08-3-LPDual

Oct 04 ISMT162/Weixin Shang 1

Duality and Shadow Price


Topics

Dual problem construction Economic interpretation for dual problem,

shadow price Duality


Example 1 Suppose that a firm produces product A and B, each

of them generates an income of $300 and $500, respectively

One unit of product A requires one unit of resource aand 3 units of c while one unit of product B needs 2 units of b and 2 units of c

The firm has 40, 120, and 180 units of resources a, b, c, respectively.

How many of A and B should the firm produce? Or what is the optimal allocation of resources to A

and B?


The Firm 1s Production Problem LP formulation

max 300x1+500x2s. t. x1 40 (1)

2x2 120 (2) 3x1+2x2 180 (3) x1, x2 0


Upper Bound: Case 1 Multiplying both sides constraint (3) by 250, we have

750x1+ 500x2 45,000 Comparing with the objective function

z = 300x1 +500x2 750x1+500x2 45,000 We can conclude 45,000 is an upper bound of the

objective function, i.e., with the availableresources the firmcannot generate morethan $45,000 in income

max 300x1+500x2s. t. x1 40 (1)

2x2 120 (2) 3x1+2x2 180 (3) x1, x2 0


Case 2 Multiplying both sides of constraint (1) by 300

and (2) by 250, we have300x1 12,000 and 500x2 30,000

Adding the two inequalities, we have z =300x1+500x242,000

The new income upperbound is 42,000

This is a tighter bound

max 300x1+500x2s. t. x1 40 (1)

2x2 120 (2) 3x1+2x2 180 (3) x1, x2 0


Case 3 Multiplying both sides of constraint (2) by 150

and (3) by 100, we have 300x2 18,000, 300x1+ 200x2 18,000

Adding them together, we havez = 300x1+500x236,000an even better upper bound

Can we do better than this? What do these bounds tell us?

max 300x1+500x2s. t. x1 40 (1)

2x2 120 (2) 3x1+2x2 180 (3) x1, x2 0


The General Case Multiplying both sides of constraints (1), (2) and (3)

by y1, y2 and y3, respectively (y1, y2, y3 0), we havey1 x1 40y1

2y2 x2 120y23y3x1+ 2y3 x2 180y3

We then have(y1+3y3)x1+(2y2+2y3)x2 40y1+120y2+180y3

- Case 1: y1= 0, y2= 0, y3=250, upper bound = 45,000;- Case 2: y1=300, y2=250, y3= 0, upper bound = 42,000;- Case 3: y1= 0, y2=150, y3=100, upper bound = 36,000


The Upper Bound Problem Similar to the three cases, we have

z=300x1+500x2 (y1+3y3)x1+(2y2+2y3)x240y1+120y2+180y3

We have: y1+3y3 300 , 2y2+2y3 500 (4) and (5)

and a smaller40y1+120y2+180y3 (6)

gives us a better (tighter) upper bound Clearly, different yis lead to different upper bounds We see a new optimization problem, i.e., finding yis

to minimize the upper bound (6), subject to (4) and (5)


The Dual Problem

(D) min 40y1+120y2 +180y3s.t. y1 +3y3 300 (4)

2y2 +2y3 500 (5)y1, y2, y3 0

(P) max 300x1+500x2 s.t. x1 40 (1)

2x2 120 (2) 3x1+2x2 180 (3) x1, x2 0

The optimal solution for problem (D) gives the Values of ys that lead to the best upper bound of the objective function value of the original problem. Obviously, it is also the

optimal objective valueof the original problem

Since solving (D) is equivalent to solving(P), we call (D) thedual problem of (P)


Dual Construction: Example 2 Following the upper (or lower) bound idea, we

can construct the dual problem for any LP; The following is a minimization problem:

This time, we need to find a tight lower bound!

