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### IEOR162_hw06

Course: IEOR 162, Spring 2010
School: Berkeley
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162, IEOR Spring 2010 Suggested Solution to Homework 06 Problem 4.1.1 The standard form is max 3x1 + 2x2 s.t. 2x1 + x2 + x3 = 100 x1 + x2 + x4 = 80 x1 + x5 = 40 xj 0 j = 1, ..., 5. Problem 4.1.2 The standard form is min 50x1 + 100x2 s.t. 7x1 + 2x2 x3 = 28 2x1 + 12x2 x4 = 24 xj 0 j = 1, ..., 4. Problem 4.1.3 The standard form is min 3x1 + x2 s.t. x1 x3 = 3 x1 + x2 + x4 = 4 2x1 x2 = 3 xj 0 j = 1, ...,...

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162, IEOR Spring 2010 Suggested Solution to Homework 06 Problem 4.1.1 The standard form is max 3x1 + 2x2 s.t. 2x1 + x2 + x3 = 100 x1 + x2 + x4 = 80 x1 + x5 = 40 xj 0 j = 1, ..., 5. Problem 4.1.2 The standard form is min 50x1 + 100x2 s.t. 7x1 + 2x2 x3 = 28 2x1 + 12x2 x4 = 24 xj 0 j = 1, ..., 4. Problem 4.1.3 The standard form is min 3x1 + x2 s.t. x1 x3 = 3 x1 + x2 + x4 = 4 2x1 x2 = 3 xj 0 j = 1, ..., 4. Problem 4.4.1 Since in the standard form we have ve variables and three constraints, there should be three basic variables and two nonbasic variables in a basic solution. The ten possible ways to choose two (nonbasic) variables to be 0 are listed in Table 1. For each possibility, we try to solve the remaining three basic variables. Note that when x1 and x5 are chosen to be nonbasic, we can not nd any solution that satises constraint 3 (the entries N/S means no solution). This happens because the constraint x1 40 is parallel to the constraint x1 0. The last column indicates the corresponding extreme point of the feasible region according to 2 Figure in Chapter 3, if the basic solution is feasible. Problem 4.4.2 Since in the standard form we have four variables and two constraints, there should be two basic variables and two nonbasic variables in a basic solution. The six possible ways to choose two (nonbasic) variables to 1 x1 0 0 0 0 50 80 40 20 40 40 x2 0 100 80 N/S 0 0 0 60 20 40 x3 100 0 20 N/S 0 60 20 0 0 20 x4 80 20 0 N/S 30 0 40 0 20 0 x5 40 40 40 0 10 40 0 20 0 0 Extreme Point H (Infeasible) D (No solution) (Infeasible) (Infeasible) E G F (Infeasible) Table 1: Basic feasible solutions and extreme points for Problem 4.4.1 be 0 are listed in Table 2. For each possibility, we try to solve the remaining two basic variables. The last column indicates the corresponding extreme point of the feasible region according to Figure 4 in Chapter 3, if the basic solution is feasible. x1 0 0 0 4 12 3.6 x2 0 14 2 0 0 1.4 x3 28 0 24 0 56 0 x4 24 144 0 16 0 0 Extreme Point (Infeasible) B (Infeasible) (Infeasible) C E Table 2: Basic feasible solutions and extreme points for Problem 4.4.2 2
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