min 4x 1 5x 2 + 3x 3 s.t. x 1 + 2x 2 + x 3 = 10 x 1 x 2 6 x 1 + 3x 2 + x 3 14

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1 The exam is three hours long and consists of 4 exercises. The exam is graded on a scale 0-25 points, and the points assigned to each question are indicated in parenthesis within the text. If necessary, use separate sheets as scratch papers, but write your final answers in the exam paper clearly, and explain your solutions. Problem 1 Solve the following problem using the Two-Phase Simplex method. Use Bland s rule as a pivoting rule. Report the tableau at each iteration for both phases and clearly indicate entering and leaving variables. (6pt) Is the optimal primal solution degenerate? (1pt) Problem 2 min 4x 1 5x 2 + 3x 3 s.t. x 1 + 2x 2 + x 3 = 10 x 1 x 2 6 x 1 + 3x 2 + x 3 14 x 1, x 2, x 3 0. Consider a minimization linear programming problem P in standard form defined by an m n constraint matrix A, a cost vector c, and a right hand side vector b. Assume that the rows of A are linearly independent. Are the following statements true or false? Justify your answers. (5pt) (a) Let x be a basic feasible solution of P, and let p be the associated basic dual solution. Then, p and x satisfy the complementary slackness conditions. (b) If the polyhedron {x R n Ax = b, x 0} is unbounded, then the optimal cost of P is. (c) Let x be a basic feasible solution. Suppose that for every basis corresponding to x, the associated basic dual solution is infeasible. Then, the optimal cost must be strictly less than c T x. (d) Let p i be the dual variable associated with the i-th equality constraint of P. Eliminating the i-th equality constraint of P is equivalent to introducing the additional constraint p i = 0 in the dual of P. (e) If P is solved by the Simplex method, the value of the objective function strictly decreases at each iteration of the Simplex algorithm. Page 1 / 8

, you can use complementary slackness, or note that in this case the duals can be derived directly from the tableau). Explain how and why.

2 Problem 3 Consider the following LP and its optimal tableau below: min 2x 1 x 2 + x 3 s.t. x 1 + 2x 2 + x 3 8 x 1 + x 2 2x 3 4 x 1, x 2, x 3 0. x 1 x 2 x 3 x 4 x x 1 = x 5 = (a) Write the dual and obtain the optimal dual variables (e.g., you can use complementary slackness, or note that in this case the duals can be derived directly from the tableau). Explain how and why. (2pt) (b) If you were to choose between increasing of 1 unit the right hand side of the first and second constraint, which one would you choose, and why? What is the effect of the increase on the optimal cost? (1pt) (c) Suppose the constraint x 1 + 2x 2 + x 3 8 is replaced by x x 2 + x 3 8. Does the current basis remain optimal? Justify. (1pt) (d) Suppose the following constraint is added to the problem: x 2 + 2x 3 3. Find a new optimal solution starting from the tableau above. (3pt) Problem 4 Consider the uncapacitated min cost flow problem defined by the graph below. The number on each arc indicates its cost, and node supplies are indicated by the bold arrows. (a) Solve the problem using the Network Simplex. Start from the tree solution defined by arcs T = {(1, 3), (3, 6), (6, 7), (7, 4), (4, 2), (2, 5)}. Explain all the steps. At each iteration, draw the graph with the updated flows, indicate the set T, and report the dual variables. (5pt) (b) Suppose the supply at node 1 is changed to 4, and the supply at node 7 is changed to -6. Is the optimal basis at the end of point (a) still optimal after the change? Justify. (1pt) Page 2 / 8

3 Solutions Problem 1 (a) The problem in standard form is: Phase I: The auxiliary problem is: min 4x 1 5x 2 + 3x 3 s.t. x 1 + 2x 2 + x 3 = 10 x 1 x 2 x 4 = 6 x 1 + 3x 2 + x 3 + x 5 = 14 x 1, x 2, x 3, x 4, x 5 0. min y 1 + y 2 + y 3 s.t. x 1 + 2x 2 + x 3 + y 1 = 10 x 1 x 2 x 4 + y 2 = 6 x 1 + 3x 2 + x 3 + x 5 + y 3 = 14 x 1, x 2, x 3, x 4, x 5, y 1, y 2, y 3 0. The starting basic feasible solution for Phase I is x i = 0, i = 1,..., 5, y 1 = 10, y 2 = 6, y 3 = 14 (note that in this case the artificial variable y 3 is unnecessary, it could be omitted and replaced by x 5 in the starting basic solution). The reduced cost vector is computed as: c T = c T c T B B 1 A = [ ] [ ] = [ ] [ ] = [ ]. Phase I tableaux are the following (pivot elements are marked by an ): x 1 x 2 x 3 x 4 x 5 y 1 y 2 y y y y x 1 x 2 x 3 x 4 x 5 y 1 y 2 y y x y Page 3 / 8

