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15.053 Thursday, February 28
• Sensitivity Analysis 2
– More on pricing out
– Effects on final tableaus
Handouts: Lecture Notes
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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 minratio 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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Shadow Price
• 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 1, gadgets on hand?
• This is the value of an extra gadget on hand.
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Shadow price vs slack variable
Claim: increasing the 6 to a 7 is mathematically equivalentto replacing “x3 ≥ 0” with “x3 ≥ -1”. This is also thereduced cost for variable x3.
Reason 1. Permitting Sarah to have 7 thousand gadgets isequivalent to giving her 6 thousand and letting her use 1thousand more than she has (at no cost).
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Shadow price vs slack variable
Claim: increasing the 6 to a 7 is mathematically equivalentto replacing “x3 ≥ 0” with “x3 ≥ -1”. This is also thereduced cost for variable x3.
Reason 2. Any solution to the original problem can betransformed to a solution with RHS 7 by subtracting 1from x3.x1 = 0, x2 = 1, x3 = 3, x4 = 0 => x1 = 0, x2 = 1, x3 = 2, x4 = 0
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Shadow pricevs slack variable
Looking at theslack variablein the finaltableau revealsshadow prices.
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
• To do with your partner: what is the shadow price for the 2nd constraint (widgets on hand)?
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The cost row inthe final tableauis obtained byadding multiplesof originalconstraints to theoriginal cost row.
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How arethe reducedcosts in the2nd tableaubelowobtained?
Take the initialcost coefficients.
Then subtract 1/3of constraint 1.
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Next: subtract½ of constraint2 from thesecosts.
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How arethe reducedcosts in the2nd tableaubelowobtained?
Subtract 1/3 ofconstraint 1and ½ ofconstraint 2from the initialcosts.
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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 theeffect ofadding Δ tothe costcoefficientfor x3?
FACT: Adding Δ tothe cost coefficient inan initial tableau alsoadds Δ to the samecoefficient insubsequent tableaus
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What is theeffect ofadding Δ tothe costcoefficientfor x2?
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Subtract Δtimes row 3from row 1 toget it back incanonical form.
How large canΔ be?
Δ ≤ 1 for thetableau toremain optimal.Bound onchanges in costcoefficients.
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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 thatwe addanothervariable, sayx5. Should weproduce x5?What is ?
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= 3/2 - 2*1/3 – 1*1/2 = 1/3
FACT: We cancompute thereduced cost of anew variable. Ifthe reduced costis positive, itshould be enteredinto the basis.
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More on Pricing Out
• Every tableau has “prices.” These are usually called simplex multipliers.
• The prices for the optimal tableau are the shadow prices.
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Simplex Multipliers
π1=1/3
π2= 1/2
FACT: x2 is abasic variableand 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 th
e 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• Shadow prices and simplex multipliers• Changes in cost coefficients• Linear relationships between columns in the• original tableau are preserved in the final tabl
eau.• Determining upper and lower bounds on Δ so
that the shadow price remains valid.