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 the following linear programming problem P. (P ) Minimize 4x 1 + 6x 2 + 5x 3 s.t. 2x 1 3x 3 3 3x 2 2x 3 6 x 1, x 2, x 3 0. (a) Solve P by using the dual Simplex method. Show the tableau at each iteration and indicate the pivot elements. Report both the primal and dual optimal solutions x and p associated with the optimal basis. (4pt) (b) Write the dual D of P. Use duality to verify that x and p are optimal. (3pt) Problem 2 Are the following statements true or false? Justify your answers. (5pt) (a) Consider a minimization linear programming problem P in standard form and let B be a basis matrix. If both the primal and dual basic solutions corresponding to B are feasible, then they are both optimal. (b) In solving an LP by the Simplex Method, a different basic feasible solution is generated after each pivoting operation. (c) The dual of a standard form minimization linear programming problem P can be either unbounded or infeasible if P has optimal cost. Problem 3 Consider the LP (P ) Minimize 2x 1 + x 2 s.t. 2x 1 + x 2 4x 3 2 x 1 x 2 + 2x 3 0 x 1 2 x 1, x 2, x 3 0. Page 1 / 6
The following is an optimal tableau for P where x 4, x 5, x 6 are slack variables. z x 1 x 2 x 3 x 4 x 5 x 6 4 0 1 0 0 0 2 2 1 0 0 0 0 1 1 0 1/2 0 1/2 1 0 1/2 0 1/4 1-1/4 0 1/2 Let x be the basic optimal solution associated with this tableau. Answer the following questions by using the optimal tableau above. Each of the questions is independent from the others. (8pt) (a) What is the optimal basis? What is the corresponding optimal solution to the dual of P? Explain why this solution can be obtained from the tableau. (b) Is x the unique optimal solution? Why? If it is not unique, find another optimal solution. (c) Suppose that the cost of variable x 3 is changed to 0 + δ. Find the range of values of δ for which the current basis remains optimal? Explain your work. How could you find a new optimal basis without resolving from scratch if δ is outside this range? (you don t have to actually find it) (d) Suppose that the cost of variable x 2 is changed to 1. After the change: Explain why the optimal cost becomes. Explain how to construct a feasible solution with cost M for any M > 0. Problem 4 Consider the uncapacitated minimum cost flow problem defined by the graph below. The number on each arc indicates its cost, and node supplies are indicated by the arrows. (a) Execute one iteration of the Network Simplex algorithm starting from the tree solution defined by arcs T = {(1, 4), (4, 2), (2, 5), (4, 3), (3, 6), (6, 7)}: (i) indicate the set T and the corresponding tree solution before and after the iteration, (iii) report the dual variables before and after the iteration, (iv) explain how the leaving and entering variables are selected and how the flows are updated. (4pt) (b) Is the dual solution that you obtained at the end of point (a) degenerate? Is it optimal? Justify the answers. (1pt) Page 2 / 6
Solutions Problem 1 (a) The problem in standard form is: min 4x 1 + 6x 2 + 5x 3 s.t. 2x 1 3x 3 + x 4 = 3 3x 2 2x 3 + x 5 = 6 x 1, x 2, x 3, x 4, x 5 0. A basis matrix B is formed by the columns corresponding to the slack variables x 4, x 5. The corresponding reduced cost vector is c = c c B B 1 A = c = [ 4 6 5 0 0 ] as c B = [ c 4 c 5 ] = [ 0 0 ]. Since c 0, B can be used to start the dual Simplex. The corresponding (primal) basic solution is x 4 = 3, x 5 = 6, x 1 = x 2 = x 3 = 0 with cost 0. The dual Simplex tableaux are the following (pivot elements are marked by an ): x 1 x 2 x 3 x 4 x 5 0 4 6 5 0 0 x 4 3 2 0 3 1 0 x 5 6 0 3 2 0 1 x 1 x 2 x 3 x 4 x 5 5 2/3 6 0 5/3 0 x 3 1 2/3 0 1 1/3 0 x 5 4 4/3 3 0 2/3 1 x 1 x 2 x 3 x 4 x 5 13 10/3 0 0 1/3 2 x 3 1 2/3 0 1 1/3 0 x 2 4/3 4/9 1 0 2/9 1/3 The optimal primal solution is x = (0, 4/3, 1, 0, 0) of cost 13. The corresponding optimal dual solution p = (p 1, p 2 ) can be obtained directly from the tableau. Since x 4, x 5 are slack variables with associated columns A 4 = e 1 and A 5 = e 2 in the constraint matrix A we have p 1 = c 4 = 1/3, p 2 = c 5 = 2. (b) The dual is max 3p 1 6p 2 s.t. 2p 1 4 3p 2 6 3p 1 2p 2 5 p 1, p 2 0. Page 3 / 6
