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Shadow Prices - Introduction to Operations Research - Lecture Slides - Document's extract

Slides, Operations Research

Post: January 9th, 2013
Extract
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?

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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
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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 ......
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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.

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More generally,  y = cBB-1 is the “dual variable”, and for the last tableu, y is an optimal solution for the dual.

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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.

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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.
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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
m

si >= 0, i=1,2,...,m.

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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

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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

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s i y i = 0 , i = 1,2,..., m tj x j = 0 , j = 1,2,..., n

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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

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Clue ...
Eq. # --Z

x1 t1

-------

xn tn

s1 y1

-------

sm

RHS (7.74) yb

ym

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