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Chapter 4 The Simplex Algorithm – Part II Based on Introduction to Mathematical Programming: Operations Research, Volume 1 4th edition, by Wayne L. Winston and Munirpallam Venkataramanan Lewis Ntaimo L. Ntaimo (c) 2005 INEN420 TAMU 1 So Far … • • • • Modeling LPs Solving 2-variable LPs graphically Simplex method - Max LPs Simplex method - Min LPs – Method 1: Multiply objective function by -1 – Method 2: Modify Step 3 of simplex algorithm for Max LPs • Simplex method – Special cases: – Alternative optimal solutions – Unbounded LPs L. Ntaimo (c) 2005 INEN420 TAMU 2 Degeneracy and Convergence of the Simplex Algorithm L. Ntaimo (c) 2005 INEN420 TAMU 3 4.11 Convergence of the Simplex Method Recall: Max z = cT x s.t. Ax = b x≥0 Where, ⎡ x1 ⎤ ⎢x ⎥ ⎢ 2⎥ ⎢.⎥ x=⎢ ⎥ ⎢.⎥ ⎢.⎥ ⎢ ⎥ ⎢⎣ xn ⎥⎦ ⎡ c1 ⎤ ⎢c ⎥ ⎢ 2⎥ ⎢.⎥ c=⎢ ⎥ ⎢.⎥ ⎢.⎥ ⎢ ⎥ ⎢⎣cn ⎥⎦ ⎡ a11 ⎢a ⎢ 21 ⎢ . A=⎢ ⎢ . ⎢ . ⎢ ⎣⎢am1 a12 . . . a22 . . . . . . . . . am 2 a1n ⎤ a2 n ⎥⎥ . ⎥ ⎥ . ⎥ . ⎥ ⎥ amn ⎦⎥ ⎡ b1 ⎤ ⎢b ⎥ ⎢ 2⎥ ⎢.⎥ b=⎢ ⎥ ⎢.⎥ ⎢.⎥ ⎢ ⎥ ⎢⎣bm ⎥⎦ Assumption: Constraint matrix A has n columns and m linearly independent rows. So we can form an m x m square matrix called the basis by setting (n - m) variables to zero (nonbasic variables or nbv’s). The remaining variables are the basic variables or bv’s. L. Ntaimo (c) 2005 INEN420 TAMU 4 4.11 Convergence of the Simplex Method Recall that for any LP with m constraints, two bfs are said to be adjacent if their sets of bv’s have (m - 1) bv’s in common. e.g. bfs of two immediate simplex tableaus are adjacent bfs. If an LP in standard form has m constraints and n variables, then there may be a basic solution for each choice of nonbasic variables. • From n variables, in how many different ways can a set of (n - m) nonbasic variables (or equivalently, m basic variables) be chosen? • ⎛n⎞ n! ⎜⎜ ⎟⎟ = ⎝ m ⎠ (n − m)!m! Thus an LP can have at most ⎛n⎞ ⎜⎜ ⎟⎟ ⎝ m⎠ basic solutions! L. Ntaimo (c) 2005 INEN420 TAMU 5 4.11 Convergence of the Simplex Method • For n = 20 and m = 10 we have: ⎛ 20 ⎞ 20! ⎜⎜ ⎟⎟ = =? 10 ( 20 − 10 )! 10 ! ⎝ ⎠ 184,756 bfs! Assuming that no bfs is repeated the simplex algorithm will find the optimal bfs after a finite number of iterations. Vast experience with the simplex algorithm indicates that an optimal solution is found after examining fewer than 3m bfs (e.g. 3(10) = 30). • Compared with examining 184,756 bfs, the simplex algorithm is quite efficient! • L. Ntaimo (c) 2005 INEN420 TAMU 6 4.11 Degeneracy and Convergence of the Simplex Method Definition: An LP is degenerate if it has at least one bfs in which a basic variable is equal to zero. • We say that an LP is nondegenerate if all bfs are positive. Theoretically, the simplex algorithm (as you have learned it so far) can fail to find the optimal solution to an LP. • However, LPs arising from actual applications seldom exhibit this behavior. • L. Ntaimo (c) 2005 INEN420 TAMU 7 4.11 Degeneracy and Convergence of the Simplex Method • Consider the following relationship for a max LP: z-value for new bfs = z-value for current bfs – (value of entering variable in new bfs)*(coefficient of entering variable in row 0 of current bfs). Recall: (coefficient of