In order to check that the solutions x = (0, 4/3, 1, 0, 0) and p = ( 1/3, 2) obtained in point (a) are optimal we first verify that they are both feasible: 2x 1 3x 3 + x 4 = 2 0 3 1 + 0 = 3 3x 2 2x 3 + x 5 = 3 4/3 2 1 + 0 = 6 x 1 = 0 0, x 2 = 4/3 0, x 3 = 1 0, x 4 = 0 0, x 5 = 0 0 and thus x satisfies all primal constraints; 2p 1 = 2 1/3 = 2/3 4 3p 2 = 3 2 = 6 6 3p 1 2p 2 = 3 1/3 2 2 = 5 5 p 1 = 1/3 0, p 2 = 2 0 and thus p satisfies all dual constraints. Now, since c x = p b = 13 from the corollary to the weak duality theorem we have that both x and p are optimal (it can also be checked that they satisfy the complementary slackness conditions). Problem 2 (a) True. Since x is a basic primal solution associated with B its cost is c B x B = c B B 1 b where x B and c B are the vectors formed by the basic variables and their costs, respectively. Since p is a basic dual solution associated with B its cost is p b = c B B 1 b. Thus, x and p are both feasible and have the same cost and are optimal by the corollary to the weak duality theorem. It can also be observed that x and p satisfy the complementary slackness conditions and are both feasible thus optimal. (b) False. If the basic solution before pivoting is degenerate and the leaving variable has value 0, then the pivoting operation will not change the value of any variable and after pivoting we have the same basic solution. (c) False. If the optimal cost of the primal is then by weak duality the dual can only be infeasible. Problem 3 (a) The optimal solution shown in the tableau has basic variables x 1, x 5, and x 3. The optimal basis is formed by the corresponding columns A 1, A 5, and A 3 of the constraint matrix A. The optimal basis matrix is B = [ ] 2 0 4 A 1 A 5 A 3 = 1 1 2 1 0 0 The optimal dual solution is p = (0, 0, 2). It can be obtained from the tableau since x 4, x 5, x 6 are slack variables with associated columns A 4 = e 1, A 5 = e 2, A 6 = e 3 : 0 = c 4 = c 4 c B B 1 A 4 = c 4 p A 4 = c 4 p e 1 = 0 p 1 p 1 = 0; 0 = c 5 = c 5 c B B 1 A 5 = c 5 p A 5 = c 5 p e 2 = 0 p 2 p 2 = 0; 2 = c 6 = c 6 c B B 1 A 6 = c 6 p A 6 = c 6 p e 3 = 0 p 3 p 3 = 2; Page 4 / 6
(b) The solution x is not the unique optimal solution. We note that the reduced cost c 4 of the non-basic variable x 4 is zero. If we let x 4 enter the basis then the basic variable x 5 must leave the basis, and since x 5 > 0 and c 4 = 0 we obtain a different basic feasible solution with the same cost as x. To compute the new solution we make a pivoting operation on row 2 and column 4: z x 1 x 2 x 3 x 4 x 5 x 6 4 0 1 0 0 0 2 2 1 0 0 0 0 1 1 0 1/2 0 1/2* 1 0 1/2 0 1/4 1-1/4 0 1/2 z x 1 x 2 x 3 x 4 x 5 x 6 4 0 1 0 0 0 2 2 1 0 0 0 0 1 2 0 1 0 1 2 0 1 0 1/2 1 0 1/2 1/2 (c) The current basis remains optimal if all reduced costs remain non-negative after the cost change. Since x 3 is a basic variable and c 3 is changed to δ, the basic cost vector c B = (c 1, c 5, c 3 ) becomes c B = ( 2, 0, δ). The reduced costs of non-basic variables become: c 2 = c 2 c B B 1 A 2 = 1 [ 2 0 δ ] 0 1/2 = 1 + 1/4δ; 1/4 c 4 = c 4 c B B 1 A 4 = 0 [ 