Mr. X is considering three universities for admission and must select based on location, academic reputation, and political violence. He analyzes each university using a decision tree that weights reputation most heavily. The analysis shows University B has the best overall score. Decision analysis allows ranking alternatives based on weighted criteria to select the optimal choice.

1. 1
Decision Theory
Definition of Decision Analysis: Decision analysis is defined as the use of a rational process/technique
for selecting the best of several alternatives. The “goodness” of a selected alternative depends on the
quality of the data used in describing decision situation.
Example: Mr X is a bright student, has received full academic scholarships from three institutions: U of
A, U of B and U of C. To select a university Mr. X lists two three main criteria: location, academic
reputation and political violence. Being the excellent student he is, he judges academic reputation to be
five times as important as location and less political violence, which gives a weight of approximation
80% to reputation, 15% to location and 5% to political violence. He then use a systematic analysis to
rank the three universities from the standard point of location, reputation and political violence. The
analysis produces the following estimates
Location: U of A (13%) U of B (28%) U of C (59%)
Reputation U of A (54%) U of B (27%) U of C (19%)
Political Violence U of A (25%) U of B 45% U of C (30%)
The structure of the decision problem is summarized below in Figure.
Select a
University
Location
(0.15)
Reputation
(0.80)
Political
Violence 0.05
U of A
0.13
U of B
0.28
U of C
0.59
U of A
0.54
U of B
0.27
U of C
0.19
U of A
0.25
U Of B
0.45
U of C
0.30
Sazzad Hossain
128

2. 2
To select the university here we use the technique of decision analysis.
Classification of Decision Situation
Decision making process can be classified into the following three broad categories
(1) Decision making under certainty in which the data are known deterministically
(2) Decision making under risk in which the data can be described by probability distributions.
3. Decision making under uncertainty in which the data cannot be assigned relative weights that
represent their degree of relevance in the decision process.
Decision Making Under Certainty:
Linear programming and transportation models provide an example of decision making under
certainty. Here only one state of nature exists, there is complete certainty about the future. The
decision maker simply finds the best payoff in that one column and chooses the associated
alternative. Few complex managerial decision making problems, however ever enjoy the luxury of
having complete information about the future and hence decision making under uncertainty is of
little consequential interest. This section presents a different approaches for the situation in which
ideas, feelings and emotions are quantified to provide a numeric scale for prioritizing decision
alternatives. The approach is known as the
Decision Making Under Risk
Here, more than one states of nature exist and decision maker has sufficient information to assign
probabilities to each of these states. For this reason decision making under risk is usually based on
the expected value criteria in which decision alternatives are compared based on the maximization of
expected profits or the minimization of expected cost. This approach has limitations in the sense that
it may not be applicable to certain situations
Expected Value Criteria:
It is also known as the expected monetary value (EMV) criteria, it consists of the following steps.
(i) Construct a payoff table listing the alternative decisions and the various states of nature. Enter the
conditional profit for each decision-event combination along with the associated probabilities. In
general, a decision problem may include n states of nature and m alternatives. If jp is the
probability of occurrence for state of nature j and let ija is the payoff of alternative i
given state of nature j (i= 1, 2,……,m; j = 1, 2,……,n)

3. 3
State j
Probability( jp )
Alternatives
1, 2, ……………. i………. ….. m
1
2
.
.
.
j
.
.
n
1p
2p
.
.
.
jp
.
.
np
11a 21a ………………. i1a ……………. m1a
12a 22a ………………… i2a …………… m2a
. ………………………………………..
. ………………………………………..
. ………………………………………..
1ja 2ja ………………….. ija ……………… mja
. …………………………………………….
. ………………………………………..
1na 2na …………………… ina ……………… mna
(ii) Calculate the EMV for each decision alternative by multiplying the conditional profits by the
assigned probabilities and adding the resulting conditional values. The expected monetary value
or expected payoff for alternative I is computed as
i ij j
1
EMV = a p ; i = 1, 2,.....,m
n
j

