Download 10.3

Section 10.3
A little background for this section: There are four (4) types of people (or
decision making) that is addressed in this section…
(1) Optimist (or maximax) is the one where they are sure that the best will
always happen, so for the Optimist you choose the one with the largest
value of any that are there.
(2) Pessimist (maximin) is the one who where they know the worst will
happen to them, so the Pessimist looks at each choice and “knows” that the
worse will happen, so chooses the smallest value for each choice; but of
those choices chooses the largest number (a later problem will better
illustrate what is being said).
(3) Regretter (or Avoiding Regret or minimax) is the one where they are
afraid of making the wrong decision, because they will have regrets, so this
avoids regrets. For this you “build a regret table” by looking at each
situation (column, not choices) and taking the largest value in the column
and making that zero, and then all the rest of the spaces become the
difference between the value you zeroed and the value in that space; then
you choose the smallest value.
(4) Bayesian (or expected value) is the one where the numbers and
probabilities tell you how to decide, by doing an expected value for each
choice; and choosing the largest expected value.
(21)
The Optimist is the choice of the largest value, so since “90” is the
largest value, then the Optimist would choose B.
(22)
The Pessimist looks at A and chooses “40” (being the lowest in that
row), then looks at B and chooses “-20” (being the lowest in that row). The
final choice is the largest between those two earlier choices, which is “40”,
which means the Pessimist chooses A.
(23)
To make a regret table, you look at the first column and see that the
“70” is the largest number, so you make that a zero in the table; and in the
other box in that column you place the difference between 70 and -20
(70 – (-20) = 90). Likewise with the other column, seeing that the “90” is
the largest number, so you make that a zero in the table; and in the other
box in that column you place the difference between 90 and 40
(90 – 40 = 50). Giving the following table:
A
0
50
B
90
0
So now it is a matter of choosing the smallest value (other than zero);
which in this case is “50”…so the Regretter will choose A.
(24)
First note that you only have one probability; the other must make
the probabilities add up to one, so the blank probability must be 0.4. Now
with that expected value for A is the values times their probabilities (all
added together): 70*0.4 + 40*0.6 = 28 + 24 = 52
Now with that expected value for B is the values times their probabilities
(all added together): -20*0.4 + 90*0.6 = -8 + 54 = 46
(25)
Using the two results from #24, you choose the largest value, which
is “52”, so the Bayesian choose A.
(26)
The maximin or Pessimist looks at C and chooses “-5” (being the
lowest in that row), then looks at D and chooses “2” (being the lowest in
that row). The final choice is the largest between those two earlier choices,
which is “2”, which means the Pessimist chooses D.
(27)
The maximax or Optimist is the choice of the largest value, so since
“8” is the largest value, then the Optimist would choose C.
(28)
To make a regret table, you look at the first column and see that the
“2” is the largest number, so you make that a zero in the table; and in the
other box in that column you place the difference between 2 and -5
(2 – (-5) = 7). Likewise with the other column, seeing that the “8” is the
largest number, so you make that a zero in the table; and in the other box
in that column you place the difference between 8 and 6 (8 – 6 = 2). Giving
the following table:
C
D
7
0
0
2
So now it is a matter of choosing the smallest value (other than zero);
which in this case is “2”…so the Regretter will choose D.
(29)
First note that you only have one probability; the other must make
the probabilities add up to one, so the blank probability must be 0.7. Now
with that expected value for C is the values times their probabilities (all
added together): -5*0.3 + 8*0.7 = -1.5 + 5.6 = 4.5
Now with that expected value for D is the values times their probabilities
(all added together): 2*0.3 + 6*0.7 = 0.6 + 4.2 = 4.8
(30)
Using the two results from #29, you choose the largest value, which
is “4.8”, so the Bayesian choose D.
(31)
First note that you are missing one probability; the probabilities must
add up to one, so the blank probability must be 0.3. Now with that
expected value for E is the values times their probabilities (all added
together): 25*0.5 + 0*0.3 + -5*0.2 = 12.5 + 0 + -1 = 11.5
Now with that expected value for F is the values times their probabilities
(all added together): 10*0.5 + 20*0.3 + -10*0.2 = 5 + 6 + -2 = 9
(32)
Using the two results from #31, you choose the largest value, which
is “11.5”, so the Bayesian choose E.
(33)
The Pessimist looks at E and chooses “-5” (being the lowest in that
row), then looks at F and chooses “-10” (being the lowest in that row). The
final choice is the largest between those two earlier choices, which is “-5”,
which means the Pessimist chooses E.
(34)
This is the same question as #33, so has the same answer…E.
(35)
The other name for an Optimist is “maximax”.
