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Lecture 4

Lecture 4. Duality and game theory. Knapsack problem – duality illustration. A thief robs a jewelery shop with a knapsack He cannot carry too much weight He can choose among well divisible objects ( gold , silver , diamond sand )

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Lecture 4

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  1. Lecture 4 Duality and gametheory

  2. Knapsack problem – dualityillustration • A thiefrobs a jewelery shop with a knapsack • He cannotcarrytoo much weight • He canchooseamongwelldivisibleobjects (gold, silver, diamondsand) • A thiefwants to takethe most valuablegoodswithhim

  3. The model • Parameters: W – knapsackmaximalweight N – number of goodsin a shop wi – goodi’sweight vi – goodi’svalue • Decisionvariables: xi – share of totalamount of good i taken to theknapsack • Objectivefunction: Maximizethetotalvalue • Constraints: • Cannottakemorethanavailable • Cannottakemorethantheknapsackcapacity • Cannottakethenegative (ifheis a thiefindeed)

  4. The model • Formulate as an LP: Max

  5. Problem of a thief (a primal problem) • Substitute N=3, W=4, w=(2,3,4) i v=(5,20,3) gold, diamondsand and silver. max p.w. A thief problem solution: (x1,x2,x3)=(0.5, 1, 0) Objectivefunctionvalue: 22.5

  6. Analysis • Only one good will be takenpartially (gold). Itis a general ruleinallknapsackproblemswith N divisiblegoods. • Intuition: • Theoptimalsolutionisunique. • In order to uniquelydetermine3 unknowns, we need 3 independent linearequations. • So atleast3 constraintsshould be satisfiedas equalities. • One constraintistheknapsackweight, but anothertwoareaboutgoodsquantities0≤xi≤1. • Henceonly one goodmay be takeninfractionalamountinthe optimum.

  7. Crime syndicate buys out thethief’s business • Thecrime syndicate wants to buy out thegoodsfromthethieftogetherwith his thief business (equipment etc. here: knapsack). • Theyproposeprices y1for gold, y2for diamondsand, y3for silverand y4for 1 kg knapsackcapacity. • But thethiefmayuse 2 kg knapsackcapacity and allthegold to generate 5 units of profit , so thepriceoffered for gold2y4+y1should be atleast5. Similarlywithothergoods. • The syndicate wants to minimizetheamountithas to paythethiefy1+y2+y3+4y4 • Thepricesshould not be negative, otherwisethethief will be insulted.

  8. The syndicate problem (a dual problem) • The syndicate problem may be formulated as follows: min p.w. Syndicate problem solution: (y1,y2,y3,y4)=(0,12.5,0,2.5) Objectivefunctionvalue: 22.5

  9. Thethief problem Isequivalent Becausee.g. Transforming: Becausee.g. Itisequivalent to thesybdicate problem

  10. Syndicate problem optimalsolution: (y1,y2,y3,y4)=(0,12.5,0,2.5) dual prices Optimalobjectuvefunctionvalue: 22.5 Thief problem optimalsolution: (x1,x2,x3)=(0.5, 1, 0) Optimalobjectivefunctionvalue: 22.5

  11. Rozwiązanieproblemusyndyka: (y1,y2,y3,y4)=(0,12.5,0,2.5) cenydualne Optymalnawartośćfunkcjicelu: 22.5 Rozwiązanieproblemuzłodzieja: (x1,x2,x3)=(0.5, 1, 0) Optymalnawartośćfunkcjicelu: 22.5

  12. Matching/assignment http://mathsite.math.berkeley.edu/smp/smp.html

  13. Individualdecisiontheoryvsgametheory

  14. Zero-sum games • In zero-sum games, payoffsineachcell sum up to zero • Movement diagram

  15. Zero-sum games • Minimax = maximin = value of thegame • Thegamemayhavemultiplesaddlepoints

  16. Zero-sum games • Or itmayhave no saddlepoints • To findthevalue of suchgame, considermixedstrategies

  17. Zero-sum games • Ifthereismorestrategies, youdon’tknowwhich one will be part of optimalmixedstrategy. • LetColumnmixedstrategy be (x,1-x) • Then Raw will try to maximize

  18. Zero-sum games • Column will try to choose x to minimizetheupperenvelope

  19. Zero-sum games • TranformintoLinearProgramming

  20. Fishing on Jamaica • In the fifties, Davenport studied a village of 200 people on thesouthshore of Jamaica, whoseinhabitantsmadetheirliving by fishing.