(iii) The best alternative is the one associated with
*
i i iEMV =max [EMV ] or
*
i i iEMV =min [EMV ]
depending, respectively, on whether the payoff of the problem represents profit (income) or loss
(expense)
Problem: A newspaper boy has the following probabilities of selling a sports magazine
No. of Copies Sold Probabilities
10 0.01
11 0.15
12 0.20
13 0.25
14 0.30
Cost of a copy is TK 30 and sale price is TK 50. He cannot return unsold copies. How many
copies should ne order?
Solution: The numbers of copies for purchases and for sales are 10, 11, 12, 13 and 14. These are
his sales magnitudes. There is no reason for him to buy less than 10 or more than 14 copies.
Stocking 10 copies each day will always result in profit TK200 irrespective of demand. For
instance, even if the demand on some day is more than 10 copies, he can sell only 10 and hence
his conditional profit is TK200. When he stocks 11 copies, his conditional profits is TK220 on
days buyers request 11, 12, 13 or 14 copies. But on days when he has 11 copies on stock and
buyers buy only 10 copies, his profit will be TK 170. The profit of TK 200 on the 10 copies sold

4. 4
must be reduced by TK 30, the cost of one copy left unsold. The same will be true when he
stocks 12, 13, or 14 copies. The conditional profit is given below;
Payoff = (20 copies sold-30 copies unsold ) TK 
The payoff table is constructed below;
Table 1: Payoff Table: Conditional profit in TK
Possible Demand
No. of Copies
Prob. Possible stock action
10 Copies 11 Copies 12 Copies 13 Copies 14 Copies
10
11
12
13
14
0.10
0.15
0.20
0.25
0.30
200
200
200
200
200
170
220
220
220
220
140
190
240
240
240
110
160
210
260
260
80
130
180
230
280
Now the expected value of each decision alternative is obtained by adding the multiplying its
conditional profit and the associated probability. The is shown in Table below;
Table: Expected profit
Possible Demand
No. of Copies
Prob. Expected profit from stocking in TK
10 Copies 11 Copies 12 Copies 13 Copies 14 Copies
10
11
12
13
14
0.10
0.15
0.20
0.25
0.30
20
30
40
50
60
17
33
44
55
66
14
28.5
48
60
72
11
24
42
65
78
8
19.5
36
57.5
84
Total Expected Profit 200 215 222.5 220 205
Thus the newsboy must order 12 copies to earn highest possible average daily profit. This stocking
will maximize the total profits over a period of time. Of course there is no guarantee that he will make
a profit of TK222.5 tomorrow. However, if he stocks 12 copies each day under the condition given, he
will have average profit of TK 222.5 per day. This is the best he can do because choice of any of the
other four possible stock actions result in a lower daily profit.
Expected Opportunity Loss (EOL) Criterion
This approach is an alternative to EMV. EOL or expected value of regrets is the amount by which
maximum possible profit will be reduced under various possible actions. The course of action that
minimize these losses is the optimal decision alternative. It consists of the following steps:
Step 1: Construct the conditional profit table for each decision-event combination and write the
associated probabilities

