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Thursday, February 28
• Sensitivity Analysis 2
– More on pricing out
– Effects on final tableaus
Handouts: Lecture Notes
1
Partial summary of last lecture
• The shadow price is the unit change in the optimal
objective value per unit change in the RHS.
• The shadow price for a “≥ 0” constraint is called the
reduced cost.
• Shadow prices usually but not always have
economic interpretations that are managerially
useful.
• Shadow prices are valid in an interval, which is
provided by the Excel Sensitivity Report.
• Reduced costs can be determined by pricing out
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Running Example (from lecture 4)
• Sarah can sell bags consisting of 3
gadgets and 2 widgets for \$2 each.
• She currently has 6000 gadgets and
2000 widgets.
• She can purchase bags with 3 gadgets
and 4 widgets for \$3.
• Formulate Sarah’s problem as an LP
and solve it.
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Shadow Prices can be found by
examining the initial and final tableaus!
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The Initial Basic Feasible Solution
Apply the min
ratio rule
min
(6/3, 2/2).
The basic feasible solution is x1 = 0, x2 = 0, x3 = 6, x4 = 2
What is the entering variable? x2
What is the leaving variable? x4
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The 2nd Tableau
The basic feasible solution is x1 = 0, x2 = 1, x3 = 3, x4 = 0, z = 2
What is the next entering variable? x1
What is the next leaving variable?
x3
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The 3rd Tableau
The optimal basic feasible solution is
x1 = 1, x2 = 3, x3 = 0, x4 = 0, z = 3
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• The shadow price of a constraint is the
increase in the optimum objective value
per unit increase in the RHS coefficient,
all other data remaining equal.
• What is the shadow price for constraint
• This is the value of an extra gadget on
hand.
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Claim: increasing the 6 to a 7 is mathematically equivalent
to replacing “x3 ≥ 0” with “x3 ≥ -1”. This is also the
reduced cost for variable x3.
Reason 1. Permitting Sarah to have 7 thousand gadgets is
equivalent to giving her 6 thousand and letting her use 1
thousand more than she has (at no cost).
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Claim: increasing the 6 to a 7 is mathematically equivalent
to replacing “x3 ≥ 0” with “x3 ≥ -1”. This is also the
reduced cost for variable x3.
Reason 2. Any solution to the original problem can be
transformed to a solution with RHS 7 by subtracting 1
from x3.
x1 = 0, x2 = 1, x3 = 3, x4 = 0 => x1 = 0, x2 = 1, x3 = 2, x4 = 0
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vs slack variable
Looking at the
slack variable
in the final
tableau reveals
What is the optimal solution if x3 ≥ 0?
What is the optimal solution if x3 ≥ -1?
What is the shadow price for constraint 1?
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Quick Summary
• Connection between shadow prices and
reduced cost. If xj is the slack variable for a
constraint, then its reduced cost is the
negative of the shadow price for the
constraint.
• The reduced cost for a variable is its cost
coefficient in the final tableau
price for the 2nd constraint (widgets on
hand)?
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The cost row in
the final tableau
is obtained by
of original
constraints to the
original cost row.
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How are
the reduced
costs in the
2nd tableau
below
obtained?
Take the initial
cost coefficients.
Then subtract 1/3
of constraint 1.
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Next: subtract
½ of constraint
2 from these
costs.
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How are
the reduced
costs in the
2nd tableau
below
obtained?
Subtract 1/3 of
constraint 1
and ½ of
constraint 2
from the initial
costs.
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Implications of Reduced Costs
• Implication 1: increasing the cost
coefficient of a non-basic variable by Δ
leads to an increase of its reduced cost
by Δ.
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What is the
effect of
the cost
coefficient
for x3?
the cost coefficient in
an initial tableau also
coefficient in
subsequent tableaus
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What is the
effect of
the cost
coefficient
for x2?
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Subtract Δ
times row 3
from row 1 to
get it back in
canonical form.
How large can
Δ be?
Δ ≤ 1 for the
tableau to
remain optimal.
Bound on
changes in cost
coefficients.
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Implications of Reduced Costs
• Implication 2: We can compute the
reduced cost of any variable if we know
the original column and if we know the
“prices” for each constraint.
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Suppose that
another
variable, say
x5. Should we
produce x5?
What is
?
23
= 3/2 - 2*1/3 – 1*1/2 = 1/3
FACT: We can
compute the
reduced cost of a
new variable. If
the reduced cost
is positive, it
should be entered
into the basis.
24
More on Pricing Out
• Every tableau has “prices.” These are
usually called simplex multipliers.
• The prices for the optimal tableau are
25
Simplex Multipliers
π1=1/3
π2= 1/2
FACT: x2 is a
basic variable
and so
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A useful fact from linear algebra
• If column j in the initial tableau is a
linear combination of the other
columns, then it is the same linear
combination of the other columns in
the final tableau.
• e.g., if A.3 = A.2 + 2 A.1 , then
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On varying the RHS
• Suppose one adds Δ to b1.
– This is equivalent to adding Δ times
the column corresponding to the first
slack variable
– One can compute the shadow price
and also the effect on
• This transformation also provides
upper and lower bounds on the interval
for which the shadow price is valid.
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Summary of Lecture
•
•
•
•
•
Using tableaus to determine information
Changes in cost coefficients
Linear relationships between columns in the
original tableau are preserved in the final
tableau.
• Determining upper and lower bounds on Δ so
that the shadow price remains valid.
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