HW2-solutions

IEOR 4004: Optimization Models and Methods Homework 2 Solutions Instructor: Shipra Agrawal 1. (10 points) Convert the following linear program (LP) to standard form: min x 1 + x 2 + 8 x 3 + x 4 x 1 + x 2 + x 3 + x 4 ≤ 12 x 1 + x 2 + x 3 + x 4 ≥ 1 x 1 + x 2 − x 3 + 5 x 4 = 10 x 1 ≥ 0 , x 2 ≤ 0 , x 3 ≥ 0 , x 4 unrestricted SOLUTION. The standard form is max c ⊤ x such that Ax = b and x ≥ 0. Converting the above LP into standard form: We transform the minimization to a maximization. Add slack variables s 1 to the ≤ constraint, and a surplus variable s 2 to the ≥ constraint. Next, we let − x 2 = y 2 and x 4 = x + 4 − x − 4 such that y 2 , x + 4 , x − 4 ≥ 0 max − x 1 + y 2 − 8 x 3 − x + 4 + x − 4 x 1 − y 2 + x 3 + x + 4 − x − 4 + s 1 = 12 x 1 − y 2 + x 3 + x + 4 − x − 4 − s 2 = 1 x 1 − y 2 − x 3 + 5 x + 4 − 5 x − 4 = 10 x 1 ≥ 0 , y 2 ≥ 0 , x 3 ≥ 0 , x + 4 ≥ 0 , x − 4 ≥ 0 2. (15 points) Consider the following linear program (LP): max 12 x 1 + 11 x 2 + 8 x 3 + 6 x 4 8 x 1 + 8 x 2 + 6 x 3 + 5 x 4 ≤ 12 x i ≥ 0 , i = 1 , 2 , 3 , 4 (a) Convert the above LP to standard form. (b) Perform one iteration of the simplex method starting with the bfs where x 1 = x 2 = x 3 = x 4 = 0. Show your work: show the dictionary, identify the entering and leaving variable, write the new bfs, transform the constraints (so that the columns of basic variables in the new bfs form the standard basis) and objective (so that it is written in terms of non-basic variables in the new bfs). (c) Is the new bfs optimal? Why? SOLUTION. (a) We just need to add a slack variable to the ≤ constraint. max 12 x 1 + 11 x 2 + 8 x 3 + 6 x 4 8 x 1 + 8 x 2 + 6 x 3 + 5 x 4 + s 1 = 12 x i ≥ 0 , i = 1 , 2 , 3 , 4 s 1 ≥ 0 1

(b) At the basic feasible solution x 1 = x 2 = x 3 = x 4 = 0, we have s 1 = 12. Using the greedy rule, we pick x 1 to be the entering variable and s 1 will be the leaving variable. The transformed LP is: max 18 − x 2 − x 3 − 3 / 2 x 4 − 3 / 2 s 1 x 1 + x 2 + 3 / 4 x 3 + 5 / 8 x 4 + 1 / 8 s 1 = 3 / 2 x i ≥ 0 , i = 1 , 2 , 3 , 4 s 1 ≥ 0 (c) The new bfs is optimal as the coefficients of all non-basic variables in the objective are negative. 3. (20 points) Consider the following LP: max 10 x 1 + x 2 s.t. x 1 ≤ 1 20 x 1 + x 2 ≤ 100 x 1 , x 2 ≥ 0 (a) Convert above LP to standard form. (b) Enumerate all the basic feasible solutions (bfs) for the standard form LP. Hint: For an LP in standard form with m constraints and n variables, there are at most ( n m ) basic feasible solutions. To find all bfs, you can check all the ( n m ) potential ways of setting n − m out of the n variables to 0. (c) Show the feasible region ( x 1 , x 2 ) graphically, and for each bfs found in part (b), plot the point ( x 1 , x 2 ) on this graph. (d) For each bfs found in part (b), transform the LP to be in ”tableau form”. That is, transform the constraints so that the columns for basic variables to form standard basis; and substitute the basic variables in the objective so that it is written in terms of the non-basic variables only. You may use a matrix inversion calculator or row reductions to perform this transform. (e) Use simplex method to solve the LP. For every bfs solution you find in the simplex method iterations, you may look up and reuse the tableau form you computed in part (d). SOLUTION. (a) max 10 x 1 + x 2 s.t. x 1 + s 1 = 1 20 x 1 + x 2 + s 2 = 100 x 1 , x 2 , s 1 , s 2 ≥ 0 (b) The basic feasible solutions are (0 , 0 , 1 , 100) , (1 , 0 , 0 , 80) , (0 , 100 , 1 , 0) , (1 , 80 , 0 , 0) (c) See image below (d) For bfs (0 , 0 , 1 , 100): max 10 x 1 + x 2 s.t. x 1 + s 1 = 1 20 x 1 + x 2 + s 2 = 100 x 1 , x 2 , s 1 , s 2 ≥ 0 2