entering variable in row 0) < 0 and (value of entering variable in new bfs) >= 0. Two Facts: (1) If (value of entering variable in new bfs) > 0), then (z-value for new bfs) > (z-value for current bfs). (2) If (value of entering variable in new bfs) = 0), then (z-value for new bfs) = (z-value for current bfs). L. Ntaimo (c) 2005 INEN420 TAMU 8 4.11 Degeneracy and Convergence of the Simplex Method Fact 1 implies that each iteration of the simplex algorithm will increase z for a nondegenerate LP. This means that it is impossible to encounter the same bfs twice. So if we use the simplex algorithm to solve a nondegenerate LP, we are guaranteed to find the optimal solution in a finite number of iterations. Fact 2 implies that the simplex algorithm may fail for a degenerate LP. In this case the algorithm will encounter the same bfs at least twice. This occurrence is called cycling. If cycling occurs then the algorithm can loop, or cycle, forever among a set of bfs and never get to the optimal solution! L. Ntaimo (c) 2005 INEN420 TAMU 9 4.11 Degeneracy and Convergence of the Simplex Method Example: Solve the following LP using the simplex tableau method. Max z = 5 x1 + 2 x2 s.t. x1 + x2 ≤ 6 x1 − x2 ≤ 0 x1 , x2 ≥ 0 Converting to standard form: Max z = 5 x1 + 2 x2 s.t. x1 + x2 + s1 x1 − x2 =6 + s2 = 0 x1 , x2 , s1 , s2 ≥ 0 L. Ntaimo (c) 2005 INEN420 TAMU 10 4.11 Degeneracy and Convergence of the Simplex Method Example: Solve using the simplex tableau method Initial Tableau Row 0 1 2 Solution z 1 0 0 x1 -5 1 1 x2 -2 1 -1 s1 0 1 0 s2 0 0 1 RHS 0 6 0 BV z= s1 = s2 = Second Tableau z Row 1 0 0 1 0 2 x1 0 0 1 x2 -7 2 -1 s1 0 1 0 s2 5 -1 1 RHS 0 6 0 BV z= s1 = x1 = MRT 0 6 0 6 0* z=0 x1 = 0 x2 = 0 MRT 0 6 0 3* None z=0 x1 = 0 x2 = 0 Cycling: We encounter same bfs twice! Third (Optimal) Tableau z Row 1 0 0 1 0 2 x1 0 0 1 x2 0 1 0 s1 3.5 0.5 0.5 s2 1.5 -0.5 0.5 RHS 21 3 3 BV z= x2 = x1 = MRT 21 3 3 z = 21 x1 = 3 x2 = 3 Optimal Solution: z = 21, x1 = 3, x2 =3 L. Ntaimo (c) 2005 INEN420 TAMU 11 4.11 Degeneracy and Convergence of the Simplex Method If an LP has many degenerate bfs (or a bfs with many bv’s equal to zero), then the simplex algorithm is often very inefficient Example: Solve the previous example using the graphical method The extreme points of the feasible region are B,C and D. Three sets of basic variables correspond to extreme point C. x2 6 B x1 − x2 ≤ 0 4 D x1 + x2 ≤ 6 2 C A 2 4 z = 10 ⇒ 5 x1 + 2 x2 = 10 6 z = 21 x1 BV's x1, x2 x1, s1 x1, s2 x2, s1 x2, s2 s1, s2 BFS x1 = x2 = 3, s1 = s2 = 0 x1 = 0, s1 = 6, x2 = s2 = 0 x1 = 6, s2 = -6, x2 = s1 = 0 x2 = 0, s1 = 6, x1 = s2 = 0 x2 = 6, s2 = 6, s1 = x1 = 0 s1 = 6, s2 = 0, x1 = x2 = 0 L. Ntaimo (c) 2005 INEN420 TAMU Extreme Point D C Infeasible C B C 12 4.11 Degeneracy and Convergence of the Simplex Method For an LP with n decision variables to be degenerate, (n+1) or more of the LP’s constraints (including the sign restrictions xi ≥ 0 as constraints) must be binding at an extreme point. Fortunately, the simplex method can be modified to ensure that cycling will never occur. Apply the so called anti-cycling rules. L. Ntaimo (c) 2005 INEN420 TAMU 13 4.11 Degeneracy and Convergence of the Simplex Method