4.1 Graphical solution of a linear program and standard form
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1 4.1 Graphical solution of a linear program and standard form Consider the problem min c T x Ax b x where x = ( x1 x ) ( 16, c = 5 ), b = 4 5 9, A = Solve the problem graphically and determine the basic and nonbasic variables of the optimal solution.. Put the problem in standard form w.r.t. the optimal basis (identify B, N, and the corresponding partition of the cost vector). 4. Geometry of linear programming Consider the linear program max z = x 1 +x x 1 +x 4 (1) x 1 +x () x 1 x 1 () x 1,x 1. Solve it graphically and indicate the optimal solution and the corresponding objective function value.. Determine the basic feasible solutions (specifying the basic and nonbasic variables) corresponding to all vertices of the feasible region (polyhedron).. Indicate the sequence of basic feasible solutions that are visited by the simplex algorithm (let x 1 be the first variable to enter the basis) when starting from the initial basic feasible solution in which all slack variables are basic variables. 4. Determine the reduced costs for the basic feasible solutions associated to the vertices ((eq. 1) (eq. )) and ((eq. 1) (eq. )), where (eq. i) is obtained from (i), by substituting with =. 5. Show, geometrically, that the gradient of the objective function is a conic combination (i.e. a linear combination with nonnegative coefficients) of the gradients of the constraints which are active at an optimal vertex. Indicate the value taken by the objective function in that vertex. Note: All the constrains must be in form, since the problem is a maximization one (e.g., x 1 must be rewritten as x 1 ). Document prepared by L. Liberti, S. Bosio, S. Coniglio, and C. Iuliano. Translation to English by S. Coniglio 1
2 6. Determine the range of values for the right hand side b 1 of constraint (1) for which the optimality of the basis solution is preserved. 7. Indicate for which values of the objective function coefficients ((x 1 = ) (eq. )) is an optimal vertex. 8. Determine the range of values for the right hand side b of constraint () for which the feasible region is (a) empty, (b) contains a single point. 9. Indicate the range of values for c 1 for which there are multiple optimal solutions. 4. Simplex algorithm with Bland s rule Solve the following linear program min z = x 1 x (4) x 1 +x = 1 (5) x 1 +x x = 5 (6) x 1,x,x (7) by applying the two-phase Simplex algorithm with Bland s rule. Document prepared by L. Liberti, S. Bosio, S. Coniglio, and C. Iuliano. Translation to English by S. Coniglio
3 Solution 4.1 Graphical resolution and standard form 1. The equations associated to the system are Ax = b x 1 +7x = 4 (8) x 1 +5x = 5 (9) x 1 +x = 9 (1) The corresponding lines are shown in Figure 1. The level curves of the objective function x x 1 +x = 9 z z = 16x 1 +5x 1 x 1 +5x = 5 x 1 +7x = x 1 Figure 1: The polyhedron is unbounded arez = 16x 1 +5x, or, equivalently, x = 16 5 x 1+ z 5. InspectingFigure, weobservethat vertex R is the unique optimal solution. Since vertex R is the intersection of equations (9) and (1), we obtain it by solving obtaining x 1 = 7,x = 1 7. x 1 +5x = 5 x 1 +x = 9. To express the problem in standard form, we introduce slack variables, s 1,s,s, one per Document prepared by L. Liberti, S. Bosio, S. Coniglio, and C. Iuliano. Translation to English by S. Coniglio
4 eq. (1) eq. (9) eq. (8) x = U R valori decrescenti funzione obiettivo Q S P T Figure : Graphical resolution: R is an optimal solution constraint. The new problem is min c T x A x = b x where x = (x 1,x,s 1,s,s ) T, c = (16,5,,,), and A = = (A I). 1 Since R is obtained by intersecting (8) and (9), the slack variables of the corresponding constraints are null in R. By letting x B = (x 1,x,s 1 ) and x N = (s,s ), we partition the variables into x = (x B x N ). The corresponding partition of matrix A is A = = (B N). Document prepared by L. Liberti, S. Bosio, S. Coniglio, and C. Iuliano. Translation to English by S. Coniglio 4
