University Decision Support Tools Assignment: Analysis and Methods

This assignment solution explores various decision support tools and techniques. It begins with a detailed explanation of the decision-making process, including identifying alternatives, states of nature, and the use of different decision-making criteria such as optimist, pessimist, Laplace, and regret matrix approaches. The solution then delves into the concept of the value of information, calculating expected monetary value and expected value of perfect information. The assignment continues with a simulation analysis of a hotel's room booking system, determining the optimal number of overbooked rooms to minimize costs. Finally, the solution concludes with a regression analysis, exploring the relationship between price and mileage/age of a product, including the creation of regression models and the interpretation of results. The document covers topics like decision analysis, simulation, regression analysis, CVP analysis, and their application in business scenarios.

Running head: DECISION SUPPORT TOOLS
DECISION SUPPORT TOOLS
Name of Student
Name of University
Author Note

1DECISION SUPPORT TOOLS
Table of Contents
Question 1: Decision Analysis.........................................................................................................2
Question 2: Value of Information....................................................................................................6
Question 3: Simulation....................................................................................................................7
Question 4: Regression Analysis...................................................................................................10
Question 5: CVP Analysis.............................................................................................................12

2DECISION SUPPORT TOOLS
Question 1: Decision Analysis
a)
The five steps in a decision making process involve, identifying the kind of decision that
is to be made and the end goals that the decision maker wishes to achieve. Consequently, it is
required to identify the relevant information, collecting the necessary data and hence identify the
alternative course of actions available to the decision maker. Upon identifying the options
available to act upon, it is required to consider the consequences of each action in terms of the
loss or gain that might be experience upon carrying them through. Then in light of those
consequences the actions are prioritised or ranked and the action coming out on top is considered
the best choice. The final step is to validate and verify whether the action meets the goals that
have been set at the beginning by the decision maker and if it does not the entire process is
repeated till an action is identified which meets all the requirements.
b)
Alternatives are course(s) of action(s) which the decision maker could opt to suggest or
agree upon as the appropriate solution to the decision making problem. An alternative is
associated with certain risks or consequences which the decision maker is required to gauge and
based on the severity of loss or gain associated with the action, the decision maker decides upon
the best course of action keeping in mind the end goals and constraints of the problem. For
example, consider the problem of a candy maker deciding upon how many units of the candy A
he makes, he should produce in a month during winter. Then given his limitations in terms of
resources and with the goal to maximize profits, each different consideration of the number of
units is an alternative, which could be greater than say 10,000 units per month or between 10,000
to 20,000 units.

3DECISION SUPPORT TOOLS
States of nature are the possible consequences that could be faced in response to the
alternative actions being taken into consideration. These are beyond the control of the decision
maker. Relating to decision problem mentioned above, the possible states of nature would be the
profits or loss that would be experienced by the vendor as consequence to the alternative actions
being considered.
c)
1. Given that cost price (C.P.) of fish is $15 per kg, selling price (S.P.) is $30 per kg and
also given that the vendor sells whatever fish is leftover for the day at $10 per kg(P.L.) to the
local proprietor, the profit is then computed using:
Profit= ( Weight bought−Weight Sold ) × P . L .+Weight sold × S . P.−Weight bought × C.P.
The following table gives the conditional profits for each combination of alternative and state of
nature:
Alternatives
Items Sold\
Items
Bought 10 15 20 25 20
States
of
Nature
10 $ 150.00 $ 75.00 $ - $ -75.00 $ -
15 $ 350.00 $ 225.00 $ 150.00 $ 75.00 $ 150.00
20 $ 550.00 $ 425.00 $ 300.00 $ 225.00 $ 300.00
25 $ 750.00 $ 625.00 $ 500.00 $ 375.00 $ 500.00
20 $ 550.00 $ 425.00 $ 300.00 $ 225.00 $ 300.00
Table 1
2.
OPTIMIST APPROACH Alternatives
Items Sold\Items Bought 10 15 20 25 20
States
of
Nature
10 $ 150.00 $ 75.00 $ - $ -75.00 $ -
15 $ 350.00 $ 225.00 $ 150.00 $ 75.00 $ 150.00
20 $ 550.00 $ 425.00 $ 300.00 $ 225.00 $ 300.00
25 $ 750.00 $ 625.00 $ 500.00 $ 375.00 $ 500.00
20 $ 550.00 $ 425.00 $ 300.00 $ 225.00 $ 300.00
Maximum $ 750.00 $ 625.00 $ 500.00 $ 375.00 $ 500.00