Bland’s Rules for Anti-Cycling: Smallest Subscript Pivoting Rule (Assume that slack and excess variables are numbered xn+1, xn+2, ...). Step 1. Choose as the entering variable (in a max problem) the variable with a negative coefficient in row 0 that has the smallest subscript. Step 2. If there is a tie in the minimum ratio test, then break the tie by choosing the winner of the ratio test so that the variable leaving the basis has the smallest subscript. L. Ntaimo (c) 2005 INEN420 TAMU 14 4.11 Degeneracy and Convergence of the Simplex Method Additional homework problems have been posted on the course website on degeneracy and convergence of the simplex method. L. Ntaimo (c) 2005 INEN420 TAMU 15 The Big M Method L. Ntaimo (c) 2005 INEN420 TAMU 16 4.12 The Big M Method Recall that the simplex method algorithm requires a starting bfs. In all the LPs we have considered so far, we have found starting bfs by using the slack variables as our basic variables. • If an LP has ≥ or = constraints, however, a starting bfs may not be readily apparent. In such a case, we need another method to solve the problem. • L. Ntaimo (c) 2005 INEN420 TAMU 17 4.12 The Big M Method Consider the following problem: Bevco manufactures an orange-flavored soft drink called Oranj by combining orange soda and orange juice. Each orange soda contains 0.5 oz of sugar and 1 mg of vitamin C. Each ounce of orange juice contains 0.25 oz of sugar and 3 mg of vitamin C. It costs Bevco 2¢ to produce an ounce of orange soda and 3¢ to produce an ounce of orange juice. Bevco’s marketing department has decided that each 10oz bottle of Oranj must contain at least 20 mg of vitamin C and at most 4 oz of sugar. Use linear programming to determine how Bevco can meet the marketing department’s requirements at minimum cost. L. Ntaimo (c) 2005 INEN420 TAMU 18 4.12 The Big M Method Decision variables: x1 = number of ounces of orange soda in a bottle of Oranj x2 = number of ounces of orange juice in a bottle of Oranj LP: Min s.t. z = 2x1 + 3x2 0.5x1 + 0.25x2 ≤ 4 x1 + 3x2 ≥ 20 x1 + x2 = 10 (sugar constraint) (vitamin C constraint) (10 oz in 1 bottle of Oranj) x1, x2 ≥ 0 L. Ntaimo (c) 2005 INEN420 TAMU 19 4.12 The Big M Method The LP in standard: Min z = 2x1 + s.t. 3x2 0.5x1 + 0.25x2 + s1 = 4 x1 + 3x2 - e2 = 20 x1 + x2 = 10 x1, x2, s1, e2 ≥ 0 The LP in standard form has z and s1 which could be used for BVs but row 2 would violate sign restrictions and row 3 has no readily apparent basic variable. L. Ntaimo (c) 2005 INEN420 TAMU 20 4.12 The Big M Method In order to use the simplex method, a bfs is needed. The idea of the Big M method is to create artificial variables to provide an initial bfs and then eliminate them from the final solution. The variables will be labeled ai according to the row i in which they are used as follows. Row 0: Row 1: z - 2x1 - 3x2 0.5x1 + 0.25x2 + s1 Row 2: x1 + 3x2 Row 3: x1 + x2 = 4 - e2 + a2 = 20 + a3 = 10 L. Ntaimo (c) 2005 INEN420 TAMU 21 4.12 The Big M Method In the optimal solution, all artificial variables must be set equal to zero. To accomplish this, in a min LP, a term Mai is added to the objective function for each artificial variable ai. For a max LP, the