s02lec05

# s02lec05 - 15.053 Todays Lecture February 21, 2002 Simplex...

This preview shows pages 1–3. Sign up to view the full content.

1 15.053 February 21, 2002 z Simplex Method Continued Handouts: Lecture Notes Note: This lecture is intended to be viewed as a slide show 2 Today’s Lecture z Review of the simplex algorithm. z Formalizing the approach z Degeneracy and Alternative Optimal Solutions z Is the simplex algorithm finite? (Answer, yes, but only if we are careful) 3 -3 3 -4 2 0 0 -3 2 1 LP Canonical Form = LP Standard Form + Jordan Canonical Form = = 2 6 x 1 x 2 -z = 0 The basic feasible solution is x 1 = 0, x 2 = 0, x 3 = 6, x 4 = 2 The basic variables are x 3 and x 4 . The non-basic variables are x 1 and x 2 . z is not a decision variable 1 0 0 1 0 0 x 4 x 3 4 Basic Variables x 1 x 2 x m+1 x m 1 0 1 0 -z 0 0 x r x n 0 a 1,m+1 a 2,m+1 0 0 0 a 1,n a 2,n b 1 b 2 = = x s a 1,s a 2,s 0 0 0 a r,m+1 0 1 a r,n b r = a r,s 0 0 a m,m+1 1 0 a m,n b m = a m,s 0 0 1 c m+1 0 0 c n - z 0 = c s 0 CV m constraints, n variables Non-basic Variables 5 Notation z n number of variables z m number of constraints z s index of entering variable z r index of pivot row note: the r th basic variable leaves the basis z The original data is c j , a ij , b i z After performing pivots we represent the revised coefficients as c j , a ij , b i 6 The basic feasible solution z The current values are all non-negative. This is needed for canonical form z There is a basic variable associated with each constraint. in this case, the basic variable associated with constraint i is x i . z There are n-m nonbasic variables. In this case, the nonbasic variables are x m+1 , … x n . z The bfs is as follows: x 1 = b 1 , … , x m = b m . All other variables are 0.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
7 -3 3 1 0 -4 2 0 1 Optimality Conditions (maximization) = = 2 6 x 2 x 4 x 3 -2 -4 0 0 -z 0 0 1= -8 This basic feasible solution is optimal! What are the optimality conditions, expressed in terms of c? x 1 8 x 1 -3 0 -4 0 1 Pivoting and the min ratio rule = = x 2 x 4 x 3 -3 0 0 -z 0 0 0 1 x 1 = 0 x 2 = x 3 = 6 - 3 x 4 = 2 - 2 z = 2 x 3 = 0 when = 6/3 x 4 = 0 when = 2/2. 6 x 3 x 4 BV -z 2 is set to the min(6/3, 2/2) = min ( b 1 / a 1s , b 2 / a 2s ). The constraint with a changed basic variable is constraint r, where r = argmin ( b 1 / a 1s , b 2 / a 2s ) = 2. Min ratio rule . Express the min ratio rule using general coefficients Pivot in variable x s , where c s > 0. 2 3 2 9 x 1 Pivoting to obtain a better solution = = x 2 x 4 x 3 -z 0 0 x 1 = 0 x 2 = 1 x 3 = 3 x 4 = 0 z = 2 x 3 x 4 BV -z Pivot in variable x s , where c s > 0. Pivot out the basic variable for constraint r according to the min ratio rule. -2 1 0 1/2 1 0 0 -1 -2 3 0 1 -3/2 3 1 10 1 3 1 0 -1 2 0 1 Alternative Optima (maximization) = = 2 6 x 2 x 4 x 3 0 -4 0 0 -z 0 0 -8 This basic feasible solution is optimal!
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 12/20/2011 for the course BUS 15.053 taught by Professor Prof.jamesorlin during the Spring '05 term at MIT.

### Page1 / 6

s02lec05 - 15.053 Todays Lecture February 21, 2002 Simplex...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online