Introduction to Discrete Optimization

Transcription

1 Prof. Friedrich Eisenbrand Martin Niemeier Due Date: March 6, 9 Discussions: March 7, 9 Introduction to Discrete Optimization Spring 9 s 4 Exercise A company must deliver d i units of its product at the end of the i th month. Material produced during a month can be delivered either at the end of the same month or can be stored as inventory and delivered at the end of a subsequent month; however, there is a storage cost of c dollars per month for each unit of product held in inventory. The year begins with zero inventory. If the company produces x i units in month i and x i units in month i, it incurs a cost of c x i x i dollars, reflecting the cost of switching to a new production level. Formulate a linear programming problem whose objective is to minimize the total cost of the production and inventory schedule over a period of twelve months. Assume that inventory left at the end of the year has no value and does not incur any storage costs. Let x i, i =,...,, reflect the units of the product produced in month i, and let x :=. Let y i, i =,...,, be the units of the product produced in month i and delivered at the end of month i. y is the amount of product left over after the end of the year. Let y :=. In each month i the company has to deliver d i many units. This constitues of the units y i on stock plus the units x i produced in month i minus the units y i+ held back for deliver next month. Thus we get the constraint y i + x i y i+ = d i for each month i =,...,. To make sure that no units of the product are stored for longer than one month, we add the constraint y i d i. The objective we want to minimize is c i= x i x i + c y i. This objective function is not linear, since. is not linear. To deal with this, we model the absolute value x i x i by introducing variables z i and the constraints x i x i z i i=

2 and Then the objective function is given as x i x i z i. c i= z i + c y i. Finally the values of x, y and z have to be nonnegative. The resulting linear program is: i= min c i= z i + c i= y i subject to y i + x i y i+ = d i i =,..., y i d i i =,..., x i x i z i i =,..., x i x i z i i =,..., x, y, z Exercise The vector x = (,,,) is an optimal solution of min (,,,) x 6 x = } {{ } =A x with x = (x, x, x, x 4 ) R 4. Use the proof of Lemma. to find another optimal solution x such that A J has full column rank with J = {i x i > }. Let J = {i x > } = {,,4}. Observe thast A i J = does not have full column rank. As seen in the proof, we find a vector d R 4 with Ad = and d(j ) = j J. Note that every vector d with these properties is contained in ( ) A 4 kern e T = span Thus choose d = (, 4,,).

3 Observe that c T d = (,,,) (, 4,,) T =. Choose ε maximal, such that + ε 4 Thus ε = /4. Then Now J = {,4} thus which has full column rank. x := x + εd = = /4 / A J = Exercise Solve the following tableau with the simplex algorithm: x x x x 4 x For each iteration give the simplex tableau, the current basis and the indices leaving/entering the basis. The initial basis is B = {x 4, x 5 }. The reduced costs of x, x and x are all negative and thus could be chosen to enter the basis. We choose x. Both entries of x s are positive, but only removing x 5 gives a feasible solution. Thus we get the new tableau x x x x 4 x with basis B = {x, x 4 }. Now only x has negative reduced costs. Thus x enters the basis and x 4 leaves the basis. The new tableau is x x x x 4 x

4 All reduced costs are nonnegative. We have found the otimal solution x = ( 7,,,,) of value. Exercise 4 Solve the following linear program by using the simplex method min ( 4 ) T x 6 x = 9 x Start with the basis B = {4,5,6}. For each iteration give the simplex tableau and the indices leaving/entering the basis. Since A B is the identity matrix, A B is the identity matrix aswell and the initial simplex tableau is given as follows: x x x x 4 x 5 x We choose x to enter the basis. Thus x 6 must leave the basis. The new tableau is: x x x x 4 x 5 x The new basis is B = {x, x 4, x 5 }. Next x enters the basis, and x 5 has to leave. x x x x 4 x 5 x The new basis is B = {x, x, x 4 }. There are no more negative reduced costs. The optimum solution is x = ( 5,,,,,) with c T x = 9. Exercise 5 Suppose you are given an oracle algorithm, which for a given polyhedron P = { x Rñ : Ã x 4

5 b} gives you a feasible solution or asserts that there is none. Show that using a single call of this oracle one can obtain an optimum solution for the LP min{c T x : x R n ; Ax = b; x }, assuming that the LP is feasible and bounded. Hint: Use duality. The LP is feasible and bounded, thus an optimum solution must exist. Strong duality tells us that the dual max{b T y A T y c} is feasible and bounded, and for optimal solutions x of the primal and y of the dual we have b T y = c T x. Thus every point (x, y ) of the polyhedron c T x = b T y Ax = b A T y c x is optimal. Hence with one oracle call for the polyhedron above we get an optimal solution of the LP. 5