2 0 δ ] 0 1/2 = 0 + 1/4δ; 1/4 c 6 = c 6 c B B 1 A 6 = 0 [ 2 0 δ ] 1 0 = 0 + 2 1/2δ; 1/2 Form the above, the reduced costs remain non-negative if δ 4, δ 0, and δ 4, that is, 0 δ 4. If δ is outside this range, then the current basis remains primalfeasible, and we can apply the primal Simplex method. We would first recompute the reduced costs and update the zeroth row of the tableau, and then proceed with the primal Simplex method. (d) If the cost of x 2 is changed to 1 the reduced cost of x 2 becomes c 2 = c 2 c B B 1 A 2 = 1 [ 2 0 0 ] 0 1/2 = 1 and the current basis is not optimal anymore. Variable x 2 has negative reduced cost and must enter the basis, but all elements in the 1/4 corresponding column of the tableau are negative. This indicates that the optimal cost is. Since all the elements in the column of x 2 in the tableau are non-positive the corresponding basic direction d 2 has all non-negative components which means that we Page 5 / 6
can construct a feasible solution x + θd 2 with cost c x + θ c 2 = c x θ (since c 2 = 1) for any θ 0 starting from any feasible solution x. From the tableau, we obtain d 2 = (0, 1, 1/4, 0, 1/2, 0). Starting from x = (0, 0, 0, 2, 0, 2) which is feasible and has cost 0, we obtain x + θd 2 = (0, θ, 1/4θ, 2, 1/2θ, 2) which is feasible and has cost θ. For any θ = M 0 we have a feasible solution of cost M. Problem 4 (a) Recall that one of the flow conservation constraints of a minimum cost network flow problem is redundant. We thus assume that the flow conservation constraint of node 7 has been removed from the problem, and for convenience we adopt the convention p 7 = 0 (where p 7 would be the dual variable associated with the removed constraint). Before the iteration. (i) T = {(1, 4), (4, 2), (2, 5), (4, 3), (3, 6), (6, 7)}, and the corresponding tree solution is f 14 = 100, f 42 = 50, f 25 = 50, f 43 = 50, f 36 = 40, f 67 = 60, f ij = 0, (i, j) T. (ii) The dual variables are computed as p i p j = c ij, (i, j) T (where we assume p 7 = 0). We obtain: p 6 = 7, p 3 = 9, p 4 = 14, p 1 = 15, p 2 = 12, p 5 = 9. (iii) To find an entering variable we compute the reduced costs for the arcs not in T as c ij = c ij p i + p j (where we assume p 7 = 0), until we find one which is negative (if there is one). Since c 47 = 8 14 < 0, f 47 is the entering variable and arc (4, 7) is added to T. This closes the cycle formed by arcs (4, 7), (6, 7), (3, 6), (4, 3) among which arcs (6, 7), (3, 6), (4, 3) are backward. The maximum flow that can be pushed around this cycle is θ = min{f 67, f 36, f 43 } = 40. A flow amount θ is added to arc (4, 7) and subtracted from arcs (6, 7), (3, 6), (4, 3). The variable f 36 becomes 0 and leaves the basis, and arc (3, 6) is removed from T. After the iteration. (i) T = {(1, 4), (4, 2), (2, 5), (4, 3), (4, 7), (6, 7)}, and the corresponding tree solution is f 14 = 100, f 42 = 50, f 25 = 50, f 43 = 10, f 47 = 40, f 67 = 20, f ij = 0, (i, j) T. (ii) The dual variables are computed as p i p j = c ij, (i, j) T (where we assume p 7 = 0). We obtain: p 6 = 7, p 4 = 8, p 3 = 3, p 1 = 9, p 2 = 6, p 5 = 3. (b) The reduced costs of non-basic variables after the iteration are: c 12 = 4 9 + 6 = 1, c 13 = 6 9 + 3 = 0, c 34 = 5 3 + 8 = 10, c 36 = 2 3 + 7 = 6, c 54 = 6 3 + 8 = 11, c 57 = 9 3 + 0 = 6, and c 64 = 4 7 + 8 = 5. All reduced costs are non-negative and the optimality condition of the network Simplex algorithm is met. Both the primal and dual solutions obtained after the iteration at the end of point (a) are optimal. The dual solution is degenerate because the reduced cost of the non-basic variable f 13 is zero. Page 6 / 6