5. 5
Step 2: For each event, calculate the conditional opportunity loss (COL) by subtracting the payoff
from the maximum payoff for the event.
Step 3: Calculate the expected opportunity loss (EOL) for each decision alternative by
multiplying the COL’s by the associated probabilities and then adding the values.
Step 4. Select the alternative that yields the lowest EOL
Problem: A newspaper boy has the following probabilities of selling a sports magazine
No. of Copies Sold Probabilities
10 0.01
11 0.15
12 0.20
13 0.25
14 0.30
Cost of a copy is TK 30 and sale price is TK 50. He cannot return unsold copies. How many
copies should ne order?
Solution: The best alternative for demand 10 copies is to order 10 copies, resulting in optimal
profit of TK 200. The conditional opportunity loss for each stock action is obtained just by
subtraction the respective conditional profit from TK 200. Likewise, for demand of 11, 12, 13
and 14 copies we subtract the conditional payoff values for each of these rows from the
maximum of that row. The resulting conditional opportunity loss table is shown below;
Table 1: Conditional Loss Table in TK
Possible Demand
No. of Copies
Prob. Possible stock action
10 Copies 11 Copies 12 Copies 13 Copies 14 Copies
10
11
12
13
14
0.10
0.15
0.20
0.25
0.30
0
20
40
60
80
30
0
20
40
60
60
30
0
20
40
90
60
30
0
20
120
90
60
30
0
Now the EOL value of each decision alternative is obtained by adding the multiplying its conditional
opportunity loss and the associated probability. The is shown in Table below;
Table: Expected profit
Possible Demand
No. of Copies
Prob. Expected profit from stocking in TK
10 Copies 11 Copies 12 Copies 13 Copies 14 Copies
10
11
12
13
14
0.10
0.15
0.20
0.25
0.30
0
3
8
15
24
3
0
4
10
8
6
4.5
0
5
12
9
9
6
0
6
12
13.5
12
7.5
0
Total Expected Profit 50 35 27.5 30 45

6. 6
The optimal stock action is the one which will minimize expected opportunity losses. If the
newsboy stock 12 copies each day, the expected loss will be minimum.
Note : It may be pointed out that EMV and EOL decision criteria are completely consistent and
yield the same optimal decision alternative. However, while the decisions alternative will be
always based on finding the minimum EOL value under EOL criterion irrespective of whether the
problem is of maximization expected profit or minimizing expected loss.
Expected Value of Perfect Information (EVPI): Complete and accurate information about
the future demand referred to as perfect information, would remove all uncertainty from the
problem. With this perfect information, the decision maker would know in advance exactly about
the future demand. EVPI represents the maximum amount he would pay to get this additional
information on which may be based the decision alternative.
Problem: A dairy firm wants to determine the quantity of butter it should produce to meet the
demand. The past records have shown the following demand patterns;
Quantity Demanded (in kg) No. of Days Demand Occurred
15
20
25
30
35
40
50
6
14
20
80
40
30
10
The stock levels are restricted to the range of 15 kg to 50 kg and the butter left unsold at the end
of the day must be disposed of due to inadequate storing facilities. Butter costs TK 80 and sold
TK 110 per kg.
(i) Construct a conditional profit table
(ii) Determine the action alternative associated with the maximization of expected profit
(iii) Determine the EVPI
Solution: The dairy firm would not produce butter less than 15 kg and more than 50 kg. Form the
given data we can calculate the conditional profit for each stock action and event (demand
combination. If CP is the conditional profit, S is the quantity in stock and D is the quantity
demanded, then CP is given by;
30S; when D S
CP=
110D-80S; when D<S



The probability of demand of 15 kg is = 6/200 = 0.03. The probabilities associated with other
demand levels are given below win payoff table below:

