S2 - C&O 350 Linear Optimization Fall 2009 Solutions to...

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C&O 350 Linear Optimization – Fall 2009 – Solutions to Assignment 2 Question # Max. marks Part marks 1 6 3 3 2 8 4 4 3 4 4 4 10 2 3 2 3 5 10 2 2 2 2 2 6 (Bonus) 5 Total 38 1. 6 marks = [3+3] Solution: (a) To convert the given LP to standard inequality form, we can do the following: (i) Substitute x 3 by u 3 - v 3 , where u 3 , v 3 are nonnegative variables. (ii) Represent the equality constraint 2 x 1 + x 2 = 7 by two inequalities, 2 x 1 + x 2 7 and - 2 x 1 - x 2 - 7. (iii) Multiply the inequality - x 1 + 5 x 2 + 2 x 4 5 by - 1 to make it a constraint. Thus, the given LP in standard inequality form is: max 2 x 1 + x 2 - u 3 + v 3 + 4 x 4 s. t. 2 x 1 + x 2 7 - 2 x 1 - x 2 - 7 x 1 + 3 x 2 - 2 u 3 + 2 v 3 + x 4 2 x 1 - 5 x 2 - 2 x 4 - 5 x 1 , x 2 , u 3 , v 3 , x 4 0 (b) To convert the given LP to standard equality form, we can do the following: (i) Use the equality constraint x 1 - x 2 = 1 to eliminate the free variable x 2 . More precisely, the constraint tells us that x 2 = x 1 - 1, and we can replace all instances of x 2 by ( x 1 - 1). (ii) Multiply the objective function by - 1 to make it a maximization problem. Note that we have dropped the constant from the objective function. (iii) Introduce nonnegative variables s 1 , s 2 to turn our inequalities into equalities Then, the given LP in standard equality form is: max - 3 x 1 + 3 x 3 s. t. 2 x 1 + x 3 - s 1 = 3 x 1 - x 3 + s 2 = 2 x 1 , x 3 , s 1 , s 2 0 . Alternatively, one may start by replacing the variable x 2 by u 2 - v 2 (where u 2 , v 2 0). This is correct, but gives an LP that has more variables and constraints. 1
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2. [8 marks = 4+4] Solution: (a) The dual of the LP in problem 1a) is: min 7 y 1 + 2 y 2 + 5 y 3 s. t. 2 y 1 + y 2 - y 3 2 y 1 + 3 y 2 + 5 y 3 1 - 2 y 2 = - 1 y 2 + 2 y 3 4 y 2 0 y 3 0 . You may refer to Section 4.3 of the course notes for rules and examples of finding duals of LPs in different forms. (b) Notice that while I 1 , . . . , I n are variables, I 0 is a constant. Thus, before finding its dual, it’s easier to first rewrite the given LP as the following: min n j =1 ( c j P j + h j I j ) s. t. - I 1 + P 1 = d 1 - I 0 I j - 1 - I j + P j = d j ( j = 2 , . . . , n ) I j , P j 0 ( j = 1 , . . . , n ) Then the dual of the above LP is: max ( d 1 - I 0 ) y 1 + n k =2 d k y k s. t. y k c k ( k = 1 , . . . , n ) - y k + y k +1 h k ( k = 1 , . . . , n - 1) - y n
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