4DECISION SUPPORT TOOLS
Table 2
Following the optimist’s approach, the action of buying 10kg per week was found to
achieve maximum profit and hence most beneficial.
3.
PESSIMIST APPROACH Alternatives
Items Sold\Items Bought 10 15 20 25 20
States
of
Nature
10 $ 150 $ 75 $ 0 $ -75 $ 0
15 $ 350 $ 225 $ 150 $ 75 $ 150
20 $ 550 $ 425 $ 300 $ 225 $ 300
25 $ 750 $ 625 $ 500 $ 375 $ 500
20 $ 550 $ 425 $ 300 $ 225 $ 300
Minimum Profit $ 150 $ 75 $ 0 $ -75 $ 0
Table 3
Following the pessimist’s approach, the action of buying 10kg per week was found to
have the maximum of the minimum possible profits, thus found to be most favourable course of
action.
4.
LAPLACE's CRITERION Alternatives
Items Sold\Items
Bought 10 15 20 25 20
States of
Nature
10 $ 150.00 $ 75.00 $ - -$75.00 $ -
15 $ 350.00 $ 25.00 $150.00 $75.00 $150.00
20 $ 550.00 $425.00 $300.00 $225.00 $300.00
25 $ 750.00 $625.00 $500.00 $375.00 $500.00
20 $ 550.00 $425.00 $300.00 $225.00 $300.00
Expected Profit $ 470.00 $355.00 $250.00 $165.00 $250.00
Table 4

5DECISION SUPPORT TOOLS
Following Laplace’s criterion, the expected profit , assuming equal chance of all states of
nature, by following the alternative of buying 10kg per week was found to be maximum and thus
is was taken as the best decision.
5.
Regret Matrix Alternatives
10 15 20 25 20
States of
Nature
10 $ 600.00
$
550.00 $ 500.00 $ 450.00 $ 500.00
15 $ 400.00
$
400.00 $ 350.00 $ 300.00 $ 350.00
20 $ 200.00
$
200.00 $ 200.00 $ 150.00 $ 200.00
25 $ - $ - $ - $ - $ -
20 $ 200.00
$
200.00 $ 200.00 $ 150.00 $ 200.00
Minimum Regret $ 600.00
$
550.00 $ 500.00 $ 450.00 $ 500.00
Table 5
The alternative buying 25kg per week was found to bear the least regret, which is the
minimum of the difference between maximum profit for that alternative and the profits over the
various states of nature and hence was found to be most preferable.
6.
Items Sold\Items
Bought 10 15 20 25 20 Probability(Given)
10 $150.00 $ 75.00 $-
$ -
75.00 $ - 0.1
15 $350.00 $225.00 $150.00 $75.00 $150.00 0.2
20 $550.00 $425.00 $300.00 $225.00 $300.00 0.4
25 $750.00 $625.00 $500.00 $375.00 $500.00 0.2
20 $550.00 $425.00 $300.00 $225.00 $300.00 0.1
Expected Profit $510.00 $390.00 $280.00 $195.00 $280.00 1
Table 6

6DECISION SUPPORT TOOLS
Using the given probability of occurrence of the states of nature, the expected profit for
each alternative or expected monetary value(EMV) were computed and the one for the action of
buying 10kg per week was found to be maximum and hence most preferable.
7.
Items Sold\Items
Bought 10 15 20 25 20
Probability
of Sold
fish/week
10 $ 150.00 $ 75.00 $ - $ -75.00 $ - 0.01079
15 $ 350.00 $ 225.00 $ 150.00 $ 75.00 $ 150.00 0.04839
20 $ 550.00 $ 425.00 $ 300.00 $ 225.00 $ 300.00 0.07978
25 $ 750.00 $ 625.00 $ 500.00 $ 375.00 $ 500.00 0.04839
20 $ 550.00 $ 425.00 $ 300.00 $ 225.00 $ 300.00 0.07978
Expected Profit $ 142.62 $ 109.77 $ 79.33 $ 56.87 $ 79.33
Table 7
Assuming that the weights of the items sold or the states of nature follow a normal
distribution with men $20 and $5 standard deviation, then using the corresponding densities of
the weights sold, the expected profit for each alternative was calculated and the action of buying
10kg of fish per week was found to have maximum expected profit of $142.62 and is hence
taken to be the best decision.
Question 2: Value of Information
a. The prior probability of success is taken to be 0.3. Therefore the prior probability of
failure is 0.7. Then if the firm expects to earn $1000000 if they achieve success and
encounter a loss of $600,000 otherwise then the expected profit or expected monetary
value is given by,
EMV= 0.3 ×1000000+0.7 ×(−600000)= $-120000
Hence the firm is expected to encounter a loss and thus the product should not marketed.