7. 7
Possible
Demand
(Event in
kg) (D)
Prob. Possible stock action (Alternative ) (in kg) (S)
15 20 25 30 35 40 50
15
20
25
30
35
40
50
0.03
0.07
0.10
0.40
0.20
0.15
0.05
450
450
450
450
450
450
450
50
600
600
600
600
600
600
-350
200
750
750
750
750
750
-750
-200
350
900
900
900
900
-1150
-600
-50
500
1050
1050
1050
-1550
-1000
-450
100
650
1200
1200
-2350
-1800
-1250
-700
-150
400
1500
The expected profit for each alternative is given below
Table: Expected profit
Possible
Demand
(Event in
kg) (D)
Prob. Possible stock action (Alternative ) (in kg) (S)
15 20 25 30 35 40 50
15
20
25
30
35
40
50
0.03
0.07
0.10
0.40
0.20
0.15
0.05
13.5
31.5
45.0
180
90
67.5
22.5
1.5
42
60
240
120
90
30
-10.5
14
75
300
150
112.5
37.5
-22.5
-14
35
360
180
135
45
-34.5
-42
-5
200
210
157.5
52.5
-46.5
-70
-45
40
130
180
60
-70.5
-126
-125
-280
-30
60
75
EMV 450.0 583.5 678.5 718.5 538.5 248.5 -496.5
Since the maximum EMV is TK 718.5 for stock of 30 kg of butter, thus the dairy firm may
produce 30 kg of butter and can expect average daily profit of TK 718.5
Now the expected profit with perfect information is given below in Table ()
Table : Expected profit with perfect information
Market Demand
(in kg)
Probabilities Conditional Profit under
certainty
Expected profit
with perfect
information
15
20
25
30
35
40
50
0.03
0.07
0.10
0.40
0.20
0.15
0.05
450
600
750
900
1050
1200
1500
13.5
42
75
360
210
180
75
EPPI 955.5

8. 8
The expected value of perfect information is given by
EVPI = EPPI- max(EMV)
= 955.5-718.5
= 237 TK
Use of Incremental (Marginal ) Analysis.
Any additional unit purchased will be either sold or remain unsold. If p denotes the probability of
selling one additional unit, then (1-p) must be the probability of not selling it. If the additional
unit sold, the conditional profit will increase as a result of the profit earned from this additional
unit. This is termed as the incremental profit or marginal profit. Let the define the marginal profit
is MP. If the additional unit is not sold, the conditional profit is reduced and the amount of
reduction is called the incremental loss or marginal loss. Let us define the marginal loss is ML.
Thus the expected incremental profit will be pMP while the expected incremental loss will be
(1-p) ML. Thus the units should be stocked up to the point such that
pMp (1-p)ML
or at least p(MP+ML) = ML
or at least p=ML/(MP+ML)

Thus additional unit should be stocked so long as the probability of selling at least one additional
unit is greater than p.
Problem: A milkman buys milk at Tk 30 and sells for TK 40 per litre. Unsold milk has to be thrown
away. The daily demand has the following probability distribution
Demand (Liter) Probability
46
48
50
52
54
56
58
60
62
64
0.01
0.03
0.06
0.10
0.20
0.25
0.15
0.10
0.05
0.05
If each day’s demand is independent of previous days’ demand, how many liters should he
ordered every day.
Solution: The marginal profit is given as
MP = 10 Tk
And the marginal loss is given as

9. 9
ML = TK 30
The milkman should stock additional liters of milk so long as the probability of selling at least an
additional liter of milk is greater than p where
ML
p=
MP+ML
30
0.75
10 30
 

The value of 0.75 for p implies that in order to justify the stocking of an additional unit, there
must be at least 0.75 cumulative probability of selling that unit. The cumulative probability of
sales are computed below;
Demand (Liter) Probability Cumulative probability that sales
will be at this level or higher
46
48
50
52
54
56
58
60
62
64
0.01
0.03
0.06
0.10
0.20
0.25
0.15
0.10
0.05
0.05
1.00
0.99
0.96
0.90
0.80
0.60
0.35
0.20
0.10
0.05
The optimum number of liters of milk to be stocked is 54 liters. If the number increased to 56
liters, the cumulative probability will become 0.60 which is less than the required probability
0.75.
The expected incremental profit is given by;
EIP = pMP=0.75 10=7.5 TK
The expected incremental loss is given by;
EIL = (1-p)ML = 0.25 0.30=7.5
If the milkman stocks 56 liters of milk, the expected incremental loss will be more than expected
incremental gain. The approach give you the same decision as provided by the EMV and EOL
approaches. However, the computational effort required in this approach is much less.
Problem: A vegetable seller buys a box of tomatoes for TK 300 and sells them for TK 450 a box.
If the box is not sold on the first selling day, it is worth of TK 200 as salvage. The past records
indicate that the demand is normally distributed with a mean of 30 boxes daily and standard