term –Mai is added to the objective function for each ai. M represents some very large number. The modified Bevco LP in standard form then becomes: Min z = 2x1 + s.t. 3x2 + Ma2 + Ma3 0.5x1 + 0.25x2 + s1 x1 + 3x2 x1 + x2 = 4 - e2 + a2 = 20 + a3 = 10 x1, x2, s1, e2, a2, a3 ≥ 0 Modifying the objective function this way makes it extremely costly for an artificial variable to be positive. The optimal solution should force a2 = a3 =0. L. Ntaimo (c) 2005 INEN420 TAMU 22 4.12 The Big M Method The rows of the simplex tableau would then be: Row 0: z - 2x1 Row 1: 3x2 - Ma2 - Ma3 = 0 0.5x1 + 0.25x2 + s1 Row 2: x1 + 3x2 Row 3: x1 + x2 = 4 - e2 + a2 = 20 + a3 = 10 But observe that to have canonical form we need to eliminate a2 and a3 from row 0! Simply set new Row 0 to Row 0 + M(Row 2) + M(Row 3) z - 2x1 - M( x1 + + M( x1 + 3x2 - Ma2 - Ma3 = 0 3x2 - e2 + a2 x2 z + (2M-2)x1 +(4M-3)x2 – Me2 L. Ntaimo (c) 2005 INEN420 TAMU = 20) + a3 = 10) = 30M 23 4.12 The Big M Method Description of the Big M Method 1. Modify the constraints so that the rhs of each constraint is nonnegative. Identify each constraint that is now an = or ≥ constraint. 2. Convert each inequality constraint to standard form (add a slack variable for ≤ constraints, add an excess variable for ≥ constraints). 3. For each ≥ or = constraint, add artificial variables. Add sign restriction ai ≥ 0. 4. Let M denote a very large positive number. Add (for each artificial variable) Mai to min problem objective functions or -Mai to max problem objective functions. 5. Since each artificial variable will be in the starting basis, all artificial variables must be eliminated from row 0 before beginning the simplex. Remembering M represents a very large number, solve the transformed problem by the simplex. L. Ntaimo (c) 2005 INEN420 TAMU 24 4.12 The Big M Method If all artificial variables in the optimal solution equal zero, the solution is optimal. If any artificial variables are positive in the optimal solution, the problem is infeasible. L. Ntaimo (c) 2005 INEN420 TAMU 25 4.12 The Big M Method The Bevco example continued (Using Method 2 for Min LP): Details Iteration 1 z x1 x2 0 1.00 2M - 2 4M -3 0.50 1.00 1.00 x1 2M - 2 0.50 0.33 1.00 x1 (2M-3)/3 0.50 0.33 1.00 x1 (2M-3)/3 0.42 0.33 1.00 x1 (2M-3)/3 0.42 0.33 0.67 0.25 3 1.00 x2 4M -3 0.25 1 1.00 x2 1 2 3 ero 0 1 2 3 ero 0 1 2 3 ero 0 1 2 3 ero 0 1 2 3 1 2 3 4 z 1.00 z 1.00 z 1.00 z 1.00 0.25 1 1.00 x2 s1 a3 -1.00 s1 1.00 e2 -M a2 -0.33 0.33 1.00 a3 1.00 s1 e2 a2 (M-3)/3 (3-4M)/3 1.00 a3 1.00 -0.33 s1 s1 rhs ratio 30M 1.00 1.00 1 a2 -M 1.00 1 1.00 x2 e2 0.33 e2 a2 (M-3)/3 (3-4M)/3 0.08 -0.08 -0.33 0.33 e2 a2 (M-3)/3 (3-4M)/3 0.08 -0.08 -0.33 0.33 0.33 -0.33 1.00 a3 1.00 a3 1.00 4.00 20.00 10.00 rhs 30M 4.00 6.67 10.00 rhs (60+10M)/3 4.00 6.67 10.00 rhs (60+10M)/3 2.33 6.67 10.00 rhs (60+10M)/3 2.33 6.67 3.33 ero Row 0 + M(Row 2) + M(Row 3) 16 6.67 10 L. Ntaimo (c) 2005 INEN420 TAMU ero Row 2 x 1/3 ero Row 0 - (4M-3)*(Row 2) ero Row 1 - 0.25*(Row 2) ero Row 3 - Row 2 26 4.12 The Big M Method Iteration 2 0 1 2 3 ero 1 0 1 2 3 ero 2 0 1 2 3 ero 3 0 1 2 3 ero 4 0 1 2 3 z x1 1.00 (2M-3)/3 0.42 0.33 0.67 x1 (2M-3)/3 0.42 0.33 1.00 x1 z 1.00 z 1.00 z 1.00 0.42 0.33 