4 x 1 x 2 x 3 x 4 x 5 y 1 y 2 y 3 8/ /3 1/3 1 7/3 2/3 0 x 2 4/ /3 1/3 0 1/3 1/3 0 x 1 22/ /3 2/3 0 1/3 2/3 0 y 3 8/ /3 1/3 1 4/3 1/3 1 x 1 x 2 x 3 x 4 x 5 y 1 y 2 y x 2 4/ /3 1/3 0 1/3 1/3 0 x 1 22/ /3 2/3 0 1/3 2/3 0 x 5 8/ /3 1/3 1 4/3 1/3 1 The optimal cost for Phase I is 0 and all artificial variables are non-basic: We have a basic feasible solution for the original problem with basic variables x 2, x 1, x 5 and corresponding basis B = [A 2 A 1 A 5 ]. The reduced costs of non-basic variables for Phase II are: c 3 = c 3 c T B B 1 A 3 = 3 [ ] 1/3 1/3 = 6 1/3 c 4 = c 4 c T B B 1 A 4 = 0 [ ] 1/3 2/3 = 1 1/3 The reduced cost of x 4 is negative, and we need to proceed with Phase II. Phase II tableaux are the following (pivot elements are marked by an ): x 1 x 2 x 3 x 4 x x 2 4/ /3 1/3 0 x 1 22/ /3 2/3 0 x 5 8/ /3 1/3 1 x 1 x 2 x 3 x 4 x x x x The optimal solution is x 2 = x 3 = 0, x 1 = 10, x 4 = x 5 = 4 of cost -40. (b) The optimal primal solution is not degenerate since all basic variables are non-zero. Problem 2 (a) True. P is in standard form and, and x and p are a pair of basic primal and dual solutions determined by the same basis B. Therefore they satisfy complementary slackness conditions. Page 4 / 8

5 The following is a more detailed explanation. Let x and p be basic primal and dual solutions determined by the same basis B. The compl. slackness conditions are (c j p T A j )x j = 0, j = 1,..., n and p i (a T i x b i ) = 0, i = 1,..., m For each basic variable x B(i), i = 1,..., m associated with column A B(i) we have: p T = c T B B 1 (c B(i) p T A B(i) )x B(i) = (c B(i) c T B B 1 A B(i) )x B(i) = 0 since B 1 A B(i) is the i-th unit vector e i and c T B e i = c B(i) ; For each non basic variable x j associated with column A j we have: x j = 0 (c j p T A j )x j = 0; Finally, since x is basic and P is in standard form, we have Ax = b p i (a T i x b i ) = 0, i = 1,..., m. (b) False. The following is an LP in standard form with optimal cost 2 and unbounded feasibility region: min x 2 s.t. x 1 + x 2 = 2 x 1, x 2, 0. (c) True. x is a basic feasible solution. If the optimal cost was c T x, then there would be an optimal basis B such that x B = B 1 b and c T B B 1 A j c j, j = 1,..., n, (i.e., such that all reduced costs with respect to B are non-negative), that is, the corresponding basic dual solution p T = c T B B 1 would be dual feasible. (d) True. The dual constraints are i=1,...,m p ia ij c j, j = 1,..., n, where a ij is the i-th element of column A j. Eliminating the i-th constraint removes the term p i a ij from all dual constraints and the term p i b i from the objective function of the dual. This is equivalent to impose p i = 0. (e) False. In presence of degeneracy, the Simplex algorithm can make a change of basis without changing the solution nor the value of the objective function (degenerate pivot). Problem 3 (a) The dual is: max 8p 1 + 4p 2 s.t. p 1 p 2 2 2p 1 + p 2 1 p 1 2p 2 1 p 1, p 2 0. Denoting B the optimal basis, the optimal dual solution is p T = c T B B 1. Note that the columns A 4, A 5 form the identity and the cost of the corresponding variables is 0. Therefore, from the reduced costs of x 4, x 5 we obtain: Page 5 / 8

Therefore, since p 1 = 2, p 2 = 0, and B remains feasible after adding one unit to the first rhs, it is profitable to increase the rhs of the first constraint to obtain an optimal cost of -18.