  21. Twenty-six fishing crews in sailing, dugout canoes fish this area [fishing grounds extend outward from shore about 22 miles] by setting fish pots, which are drawn and reset, weather and sea permitting, on three regular fishing days each week … The fishing grounds are divided into inside and outside banks. The inside banks lie from 5-15 miles offshore, while the outside banks all lie beyond … Because of special underwater contours and the location of one prominent headland, very strong currents set across the outside banks at frequent intervals … These currents are not related in any apparent way to weather and sea conditions of the local region. The inside banks are almost fully protected from the currents. [Davenport 1960]

  22. Jamaica on a map

  23. Strategies • Therewere 26 woodencanoes. Thecaptains of thecanoesmightadopt 3 fishingstrategies: • IN – putallpots on theinside banks • OUT – putallpots on theoutside banks • IN-OUT) – putsomepots on theinside banks, somepots on theoutside

  24. Advantages and disadvantages of fishingintheopensea Disadvantages • Ittakesmore time to reach, so fewerspotscan be set • Whenthecurrentisrunning, itisharmful to outsidepots • marksaredraggedaway • potsmay be smashedwhilemoving • changesintemeperaturemaykillfishinsidethepots Advanatages • Theoutside banks producehigherqualityfishbothinvariaties and insize. • If many outsidefishareavailable, theymaydrivetheinsidefishoffthe market. • The OUT and IN-OUT strategiesrequirebettercanoes. • Theircaptainsdominatethe sport of canoe racing, whichisprestigious and offerslargerewards.

  25. Collecting data • Davenport collected the data concerning the fishermenaveragemonthly profit depending on the fishingstrategiestheyused to adopt.

  26. OUT Strategy

  27. Zero-sum game? Thecurrent’s problem • Thereis no saddle point • Mixedstrategy: • Assumethatthecurrentisvicious and playsstrategy FLOW withprobability p, and NO FLOW withprobability 1-p • Fishermen’sstrategy: IN with prob. q1, OUT with prob. q2, IN-OUT with prob. q3 • For every p, fishermenchoose q1,q2 and q3 thatmaximizes: • And theviciouscurrentchooses p, so thatthefishermenget min

  28. Graphicalsolution of thecurrent’s problem Solution: p=0.31 Mixedstrategy of thecurrent

  29. Thefishermen’s problem • Similarly: • For everyfishermen’sstrategy q1,q2 and q3, theviciouscurrentchooses p so thatthefishermenearntheleast: • Thefishermen will try to choose q1,q2 and q3 to maximizetheirpayoff:

  30. Maximin andminimax Optimalstrategy for thefishermen Value of thegame Optimalstrategy for thecurrent

  31. Minimaxsensitivity report

  32. Maximinsensitivity report

  33. Forecast and observation Gametheorypredicts Observationshows No fishermenrisksfishingoutside Strategy69% IN, 31% IN-OUT [Payoff: 13.38] Current’s „strategy”: 25% FLOW, 75% NO FLOW • No fishermenrisksfishingoutside • Strategy67% IN, 33% IN-OUT [Payoff: 13.31] • Optimalcurrent’sstrategy31% FLOW, 69% NO FLOW The similarity is striking Davenport’s finding went unchallenged for several years Until …

  34. Currentis not vicious • Kozelka 1969 and Read, Read 1970 pointed out a seriousflaw: • The currentis not a reasoningentityand cannotadjust to fishermenchangingtheirstrategies. • HencefishermenshoulduseExpected Value principle: • Expectedpayoff of the fishermen: • IN: 0.25 x 17.3 + 0.75 x 11.5 = 12.95 • OUT: 0.25 x (-4.4) + 0.75 x 20.6 = 14.35 • IN-OUT: 0.25 x 5.2 + 0.75 x 17.0 = 14.05 • Hence, all of the fishermenshouldfishOUTside. • Maybe, theyare not welladaptedafterall

  35. Currentmay be viciousafterall • The currentdoes not reason, but itisveryrisky to fishoutside. • Evenif the currentruns 25% of the timeON AVERAGE, itmight run considerablymoreor less in the short run of a year. • Suppose one yearit ran 35% of the time. Expectedpayoffs: • IN: 0.35 x 17.3 + 0.65 x 11.5 = 13.53 • OUT: 0.35 x (-4.4) + 0.65 x 11.5 = 11.85 • IN-OUT: 0.35 x 5.2 + 0.65 x 17.0 = 12.87. • By treating the current as theiropponent, fishermenGUARANTEEthemselvespayoff of atleast13.31. • Fishermenpay 1.05 pounds as insurancepremium

  36. Decisionmaking under uncertainty

  37. Decisionmaking under uncertainty

  38. Decisionmaking under uncertainty Regretmatrix

  39. Decisionmaking under uncertainty Regretmatrix

  40. Decisionmaking under uncertainty

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