1.00 x1 x2 s1 1.00 1 x2 s1 1.00 1 x2 s1 1.00 1.00 x2 s1 1.00 z 1.00 0.33 1.00 x1 1.00 x2 s1 1.00 1.00 1.00 e2 (M-3)/3 0.08 -0.33 0.33 e2 (M-3)/3 0.08 -0.33 0.50 e2 -0.50 0.08 -0.33 0.50 e2 -0.50 -0.13 -0.33 0.50 e2 -0.50 -0.13 -0.50 0.50 a2 a3 (3-4M)/3 -0.08 0.33 -0.33 1.00 a2 a3 (3-4M)/3 -0.08 0.33 -0.50 1.50 a2 a3 (1-2M)/2 (3-2M)/2 -0.08 0.33 -0.50 1.50 a2 a3 (1-2M)/2 (3-2M)/2 0.13 -0.63 0.33 -0.50 1.50 a2 a3 (1-2M)/2 (3-2M)/2 0.13 -0.63 0.50 -0.50 -0.50 1.50 rhs (60+10M)/3 2.33 6.67 3.33 rhs (60+10M)/3 2.33 6.67 5.00 rhs 25.00 2.33 6.67 5.00 rhs 25.00 0.25 6.67 5.00 rhs 25.00 0.25 5.00 5.00 ratio 5.60 20.00 5.00 ero (Row 3)*(3/2) ero Row 0 + (3-2M)*(Row 3)/3 ero Row 1 - (5/12)*Row 3) ero Row 2 -(1/3)*Row 3 Optimal Solution: z = 25, x1 = 5, x2 =5 L. Ntaimo (c) 2005 INEN420 TAMU 27 4.12 The Big M Method The Bevco example continued (Using Method 2 for Min LP): Initial Tableau Row 0 1 2 3 z 1 0 0 0 Second Tableau z Row 1 0 0 1 0 2 0 3 x1 2M-2 0.5 1 1 x2 4M-3 0.25 3 1 s1 0 1 0 0 e2 -M 0 -1 0 a2 0 0 1 0 a3 0 0 0 1 RHS 30M 4 20 10 x1 (2M-3)/3 5/12 1/3 2/3 x2 0 0 1 0 s1 0 1 0 0 e2 (M-3)/3 1/12 -1/3 1/3 a2 (3-4M)/3 -1/12 1/3 -1/3 a3 0 0 0 1 RHS (60+10M)/3 7/3 20/3 10/3 x2 0 0 1 0 s1 0 1 0 0 e2 -1/2 -1/8 -1/2 1/2 a2 (1-2M)/2 1/8 1/2 -1/2 a3 (3-2M)/2 -5/8 -1/2 3/2 RHS 25 1/4 5 5 Third (Optimal) Tableau z Row 1 0 0 1 0 2 0 3 x1 0 0 0 1 BV z= s1 = a2 = a3 = MRT 30M 4 20 10 16 6.67* 10 MRT BV z = (60+10M)/3 s1 = 7/3 5.6 x2 = 20/3 20 a3 = 10/3 5* BV z= s1 = x2 = x1 = MRT 25 1/4 5 5 Optimal Solution: z = 25, x1 = 5, x2 =5 L. Ntaimo (c) 2005 INEN420 TAMU 28 4.12 The Big M Method Issues with the Big M Method 1) It is difficult to determine how large M should be Generally chosen to be at least 100 times the largest coefficient in the objective function. 2) Introduction of such large numbers cause round off errors and other computational difficulties (numerical instability). Read section on Scaling of LPs on page 167. This motivates the need for yet another method! L. Ntaimo (c) 2005 INEN420 TAMU 29 4.12 The Big M Method Problem (10 pts). Get with your team and solve the following LP using the Big M method : Min s.t. z = 3 x1 − 2 x1 − x2 ≤ −6 x1 + 2 x2 = 3 x1 , x2 ≥ 0 L. Ntaimo (c) 2005 INEN420 TAMU 30 4.12 The Big M Method Solution to Problem (10 pts): LP is standard form: Min z = 3x1 s.t. 2 x1 + x2 − e1 = 6 x1 + 2 x2 =3 x1 , x2 , e1 ≥ 0 Big M Formulation: Min z = 3x1 + Ma1 + Ma2 s.t. 2 x1 + x2 − e1 + a1 x1 + 2 x2 =6 + a2 = 3 x1 , x2 , e1 , a1 , a2 ≥ 0 L. Ntaimo (c) 2005 INEN420 TAMU 31 4.12 The Big M Method Eliminate a1 and a2 from row 0. Simply set new Row 0 to Row 0 + M(Row 2) + M(Row 3): z - 3x1 M( 2x1 + + M( x1 + - Ma1 - Ma2 = 0 x2 - e1 + 2x2 z + (3M-3)x1 + (3M)x2 – Me1 a1 = 6) + a2 = 3) = 9M L. Ntaimo (c) 2005 INEN420 TAMU 32 4.12 The Big M Method Using Method 2 of the simplex algorithm for min LPs: 2 pts 3 pts 3 pts Initial Tableau Row 0 1 2 z 1 0 0 Second Tableau z Row 1 0 0 1 0 2 x1 3M-3 2 1 x2 3M 1 2 e1 -M -1 0 a1 0 1 0 a2 0 0 1 RHS 9M 6 3 x1 (3M-6)/2 1.5 0.5 x2 0 0 1 e1 -M -1 0 a1 0 1 0 a2 -3/2M -1 0.5 RHS 4.5M 4.5 1.5 BV z= a1 = x2 = 4.5M 4.5 1.5 x2 0 0 1 e1 -2 -2/3 0.5 a1 2-M 2/3 -0.5 a2 -(4+M)/2 -2/3 1 RHS 9 3 0 BV z= x1 = x2 = 9 3 0 Third (Optimal) Tableau z Row 1 0 0 1 0 2 x1 0 1 0 Optimal Solution: z = 9, x1 = 3, x2 = 0 L. Ntaimo (c) 2005 INEN420 TAMU BV z= a1 = a2 = MRT 9M 6 3 6 3/2* MRT 3* 3 MRT 2 pts 33 The Two-Phase Simplex Method L. Ntaimo (c) 2005 INEN420 TAMU 34 4.13 The Two-Phase Simplex Method • When an initial basic feasible solution is not available, the twophase simplex method can be used as an alternative to the Big M method. • In this method we can add artificial variables, and then focus entirely on obtaining an initial basic feasible solution, any basic feasible solution (This is called Phase I). • We can then start the simplex algorithm with the basic feasible solution we have found (This is called Phase II) L. Ntaimo (c) 2005 INEN420 TAMU 35 4.13 The Two-Phase Simplex Method • Observe that if all we want is a basic feasible solution, then we can select any objective function. • Once we find a basic feasible solution, we can reconsider the original cost coefficients. • Also note that it is easy to bring cost coefficients into canonical form. • It is desirable to choose an objective function such that solving the Phase I LP will force the artificial variables to be zero. • Minimize the sum of artificial variables! L. Ntaimo (c) 2005 INEN420 TAMU 36 4.13 The Two-Phase Simplex Method Consider solving the Bevco problem using the two-phase simplex method: Decision variables: x1 = number of ounces of orange soda in a bottle of Oranj x2 = number of ounces of orange juice in a bottle of Oranj LP: Min s.t. z = 2x1 + 3x2 0.5x1 + 0.25x2 ≤ 4 x1 + 3x2 ≥ 20 x1 + x2 = 10 (sugar constraint) (vitamin C constraint) (10 oz in 1 bottle of Oranj) x1, x2 ≥ 0 L. Ntaimo (c) 2005 INEN420 TAMU 37 4.13 The Two-Phase Simplex Method The LP in standard: Min z = 2x1 + s.t. 3x2 0.5x1 + 0.25x2 + s1 = 4 x1 + 3x2 - e2 = 20 x1 + x2 = 10 x1, x2, s1, e2 ≥ 0 Note again that the LP in standard form has z and s1 which could be used for BVs but row 2 would violate sign restrictions and row 3 has no readily apparent basic variable. L. Ntaimo (c) 2005 INEN420 TAMU 38 4.13 The Two-Phase Simplex Method Adding artificial variables we get the following: Min z = 2x1 + s.t. 3x2 0.5x1 + 0.25x2 + s1 x1 + 3x2 x1 + x2 = 4 - e2 + a2 = 20 + a3 = 10 x1, x2, s1, e2, a2, a3 ≥ 0 Then the Phase I LP is: Min w’ = s.t. a2 + a3 0.5x1 + 0.25x2 + s1 x1 + 3x2 x1 + x2 = 4 - e2 + a2 = 20 + a3 = 10 x1, x2, s1, e2, a2, a3 ≥ 0 L. Ntaimo (c) 2005 INEN420 TAMU 39 4.13 The Two-Phase Simplex Method Note that row 0 for this tableau contains the basic variables a2 and a3: Row 0: w’ Row 1: - a 2 - a3 0.5x1 + 0.25x2 + s1 Row 2: x1 + 3x2 Row 3: x1 + x2 =0 = 4 - e2 + a2 = 20 + a3 = 10 As in the Big M method, we need to eliminate a2 and a3 from row 0 to get a canonical form. Simply add Row 2+ Row 3 to Row 0: - a 2 - a3 = 0 w’ x1 + 3x2 - e2 + a2 + x1 + x2 w’ + 2x1 + 4x2 – e2 = 20 + a3 = 10 = 30 L. Ntaimo (c) 2005 INEN420 TAMU 40 4.13 The Two-Phase Simplex Method Description of the Two-Phase Simplex Method 1. Modify the constraints so that the rhs of each constraint is nonnegative. Identify each constraint that is now an = or ≥ constraint. 2. Convert each inequality constraint to standard form (add a slack variable for ≤ constraints, add an excess variable for ≥ constraints). 