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Lesson 30 (Section 19.2–3)
           Duality in Linear Programming

                          Math 20


                     December 3, 2007


Announcements
   Problem Set 11 on the WS. Due December 5.
   next OH: Monday 1–2 (SC 323)
   next PS: Sunday 6–7 (SC B-10)
   Midterm II review: Tuesday 12/4, 7:30-9:00pm in Hall E
   Midterm II: Thursday, 12/6, 7-8:30pm in Hall A
Outline



   Recap


   Example


   Shadow Prices


   The Dual Problem
Definition
A linear programming problem is a constrained optimization
problem with a linear objective function and linear inequality
constraints.
Definition
A linear programming problem is a constrained optimization
problem with a linear objective function and linear inequality
constraints.

Definition
An LP problem is in standard form if it is expressed as

                 max z = c1 x1 + c2 x2 + · · · + cn xn

subject to the constraints

                  a11 x1 + a12 x2 + · · · + a1n xn ≤ b1
                  a21 x1 + a22 x2 + · · · + a2n xn ≤ b2
                                                 ..
                                                 ..
                                                 ..
                am1 x1 + am2 x2 + · · · + amn xn ≤ bm
                                  x1 , x2 , . . . , xn ≥ 0
In vector notation, an LP problem in standard form looks like

                           max z = c · x

subject to constraints

                  Ax ≤ b                   x≥0
Theorem of the Day for Friday




   Theorem (The Corner Principle)
   In any linear programming problem, the extreme values of the
   objective function, if achieved, will be achieved on a corner of the
   feasibility set.
Outline



   Recap


   Example


   Shadow Prices


   The Dual Problem
Example



  Example
  We are starting a business selling two Harvard insignia products:
  sweaters and scarves. The profits on each are $35 and $10,
  respectively. Each has a pre-bought embroidered crest sewn on it;
  we have 2000 crests on hand. Sweaters take four skeins of yarn
  while scarves only take one, and there are 2300 skeins of yarn
  available. Finally, we have available storage space for 1250 scarves;
  we could use any of that space for sweaters, too, but sweaters take
  up half again as much space as scarves.
  What product mix maximizes revenue?
Formulating the problem


   Let x be the number of sweaters and y the number of scarves
   made. We want to
                         max z = 35x + 10y
   subject to

                             x + y ≤ 2000
                            4x + y ≤ 2300
                           3x + 2y ≤ 2500
                               x, y ≥ 0
Finding the corners

  2300

  2000

                                                       Notice one constraint is
           4x +



                                                       superfluous!
                                x
                y =2




                                    +                            z(0, 0) = 0
  1250                                  y
                                            =                  z(575, 0) = 20, 125
                     300




                                                20
                                                  00        z(0, 1250) = 12, 500
                                                           z(420, 620) = 20, 900
                    3x




                   •
                           +




   (420, 620)
                           2y
                               =
                                25
                                   00




                       575 833 1                        2000
                                     3
Answer




  We should make 420 sweaters and 620 scarves.
Outline



   Recap


   Example


   Shadow Prices


   The Dual Problem
Suppose our business were suddenly given
    one additional crest patch?
    one additional skein of yarn?
    one additional unit of storage space?
How much would profits change?
One more patch

2300                                                    Since we weren’t “up against”
                                                        this constraint in the first place,
2000                                                    one extra doesn’t change our
                                                        optimal product mix.
         4x +




                                                        At this product mix, the
                                                        marginal profit of patches is 0.
                                 x
              y =2




                                     +
1250                                     y
                                             =
                                                 20
                   300




                                                   01
                  3x




                 •
                         +




 (420, 620)
                         2y
                             =
                              25
                                 00




                     575 833 1                          2000
                                     3
One more skein

 2300                                                  We’ll make a little more sweater
                                                       and less scarf
 2000
                                                       The marginal profit is
           4x +




                                                       ∆z = 35(0.4) + 10(−0.6) = 8
                y =2




                                x
                                    +
 1250                                   y
                                            =
                     301




                                                20
                                                  00
                    3x




                    •
                           +
                           2y




(420.4, 619.4)
                               =
                                25
                                   00




                      575 833 1                        2000
                                     3
One more storage unit

2300                                            We’ll make a little less sweater
                                                and more scarf
2000
                                                The marginal profit is

                                                ∆z = 35(−0.2) + 10(0.8) = 1

                         x
                             +
1250                             y
                                     =
                                         20
                                           00
             •
                 3x




(419.8, 620.8)
                    +
                    2y
                        =
                         25
                             01




                 575 833 1                      2000
                             3
Shadow Prices


   Definition
   In a linear programming problem in standard form, the change in
   the objective function obtained by increasing a constraint by one is
   called the shadow price of that constraint.
Shadow Prices


   Definition
   In a linear programming problem in standard form, the change in
   the objective function obtained by increasing a constraint by one is
   called the shadow price of that constraint.