(P) min 2x1 + x2 + x3s.t. x1 x2 + x3 2 (7)

x1 + x2 3 (8)x1, x2, x3 0


Example 2 Multiplying constraint (7) by y1 and (8) by y2:

y1x1 y1x2 + y1x3 2y1y2x1 + y2x2 3y2

Then (y1+y2)x1+(-y1+y2) x2 +y1x3 2y1+3y2, or2x1+x2+x3 (y1+y2)x1+(-y1+y2)x2+y1x32y1+3y2

(P) min 2x1 + x2 + x3s.t. x1 x2 + x3 2 (7)

x1 + x2 3 (8)x1, x2, x3 0


Example 2 We require that

y1+y2 2, y1+y2 1, y1 1 The dual problem: (D) max 2y1+ 3y2

subject to y1 + y2 2 y1+ y2 1y1 1y1, y2 0(P) min 2x1 + x2 + x3

s.t. x1 x2 + x3 2 (7)x1 + x2 3 (8)

x1, x2, x3 0Each dual variable is associated to a constraint of the prime problem


Economic Interpretation of Dual Consider example 1. Suppose that another firm, firm

2, wants to buy the resources from firm 1. What are the fair prices for the resources?

Let y1, y2, and y3 be the prices that firm 2 offers to firm 1. The total amount firm 2 pays is

40y1+120y2+180y3 Of course, firm 2 wants to pay as little as possible, i.e.

minimize 40y1+120y2+180y3


What Can Firm 1 Accept? The prices must satisfy firm 1 For firm 1, using one unit of resource a and 3

units of resource c, it could produce one unit product A for a profit of $300.

Thus, the prices must satisfy the following constraints:

y1 + 3y3 300 (for A)2y2+2y3 500 (for B)


The Optimal Offer Prices Firm 2 faces the optimal pricing problem

The LP for firm 2 is the dual problem of the LP for firm 1

(D) min 40y1+120y2 +180y3s.t. y1 +3y3 300 (4)

2y2 +2y3 500 (5)y1, y2, y3 0


Shadow Price The optimal prices y1*, y2* and y3* are called

shadow prices For the seller, shadow prices are fair prices

because the total payment received from selling the resources equals the maximum income that the seller can earn if she uses the resources to produce herself. Shadow prices are the best prices she can get under normal conditions

For the purchaser, shadow prices yield the minimum acceptable total acquisition payment. It is also the maximum amount he is willing to pay normally


Relations Between Primal and Dual

PrimalDual

Have an Opt.Solution Unbounded Infeasible

Have an Opt.Solution

Unbounded

Infeasible


Complementary Slackness Theorem If a constraint is inactive (strictly > or


Using Complementary Slackness The optimal decision for firm 1 is x1*=20, x2*=60,

what are the fair prices for the resources? x1*, x2*>0, so (4) and (5) in dual are active; (1) is

inactive, so y1*=0; Therefore: y1*=0; y1*+3y3*=300; 2y2*+2y3*=500

Shadow price: y1*=0; y3*=100; y2*=150 (P) max 300x1+500x2

s.t. x1 40 (1)2x2 120 (2)

3x1+2x2 180 (3)x1, x2 0

(D) min 40y1+120y2 +180y3s.t. y1 +3y3 300 (4)

2y2 +2y3 500 (5)y1, y2, y3 0


Marginal Value of A Resource The shadow price of a resource also

represents the marginal value of this resource under the firms current productivity and resource conditions

If the shadow price is zero, increasing resource a will not lead to an increase in objective value

An increase of resource b by one unit, the objective value will increase by $150


Summary

We learn how to construct a dual problem based on the idea of getting a (lower or upper) bound of the objective function

The relations between the primal and dual LP Important economic interpretation: shadow

price


Review Problem ABC Manufacturing Company produces brass bowls

and trays. The table below summarizes the needs and profits of the products:

The supply of brass is limited by 400 lbs per week, while the supply of the labor is limited by 1200 hours per week. Besides, the marketing research indicates that the demand for trays is limited by 150 units per week. However, the company can sell as many bowls as it produces.

The company needs to decide the numbers of bowls and trays to produce each week.

Product Brass required(lbs)

Labor required(hours)

Unit profit

Bowl 1 4 $7

Tray 2 3 $6


Review Problem

Formulate an LP for ABC Use graphic method to get the optimal

decision Write down the dual problem Derive the shadow prices

Duality and Shadow PriceTopicsExample 1 The Firm 1s Production Problem Upper Bound: Case 1Case 2Case 3The General CaseThe Upper Bound Problem The Dual Problem Dual Construction: Example 2Example 2Example 2Economic Interpretation of DualWhat Can Firm 1 Accept?The Optimal Offer PricesShadow Price Relations Between Primal and DualComplementary Slackness TheoremUsing Complementary SlacknessMarginal Value of A ResourceSummaryReview ProblemReview Problem

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08-3-LPDual