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Making Decisions Against an Opponent: An application of Mathematical Optimization

Making Decisions Against an Opponent: An application of Mathematical Optimization. ISYE 2030 Monday, October 2, 2000 Dr. Joel Sokol jsokol@isye.gatech.edu. Making Decisions Against an Opponent. Two people/teams in competition. Outcome depends on both sides’ simultaneous decisions.

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Making Decisions Against an Opponent: An application of Mathematical Optimization

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  1. Making Decisions Against an Opponent:An application ofMathematical Optimization ISYE 2030 Monday, October 2, 2000 Dr. Joel Sokol jsokol@isye.gatech.edu

  2. Making Decisions Against an Opponent • Two people/teams in competition • Outcome depends on both sides’ • simultaneous decisions • Many examples in business, sports, • economics, and even everyday life

  3. Simpsons Decision-Making vs. Bart Lisa

  4. Lisa’s Optimal Choice Poor, predictable Bart…always chooses rock. Rock-Paper-Scissors Good old rock… You can always count on rock. Rock Paper Scissors Rock 0 -11 Paper 1 0 -1 Scissors -11 0 1 = Bart wins -1 = Lisa wins

  5. ? Good old rock… 60% of the time, you can count on rock. What if…? Rock Paper Scissors Rock 0 -11 Paper 1 0 -1 Scissors -11 0 1 = Bart wins -1 = Lisa wins

  6. Paper! Good old rock… 60% of the time, you can count on rock. What if…? Rock Paper Scissors 60% Rock 0 -11 20% Paper 1 0 -1 20% Scissors -11 0 Average = 0 -0.4 +0.4 1 = Bart wins -1 = Lisa wins

  7. So, what’s the outcome? Rock Paper Scissors Average = 0 -0.4 +0.4 • Lisa should choose the column with the • best results (paper), and wins 40% more • In general, Bart’s overall result equals the • value of Lisa’s best (lowest) column. • But… if Bart knows Lisa will choose paper, • he should choose scissors… and so on

  8. Bart’s Mathematical Model x = probability that Bart chooses rock, y = probability that Bart chooses paper, z = probability that Bart chooses scissors v = total value (smallest of column totals) Rock Paper Scissors (x) Rock 0 -11 (y) Paper 1 0 -1 (z) Scissors -11 0 Average = y - z z - x x - y • Let’s use a Linear Programming Model

  9. Linear Programming Models • All linear functions • No powers (x2) • No square roots, trig functions, logs, etc. • No variables multiplied by each other • Find values for the variables that • satisfy all constraints • give the best (optimal) objective function value • “Objective Function” - minimize or maximize • “Constraints” - mathematical restrictions

  10. Bart’s Linear Program Rock Paper Scissors (x) Rock 0 -11 (y) Paper 1 0 -1 (z) Scissors -11 0 Subject to v  y - z (Rock) v  z - x (Paper) v  x - y (Scissors) x + y + z = 1 x  0, y  0, z  0 Maximize v Average = y - z z - x x - y

  11. Lisa’s Mathematical Model a = probability that Lisa chooses rock, b = probability that Lisa chooses paper, c = probability that Lisa chooses scissors d = total value (largest of row totals) • Another Linear Programming Model Rock(a) Paper(b) Scissors(c) Avg. Rock 0 -11 c - b Paper 1 0 -1 a - c Scissors -11 0 b - a

  12. Lisa’s Linear Program Rock(a) Paper(b) Scissors(c) Avg. Rock 0 -11 c - b Paper 1 0 -1 a - c Scissors -11 0 b - a Minimize d Subject to d  b - c (Rock) d  c - a (Paper) d  a - b (Scissors) a + b + c = 1 a  0, b  0, c  0

  13. Solution: x = y = z = 1/3 Value = 0 Lisa’s Model Two Optimization Models • Maximize outcome • Outcome = lowest column average • Bart’s Model • Minimize outcome • Outcome = highest row average • Solution: a = b = c = 1/3 • Value = 0 Note: The two values will always be equal

  14. A More Interesting Example:2-point conversions in football • Offense gets the ball at 3 yard line, has one play to score • Many offensive play choices • Many defensive strategies So, let’s simplify a little bit…

  15. Simplified Model: Run or Pass • Offense can choose to run or to pass • Defense can choose run defense or pass defense Run Pass Defense Defense Run 30% 50% Pass 80% 40% Success Rates What do you expect?

  16. Offense Maximize v Subject to v  .3x + .8y v  .5x + .4y x + y = 1 x  0, y  0 Defense Minimize d Subject to d  .3a + .5b d  .8a + .4b a + b = 1 a  0, b  0 Two Linear Programs Run (a) Pass (b) Defense Defense Run (x) 30% 50% Pass (y) 80% 40%

  17. d a=1/6, d= .47 1 .5 0 d  .4 + .4a d  .5 – .2a a 0 .5 1 Defense’s Optimization Minimize d Subject to d  .3a + .5b d  .8a + .4b a + b = 1 a  0, b  0 Minimize d Subject to d  .5 – .2a d  .4 + .4a 0  a  1 • Substitute b = 1 – a

  18. v 1 .5 0 v  .8 – .5x v  .4 + .1x x=2/3, v= .47 x 0 .5 1 Offense’s Optimization Maximize v Subject to v  .3x + .8y v  .5x + .4y x + y = 1 x  0, y  0 • Substitute y = 1 – x Maximize v Subject to v  .8 – .5x v  .4 + .1x 0  x  1

  19. Summary • Start with a complex problem (making decisions against an opponent) • Create a simplified mathematical model (Optimization/linear programming) • Use the model to suggest a strategy (in this case, a surprising strategy!) Slides at www.isye.gatech.edu/~jsokol/2030.ppt

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