   Example
   In our example problem,
       The shadow price of patches is zero
       The shadow price of yarn is 8
       The shadow price of storage is 1
   We should look into getting more yarn!
Outline



   Recap


   Example


   Shadow Prices


   The Dual Problem
Question
Suppose an entrepreneur wants to buy our business’s resources.
What prices should be quoted for each crest? skein of yarn? unit
of storage?
Question
Suppose an entrepreneur wants to buy our business’s resources.
What prices should be quoted for each crest? skein of yarn? unit
of storage?
Answer.
Suppose the entrepreneur quotes p for each crest patch, q for each
skein of yarn, and r for each storage unit.
    Each sweater takes one patch, 4 skeins, and 3 storage units,
    so effectively p + 4q + 3r is bid per sweater
    Likewise, p + q + 2r is bid per scarf.
So we must have

                         p + 4q + 3r ≥ 35
                          p + q + 2r ≥ 10

for us to sell out. The entrepreneur’s goal is to minimize the total
payout
                     w = 2000p + 2300q + 2500r
Definition
Given a linear programming problem in standard form, the dual
linear programming problem is

                    min w = b1 y1 + · · · + bm ym

subject to constraints

                 a11 y1 + a21 y2 + · · · + am1 ym ≥ p1
                 a12 y1 + a22 y2 + · · · + am2 ym ≥ p2
                                                 ..
                                                 ..
                                                 ..
                a1n y1 + a2n y2 + · · · + amn ym ≥ pn
                                     y1 , . . . , ym ≥ 0
In fancy vector language, the dual of the problem

           max z = p · x subject to Ax ≤ b and x ≥ 0

is
           min w = b · y subject to A y ≥ p and y ≥ 0
Solving the Dual Problem

      The feasible set is unbounded (extending away from you)




                  r

                                           p
        q
Solving the Dual Problem

            The feasible set is unbounded (extending away from you)




                (0, 0, 83/4)
                               •




                                   r                                         •

(0, 8, 1)           (12/3, 81/3, 0)                             (35, 0, 0)
                                                 p
            •
       •        •

                    q
  (0, 10, 0)
Solving the Dual Problem

         The feasible set is unbounded (extending away from you)




         w = 21, 875
                       •




                    r                                                   •
  w = 20, 900                                             w = 70, 000
           w = 22, 500
                                              p
         •
     •       •

w = 23, 000 q
Solving the Dual Problem

            The feasible set is unbounded (extending away from you)
            w (0, 8, 1) = 20, 900 is minimal




                (0, 0, 83/4)
                               •




                                   r                                         •

(0, 8, 1)           (12/3, 81/3, 0)                             (35, 0, 0)
                                                 p
            •
       •        •

                    q
  (0, 10, 0)
The Big Idea




      The shadow prices are the solutions to the dual problem
      The payoff is the same in both the primal problem and the
      dual problem

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Lesson 30: Duality In Linear Programming