7DECISION SUPPORT TOOLS
b. The expected value of perfect information (EVPI) is given by:
EVPI= EMV – EV|PI
The expected value given perfect information (EV|PI) was computed as 0.3×1000000.
The EMV is $120000.Thus the EVPI was computed as $420000.
c. Given that the conditional probability of getting favourable result when success was
achieved is 0.7 and unfavourable result given failure is 0.8.
Then, P (favourable, Success) = P (Success) × P (favourable| Success) = 0.3×0.7 = 0.21
P (unfavourable, failure) = P (failure) × P (unfavourable| failure) = 0.7 ×0.8 = 0.56
Then, P (favourable, failure) = P (failure) – P (unfavourable, failure) = 0.7 – 0.56 = 0.14
and,
P (favourable) = P (favourable, failure) + P (favourable, success) = 0.14+ 0.21 =0.35.
Again, P (unfavourable) = 1- P (favourable) = 1- 0.35= 0.65.
d. Using Bayes’ theorem the probability P (Success| favourable) is given by:
P(Favorable∨Success)
P (Unfavovrable∨Success)+ P( Favorable∨Success)
Then,
P (unfavourable |success) = P (Unfavourable , Success)
P( Success) =
P (Success)−P(Favourable , Success)
P( Success)
= 0.3−0.21
0.3 = 0.3
Hence, P (Favourable | Success) = 0.7
0.3+0.7 = 0.7.
e. The firm ought to pay a maximum of $420000 for the market research survey which is
the EVPI amount.

8DECISION SUPPORT TOOLS
Question 3: Simulation
a. The following section simulates the room booking records of one Heartbreak Hotel. The
hotel faces a situation where it intentionally overbooks three room during peak season to counter
the loss that guests cancelling their stay would incur. Instead if they find that they are unable to
provide a room due to overbooking they pay $125 on behalf of the guest to stay at a competing
hotel down the street. The cost of a vacant room is given as $50. The problem is to find out what
is the optimum number of over booked rooms so that their average daily cost is minimum. The
following figure gives the model for determining just that:
Figure 1
b. Model Formulae:

9DECISION SUPPORT TOOLS
Figure 2
Output:
Figure 3
c.
Rooms Average daily cost

10DECISION SUPPORT TOOLS
Overbooked
0 $ 121.67
1 $ 77.50
2 $ 62.50
3 $ 123.33
4 $ 207.50
5 $ 320.83
Table 8
d.
To,
The Manager,
Heartbreak Hotel.
Sir,
Upon analysing the patterns of vacant rooms in you hotel each day and the associated
costs of $50 per vacant room and $125 as consolation for cases when overbookings rendered
your hotel unable to provide lodgings for any guest, an entire month, for it has been determined
that the daily average cost is minimum in case you overbook 2 rooms instead of 3, leading up to
an average daily cost of $62.50.
Regards,
Name of Student
Date

11DECISION SUPPORT TOOLS
Question 4: Regression Analysis
a.
Regression Statistics
Multiple R
0.84975
7
R Square
0.72208
7
Adjusted R Square
0.68734
8
Standard Error
1532.39
3
Observations 10
Table 9
Coefficient
s
Standard
Error t Stat P-value
Intercept 17227.29 1188.417 14.496 5.02E-07
Mileage -0.09553 0.020953 -4.55916 0.001852
Table 10
The regression model or cost equation based on mileage is:
Price = 17227.29 – 0.09553 × Mileage
The regression has R-squared value of 0.72 so it explains 72% of the variation in Price. Effect of
Mileage was found to be significant.
Regression Statistics
Multiple R
0.85506
8
R Square
0.73114
2
Adjusted R Square
0.69753
4
Standard Error
1507.22
2
Observations 10
Table 11
Coefficien
ts
Standar
d Error t Stat P-value
Intercept 16226.39
971.101
9
16.7092
6
1.66E-
07