Ch 4 Test Bank Simplex 
Test Bank for Chapter 4
Problem 4-1:
Work through the simplex method (in algebraic form) step by step to solve the following problem.
Maximize Z = x1 + 2x2 + 2x3, subject to 5x1 + 2x2 + 3x3 ≤ 15 x1 + 4x2 + 2x3 ≤ 12 2x1 + x3 ≤ 8 and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0.
Solution for Problem 4-1:
We introduce x4, x5, and x6 as slack variables for the respective functional constraints. The augmented form of the problem then is Maximize Z = x1 + 2 x2 + 2 x3, subject to 5 x1 + 2 x2 + 3 x3 + x4 = 15 x1 + 4 x2 + 2 x3 + x5 = 12 2 x1 + x3 + x6 = 8 and x1 0, x2 0, x3 0, x4 0, x5 0, x6 0.
The simplex method in algebraic form is applied to this augmented form of the problem as follows.
Initialization:
Let x1, x2 and x3 be the nonbasic variables, so x1 = x2 = x3 = 0. Solving for x4, x5 and x6 from the equations in the constraints:
(1) 5 x1 + 2 x2 + 3 x3 + x4 = 15
(2) x1 + 4 x2 + 2 x3 + x5 = 12
(3) 2 x1 + x3 + x6 = 8
we obtain the initial BF solution (0, 0, 0, 15, 12, 8). The objective function is Z = 1 x1 + 2 x2 + 2 x3. The current BF solution is not optimal since we can improve Z by increasing x1 or x2 or x3.
Iteration 1:
Z = x1 + 2 x2 + 2 x3.
We choose x2 as the entering basic variable because increasing it (or x3) increases Z at the fastest rate. (Alternatively, because of the tie, x3 can be chosen instead to be the entering basic variable, in which case Iteration 2 would lead to x2 becoming the entering basic variable at that point instead of x3.) Next, we need to decide how far we can increase x2. Since we need variables x4, x5 and x6 to stay nonnegative, from the minimum ratio test, we
Problem 4-1:
Work through the simplex method (in algebraic form) step by step to solve the following problem.
Maximize Z = x1 + 2x2 + 2x3, subject to 5x1 + 2x2 + 3x3 ≤ 15 x1 + 4x2 + 2x3 ≤ 12 2x1 + x3 ≤ 8 and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0.
Solution for Problem 4-1:
We introduce x4, x5, and x6 as slack variables for the respective functional constraints. The augmented form of the problem then is Maximize Z = x1 + 2 x2 + 2 x3, subject to 5 x1 + 2 x2 + 3 x3 + x4 = 15 x1 + 4 x2 + 2 x3 + x5 = 12 2 x1 + x3 + x6 = 8 and x1 0, x2 0, x3 0, x4 0, x5 0, x6 0.
The simplex method in algebraic form is applied to this augmented form of the problem as follows.
Initialization:
Let x1, x2 and x3 be the nonbasic variables, so x1 = x2 = x3 = 0. Solving for x4, x5 and x6 from the equations in the constraints:
(1) 5 x1 + 2 x2 + 3 x3 + x4 = 15
(2) x1 + 4 x2 + 2 x3 + x5 = 12
(3) 2 x1 + x3 + x6 = 8
we obtain the initial BF solution (0, 0, 0, 15, 12, 8). The objective function is Z = 1 x1 + 2 x2 + 2 x3. The current BF solution is not optimal since we can improve Z by increasing x1 or x2 or x3.
Iteration 1:
Z = x1 + 2 x2 + 2 x3.
We choose x2 as the entering basic variable because increasing it (or x3) increases Z at the fastest rate. (Alternatively, because of the tie, x3 can be chosen instead to be the entering basic variable, in which case Iteration 2 would lead to x2 becoming the entering basic variable at that point instead of x3.) Next, we need to decide how far we can increase x2. Since we need variables x4, x5 and x6 to stay nonnegative, from the minimum ratio test, we