 0 1 1 -1 s2 0 0 1 0 0 1 Sol. 0 6 10 6 6 4 z 1 0 0 1 0 6 x1 0 1 0 2 -1 2 x2 0 0 1 -1 1 4 Note that we have got an alt. opt. soln. We understand this graphically also in the next slide. This is the optimal tableau. Since coefficient of the nonbasic variable x2=0, alt. opt. soln exists. (0,6) (0,5) Optimal BFSs (2,4) All points on this line segment (except the endpoints) are nonbasic optimal feasible solutions. (10,0) (6,0) Note that the objective function line is parallel to the first (binding) constraint line. Maximize z = 2 x1 − x2 Subject to the constraints x1 − x2 ≤ 10 2 x1 − x2 ≤ 40 x1 , x2 ≥ 0 We shall solve this problem by the Simplex method. Basic z z 1 s1 s2 z x1 s2 0 0 1 0 0 x1 -2 1 2 0 1 0 x2 1 -1 -1 -1 -1 1 s1 0 1 0 2 1 -2 s2 0 0 1 0 0 1 Sol. 0 10 40 20 10 20 z 1 0 0 0 1 40 x1 0 1 0 -1 1 30 x2 0 0 1 -2 1 20 Note that we have got the optimal tableau. Since the coefficient of the nonbasic variable s1 is zero in the objective function row, we expect an alternative optimal BFS to exist. We see that the problem indeed has an infinite number of alternative feasible solutions which are NOT basic. Could you guess them? Answer: x1= 30+t; x2 = 20+2t, where t is any nonnegative number. We can easily see this graphically also. (30,20) All points on this line from (30,20) onwards are Optimal nonbasic solutions. Problem 2 Maximize Problem Set 3.5B Page 108 z = 2 x1 − x2 + 3 x3 Subject to the constraints x1 − x2 + 5 x3 ≤ 10 2 x1 − x2 + 3 x3 ≤ 40 x1 , x2 , x3 ≥ 0 Show that the above LPP has an infinite number of Optimal (nonbasic) feasible solutions. Unbounded solutions In some LPPs, the values of some decision variables can be increased indefinitely without violating any of the constraints, meaning that the solution space is unbounded in at least one direction. As a result the objective value may increase (maximization case) or decrease (minimization case) indefinitely. Thus both the feasible solution space and the objective function are unbounded. The solution space is unbounded if in a Simplex iteration, all the entries in a column corresponding to a non-basic variable are negative (or zero) in all the constraint rows and in the objective function row, the entry is: (a) negative (or zero) in the case of a maximization problem (b) positive (or zero) in the case of a minimization problem We illustrate by an example. Problem 2 Problem Set 3.5C Page 110 Maximize z = 20 x1 + 10 x2 + x3 Subject to the constraints 3 x1 −3 x2 +5 x3 ≤ 50 x1 + x3 ≤10 x1 − x2 + 4 x3 ≤ 20 x1 , x2 , x3 ≥ 0 It is clear that the solution space is unbounded in the x2 direction. Thus the variable x2 can be increased indefinitely without violating any of the constraints and thus the solution space and the objective function are both unbounded. (You can form the Simplex tableau and verify that in x2 column all the entries in the constraint rows are negative or zero and that in the z-Row, it is negative.) An example where cycling happens 3 1 Maximize z = x1 − 20 x2 + x3 − 6 x4 4 2 Subject to the constraints 1 x1 − 8 x2 − x3 + 9 x4 ≤ 0 4 1 1 x1 −12 x2 − x3 + 3 x4 ≤ 0 2 2 x3 ≤1 x1 , x2 , x3 , x4 ≥ 0 ...
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This note was uploaded on 02/23/2011 for the course MATH 112 taught by Professor Ritadubey during the Spring '11 term at Amity University.

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