5 The basic variables in R have values x B = B 1 b. Therefore, B 1 = and x B = ( 7, 1 7, 9 7 ), where x B = (x 1,x ). 4. Geometry of linear programming 1. The equations associated to constraints (1), (), () are or, equivalently, x 1 +x = 4 (eq. 1) x 1 +x = (eq. ) x 1 x = 1 (eq. ) x = x 1 +4 x = x 1 + x = x 1 1. The level curves of the objective function are z = x 1 + x, i.e., x = x 1 + z. The polyhedron PQROS of the feasible solutions is shown in Figure. The unique optimal solution is achieved at vertex P = ( 1,), where z = 15. eq. (1) eq. () P eq. () S max z* Q O R Figure : Polyhedron of the feasible solutions Document prepared by L. Liberti, S. Bosio, S. Coniglio, and C. Iuliano. Translation to English by S. Coniglio 5
6 . To express the problem in standard form, we introduce slack variables, s 1,s,s, one per constraint. The problem is where x = x 1 x s 1 s s, c = max z = c T x A x = b x,, b = b, A = (A I) = (a) Vertex P: s 1 =,s =, where x B = (x 1,x,s ), x N = (s 1,s ), 1 1 B = 1, N = (b) Vertex Q: s 1 =,s =, where x B = (x 1,x,s ), x N = (s 1,s ), 1 1 B = 1 1, N = (c) Vertex R: x =,s =, where x B = (x 1,s 1,s ), x N = (x,s ), 1 1 B = 1, N = (d) Vertex O: x 1 =,x =, where x B = (s 1,s,s ), x N = (x 1,x ), 1 1 B = 1, N = (e) Vertex S: x 1 =,s =, where x B = (x,s 1,s ), x N = (x 1,s ), 1 1 B = 1, N = Assume that, initially, the basic solution is given by the slack variables, i.e., x B = (s 1,s,s ). Variables x 1,x are nonbasic. The solution corresponds to vertex O (the origin). Assuming that x 1 becomes basic, we are increasing the value of x 1, i.e., we are moving on segment O R. Variables s 1 and s decrease as x 1 increases. The first one to become is s. We reach vertex R, where x B = (x 1,s 1,s ). The next variable to become basic is x. We are moving on segment R Q, and obtain the next solution in Q, where x B = (x 1,x,s ). Then s becomes basic and s becomes nonbasic, i.e., we move on segment Q P, reaching the unique optimal vertex P, where x B = (x 1,x,s ),. Document prepared by L. Liberti, S. Bosio, S. Coniglio, and C. Iuliano. Translation to English by S. Coniglio 6
7 4. The vertex given by (1) () isp. That given by (1) () is Q. The reduced costs are c = c T c T B B 1 A. Observe that c is zero for basic variables, and (possibly!) nonzero for nonbasic ones. Therefore, we only consider c N = c T N ct B B 1 N. In vertex P, we have B and N as in (b)i, so B 1 N = where c B = (,,) and c N = (,). We obtain c N = ( 7 4, 1 4 ). Since both values are and the problem is maximization one, the basic solution corresponding to vertex P is optimal. In vertex Q, we have B,N as in 4.(b)ii, so B 1 N = where c B = (,,) and c N = ( 5, 1 ). Therefore, Q is not an optimal solution. 5. The gradient of the objective function is a conic combination of the gradients of the active constraints only in an optimal vertex. It means that any improving direction is infeasible, which implies that the current vertex is optimal. Consider P = ( 1,). The gradient of the objective function is f = (,). The active constraints, in P, are (1) and (). The gradients are (,1) and (,1). We are to check whether the system ( ) ( ) ( ) λ 1 +λ 1 = 1 admits a solution where λ 1 and λ. Since it is of full rank, the solution is unique. Since it amounts to λ 1 = 7 4,λ = 1 4, P satisfies the condition. See Figure 4. We verify that the condition is not satisfied in the nonoptimal vertices Q,R,O,S. Vertex Q. Active constraints (1), () with gradients (,1), (1, 1). We obtain λ 1 = 5, λ = 1 <. The condition is not satisfied. Vertex R. Active constraints (), x with gradients (1, 1), (, 1). We obtain λ 1 =,λ = 5 <. The condition is not satisfied. Vertex O. Active constraints x 1, x with gradients ( 1,), (, 1). We obtain λ 1 = <,λ = <. The condition is not satisfied. Vertex S. Active constraints x 1, () with gradients ( 1,), (,1). We obtain λ 1 = 7 <,λ =. The condition is not satisfied. 6. Geometrical solution, without sensitivity analysis. For b 1, the optimality of the optimal basis is preserved. Let S = (,s). If b 1 decreases, while b 1 > s, the optimality of the basis is preserved. If b 1 = s, we have a degenerate solution which can be expressed both as x B = (x 1,s,x ) or x B = (x 1,s,s ). For < b 1 < s, x 1 = becomes nonbasic and s becomes basic, since the corresponding constraints () is no more active. For b 1 =, the problem has a single feasible point, (,), and for b 1 < the feasible region is empty. 7. S is an optimal solution for any objective function with a gradient which is a conic combination of ( 1,) T and (,1) T. Document prepared by L. Liberti, S. Bosio, S. Coniglio, and C. Iuliano. Translation to English by S. Coniglio 7