10. 10
deviation is 9. How many boxes should be stock?
Decision Trees Analysis:
A decision tree is a graphical representation of the decision process indicating decision
alternatives, states of nature, probabilities attached to the state of nature and conditional benefits
and losses. It consists of a network of nodes and branches. Two types of nodes are used: (i)
decision node represented by a square and state of nature (chance or event) node represented by a
circle. Alternative courses of action (strategies) originate from the decision nodes as main
branches (decision branches). At the end of each decision branch, there is a state of nature node
from which emanate chance events in the form of sub-branches (chance branches). The
respective payoff and the probabilities associated with alternative courses and the chance events
are shown alongside these branches. At the terminal of the chance branches are shown the
expected value of the outcome.
The general approach used in decision tree analysis is to work backward through the tree from
right to left, computing the expected value of each chance node. We then choose the particular
branch leaving a decision node which leads to the chance node with the highest expected value.
This approach is known as roll back or fold back process.
Example: A client asks an estate agent to sell three properties A, B, and C for him and agrees to
pay him 5% commission on each sale. He specifies certain conditions. The estate agent must sell
property A first, and this he must do within 60 days. If and when A is sold the agent receives his
5% commission on that sale. He can then either back out at this stage or nominate and try to sell
one of the remaining two properties within 60 days. If he does not succeed in selling the
nominated property in that period, he is not given the opportunity to sell the third property on the
same conditions. The prices, selling costs( incurred by the estate agent whenever a sale is made)
and the estate agent’s estimated probabilities of making
a sale are given below;
Property Price of Property Selling Cost Probability of Sale
A 125000 5000 0.70
B 175000 4000 0.60
C 225000 6000 0.50
(i) Draw up an appropriate decision tree for the state agent
(ii) What is the estate agent’s best strategy under EMV approach?
Solution: The state agent’s gets 5% commission if he sells the properties and satisfies the
specified conditions. The amount he receives as commission on sale of properties A, B and C will
be TK 6250, TK 8750, and TK 11250 respectively. Since the selling costs incurred by his are TK
5000, TK 4000 and TK 6000 respectively and his conditional profits are TK 1250, TK 4750 and TK

11. 11
5250 respectively. The decision tree for the problem is shown below;
EMV of Node D = TK [(5250 0.50)+(0 0.50)]=TK2625 
EMV of Node E = TK [(4750 0.60)+(0 0.40)]=TK2850 
EMV of Node 3 = Max of TK [ 2625, 0] = TK2625
EMV of Node 4 = Max of TK [ 2850, 0] = TK 2850
EMV of Node B = TK [(4750+2625) 0.60)+(0 0.40)]=TK4425 
EMV of Node C = TK [(5250+2850) 0.50)+(0 0.50)]=TK4050 
EMV of Node 2 = Max of [ 4425, 4050] = TK 4425
EMV of Node A = TK [(1250+4425) 0.70)+(0 0.30)]=TK3972.5 
EMV of Node 1 = TK 3972.5
Thus the optimal strategy for the estate agent is to sell A, if he sells A, then try to sell B and if he sells
B and then try to sell C to get an optimum amount of TK 3972.5
Sells C, 5250. 0.5
Takes c
Stops, TK 0 Sells Does Not C, TK 0
4750, 0.60 Stop, TK0 4750, 0.6
Sells A, Takes B Does Not Sell B Sells B
TK 0, 0.40 Takes B Does
Accepts TK1250, 0.70 Take C Sells, 5250, 0.50 Not Sell
Does not Sell, TK 0 Does Not Sell C Stops, TK0 TK0
0.30 0.50, TK 0
1 A
2
B
C
3
4
D
E