  • 1. Lesson 30 (Section 19.2–3) Duality in Linear Programming Math 20 December 3, 2007 Announcements Problem Set 11 on the WS. Due December 5. next OH: Monday 1–2 (SC 323) next PS: Sunday 6–7 (SC B-10) Midterm II review: Tuesday 12/4, 7:30-9:00pm in Hall E Midterm II: Thursday, 12/6, 7-8:30pm in Hall A
  • 2. Outline Recap Example Shadow Prices The Dual Problem
  • 3. Definition A linear programming problem is a constrained optimization problem with a linear objective function and linear inequality constraints.
  • 4. Definition A linear programming problem is a constrained optimization problem with a linear objective function and linear inequality constraints. Definition An LP problem is in standard form if it is expressed as max z = c1 x1 + c2 x2 + · · · + cn xn subject to the constraints a11 x1 + a12 x2 + · · · + a1n xn ≤ b1 a21 x1 + a22 x2 + · · · + a2n xn ≤ b2 .. .. .. am1 x1 + am2 x2 + · · · + amn xn ≤ bm x1 , x2 , . . . , xn ≥ 0
  • 5. In vector notation, an LP problem in standard form looks like max z = c · x subject to constraints Ax ≤ b x≥0
  • 6. Theorem of the Day for Friday Theorem (The Corner Principle) In any linear programming problem, the extreme values of the objective function, if achieved, will be achieved on a corner of the feasibility set.
  • 7. Outline Recap Example Shadow Prices The Dual Problem
  • 8. Example Example We are starting a business selling two Harvard insignia products: sweaters and scarves. The profits on each are $35 and $10, respectively. Each has a pre-bought embroidered crest sewn on it; we have 2000 crests on hand. Sweaters take four skeins of yarn while scarves only take one, and there are 2300 skeins of yarn available. Finally, we have available storage space for 1250 scarves; we could use any of that space for sweaters, too, but sweaters take up half again as much space as scarves. What product mix maximizes revenue?
  • 9. Formulating the problem Let x be the number of sweaters and y the number of scarves made. We want to max z = 35x + 10y subject to x + y ≤ 2000 4x + y ≤ 2300 3x + 2y ≤ 2500 x, y ≥ 0
  • 10. Finding the corners 2300 2000 Notice one constraint is 4x + superfluous! x y =2 + z(0, 0) = 0 1250 y = z(575, 0) = 20, 125 300 20 00 z(0, 1250) = 12, 500 z(420, 620) = 20, 900 3x • + (420, 620) 2y = 25 00 575 833 1 2000 3
  • 11. Answer We should make 420 sweaters and 620 scarves.
  • 12. Outline Recap Example Shadow Prices The Dual Problem
  • 13. Suppose our business were suddenly given one additional crest patch? one additional skein of yarn? one additional unit of storage space? How much would profits change?
  • 14. One more patch 2300 Since we weren’t “up against” this constraint in the first place, 2000 one extra doesn’t change our optimal product mix. 4x + At this product mix, the marginal profit of patches is 0. x y =2 + 1250 y = 20 300 01 3x • + (420, 620) 2y = 25 00 575 833 1 2000 3
  • 15. One more skein 2300 We’ll make a little more sweater and less scarf 2000 The marginal profit is 4x + ∆z = 35(0.4) + 10(−0.6) = 8 y =2 x + 1250 y = 301 20 00 3x • + 2y (420.4, 619.4) = 25 00 575 833 1 2000 3
  • 16. One more storage unit 2300 We’ll make a little less sweater and more scarf 2000 The marginal profit is ∆z = 35(−0.2) + 10(0.8) = 1 x + 1250 y = 20 00 • 3x (419.8, 620.8) + 2y = 25 01 575 833 1 2000 3
  • 17. Shadow Prices Definition In a linear programming problem in standard form, the change in the objective function obtained by increasing a constraint by one is called the shadow price of that constraint.
  • 18. Shadow Prices Definition In a linear programming problem in standard form, the change in the objective function obtained by increasing a constraint by one is called the shadow price of that constraint. Example In our example problem, The shadow price of patches is zero The shadow price of yarn is 8 The shadow price of storage is 1 We should look into getting more yarn!
  • 19. Outline Recap Example Shadow Prices The Dual Problem
  • 20. Question Suppose an entrepreneur wants to buy our business’s resources. What prices should be quoted for each crest? skein of yarn? unit of storage?
  • 21. Question Suppose an entrepreneur wants to buy our business’s resources. What prices should be quoted for each crest? skein of yarn? unit of storage? Answer. Suppose the entrepreneur quotes p for each crest patch, q for each skein of yarn, and r for each storage unit. Each sweater takes one patch, 4 skeins, and 3 storage units, so effectively p + 4q + 3r is bid per sweater Likewise, p + q + 2r is bid per scarf. So we must have p + 4q + 3r ≥ 35 p + q + 2r ≥ 10 for us to sell out. The entrepreneur’s goal is to minimize the total payout w = 2000p + 2300q + 2500r
  • 22. Definition Given a linear programming problem in standard form, the dual linear programming problem is min w = b1 y1 + · · · + bm ym subject to constraints a11 y1 + a21 y2 + · · · + am1 ym ≥ p1 a12 y1 + a22 y2 + · · · + am2 ym ≥ p2 .. .. .. a1n y1 + a2n y2 + · · · + amn ym ≥ pn y1 , . . . , ym ≥ 0
  • 23. In fancy vector language, the dual of the problem max z = p · x subject to Ax ≤ b and x ≥ 0 is min w = b · y subject to A y ≥ p and y ≥ 0
  • 24. Solving the Dual Problem The feasible set is unbounded (extending away from you) r p q
  • 25. Solving the Dual Problem The feasible set is unbounded (extending away from you) (0, 0, 83/4) • r • (0, 8, 1) (12/3, 81/3, 0) (35, 0, 0) p • • • q (0, 10, 0)
  • 26. Solving the Dual Problem The feasible set is unbounded (extending away from you) w = 21, 875 • r • w = 20, 900 w = 70, 000 w = 22, 500 p • • • w = 23, 000 q
  • 27. Solving the Dual Problem The feasible set is unbounded (extending away from you) w (0, 8, 1) = 20, 900 is minimal (0, 0, 83/4) • r • (0, 8, 1) (12/3, 81/3, 0) (35, 0, 0) p • • • q (0, 10, 0)
  • 28. The Big Idea The shadow prices are the solutions to the dual problem The payoff is the same in both the primal problem and the dual problem