{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lecture 6

# Lecture 6 - 540: OPERATIONSRESEARCH Lecture6:Chapter4.64.12...

This preview shows pages 1–9. Sign up to view the full content.

540:311 DETERMINISTIC MODELS IN OPERATIONS RESEARCH Lecture 6: Chapter 4.6‐4.12 Class MeeDng: Thu Feb 10 th 10:20‐11:40am Prof. W. Art Chaovalitwongse

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

View Full Document
Simplex and Its Geometry max z = 3x 1 + 2x 2 s.t. -x 1 + 3x 2 12 x 1 + x 2 8 2x 1 - x 2 10 x 1 , x 2 0 max z = + 3x 1 + 2x 2 s.t. -x 1 + 3x 2 +s 1 = 12 x 1 + x 2 +s 2 = 8 2x 1 - x 2 +s 3 = 10 x 1 , x 2 0
4.5 Problem 3 Solve the following LP. Max z = 2x 1 ‐ x 2 + x 3 st. 3x 1 + x 2 + x 3 ≤ 60 x 1 ‐ x 2 + 2x 3 ≤ 10 x 1 + x 2 x 3 ≤ 20 All variables ≥ 0

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

View Full Document
4.6 – The Simplex Algorithm (min LPs) Two different ways the simplex method can be used to solve minimizaVon problems. Consider the LP shown to the right: min z = 2x 1 – 3x 2 s.t. x 1 + x 2 4 x 1 – x 2 6 x 1 , x 2 0 The optimal solution is the point (x 1 ,x 2 ) that makes z = 2x 1 – 3x 2 the smallest. Equivalently, this point makes max -z = - 2x 1 + 3x 2 the largest. This means we can find the optimal solution to the LP by solving the LP shown to the right. max -z = -2x 1 + 3x 2 s.t. x 1 + x 2 4 x 1 – x 2 6 x 1 , x 2 0 Method 1
4.6 – The Simplex Algorithm (min LPs) In solving this modified LP, use –z as the basic variable in row 0. AYer adding slack variables s1 and s2 to the constraints we obtain the iniVal tableau. Initial Tableau Row - z x1 x2 s1 s2 rhs BVs 0 1 2 -3 0 0 0 - z = 0 1 0 1 1 1 0 4 s1 =4 2 0 1 -1 0 1 6 s2 = 6 ero1 - z x1 x2 s1 s2 rhs 0 1 2 -3 0 0 0 1 0 1 1 1 0 4 2 0 2 0 1 1 10 ero 2 - z x1 x2 s1 s2 rhs BVs 0 1 5 0 3 0 12 - z = 12 1 0 1 1 1 0 4 x2 = 4 2 0 2 0 1 1 10 s2 = 10

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

View Full Document
4.6 – The Simplex Algorithm (min LPs) The opVmal soluVon (to the max problem) is ‐z = 12, x 2 = 4, s 2 = 10, x 1 = s 1 = 0. Then the opVmal soluVon to the min problem is z = ‐12, x 2 =4, s 2 = 10, x 1 = s 2 = 0. The min LP objecVve funcVon confirms this: z = 2x 1 ‐3x 2 =2(0) – 3 (4) = ‐12. In summary, multiply the objective function for the min problem by -1 and solve the problem as a maximization problem with the objective function –z. The optimal solution to the max problem will give you the optimal solution for to the min problem. Remember that (optimal z-value for the min problem) = - (optimal z-value for the max problem).
4.6 – The Simplex Algorithm (min LPs) Method 2 A simple modification of the simplex algorithm can be used to solve min problems directly. Modify Step 3 of the simplex algorithm as follows: If all nonbasic variables (NBV) in row 0 have nonpositive coefficients, the current bfs is optimal. If any nonbasic variable has a positive coefficient, choose the variable with the most positive coefficient in row 0 as the entering variable. This modification of the simplex algorithm works because increasing a nonbasic variable (NBV) with a positive coefficient in row 0 will decrease z.

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

View Full Document
Example of MinimizaVon Problem Min z=2x 1 ‐ 3x 2 st. x 1 + x 2 ≤ 4 x 1 ‐ x 2 ≤ 6 All variables ≥ 0 1 1 1 -1 0 0 -2 +3 1 = = 6 4 x 1 x 2 -z = 0 1 0 0 1 0 0 s 2 s 1 The basic feasible soluVon is x 1 = 0, x 2 = 0, s 1 = 4, s 2 = 6 The basic variables are s 1 and s 2 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}