(36)
To make a regret table, you look at the first column and see that the
“25” is the largest number, so you make that a zero in the table; and in the
other box in that column you place the difference between 25 and 10
(25 - 10 = 15). Likewise with the second column, seeing that the “20” is the
largest number, so you make that a zero in the table; and in the other box
in that column you place the difference between 20 and 0 (20 – 0 = 20).
Finally, the third column, seeing that the “-5” is the largest number, so you
make that a zero in the table; and in the other box in that column you place
the difference between -5 and -10 (-5 – (-10) = 5). Giving the following
table:
E
0
20
0
F
15
0
5
So now it is a matter of choosing the smallest value (other than zero);
which in this case is “5”…so the Regretter will choose F.
(37)
If we say “p” is the probability of the first column, then the
probability of the second column will be (1-p). Then we calculate the
expected value of A [ 70*p + 40*(1-p) ] and the expected value of B
[-20*p + 90*(1-p) ] and set them equal to one another…
70*p + 40*(1-p) = -20*p + 90*(1-p)
70p + 40 – 40p = -20p + 90 -90p
30p + 40 = -110p + 90
Now subtract 40 from both sides
30p = -110p + 50
Now add 110p to both sides
140p = 50
Divide both sides by 140
p = 5/14 = 0.3571
(38)
Easiest way to answer this is to put in a smaller probability, like 0.3
into the two equation…70*p + 40*(1-p) and -20*p + 90*(1-p)
70*0.3 + 40*(1-0.3) = 21 + 28 = 49 and
-20*0.3+ 90*(1-0.3) = -6 + 63 = 57, so since 57 is larger the answer is B.
(39)
If we say “p” is the probability of the first column, then the
probability of the second column will be (1-p). Then we calculate the
expected value of C [ -5*p + 8*(1-p) ] and the expected value of D
[2*p + 6*(1-p) ] and set them equal to one another…
-5*p + 8*(1-p) = 2*p + 6*(1-p)
-5p + 8 – 8p = 2p + 6 -6p
-13p + 8 = -4p + 6
Now subtract 6 from both sides
-13p + 2 = -4p
Now add 13p to both sides
2 = 9p
Divide both sides by 9
p = 2/9 = 0.2222
(40)
Easiest way to answer this is to put in a larger probability, like 0.3
into the two equation…-5*p + 8*(1-p) and 2*p + 6*(1-p)
-5*0.3 + 8*(1-0.3) = -1.5 + 5.6 = 4.1 and
2*0.3 + 6*(1-0.3) = 0.6 + 4.2 = 4.8 , so since 4.8 is larger the answer is D.
For problems 41-44 let’s set up the demand/probability table from the text:
Demand
3
4
5
6
Probability
26 days
50 days
76 days
48 days
Selling Prob.
As you can see days do not equal probability, but if you divide the days by
the total number of days (200 days) then you will have probabilities:
Demand
3
4
5
6
Probability
0.13
0.25
0.38
0.24
Selling Prob.
And sell probability is calculated by adding up everything from right to left
until you get to the probability. For example, if you want the probability of
5, you add 0.24 and 0.38.
Demand
3
4
5
6
Probability
0.13
0.25
0.38
0.24
Selling Prob.
1.00
0.87
0.62
0.24
Now for the specific questions...
(41)
The pessimist would choose the smallest probability, which is for 3
cranes.
(42)
The optimist would choose the largest probability, which is for 5
cranes.
(43)
To maximize the profit you have to know how much profit you will
make on a given sale; and in this case is would be $1200 - $800 = $400; also
how much loss if you rent the cranes and do not rent them out: -$800
Now it is a matter of using the sell/no sell table below:
SELL
$400
P
NO SALE
-$800
1-p
The p is the selling probability for the best long term in the above table
(which is what we need to know to answer the question). Next the
expected value: 400*p + (-800)*(1-p) and set this equal to zero…
400p – 800 + 800p = 0
1200p - 800 = 0
add 800 to both sides
1200p = 800
divide both sides by 1200
p = 800/1200 = 0.6667
Now going back to the table with the selling probability, starting at the right
and moving left until you reach the first selling probability that is equal to
or greater than 0.6667, which is at 4 cranes.
(44)
The largest probability is at 5 cranes, which in a regret table would be
zero, and the next largest is at 4 cranes which means the difference
between that probability and the next will be the smallest (or the smallest
amount of regret).
First a table to represent the information given in the question:
Lose Sailboat
$100,000
0.1
Cost is $3800
Not Lose Sailboat
0
0.9
(45)
The sailboat owner is definitely favored since in 1 in every 10 cases
the sailboat will be lost and the insurance company will have to pay out
$100,000 versus only receiving 10*3800 = $38,000. The offset is that this
does not happen very often, so it is more likely that the insurance company
will just receive its premium and not have to pay anything out.
