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# opgaven - Advanced Linear Programming The Exercises The...

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Unformatted text preview: Advanced Linear Programming: The Exercises The answers are sometimes not written out completely. 1.5 a) min c T x + d T y s . t . Ax + By ≤ b y = | x | (1) First reformulation, using z smallest number satisfying x ≤ z and- x ≤ z : min c T x + d T z s . t . Ax + Bz ≤ b x ≤ z- x ≤ z. (2) Second reformulation, using x = x +- x- and | x | = x + + x- : min c T x +- c T x- + d T x + + d T x- s . t . Ax +- Ax- + Bx + + Bx- ≤ b x + ≥ x- ≥ . (3) 1.5 b) Suppose (1) has feasible solution x (hence y = | x | ), then choosing x + i = x i and x- i = 0 if x i ≥ 0 and x + i = 0 and x- i =- x i if x i ≤ 0 gives a feasible solution of (3). Moreover, the solution values coincide, implying that the optimal value of (3) is at most that of (1). Suppose (3) has feasible solution x + ,x- with the property that for all i = 1 ,...,n , x + i > ⇒ x- i = 0 and x- i > ⇒ x + i = 0. Then setting x i = x + i- x i and z = x + i + x- i gives a feasible solution for (2) with equal value, implying that the optimal value of (2) is at most that of (3). If (3) has a feasible so- lution that does not have the property then there are variables x + i > 0 and x- i > 0. Setting ˆ x + i = x + i- min { x + i ,x- i } and ˆ x- i = x- i- min { x + i ,x- i } gives c i (ˆ x + i- ˆ x- i ) = c i ( x + i- x- i ) and, since d i ≥ 0 and the i-th row of B is non- negative, we obtain again a feasible solution and d i (ˆ x + i- ˆ x- i ) < d i ( x + i- x- i ). Hence a better feasible solution. Suppose (2) has feasible solution x,z satisfying for all i = 1 ,...,n that z i = x i or z i =- x i . Then, setting y = z and keeping x as it is is a feasible solution of (1) with equal value, implying that the optimal value of (1) is at most that of (2). If (2) has a feasible solution that does not have the property then there exist variables z i > x i and z i >- x i . Clearly, since d i ≥ 0 and the i-th row of B is non-negative, diminishing the value of z i yields another feasible solution with better objective value. 1 1.5 c) The set { x ∈ IR | x- 2 | x | ≤ - 1 , | x | ≤ 1 } is not even connected, let alone convex. Minimizing x gives local optimum x = 1, whereas x =- 1 is the global optimum. ———————- 1.15 a) min (9- 1 . 2) x 1 + (8- . 9) x 2 s . t . 1 4 x 1 + 1 3 x 2 ≤ 90 1 8 x 1 + 1 3 x 2 ≤ 80 x 1 ≥ , x 2 ≥ (4) 1.15 b(i)) min (9- 1 . 2) x 1 + (8- . 9) x 2- 7 x 3- 7 x 4 s . t . 1 4 x 1 + 1 3 x 2- x 3 ≤ 90 1 8 x 1 + 1 3 x 2- x 4 ≤ 80 x 3 + x 4 ≤ 50 x 1 ≥ , x 2 ≥ , x 3 ≥ , x 4 ≥ (5) 1.15 b(ii)) The easiest way is to solve the LP with the discounted price for raw material. If the optimal solution yields a total bill for raw materials of at least of \$300 then this is the best decision. If not, solve the LP as in a) and solve the LP with the extra restriction that the total raw material cost is \$300. Choose the best of these two solutions....
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opgaven - Advanced Linear Programming The Exercises The...

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