6 2 = c 4 = c 4 c T B B 1 A 4 = c 4 p T A 4 = 0 p 1 p 1 = 2; 0 = c 5 = c 5 c T B B 1 A 5 = c 5 p T A 5 = 0 p 2 p 2 = 0; (b) The optimal solution is not degenerate: ɛp i represents the change in the optimal cost if the rhs of the i-th constraint is increased by ɛ, as long as the current optimal basis B remains feasible. Therefore, since p 1 = 2, p 2 = 0, and B remains feasible after adding one unit to the first rhs, it is profitable to increase the rhs of the first constraint to obtain an optimal cost of -18. [ ] 1 (c) The column A 2 becomes A 2 + δ = A δe 1 where δ = 11/6. The reduced cost c 2 of x 2 is changed as: c 2 c T B B 1 (A 2 + δe 1 ) = c 2 p 1 δ = 3 11/3 < 0. Therefore, the basis is not optimal after the change. (d) The row of the new constraint in the constraint matrix is: a T 3 = [ ], and the vector formed by the components of a 3 associated with basic variables is a T = [ 0 0 ]. We add the slack variable x 6 to put the constraint in equality form (x 2 + 2x 3 x 6 = 3), and derive the corresponding tableau row as [ a T B 1 A a T 3 1 ] = [ ]. The updated tableau with the new constraint is: x 1 x 2 x 3 x 4 x 5 x We can use the dual Simplex to re-optimize. x 1 x 2 x 3 x 4 x 5 x 6 23/2 0 3/ /2 13/2 1 3/ /2 27/2 0 7/ /2 3/2 0 1/ /2 The new optimal solution is x 1 = 13/2, x 2 = 0, x 3 = 3/2, x 4 = 0, x 5 = 27/2, x 6 = 0. The optimal cost is 23/2. Problem 4 (a) Iteration 1. Page 6 / 8

We then compute the reduced costs for the arcs not in T as c ij = c ij p i + p j : c 41 = 0+1+10 = 11, c 34 = 2 8 1 = 7, c 64 = 1 4 1 = 4, c 21 = 5+3+10 = 18, c 57 = 2 + 5 = 7.

7 T = {(1, 3), (3, 6), (6, 7), (7, 4), (4, 2), (2, 5)} (arcs in T are represented as dashed lines in the figures). Dual variables are computed as p i p j = c ij, (i, j) T after setting p 7 = 0. We obtain: p 7 = 0, p 4 = 1, p 6 = 4, p 5 = 5, p 2 = 3, p 3 = 8, p 1 = 10. We then compute the reduced costs for the arcs not in T as c ij = c ij p i + p j : c 41 = = 11, c 34 = = 7, c 64 = = 4, c 21 = = 18, c 57 = = 7. Variable f 34 has negative reduced cost and is chosen to enter the basis. Adding the corresponding arc (3, 4) to T forms the cycle 3, 4, 7, 6, 3 with backward arcs (7, 4), (6, 7), (3, 6). The maximum flow that can be pushed around this cycle is θ = min{f 74, f 67, f 36 } = 5. A flow θ is added to arc (3, 4) and subtracted from arcs (7, 4), (6, 7), (3, 6), f 36 becomes 0 and leaves the basis. Iteration 2. T = {(1, 3), (3, 4), (6, 7), (7, 4), (4, 2), (2, 5)}. Dual variables are p 7 = 0, p 4 = 1, p 6 = 4, p 5 = 5, p 2 = 3, p 3 = 1, p 1 = 3. Variable f 64 has negative reduced cost c 64 = = 4 and is chosen to enter the basis. Adding (6, 4) to T closes the cycle 6, 4, 7, 6 and θ = min{f 74, f 67 } = 3. Variable f 74 drops to 0 and leaves the basis. Iteration 3. T = {(1, 3), (3, 4), (6, 4), (6, 7), (4, 2), (2, 5)}. Dual variables are p 7 = 0, p 6 = 4, p 4 = 3, p 3 = 5, p 1 = 7, p 2 = 1, p 5 = 1. Reduced costs are c 21 = > 0, c 57 = 2+1 > 0, c 74 = > 0, c 41 = > 0, c 36 = > 0. All reduced costs are non-negative and the solution is optimal. The optimal cost is 55. Page 7 / 8

8 (b) Changing the supplies does not change the dual solution nor the reduced costs if the basis is not changed. To see if the basis remains optimal we just have to check if it remains feasible after the change. After changing the supplies, the arcs in T determine the solution f 13 = 4, f 34 = 6, f 64 = 2, f 67 = 6, f 42 = 8, f 25 = 0 which is feasible. Therefore, the basis remains optimal. Page 8 / 8

(P ) Minimize 4x 1 + 6x 2 + 5x 3 s.t. 2x 1 3x 3 3 3x 2 2x 3 6

(P ) Minimize 4x 1 + 6x 2 + 5x 3 s.t. 2x 1 3x 3 3 3x 2 2x 3 6 The exam is three hours long and consists of 4 exercises. The exam is graded on a scale 0-25 points, and the points assigned to each question are indicated in parenthesis within the text. Problem 1 Consider