(46)
The fair price would equal the gross expected value of the “game” as
illustrated in the boxes above…100000*0.1 + 0*0.9 = $10,000
Insure
Don’t Insure
(47)
Lose Sailboat
$100,000 -$3800
-$100,000
Not Lose Sailboat
-$3800
$3800
The regret table for the above looks like:
Lose Sailboat
Not Lose Sailboat
Insure
0
$7200
Don’t Insure
$196,200
0
On this basis it is better to insure, because the least regret is $7200.
(48)
Better to lose a little than lose a lot.
(49)
Rewriting the above table without the probabilities:
Lose Sailboat
Not Lose Sailboat
$100,000
0
p
1-p
Cost is $3800
The expected value for a “fair game” would be:
100000*p + 0*(1-p) = 3800
100000p = 3800
divide both sides by 100000
P = 3800/100000 = 0.038
Figure 10.11
Stock
Level
14
14
4.2
15
3.7
16
3.2
17
2.7
18
2.2
19
1.7
Prob.
0.12
15
4.2
4.5
4.0
3.5
3.0
2.5
0.24
Possible Demands
16
17
4.2
4.2
4.5
4.5
4.8
4.8
4.3
5.1
3.8
4.6
3.3
4.1
0.14
0.20
18
4.2
4.5
4.8
5.1
5.4
4.9
0.24
19
4.2
4.5
4.8
5.1
5.4
5.7
0.06
Expected
Profit
$4.20
$4.40
$4.42
$4.32
$4.06
$3.60
(50)
Now making this into a regret table…regrets based on probability,
regrets based on expected profits and regrets based on individual profits…
Stock
Level
14
15
16
17
18
19
Prob.
14
1.5
2.0
2.5
3.0
3.5
4.0
0.12
15
1.5
1.3
1.7
2.2
2.7
3.2
0
Possible Demands
16
17
1.5
1.5
1.3
1.3
0.9
0.9
1.4
0.6
1.9
1.1
2.4
1.6
0.10
0.04
18
1.5
1.3
0.9
0.6
0.3
0.8
0
19
1.5
1.3
0.9
0.6
0.3
0
0.18
Expected
Profit
0.22
0.02
0
0.10
0.36
0.80
No specific level shows up as a definite choice…if you based it on
probability you would choose 16 (the smallest amount of regret that is nonzero … though you may want to select either 15 or 18 since there is zero
regret when you base the selection on the other). And if you choose the
expected profit you would select 15; and if you select based on individual
profits you would select either 18 or 19…SO DEFINITE SELECTION stands
out!
Demand
14
15
16
17
18
19
Probability
0.12
0.24
0.14
0.20
0.24
0.06
Selling Prob.
1.00
0.88
0.64
0.50
0.30
0.06
Cost: 50 cents
Sale Price: 80 cents
This table is to be used for problems 51-55. (NOTE: Remember that the
selling probability is just adding up everything from right to left.)
(51)
The profit for selling is 80 – 50 = 30 cents; and with a “salvage” (being
paid for leftovers) being 35 cents, then the possible loss is 50 – 35 = 15
cents.
Sell
Don’t Sell
30 cents
-15 cents
P
1-p
Based on the above table, the expected value (set to zero maximum profit)
is: 30*p + (-15)*(1 – p) = 0
30p -15 +15p = 0
45p – 15 = 0
45p = 15
p = 15/45 = 0.33
add 15 to both sides
divide both sides by 45
Now looking at the above chart with selling probability, starting at the right on
going left until the selling probability is equal to or greater than 0.33… and you
get 17 (which is the answer).
(52)
The profit for selling is 80 – 50 = 30 cents; and with a “salvage” (being
paid for leftovers) being 35 cents, then the possible loss is 50 – 45 = 5 cents.
Sell
Don’t Sell
30 cents
-5 cents
P
1-p
Based on the above table, the expected value (set to zero maximum profit)
is: 30*p + (-5)*(1 – p) = 0
30p -5 +15p = 0
45p – 5 = 0
45p = 5
p = 5/45 = 0.11
add 15 to both sides
divide both sides by 45
Now looking at the above chart with selling probability, starting at the right on
going left until the selling probability is equal to or greater than 0.11… and you
get 18 (which is the answer).
(53)
No he would NOT, because all he is looking to do is to lose
money…though on some rare occasions he would make some more if he
stocked 19, but he always lose money on items 20 and on.
(54)
He should buy as many Rutabagas as possible, since he will profit on
every one he gets…and more profit is all good.
(55)
ZERO, since there is no profit to be made, only loss.
(56)
ZERO, since there is no profit to be made, only loss.