3. For each ≥ or = constraint, add artificial variables. Add sign restriction ai ≥ 0. 4. For now, ignore the original LP’s objective function. Instead solve an LP whose objective function is min w’ = (sum of all artificial variables). This is the Phase I LP. Solving the Phase I LP will force the artificial variables to be zero. L. Ntaimo (c) 2005 INEN420 TAMU 41 4.13 The Two-Phase Simplex Method Description of the Two-Phase Simplex Method cont… Since each ai ≥ 0, solving the Phase I LP will result in one of the following three cases: Case 1: The optimal value of w’ > 0. In this case the original LP is infeasible. Case 2: The optimal value of w’ = 0 and no artificial variables are in the optimal Phase I basis. Drop all columns in the optimal Phase I tableau that correspond to the artificial variables. In this case, combine the original objective function with the constraints from the optimal Phase I tableau. This yields the Phase II LP. The optimal solution to the Phase II LP is the optimal solution to the original LP. Case 3: The optimal value of w’ = 0 and at least one artificial variable is in the optimal Phase I basis. In this case, we can find the optimal solution to the original LP if at the end of Phase I we drop from the optimal Phase I tableau all nonbasic artificial variables and any variable from the original problem that has a negative coefficient in row 0 of the optimal Phase I tableau. L. Ntaimo (c) 2005 INEN420 TAMU 42 4.13 The Two-Phase Simplex Method The Bevco example continued (Using Method 2 for Min LP): Initial Phase I Tableau w' Row 0 1 1 0 2 0 3 0 x1 2 0.5 1 1 x2 4 0.25 3 1 s1 0 1 0 0 e2 -1 0 -1 0 a2 0 0 1 0 a3 0 0 0 1 RHS 30 4 20 10 BV w' = s1 = a2 = a3 = Second Phase I Tableau w' x1 Row 1 2/3 0 0 5/12 1 0 1/3 2 0 2/3 3 x2 0 0 1 0 s1 0 1 0 0 e2 1/3 1/12 -1/3 1/3 a2 -4/3 -1/12 1/3 -1/3 a3 0 0 0 1 RHS 10/3 7/3 20/3 10/3 BV w' = s1 = x2 = a3 = Third (Optimal) Phase I Tableau w' x1 Row 1 0 0 0 0 1 0 0 2 0 1 3 x2 0 0 1 0 s1 0 1 0 0 e2 0 -1/8 -1/2 1/2 a2 -1 1/8 1/2 -1/2 a3 -1 -5/8 -1/2 3/2 RHS 0 1/4 5 5 BV w' = s1 = x2 = x1 = MRT 30 4 20 10 16 6.67* 10 MRT 10/3 7/3 20/3 10/3 28/5 20.00 5* MRT 0 1/4 5 5 We will now drop columns a2 and a3. L. Ntaimo (c) 2005 INEN420 TAMU 43 4.13 The Two-Phase Simplex Method We begin Phase II with the following set of equations: Row 0: z - 2x1 – 3x2 =0 s1 - 1/8e2 = 1/4 Row 1: Row 2: x2 Row 3: - 1/2e2 = 5 + 1/2e2 = 5 x1 We need to eliminate x1 and x2 from row 0 to get a canonical form. Add 3(Row 2) + 2(Row 3) to Row 0: z - 2x1 – 3x2 3( x2 + 2( x1 z =0 - 1/2e2 = 5) + 1/2e2 = 5) - 1/2e2 = 25 L. Ntaimo (c) 2005 INEN420 TAMU 44 4.13 The Two-Phase Simplex Method Initial (Optimal) Phase II Tableau z x1 Row 1 0 0 0 0 1 0 0 2 0 1 3 x2 0 0 1 0 s1 0 1 0 0 e2 -1/2 -1/8 -1/2 1/2 RHS 0 1/4 5 5 BV MRT z = 25 s1 = 1/4 x2 = 5 x1 = 5 In this case Phase II requires no pivots to find an optimal solution. If the Phase II row 0 does not indicate an optimal tableau, then simply continue with the simplex until an optimal row 0 is obtained. The optimal Phase II tableau shows that the optimal solution to the Bevco problem is: Optimal Solution: z = 25, x1 = 5, x2 = 5 Note that we