asst2sol

asst2sol - C&O 350 Linear Optimization – Winter 2009 –...

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C&O 350 Linear Optimization – Winter 2009 – Solutions to Assignment 2 Question # Max. marks Part marks 1 9 3 3 3 2 5 2 3 3 6 3 3 4 3 5 6 2 2 2 6 13 2 4 3 3 1 Total 42 1. 9 marks = 3+3+3 1 point for each part (a), (b), (c) This question addresses mathematical programming problems (abbreviated MP problems) that are gen- eralizations of LP problems, where we may have any function (rather than a linear function) in the objective function as well as in the constraints, and moreover, we allow strict inequalities in the con- straints. A feasible solution to an MP problem is a vector x that satisfies all of the constraints, and an MP problem is called infeasible if it has no feasible solution. An optimal solution to an MP problem is a feasible solution that has the best objective value, and that objective value is called the optimal value. For each of the following MP problems (a) explain why it is not an LP problem; (b) state whether or not it is infeasible, and if not then write down a feasible solution; (c) state whether or not it has an optimal solution, and if yes then write down an optimal solution and the optimal value. (P1) max x 1 subject to x 1 + x 2 < 1 (P2) max x 1 subject to x 2 1 + x 2 2 1 (P3) max x 1 subject to ( x 1 + x 2 ) 2 1 Solution : (P1) (a) This MP is not an LP problem because its single constraint is a strict inequality. (b) This problem has a feasible solution. The vector (0 , 0) satisfies 0 + 0 < 1. (c) This problem has no optimal solution. If ( x 1 , x 2 ) is a feasible solution, then we can find a new feasible solution ( x 1 , x 2 ) with greater objective value by taking any x 1 < x 1 < 1 - x 2 . (P2) (a) This MP is not an LP problem because its single constraint is not a linear inequality. (b) This problem has a feasible solution. The vector (0 , 0) satisfies 0 2 + 0 2 1. (c) This problem has an optimal solution. The vector (1 , 0) has objective value 1. Moreover the constraint implies that since x 2 2 0, x 2 1 1, and hence if the problem has a feasible solution, then its objective value is bounded above by 1. (P3) (a) This MP is not an LP problem because its single constraint is not a linear inequality. (b) This problem has a feasible solution. The vector (0 , 0) satisfies (0 + 0) 2 1. 1

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(c) This problem has no optimal solution. Note that for any positive number x , the vector ( x, - x ) is feasible. So for any strictly increasing sequence { x i } of positive numbers, we have a sequence { ( x i , - x i ) } of feasible solutions, each with greater objective value than the last. In particular if we take { x i } to be an unbounded sequence (e.g. the set of natural numbers) we have the result. 2. 5 marks = 2+3 Exercises 3.3.1(b) and 3.3.2 from the Course Notes, stated below: Convert the following LP problems to standard equality form: (a) minimize - x 1 + x 2 subject to 5 x 1 - 2 x 2 4 2 x 1 + x 2 3 x 1 , x 2 0 (b) minimize 2 x 1 + x 2 - 3 x 3 subject to x 1 + x 2 + x 3 2 x 1 - x 2 = 1 x 1 - x 3 2 x 1 , x 3 0 Solution : (a) First we convert the “min - x 1 + x 2 ” to “max x 1 - x 2 ” to get the following LP: max x 1 - x 2 subject to 5 x 1 - 2 x 2 4 2 x 1 + x 2 3 x 1 , x 2 0
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