Managerial

Managerial

Economics

Economics

Jarek Neneman

[e-mail]

601305093

Lecture 6
Optimal search
(ch. 9, Samuelson & Marks)

Optimal search

If at first you don’t succeed, try,

try again. Then stop. No use

being a damn fool about it.

W.C. Fields

Optimal search

A exercise to start with – Finding the

best item

Secretary problem, hotel problem.
Suppose that you will be shown three "prizes" in

order. Ahead of time, you know absolutely

nothing about how valuable the prizes might

be. Only after viewing all three can you

determine which you like best. You are shown

the prizes in order and are allowed to select

one. However, there is no "going back.„

You must select a prize immediately after

seeing it, and before seeing any subsequent

prize.

Optimal search

A exercise to start with – Finding the

best item

 Your sole objective is to obtain the best of the

three prizes. (Second best doesn’t count.)

 A random selection provides a one-third

chance of getting the best prize.

 Find a strategy that provides a strictly

greater chance (and compute the actual

chance).

Optimal search

A exercise to start with – Finding the best

item - solution

Winning strategy:
 Observe but bypass the first prize.
 Select the second prize only if it is better

than the first;

 otherwise go on and select the third prize.
Why?

Optimal search

A exercise to start with – Finding the best

item - solution

Six distinct (equally likely) orderings

of the prizes

First Prize

best

best

2nd best

2nd best worst

worst

Second Prize

2nd best

worst

best

worst

best

2nd

best

Third Prize

worst

2nd best

worst

best

2nd best

best

Prices selected with this strategy are

shown in

red

. The likelihood of

selecting best prize is 50% instead of

1/3 if selected by random

Optimal search

Optimal search

 Management search and uncover a number

of uncertain opportunities.

 Options are explored (or search) in sequence
 Management tasks is:

- to find the best strategy search (order of

pursuing options), and
- to find when to stop

Optimal search

Optimal stopping – escalating

investment in R&D

 An electronic firm can initiate an R&D

program by making $3million investment

 The chance for immediate success is 1/5 and

return is: $10 m, net profit is: 10-3=$7m

 If success does not come, the firm can invest

another $3m with the chance for success

now ¼

 The investment cost is $3m for each stage.
 The chances of success are: 1/5, ¼, 1/3, ½

and 1

Optimal search

Optimal stopping – escalating

investment in R&D

 Should risk-neutral firm pursue this program?
 At what stage should it stopped?
 What is you opinion?

Optimal search

Optimal stopping – escalating

investment in R&D

 Typically:

- 20% of students – no investment at all
- 30% of students – invest till the end
- 50% of students – start investment and stop

(2/3 decided to stop after 3rd failure.

 Does stopping after 3rd failure make any

sense? Compare cost and profits

Optimal search

Optimal stopping – escalating

investment in R&D

 Notice, that after failure investment cost is

sunk cost and the firm faces the same

problem under nearly the same conditions as

before.

 The only difference is the probability of

success, but this is increasing, therefore

 If it is worth investing initially, then it is worth

continuing to invest.

 The question is: should the firm start

investing?

Optimal search

Optimal stopping – escalating

investment in R&D

 It is high time to draw a decision tree

Third

investment

Fourth

investment

Fifth

investment

-2

-3.5

-3.5

-5

- 5

1/3

1

Quit

- 9

1/2

- 2

Quit

- 12

1/2

2/3

First

investment

Second

investment

1

1

-1/2

-2

-1/2

Don't
invest

0

1/5

7

Quit

- 3

1/4

4

Quit

- 6

3/4

4/5

Optimal search

Optimal stopping – escalating

investment in R&D

 Usually however, there is a risk that the

program may not succeed at all.

 With declining probabilities of success, the

firm should give up investment (irrespective

of sunk cost) when revised probability of

success falls sufficiently low

 Cut-off value of probability must satisfy zero-

profit condition:

π

*

Π – c = 0, hence

π

*

= c/Π

Optimal search

Optimal stopping – escalating

investment in R&D



π

*

Π – c = 0, hence

π

*

= c/Π

 Example
 Assume:

Π = $20m., c = $3m.,
π = 0.25, 0.21, 0.17, 0.13, 0.7, 0.01
When to stop investing?

Optimal search

Optimal stopping – escalating

investment in R&D

 Cutoff value

π

*

= 3/20 = 0.15

 The firm should invest 3 times and

then abandon

Optimal search

Optimal sequencial decisions

 Assume:

- several methods of developing a new

product
- predictable profit (Π) does not

depend on the method
- cost (c) of each method is different
- probability of success (π) of each

method is different

What is the best order of pursuing the

methods?

Optimal search

Optimal sequencial decisions

 What is the best order of pursiung the

methods?

 The intuitive answer is correct:

the program with greatest probability-

to-cost ratio should be tried first

Optimal search

Optimal sequencial decisions

 If there are only two methods, and if

firm starts with A first, then the

expected net benefit is:

π

A

Π – c

A

+ (1 – π

A

)(π

B

Π – c

B

)

=

π

A

Π + π

B

Π – π

A

π

B

Π - c

A

– c

B

+ π

A

c

B

 The first three terms are expected

gross profit,

 the cost are: c

A

+ c

B

, and last term is

 saving of investment cost (c

B

) if A is

successful, but this happens with

probability π

A

Optimal search

Optimal sequencial decisions

 If method B is the first one, the

expected benefit is almost the same:
π

A

Π + π

B

Π – π

A

π

B

Π - c

A

– c

B

+ π

B

c

A

,

therefore

 Starting from A is more profitable than

starting form B, only iff:
π

A

c

B

> π

B

c

A

or equivalently:

π

A

/c

A

> π

B

/c

B

Optimal search

Optimal search – summary

 A risk neutral firm should:
1. continue to invest as long as π > c/Π
2. Determine the sequence of

investments in descending order of π/c

Optimal search

The value of additional alternatives – searching for the

best price

 Is more options available better for the

decision maker?

 What if additional option can be

obtained at a cost?

 Example of the sale of the bank

division

Optimal search

The value of additional alternatives – searching for the

best price

 Seller believes that the potential

buyers offers are uniformly distributed

in the range form $50-64m

 What is the best price the firm can

obtain from from contracting outside

buyers?

 What is expected price from single

buyer?

 The average is $52m
 What if there are two potential buyers

contacted?

 The average of two is: $56m

Optimal search

The value of additional alternatives – searching for the

best price

 What is the best price the firm can

obtain from contracting outside

buyers?

 Generally, the more the outside buyers

conntacted, the higher the average

price.

 The expected maximum value:

E(V

max

) = [1/(n+1)]L + [n/(n+1)]U

L – the lowest possible value,
U- the greatest possible value

Optimal search

The value of additional alternatives – searching for the

best price

Expected maxium price ($

millions)

No of buyers Uniform distrib. Normal

distrib.

1

52.0

52.0(0)

2

56.0

56.5(0.56)

3

58.0

58.5(0.85)

4

59.2

60.2(1.03)

5

60.0

61.3(1.16)

6

60.6

62.2(1.27)

7

61.0

62.8(1.35)

8

61.3

63.4(1.42)

9

61.6

63.9(1.49)