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Shadow Prices

**Economic interpretation?
**

*a x a x a x b
a x a x a x b
*

*a x a x a x b
x x x
*

*n n
*

*n n
*

*m m mn n m
*

*n
*

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

1 2 0

+ + + ≤ + + + ≤

+ + + ≤ ≥

. .. . ..

. .. .. . . .. .. .

. .. .. . . .. .. . . ..

, , .. .,

max
*x
*

*j j
j
*

*n
Z c x*= ∑

=1

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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

*a x a x a x s b
a x a x a x s b
*

*a x a x a x s b
*

*x j n m
*

*n n
*

*n n
*

*m m mn n m m
*

*j
*

11 1 12 2 1 1 1

21 1 22 2 2 2 2

1 1 2 2

0 1

+ + + + =

+ + + + =

+ + + + =

≥ = +

.. . .. .

. .. .. . .. . . .. . ..

. .. .. . .. . . .. . .. . . .

, ,. . .,

max ... .. .
,*x s n n m
*

*Z c x c x s s s*= + + + + + +1 1 1 20 0

**Correction: Add si >= 0, i=1,2,...,m.
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*a y a y a y t c
a y a y a y t c
*

*a y a y a y t c
y y y t t t
*

*m m
*

*m m
*

*n n mn m n n
*

*m n
*

11 1 21 2 1 1 1

12 1 22 2 2 2 2

1 1 2 2

1 2 1 2 0

+ + + − = + + + − =

+ + + − = ≥

. . . . ..

.. . . . . . . . . . . . . .

.. . . . . . . . . . . . . . . . .

, , . . . , , , , . . . ,

min . . . . . .
, *y t m m
*

*w b y b y t *= + + + + + 0tn 1 1 1 0

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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 non-

negative, this is equivalent to

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*s y i mi i *= =0 1 2, , , .. . ,

*t x j nj j *= =0 1 2, , , . . .,

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Example 7.5.2 In example 7.4.2 we have x = (12,0,0) ; s=(4,0) t = (0,4,14) ; y =(0,16)

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Clue ...

Eq. # *x*1 --- *xn s*1 --- *sm *RHS
--- --- --- (7.74)
Z *t*1 --- *tn y*1 --- *ym *yb

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