Name: Section: ISyE 3833: Engineering Optimization Final May 3rd, 2016 6:00pm to 8:50pm Remarks. • This is a closed book exam. • There are 5 problems in total. • Write your name on every sheet. Row: 1 Name: Problem 1. Section: Row: 2 (18 pts) We are given a maximization integer programming problem (IP): max s.t. cx Ax ≤ b x≥0 x integer (a) The LP relaxation of the IP is obtained by In the questions below circle the most appropriate answer. Note that always and sometimes are mutually exclusive, i.e., if an event always happens, “sometimes” is an incorrect answer. (b) The optimal value of the LP relaxation is always sometimes never strictly greater than the optimal value of the IP. (c) If the LP relaxation is infeasible, then the IP is always sometimes never infeasible. (d) If the LP relaxation has an optimal solution, then the IP always sometimes never has an optimal solution. (e) If the LP relaxation has an optimal solution in which all of the variables have integer values, then that solution is always sometimes never an optimal solution to the IP. (f) Suppose we are solving the IP by a branch-and-bound algorithm. We solve the LP relaxation at node i of the tree and the value of the optimal LP solution is larger than the value of the best solution we have to the IP. Then it will always sometimes never be necessary to create children nodes from node i to solve the IP. . Name: Section: Row: 3 (g) Suppose we are solving the IP by a branch-and-bound algorithm. We solve the LP relaxation at node i of the tree and the value of the optimal LP solution is less than or equal to the value of the best solution we have to the IP. Then it will always sometimes never be necessary to create children nodes from node i to solve the IP. (h) Suppose we are solving the IP by a branch-and-bound algorithm. We solve the LP relaxation at node i of the tree. The optimal solution to the LP relaxation is given by x1 = 3, x2 = 2.7, x3 = 0 and we need to create 2 children nodes from node i, say nodes i + p and i + p + 1. Give any additional constraints needed to define the problem at each child node: node i + p node i + p + 1 . . Name: Problem 2. Section: Row: (22 pts) In the questions below circle the most appropriate answer. We are given an LP: max cx s.t. Ax ≤ b x≥0 (a) The objective function always sometimes never strictly increases at each iteration of the simplex algorithm. (b) If the LP has a feasible solution, its dual always sometimes never has an optimal solution. (c) If the LP has an optimal solution, its dual always sometimes never has an optimal solution. (d) If the LP is infeasible, its dual is always sometimes never infeasible. (e) If the LP is unbounded, its dual is always sometimes never feasible. (f) If the dual LP is unbounded, the primal LP is always feasible. sometimes never 4 Name: Section: Row: 5 (g) Suppose the optimal solution of the LP is unique and in the optimal solution, the slack variable of the i-th constraint is positive. Then in an optimal solution to the dual, the i-th dual variable is always sometimes never positive. (h) Suppose the optimal solution of the LP is unique and in the optimal solution, the variable xj is positive. Then in an optimal solution to the dual, the j-th dual constraint is always sometimes never satisfied at equality. (i) Suppose the optimal solution of the LP is unique and in the optimal solution, the objective coefficient (reduced profit) of the slack variable of the i-th constraints equals -12. It would be profitable to purchase some amount of the i-th resource if its cost per unit was more than equal to less than 12. (j) Suppose the LP has 3 variables and 2 constraints, and that (x1 , x2 , x3 , s1 , s2 ) = (2, 0, 0, 0, 3) is feasible to the LP problem and (y1 , y2 , t1 , t2 , t3 ) = (2, 2, 1, 0, 0) is feasible to its dual, where s is the vector of primal slacks, y is the vector of dual variables and t is the vector of dual surplus (negative slack) variables. Then it is always sometimes never the case that both of these solutions are optimal. (k) Suppose the LP has 3 variables and 2 constraints, and that (x1 , x2 , x3 , s1 , s2 ) = (2, 0, 0, 0, 3) is feasible to the LP problem and (y1 , y2 , t1 , t2 , t3 ) = (2, 2, 1, 0, 0) is feasible to its dual, where s is the vector of primal slacks, y is the vector of dual variables and t is the vector of dual surplus (negative slack) variables. Then it is always sometimes never the case that at least one of these solutions is optimal. Name: Problem 3. Section: Row: 6 (15 pts) (a) (5 pts) Suppose we have two binary variables, x and y, and the nonlinear constraint x 6= y. For binary variables there is one simple linear constraint for enforcing x 6= y, i.e., excluding the points (x, y) = (0, 0) and (x, y) = (1, 1). That constraint is: . (b) (10 pts) Now suppose x and y are both integer variables between 0 and 9. Now it is harder to enforce the constraint x 6= y, but it can be done with two constraints and an additional binary variable z. One of the constraints is: x ≤ y − 1 + 10z which enforces x 6= y with z = 0 when x < y. Give the second constraint required in the variables x, y and z: . Explain why the two constraints together guarantee x 6= y. Name: Problem 4. Section: Row: (25 pts) Consider the LP: max 5x1 + 2x2 + c3 x3 s.t. x1 + 5x2 + 2x3 + s1 x1 − 5x2 − 6x3 = b1 + s2 = b2 x1 , x2 , x3 , s1 , s2 ≥ 0 where c3 , b1 and b2 are constants. Suppose that the optimal set of equations is: x1 = 30 + dx2 + ex3 − s1 s2 = 10 + f x2 + gx3 + s1 z = 150 + hx2 − 7x3 − 5s1 Note that the optimal basis matrix is: AB = 1 1 0 1 with inverse: A−1 B 1 = −1 Determine: (a) (6 pts) The values of c3 , b1 and b2 . (b) (10 pts) The values of d, e, f, g, h. (c) (5 pts) The dual of the LP. (d) (4 pts) An optimal solution to the dual. 0 . 1 7 Name: Problem 4. Section: Row: 8 Name: Problem 5. Section: Row: 9 (20 pts) A company requires at least Di machines in month i for each of the next four months, i = 1, 2, 3, 4. Depending on its quality, a machine is productive for 1, 2 or 3 months, i.e., it can have a service life of up to 3 months. A machine bought in month i with a service life of k months costs cik dollars. Naturally a machine with longer service life costs more so that cip < cis for p < s and all i. In addition, if any machines are bought in month i, there is a delivery charge of fi > 0. Let xik be the number of machines bought at the start of month i, i = 1, 2, 3, 4, with service life of k months, k = 1, 2, 3. Note that it is undesirable, with respect to cost, to buy a machine whose service life goes beyond month 4. (a) (10 pts) Formulate a linear integer program that will determine the number of machines of different service life to be bought each month to minimize total cost. Please define any additional variables you add to the problem. (b) (10 pts) Now suppose that if more than 5 machines of service life k are bought in month i, i = 1, 2, 3, 4 and k = 1, 2, 3 then each of the additional machines of this type and this month can be bought at a 10% discount. Formulate a linear integer program that will determine the number of machines of different service life to be bought each month to minimize total cost. Please define any additional variables you add to the problem. Name: Problem 5. Section: Row: 10