Shadow Prices max Z = ∑ c j x j x j =1 n a11 x1 + a12 x 2 + ... + a1n x n ≤ b1 a21 x 1 + a22 x 2 + ... + a2 n x n ≤ b2 ... ... ... ... ... ... ... .. . am1 x1 + a m2 x 2 + ... + amn x n ≤ bm x1 , x 2 ,..., x n ≥ 0 Economic interpretation? Docsity.com xj : units of activity j bi : resource level i aij : units of resource level i used per one unit of activity j cj : return/loss from unit of activity j z : total return/loss z* : optimal return/loss Docsity.com what is the “value” of resource i ? How much are we going to gain/lose if we have more from resource i ? Instead of bi we have bi + ε. As a result, instead of z* we now have z* + ∆. The value of ∆ generated by one unit of ε is called the shadow price of resource i. It is better to use the term “marginal value of resource i” ...... but ...... Docsity.com Clue ... From duality theory we know that z* = y*b where y* is the optimal solution to the dual problem. Furthermore, for problems in standard form y* is equal to the reduced costs of the slack variables. Docsity.com More generally, y = cBB-1 is the “dual variable”, and for the last tableu, y is an optimal solution for the dual. Docsity.com Recipe For problems in stadard form: The shadow price of the i-th resource is equal to the optimal value of the i-th dual variable. Docsity.com warning: The recipe “assumed” that the change in the RHS value does not change the basis itself, namely the elements of the basis are assumed to be the same after the change. If this assumption is not valid, the recipe may not be valid. Docsity.com 7.5.1 Complementary Slackness Theorem max Z = c1x 1 + ... + c n x n + 0s 1 + 0 s2 + ... + sm x ,s a11 x1 + a12 x 2 + ... + a1 n x n + s1 + s2 ... ... = b1 = b2 a21 x 1 + a22 x 2 + ... + a2 n x n ... ... ... ... ... ... .. . ... am1 x1 + am 2 x 2 + .. . + amn x n x j ≥ 0 , j = 1,..., n + m Correction: Add + s m = b

Download original documentm si >= 0, i=1,2,...,m. Docsity.com min w = b1 y1 + . . . + bm y m + 0 t1 + . .. + 0tn y,t a11 y1 + a 21 y 2 + ... + am1 y m − t 1 = c1 a12 y1 + a 22 y2 + ... + a m 2 y m − t2 = c2 .. . ... ... .. . ... ... ... . .. ... . .. a1n y1 + a 2 n y 2 + ... + amn y m y1 , y 2 ,. .., y m , t 1 , t 2 ,... , t n ≥ 0 − tn = cn Docsity.com Let (x,s) be a feasible solution to the primal and (y,t) be a feasible solution to the dual. Then a necessary and sufficient condition for optimality of both solutions is sy = 0 ; tx = 0 Observe that because all the variables are nonnegative, this is equivalent to Docsity.com s i y i = 0 , i = 1,2,..., m tj x j = 0 , j = 1,2,..., n Docsity.com In example 7.4.2 we have x = (12,0,0) ; s=(4,0) t = (0,4,14) ; y =(0,16) Example 7.5.2 Docsity.com Clue ... Eq. # --Z x1 t1 ------- xn tn s1 y1 ------- sm RHS (7.74) yb ym Docsity.com