8 (eq.1) eq. () g g 1 P f S Q O R Figure 4: Optimality of vertex P 8. The feasible region contains a single point, Q, for b such that Q is the intersection of the three lines corresponding to the three constraints, i.e., for b = 8. The feasible region is empty for b < We have multiple optimal solutions if the gradient of the objective function is parallel to that of an inequality defining a facet f of the polyhedron, i.e., if the two gradients are equivalent up to a positive multiplicative factor. 4. Simplex method with Bland s rule The problem is already in standard form. Constraints (1)-() equal to the systemax = b, where x = (x 1,x,x ) T, b = (1,5) T and A = ( 1 Since a feasible basic solution is not evident, we apply the two phases simplex method. Phase I This phase finds a feasible basic solution, if any. We solve the auxiliary problem ). min v = i y i s.t. Ax+Iy = b x y Document prepared by L. Liberti, S. Bosio, S. Coniglio, and C. Iuliano. Translation to English by S. Coniglio 8
9 where an artificial y i is introduced for each equality constraint. Observe that vector b must be nonnegative. If it is not, we need to multiply the lines where the entry is negative by 1. A feasible basic solution for the auxiliary problem is given by x B = (y 1,y ), where the objective function has a strictly positive value. The problem admits an optimal solution of value if and only if the original problem admits a feasible solution. Indeed, to achieve a value of, all variables y i must be zero, which implies that Ax = b can be satisfied. If no optimal solution of value is found, the original problem is infeasible and the algorithm stops. For the current problem, the auxiliary problem reads min { y 1 +y Ā x = b,x,y }. The initial basic solution is x B = (y 1,y ). We express the basic variables w.r.t. the nonbasic ones y 1 = 1 x 1 x y = 5 x 1 x +x so that the objective function becomes v = y 1 +y = 6 5x 1 x x. The initial tableau is x 1 x x y 1 y y y The reduced costs of the nonbasic variables x N = (x 1,x,x ) are all negative. Using Bland s rule, we pick the nonbasic variable with smallest index, x 1. There are two limitations to the growth of x 1, i.e., given by y 1 = 1 x 1 and y = 5 x 1. Therefore, the tightest limit is min { 1, 5 } = 1, given by the basic variable y 1, which leaves the basis. Pivoting is performed on coefficient in tableau position (1,1) 1. divide row 1 by. add 5 times row 1 to row. subtract times row 1 from row We obtain the tableau x 1 1 y 7 x 1 x x y 1 y which shows the current basic solution x B = (x 1,y ) T = (1/,7/) T of value 7/ (obviously x N = ). Since the only negative reduced cost is given by x, x enters the basis. The only limit to its growth is given by y = 7 x, therefore x 7 4 and y leaves the basis. Pivoting is performed on coefficient in position (,) of the previous tableau Document prepared by L. Liberti, S. Bosio, S. Coniglio, and C. Iuliano. Translation to English by S. Coniglio 9
10 1. add row to row. divide row by We obtain x 1 x x y 1 y x x Since the reduced costs for the nonbasic variables are all nonnegative, Phase I stops, proving that the feasible region of the original problem is nonempty, and yielding an initial basic solution of value x B = (x 1,x ). Phase II The objective function is x 1 x, and the initial basis is (x 1,x ), with x nonbasic. We express x 1,x w.r.t. x x 1 = 1 x x = x The objective function becomes 7x. Removing from the tableau the columns of the auxiliary variables, we obtain x 1 x x -7 1 x x Variable x, having a negative reduced cost of -7, enters the basis. Since the only limit to its growth is given by x 1 = 1 x, x 1 leaves the basis. Therefore, we perform a pivoting operation on coefficient in position (,1) 1. divide row 1 by ;. add 7 times row 1 to row. add 11 4 times row 1 to row We obtain 16 1 x 8 x x 1 x x Since the reduced costs are all nonnegative, the algorithm terminate and it yields the optimal solution (, 8, 1 ) with a value z = 16. Document prepared by L. Liberti, S. Bosio, S. Coniglio, and C. Iuliano. Translation to English by S. Coniglio 1
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