David Evans
February 12, 2019
590

# Class 9: Linear Programming

Class 9: Linear Programming
https://uvammm.github.io/class9

Markets, Mechanisms, and Machines
University of Virginia
cs4501/econ4559 Spring 2019
David Evans and Denis Nekipelov
https://uvammm.github.io/

#### David Evans

February 12, 2019

## Transcript

1. ### MARKETS, MECHANISMS, MACHINES University of Virginia, Spring 2019 Class 9:

Applications of Linear Programming 12 February 2019 cs4501/econ4559 Spring 2019 David Evans and Denis Nekipelov https://uvammm.github.io Zinedine Zidane’s penalty in 2006 World Cup Final

3. None

5. ### Possibilities 4 ) subject to ≤ Infeasible: no solution satisfies

all the inequalities
6. ### Possibilities 5 ) subject to ≤ Infeasible: no solution satisfies

all the inequalities 6 subject to 6 ≤ 5 −6 ≤ −7
7. ### Possibilities 6 ) subject to ≤ Unbounded: no limit on

maximized value 6 subject to −6 ≤ −7
8. ### Brewer’s Problem 7 Robert Sedgewick and Kevin Wayne, Princeton Course

Limited Resources Available Recipes and Outputs How much of each should we brew to maximize total profits?

10. ### 9 Robert Sedgewick and Kevin Wayne, Princeton Course Maxmimize =

13 + 23 subject to constraints: 5 + 15 ≤ 480 (corn) 4 + 4 ≤ 160 hops 35 + 20 ≤ 1190 (malt) ≥ 0, ≥ 0
11. ### 10 Robert Sedgewick and Kevin Wayne, Princeton Course Maxmimize =

13 + 23 subject to constraints: 5 + 15 ≤ 480 (corn) 4 + 4 ≤ 160 hops 35 + 20 ≤ 1190 (malt) ≥ 0, ≥ 0 Feasible Region (convex polygon)
12. ### 11 Robert Sedgewick and Kevin Wayne, Princeton Course Maxmimize =

13 + 23 subject to constraints: 5 + 15 ≤ 480 (corn) 4 + 4 ≤ 160 hops 35 + 20 ≤ 1190 (malt) ≥ 0, ≥ 0
13. ### 12 Robert Sedgewick and Kevin Wayne, Princeton Course Optimal solution

must be at an extreme point (intersection of constraints)

15. ### Dual Problem 14 Brewer Entrepreneur = 13 + 23 subject

to constraints: 5 + 15 ≤ 480 (corn) 4 + 4 ≤ 160 hops 35 + 20 ≤ 1190 (malt) ≥ 0, ≥ 0 = 480 + 160ℎ + 1190 , ℎ, = unit prices for ingredients 5 + 4ℎ + 35 ≥ 13 (ale) 15 + 4ℎ + 20 ≥ 23 beer ≥ 0, ℎ ≥ 0, ≥ 0
16. ### Dual Theorem 15 ) subject to ≤ ≥ 0 )

subject to ) ≥ ≥ 0 primal problem dual problem T = T Proof sketch: roughly – show that simplex algorithm for both problems produces same result. if both feasible:

18. ### Example: Network Flow 17 Source: Avrim Blum, Manual Blum, CMU

course: https://www.cs.cmu.edu/~avrim/451f11/lectures/lect1101.pdf
19. ### Example: Network Flow 18 Source: Avrim Blum, Manual Blum, CMU

course: https://www.cs.cmu.edu/~avrim/451f11/lectures/lect1101.pdf hi ≤ 4 hj ≤ 2 ik ≤ 3 jk ≤ 2 kj ≤ 1 kl ≤ 2 ml ≤ 4 maximize kl + ml subject to:
20. ### Example: Network Flow 19 Source: Avrim Blum, Manual Blum, CMU

course: https://www.cs.cmu.edu/~avrim/451f11/lectures/lect1101.pdf hi ≤ 4 hj ≤ 4 ik ≤ 3 jk ≤ 2 kj ≤ 1 kl ≤ 2 jm ≤ 3 ml ≤ 4 maximize kl + ml subject to: hi = ik kl + kj = ik hj + kj = jm + jk ml = jm

22. ### Minimize Cost: Network Flow 21 Source: Avrim Blum, Manual Blum,

CMU course: https://www.cs.cmu.edu/~avrim/451f11/lectures/lect1101.pdf \$2 \$2

24. ### Penalty Kicks Model Goalkeeper Direction Left Center Right Kicker Direction

Left Center Right 23
25. ### Penalty Kicks Model Goalkeeper Direction Left Center Right Kicker Direction

Left 0.20 0.90 0.90 Center 0.99 0.01 0.99 Right 0.98 0.98 0.40 24 two-player, zero-sum game
26. ### Goalkeeper Direction Left Center Right Kicker Direction Left 0.20 0.90

0.90 Center 0.99 0.01 0.99 Right 0.98 0.98 0.40 25 Kicker’s goal: maximize minimum payoff (probability to score assuming goalkeeper behaves optimally)
27. ### Goalkeeper Direction Left Center Right Kicker Direction Left 0.20 0.90

0.90 Center 0.99 0.01 0.99 Right 0.98 0.98 0.40 26 Kicker’s goal: maximize minimum payoff (probability to score assuming goalkeeper behaves optimally) Variables: opqr , sptrpu , uvwxr , v ≥ 0 and ∑ v = 1 v Objective: maximize Constraints: | v v} ≥ v

30. ### Charge If you don’t have a team for Project 3,

see us now 29