Chapter 6
Sensitivity Analysis and Duality
PART 3
Mahmut Ali GÖKÇE
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6.10 – Complementary Slackness
The Theorem of Complementary Slackness is an important result
that relates the optimal primal and dual solutions.
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6.10 – Complementary Slackness
This means that:
If a constraint in either the primal or dual is nonbinding,
then the corresponding variable in the other (or complementary)
problem must equal 0.
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6.10 – Complementary Slackness
Dakota Primal:
Dakota Dual:
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6.10 – Complementary Slackness
Optimal Primal Solution:
Optimal Dual Solution:
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6.10 – Complementary Slackness
Interpretation:
At the optimal solution;
Since s1 > 0, y1 = 0  Positive slack in the lumber constraint
implies that lumber must have a zero shadow price.
Slack in the lumber constraint means that extra lumber would not
be used, so an extra board foot of lumber should indeed be
worthless.
Since y2 > 0, s2 = 0  An extra finishing hour has some value. This
can only occur if we are using all available finishing hours.
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6.10 – Complementary Slackness
Interpretation:
At the optimal solution;
Since e2 > 0, x2 = 0  This is reasonable because
e2 = 6y1 + 2y2 + 1.5y3 - 30.
Tables are selling for a price that is less than the value of the
resources. This means that no tables should be made.
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6.10 – Complementary Slackness
Interpretation:
At the optimal solution;
Since x1 > 0, e1 = 0  For any variable xj in the optimal primal basis, the
marginal revenue obtained from producing a unit of xj must equal the marginal
cost of the resources used to produce a unit of xj.
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6.10 – Complementary Slackness
Using Complementary Slackness to solve LPs:
For example, suppose we were told that the optimal solution to the Dakota
problem was z = 280, x1 = 2, x2 = 0, x3 = 8, s1 = 24, s2 = 0, s3 = 0.
Can we use Theorem 2 to help us find the optimal solution to the Dakota dual?
• Because s1 > 0, optimal dual solution must have y1 = 0.
• Because x1 > 0 and x3 > 0, optimal dual solution must have e1 = 0, and e3 = 0. This
means that for the optimal dual solution, the first and third constraints must be
binding.
• From the dual theorem, we know also that z = w = 280.
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6.8 – Shadow Prices
By using the Dual Theorem, we can easily determine the shadow
price of the ith constraint.
The shadow prices of the Dakota problem:
The shadow price of the ith constraint of a max problem is the optimal
value of the ith dual variable.
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6.8 – Shadow Prices
For a max problem:
For a min problem:
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6.8 – Shadow Prices
The shadow price for the lumber constraint is 0; for the finishing constraint, 10; and for
the carpentry constraint, 10.
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6.8 – Shadow Prices
1 – Answer:
The shadow price for the lumber constraint is 0; for the finishing
constraint, 10; and for the carpentry constraint, 10.
In this problem, the shadow price of the ith constraint may be
thought of as the maximum amount that the company would
pay for an extra unit of the resource associated with the ith
constraint.
For example, an extra carpentry hour would raise revenue by
y3 = $10. Thus, Dakota could pay up to $10 for an extra carpentry
hour.
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6.8 – Shadow Prices
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6.8 – Shadow Prices
Example:
Leatherco manufactures belts and shoes.
• A belt requires 2 square yards of leather and 1 hour of skilled
labor.
• A pair of shoes requires 3 sq yd of leather and 2 hours of skilled
labor.
• As many as 25 sq yd of leather and 15 hours of skilled labor can be
purchased at a price of $5/sq yd of leather and $10/hour of skilled
labor.
• A belt sells for $23, and a pair of shoes sells for $40.
Leatherco wants to maximize profits (revenues - costs). Formulate
an LP that can be used to maximize Leatherco’s profits. Then find
and interpret the shadow prices for this LP.
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6.8 – Shadow Prices
Solution:
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6.8 – Shadow Prices (Solution cont.)
If one more square yard of leather were available, then Leatherco’s
objective function (profits) would increase by $1.
If we purchase another square yard of leather at the current price
of $5, profits increase by y1 = $1.
So, if we purchase another square yard of leather at a price of $6 =
$5 + $1, then profits increase by $1 - $1 = $0.
Thus, the most Leatherco would be willing to pay for an extra
square yard of leather is $6 (it is not equal to the shadow price).
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6.8 – Shadow Prices
Example:
Steelco has received an order for 100 tons of steel. The order must contain at
least 3.5 tons of nickel, at most 3 tons of carbon, and exactly 4 tons of
manganese.
Steelco receives $20/ton for the order. To fill the order, Steelco can combine four
alloys, whose chemical composition is given in Table 28.
Steelco wants to maximize the profit (revenues - costs) obtained from filling the
order. Formulate the appropriate LP. Also find and interpret the shadow prices for
each constraint.
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6.8 – Shadow Prices
Solution:
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6.8 – Shadow Prices
Solution (cont.):
The nickel constraint has a negative shadow price because increasing the righthand side of the nickel constraint makes it harder to satisfy the nickel constraint.
In this problem, both equality constraints had positive shadow prices. In general,
we know that it is possible for an equality constraint’s dual variable (and
shadow price) to be negative.
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3.4. A Diet Problem (Stigler, 1939)
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My diet requires that all the food I get come from one of the four “basic
food groups”.
At present, the following four foods are available for consumption:
brownies, chocolate ice cream, cola and pineapple cheesecake.
Each brownie costs 50¢, each scoop of ice cream costs 20 ¢, each bottle of
cola costs 30 ¢, and each piece of pineapple cheesecake costs 80 ¢.
Each day, I must ingest at least 500 calories, 6 oz of chocolate, 10 oz of
sugar, and 8 oz of fat.
The nutritional content per unit of each food is given.
Formulate a linear programming model that can be used to satisfy my
daily nutritional requirements at minimum cost.
ISE203/IE251
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ISE203/IE251
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Output Interpretation (Minimization)
These are equal to the negative of the dual variable values:
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