obtained the same solution using the Big M Method. L. Ntaimo (c) 2005 INEN420 TAMU 45 4.13 The Two-Phase Simplex Method • For examples of Case 1 and Case 3 of the two-phase simplex method read Example 6 and 7 of Chapter 4. • Important note: Example 7 should be a max problem and not a min problem! L. Ntaimo (c) 2005 INEN420 TAMU 46 4.13 The Two-Phase Simplex Method Problem (10 pts). Get with your team and solve the following LP using the two-phase simplex method : Min s.t. z = 3 x1 − 2 x1 − x2 ≤ −6 x1 + 2 x2 = 3 x1 , x2 ≥ 0 L. Ntaimo (c) 2005 INEN420 TAMU 47 4.13 The Two-Phase Simplex Method Solution to Problem (10 pts): Maintaining nonegative RHS: Min z = 3 x1 s.t. 2 x1 + x2 ≥ 6 x1 + 2 x2 = 3 x1 , x2 ≥ 0 LP is standard form: Min z = 3 x1 s.t. 2 x1 + x2 − e1 = 6 x1 + 2 x2 =3 x1 , x2 ≥ 0 Phase I Formulation: Min w' = s.t. a1 + a2 2 x1 + x2 − e1 + a1 x1 + 2 x2 =6 + a2 = 3 x1 , x2 ≥ 0 L. Ntaimo (c) 2005 INEN420 TAMU 48 4.13 The Two-Phase Simplex Method Eliminate a1 and a2 from row 0. Add Row 2 + Row 3 to Row 0: w’ - a 1 - a2 = 0 2x1 + x2 - e1 + a1 + x1 + 2x2 w’ + 3x1 + 3x2 - e1 =6 + a2 = 3 =9 L. Ntaimo (c) 2005 INEN420 TAMU 49 4.13 The Two-Phase Simplex Method Using Method 2 of the simplex algorithm for min LPs: 2 pts Initial Phase I Tableau w' Row 1 0 0 1 0 2 x1 3 2 1 x2 3 1 2 e1 -1 -1 0 a1 0 1 0 a2 0 0 1 RHS 9 6 3 BV w' = a1 = a2 = 2 pts Second Phase I Tableau w' Row 1 0 0 1 0 2 x1 0 1 0 x2 3/2 1/2 3/2 e1 1/2 -1/2 1/2 a1 -3/2 1/2 -1/2 a2 0 0 1 RHS 0 3 0 BV w' = x1 = a2 = x2 0 0 1 e1 0 -5/6 1/3 a1 -1/2 7/12 -1/6 a2 -1 -1/3 2/3 RHS 0 3 0 BV w' = x1 = x2 = Third (Optimal) Phase I Tableau w' x1 Row 1 0 0 2 pts 0 1 1 0 0 2 MRT 9 6 3 3* 3 MRT 0 3 0 6 0* MRT 0 3 0 We will now drop columns a1 and a2. L. Ntaimo (c) 2005 INEN420 TAMU 50 4.13 The Two-Phase Simplex Method We begin Phase II with the following set of equations: Row 0: z - 3x1 Row 1: x1 Row 2: =0 - 5/6e1 = 3 x2 + 1/3e2 = 0 We need to eliminate x1 from row 0 to get a canonical form. Add 3(Row 1) to Row 0: =0 z - 3x1 + 3( x1 - 5/6e1 = 3) z - 5/2e1 = 9 L. Ntaimo (c) 2005 INEN420 TAMU 51 4.13 The Two-Phase Simplex Method 2 pts Initial Phase II Tableau z x1 Row 0 1 0 1 0 1 2 0 0 x2 0 0 1 e1 -5/2 -5/6 1/3 RHS 0 3 0 BV z= x1 = x2 = MRT 9 3 0 In this case Phase II requires no pivots to find an optimal solution. If the Phase II row 0 does not indicate an optimal tableau, then simply continue with the simplex until an optimal row 0 is obtained. The optimal Phase II tableau shows that the optimal solution to the problem is: Optimal Solution: z = 9, x1 = 3, x2 = 0 2 pts Note that we obtained the same solution using the Big M Method. L. Ntaimo (c) 2005 INEN420 TAMU 52 4.13 The Two-Phase Simplex Method Summary • To get started with the simplex method, add an artificial basis, but ensure that the artificial variables do not occur in an optimal solution. • Big M method: place a large cost (penalty) on each of the artificial variables • Two-phase method: In phase I minimize the sum of the artificial variables. At the end of phase I, we will have a basic feasible solution to the original problem. Use this as a starting point for Phase II, which solves the original problem. L. Ntaimo (c) 2005 INEN420 TAMU 53