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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 ith 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 ith row of B is nonnegative, 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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 Spring '10
 Bertsekas
 Trigraph, feasible solution, Frank Reinders

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