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Published byNoel nolan Monger Modified about 1 year ago

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תרגילי חזרה

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General Things to Remember Be careful to identify if we’re talking about gains or costs (because the search cost is always negative) Check carefully if the payoff cost allows you to use advance formulation of the original model

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Problem 1 You have 5 boxes on the table – Each box can generate a random number from a uniform distribution (0,100) – In order to activate a box you need to feed it with a $1 – In addition, in order to approach the table you need to pay $2 – You can approach the table up to 5 times, and activate as many boxes as you want each time – Your revenue will be the maximum prize revealed

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Problem 1 (cont.) Show that 81 is an upper bound for your expected revenue

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Solution Searching sequentially, with a search cost of c=1 and an infinite decision horizon will always yield a better expected net revenue thus this is an upper bound: Good, but not good enough…

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Parallel Search is an even tighter bound (if infinite horizon)… We know that: N=1: x=75.5 N=2: x=79.2 N=3: x=80.5 N=4: x=80.95 N=5: x=81

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בעיית המזכירה נתונה בעיית המזכירה עם n=5 מועמדים, כאשר בניגוד לבעיה המקורית, גם אם סיימנו את החיפוש ולא מצאנו את המועמד הטוב ביותר הרי שבהסתברות α הוא עדיין יהיה פנוי וניתן יהיה לשכור אותו –מהי האסטרטגיה האופטימלית לחיפוש כאשר α=0.2? –בחר ערך N ( גדול מ - 10) ו - α כרצונך ופתור שנית.

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Analysis What is the probability of selecting the best applicant? – For r=1 or r=n, 1/n – For r>1: timeline n max of r-1max of j-r> the probability that the maximum number in a sample of j-1 is one of the first r-1 numbers

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Adding α The revised equation: r1/(j-1) sum(1/(j- 1))(r-1)/5A(r-1)*a/n A

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And if you’re a wiseguy… Just choose α=1 and the probability of ending up with the best equals 1…

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Buying a Laptop Assume you are a comparison shopping agent and a buyer offers you sqrt(x) dollars for each saving of x dollars below a price of $50 for a product You have 3 stores to look at: How will you search and what is the probability of getting to store C? f(x) 0100 Store A c A =1.3 f(x) 1090 Store B c B =0.5 f(x) 2080 Store C c C =0.3

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This is a Pandora’s version! Sellerfi(x) A1/100 B1/80 C1/60 And the rule: go to store with minimal zi unless found a price below zi

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Cont. Optimal strategy: start with B, if price above then go to C and if best price is above then go to A Sellerzi A45.23 B36.98 C37.41

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Probability of getting to C

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Yet another Pandora… You go to NY for a month There are N pubs you might find attractive Each pub has a different a-priori distribution of value for you and different exploration cost Your performance will be the accumulated value What is the optimal strategy?

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Solution All search should be conducted during first day. Cost to be used is c/30

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Pandora’s Problem We have two boxes: – One with payoffs $10 with probability 0.99 and $K with probability Cost of opening is $1. – The other with probability 0.5 has a prize of 100$ and with probability 0.5 has a prize of $200. its cost for opening is c->0. – What should be the value of K so that the first box will be opened first

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Solution RV of second box is 200 RV of first box is: (K-200)*0.01=1 0.01K=3 K=300

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שאלה במערכת מחשב מרובת מעבדים בנוי כל מעבד משתי ליבות ידוע שהעומס הרגעי ( זמן המתנה ) על כל ליבה מתפלג f(x)=2x ( עבור 0

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פתרון ניתן להסתכל על כל מעבד כהזדמנות בעלת שני שלבים. נתחיל מהשלב השני : כלומר כל זמן שמצאנו ערך גדול מ נרצה לדגום גם את הליבה השניה

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פתרון כעת נסתכל על הליבה הראשונה : כלומר כל זמן שלא מצאנו ערך קטן מ נרצה לדגום את הליבה הראשונה

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