hw2_sol - Homework 2 Solutions Due: February 3, 2009...

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Homework 2 Solutions Due: February 3, 2009 Solution to Exercise 4, page 83 : Let x 11 and x 12 denote the production level (units) of product P 1 on machines M 1 and M 2, respectively. Similarly, let x 21 and x 22 denote the production level of product P 2 on machines M 1 and M 2, respectively. Note that x 11 + x 21 represents the total production of machine M 1, and x 12 + x 22 represents the production of machine M 2. The capacity constraints for machines M 1 and M 2 are, respectively, given by: x 11 + x 21 200 , x 12 + x 22 250 . The constraint balancing the production schedule on the two machines is given by: | ( x 11 + x 21 ) - ( x 12 + x 22 ) | ≤ 5 . This constraint can be represented equivalently by two inequalities, as follows: x 11 + x 21 - x 12 - x 22 5 , - x 11 - x 21 + x 12 + x 22 5 . The total production of P 1 is x 11 + x 12 , and that of P 2 is x 21 + x 22 . Given the profit of $10 per unit of P 1 and $15 per unit of P 2, the total profit is given by 10( x 11 + x 12 ) + 15( x 21 + x 22 ). Thus, the LP formulation of the problem is maximize 10 x 11 + 10 x 12 + 15 x 21 + 15 x 22 subject to x 11 + x 21 200 x 12 + x 22 250 x 11 + x 21 - x 12 - x 22 5 - x 11 - x 21 + x 12 + x 22 5 x 11 ,x 12 ,x 21 ,x 22 0 , where x 11 ,x 12 ,x 21 ,x 22 0 is a short hand notation for denoting that all listed variables are nonnegative. 1
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To transform this LP to a standard form, we need to the transform the inequality constraints into equalities by adding slack variables. Thus, the standard LP is maximize 10 x 11 + 10 x 12 + 15 x 21 + 15 x 22 subject to x 11 + x 21 + s 1 = 200 x 12 + x 22 + s 2 = 250 x 11 + x 21 - x 12 - x 22 + s 3 = 5 - x 11 - x 21 + x 12 + x 22 + s 4 = 5 x 11 ,x 12 ,x 21 ,x 22 ,s 1 ,s 2 ,s 3 ,s 4 0 . Solution to the modification of Exercise 5, page 83: The given problem is not LP, so at first, we provide its LP formulation. The first step is to eliminate the objective function given by maximization by introducing a new variable y . The equivalent formulation is maximize y subject to y ≤ | x 1 - x 2 + 3 x 3 | y ≤ | - x 1 + 3 x 2 - x 3 | x 1 ,x 2 ,x 3 0 , y is unrestricted . Note that y is a variable in the preceding problem. The next step toward LP formulation is to eliminate nonlinear constraints involv- ing absolute value. By doing so, we obtain the following LP formulation: maximize y subject to x 1 - x 2 + 3 x 3 - y 0 - x 1 + x 2 - 3 x 3 - y 0 - x 1 + 3 x 2 - x 3 - y 0 x 1 - 3 x 2 + x 3 - y 0 x 1 ,x 2 ,x 3 0 , y is unrestricted . Now, we provide the standard form of the preceding LP. To do so, we need to introduce a slack variable for each of the inequality constraints, and we need to write y as y = x + 4 - x - 4 with x + 4 0 and x - 4 0 (since y is unrestricted). Thus, the standard form of the preceding LP is: maximize
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hw2_sol - Homework 2 Solutions Due: February 3, 2009...

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