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Oct 04 ISMT162/Weixin Shang 1
Duality and Shadow Price
Oct 04 ISMT162/Weixin Shang 2
Topics
Dual problem construction Economic interpretation for dual problem,
shadow price Duality
Oct 04 ISMT162/Weixin Shang 3
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?
Oct 04 ISMT162/Weixin Shang 4
The Firm 1s Production Problem LP formulation
max 300x1+500x2s. t. x1 40 (1)
2x2 120 (2) 3x1+2x2 180 (3) x1, x2 0
Oct 04 ISMT162/Weixin Shang 5
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
Oct 04 ISMT162/Weixin Shang 6
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
Oct 04 ISMT162/Weixin Shang 7
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
Oct 04 ISMT162/Weixin Shang 8
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
Oct 04 ISMT162/Weixin Shang 9
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)
Oct 04 ISMT162/Weixin Shang 10
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)
Oct 04 ISMT162/Weixin Shang 11
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
Oct 04 ISMT162/Weixin Shang 12
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
Oct 04 ISMT162/Weixin Shang 13
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
Oct 04 ISMT162/Weixin Shang 14
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
Oct 04 ISMT162/Weixin Shang 15
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)
Oct 04 ISMT162/Weixin Shang 16
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
Oct 04 ISMT162/Weixin Shang 17
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
Oct 04 ISMT162/Weixin Shang 18
Relations Between Primal and Dual
PrimalDual
Have an Opt.Solution Unbounded Infeasible
Have an Opt.Solution
Unbounded
Infeasible
Oct 04 ISMT162/Weixin Shang 19
Complementary Slackness Theorem If a constraint is inactive (strictly > or
Oct 04 ISMT162/Weixin Shang 20
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
Oct 04 ISMT162/Weixin Shang 21
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
Oct 04 ISMT162/Weixin Shang 22
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
Oct 04 ISMT162/Weixin Shang 23
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
Oct 04 ISMT162/Weixin Shang 24
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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