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**– Effects on final tableaus**

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 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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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 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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**Shadow price vs slack variable**

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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**Shadow price vs slack variable**

Looking at the slack variable in the final tableau reveals shadow 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 in the final tableau is obtained by adding multiples 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 adding Δ to the cost coefficient for x3? FACT: Adding Δ to the cost coefficient in an initial tableau also adds Δ to the same coefficient in subsequent tableaus

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What is the effect of adding Δ to 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 we add another variable, say x5. Should we produce x5? What is ?

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= 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.

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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 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 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**

Shadow prices and simplex multipliers 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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