# Slide 1

### 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

### 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

### Problem 1 (cont.)

• Show that 81 is an upper bound for your expected revenue

### 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:

V ( x )

  c

 y

 x yf ( y ) dy

F ( x ) V ( x )

 x x

1

 y

100

 x

0 .

01 ydy

0 .

01 x * x

85.86

 x

Good, but not good enough… x

114 .

14

Parallel Search is an even tighter bound (if infinite horizon)…

• We know that:

V

F

N f

N

( x )

( x )

( x )

 c ( N )

 y

 x yf

N

( y ) dy

F

N

F ( x )

N 

0 .

01 x

N

N

F ( x )

N

1 f ( x )

0 .

01 *

( x ) V ( x )

 x

N

0 .

01 x

N

1

N

2

100 y

 x y 0 .

01 * N

0 .

01 y

N

1 dy

0 .

01 x

N x

 x

N=1: x=75.5

N=2: x=79.2

N=3: x=80.5

N=4: x=80.95

N=5: x=81

### הריכזמה תייעב

רשאכ , םידמעומ n=5 םע הריכזמה תייעב הנותנ

שופיחה תא ונמייס םא םג , תירוקמה היעבל דוגינב

ירה רתויב בוטה דמעומה תא ונאצמ אלו

היהי ןתינו יונפ היהי ןיידע אוה α תורבתסהבש

ותוא רוכשל

?

α=0.2

רשאכ שופיחל תילמיטפואה היגטרטסאה יהמ –

.

תינש רותפו ךנוצרכ α ו ) 10 מ לודג ( N ךרע רחב –

### Analysis

 n

• n

  best applicant?

– For r=1 or r=n, 1/n

– For r>1:

 j n 

 r

P

 j th applicant and you select is it best



 j n 

 r

1 n

 r j

1

1



 r n

1 j n 

 r

 j

1

1

 max of r-1 > max of j-r the probability that the maximum number in a sample of j-1 is one of the first r-1 numbers timeline n

• The revised equation:

A

 n

 n

1 j n 

 r

 j

1

1



 r n

1

1

2 r 1/(j-1) sum(1/(j-

1)) (r-1)/5 A (r-1)*a/n

 n

0.2

1 2.08

0.2

0.42

0.04

0.46

3 0.5

1.08

0.4

0.43

0.08

0.51

4 0.33

0.58

0.6

0.35

0.12

0.47

5 0.25

0.25

0.8

0.2

0.16

0.36

### And if you’re a wiseguy…

• Just choose α=1 and the probability of ending up with the best equals 1…

• 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: f(x) f(x) f(x)

0

Store A c

A

=1.3

100 10

Store B c

B

=0.5

90 20

Store C c

C

=0.3

• How will you search and what is the probability of getting to store C?

80

### This is a Pandora’s version!

c i

  x z i

50

 x i

 

50

 z i

 f i

  i dx i

Seller

A

B

C fi(x)

1/100

1/80

1/60

And the rule: go to store with minimal zi unless found a price below zi

c i

  x z i

### Cont.

50

 x i

 

50

 z i

 f i

  i dx i

Seller

A

B

C zi

45.23

36.98

37.41

37.41 then go to C and if best price is above

45.23 then go to A

P

P ( B

37 .

41 )

0 .

66

### 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?

### Solution

• All search should be conducted during first day.

• Cost to be used is c/30 c / 30

  z

  y

 z

   dy

### Pandora’s Problem

• We have two boxes:

– One with payoffs \$10 with probability 0.99 and \$K with probability 0.01. 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

### Solution

• RV of second box is 200

• RV of first box is:

(K-200)*0.01=1

0.01K=3

K=300 c

  z

  y

 z

   dy

### הלאש

דבעמ לכ יונב םידבעמ תבורמ בשחמ תכרעמב

תוביל יתשמ

הביל לכ לע

)

) הנתמה ןמז

למרונמ , 0<x<1

( יעגרה סמועהש עודי

רובע ( f(x)=2x גלפתמ

לש

הדימבו

הינשה

0.05

אוה דבעמב הבילל השיג ןמז

הבילל םג תשגל ןכמ רחאל םיטילחמו

0.02

– דבעמ ותוא

איבהל איה הרטמהו הדימב המיגדה תוינידמ יהמ

םומינימל ללוכה ןמזה תא

### ןורתפ

ינש תלעב תונמדזהכ דבעמ לכ לע לכתסהל ןתינ

.

םיבלש

: ינשה בלשהמ ליחתנ c

 

0 z  z

 y

   dy

0 .

02

 x

0 .

39

0 z  z

 y

2 y dy

םוגדל הצרנ 0.39

מ לודג ךרע ונאצמש ןמז לכ רמולכ

הינשה הבילה תא םג

### ןורתפ

: הנושארה הבילה לע לכתסנ תעכ • c

  y

0 .

39  z

 y

   dy

0

  y z

0 .

39 z

0 .

52 z

  w y

0 wf ( w ) dw

 y

1

F ( y )

0 .

02 f

  dy

הצרנ 0.52

מ ןטק ךרע ונאצמ אלש ןמז לכ רמולכ

הנושארה הבילה תא םוגדל