Exam Questions: Decision Analysis, Value Of Information, Simulation, Regression Analysis

Question

a)

There are five major steps in the decision making process. Firstly, it is required to clearly identify the nature of the decision based on the nature of the problem and situation giving arise to it. Secondly, it is to identify the information necessary to address the problem and make the best suited decision and hence develop and execute a data collection strategy. After gathering the necessary information, the next step involves identifying the plausible alternative course of actions that could be considered based on the gathered information and due diligenceand list them. Fourthly, consider the associated consequences of undertaking one of the many available options for action to be taken on the matter and identify the actions according to the desired goals, prioritising them as per how much each work to reach the goal(s). This is typically done via statistical techniques of optimization and inferential analysis, taking into account the relevant data. The ordering of the actions should reflect their effectiveness towards attaining the set goals. Next step involves choosing the action which is placed at the top of this ordered list, or the action which best suits the attainment of the goal or has the least loss. Finally, it is required to review whether the decision or action identified serves to satisfy the goals in step 1 and if it is discovered that it fails to do so, then the entire process is repeated till a new favourable action or decision could be identified.

b)

An “alternative” in decision theory is what is referred to as any course of action or decision or strategy that is available to the decision maker that he may choose. Consider a decision making problem where it is to be decided which locations among three, namely A, B and C an aspiring bakery ought to set up shop in so as to ensure good profits. Then the alternatives available to the decision maker would be either to set up shop at location A or at location B or at location C out of which best possible course of action is to be determined.

“State of nature” refers to an event or outcome, upon the occurrence of which, the decision maker has little to no control whatsoever. An example of such an outcome could be outcomes of random experiments such as getting a “head” as the result of a coin toss or the chance of blindly drawing an ace of spades from a pack of well shuffled cards.

c)

The cost price (CP) of fish is said to be $15 per kg and the selling price(SP) set is $30 per kg. The vendor sells the leftover stock of fish for the day to the local proprietor with the price of the leftover (PL) being $10 per kg. Hence the profit for each day is given by the formula:

CP

The following table gives the conditional profits for each strategy or the amount of fish to be bought per day corresponding to each states of nature or the items sold per day, using the above formula.

Items SoldBought(in kg)	10	15	20	25	20
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

1 Decision Analysis

Table 1

The optimist’s approach abides by the rule of choosing the decision which would lead to maximization of  maximum profit from the possible states of nature, that is, if the vendor were an optimist, he would favour the decision which corresponds with the choice of buying a certain amount of fish such that it yields maximum profit among the maximum profits from each level of fish sold. The following table gives the figure as 10 kg per day with maximum of $750 profits.

		Alternatives
	Items SoldItems 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
	Maximum	750	625	500	375	500

Table 2

The pessimist’s approach abides by the rule of choosing the decision which would lead to maximization of the minimum possible profit for the states of nature, that is, if the vendor were an optimist, he would favour the decision which corresponds with maximum of the minimum profits from each level of fish sold. It was seen that buying 10 kg per week of fish served best to maximize profit through this approach.

		Alternatives
	Items SoldItems 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	150	75	0	-75	0

Table 3

The Laplace’s criteria assumes equal chance of occurrence for all states of nature. It computes the expected profit for each alternative based on this assumption and then the maximum expect profit indicates which action would serve to ensure maximum profits. It has been seen that 10kg per week yielded maximum profit of $470.

		Alternatives
	Items SoldItems 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
	Expected Profit	470	355	250	165	250

Table 4

The criterion of regret calculates the regret metric given by the maximum profit minus the profit for that combination of outcome and decision. The action with minimum of the minimum regret measure is considered the best option. Buying 25kg this week is supposed to be the best decision.

Regret Matrix	Alternatives
		10	15	20	25	20
States of Nature	10	600	550	500	450	500
15	400	400	350	300	350
20	200	200	200	150	200
25	0	0	0	0	0
20	200	200	200	150	200
	Minimum Regret	600	550	500	450	500

Table 5

The given frequencies of the sales per week in kilograms were used to compute the probability of each sales per week. The expected profit for each alternative option for the amount to be bought per week were calculated and the maximum expected monetary value of the profit was deemed as the indicator of which alternative would serve to be the best choice.10 kg per week have maximum expected profit with $510.

Items SoldBought	10	15	20	25	20	Probability
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(EMV)	$ 510.00	$  390.00	$  280.00	$195.00	$280.00

Table 6

Let the amount of fish sold per week follow a normal distribution with mean 20kg and standard deviation 5kg. Then the probability density at each observed sold amount was computed and the expected profit per week for each alternative amount bought per week was computed. The alternative with maximum expected profit was deemed to be the best. The option of buying 10kg per week showed maximum expected profit of $142.62.

Items SoldItems Bought	10	15	20	25	20	Probability of Sold fish/week
10	$  150.00	$    75.00	$   –	$   -75.00	$   –	0.010798193
15	$  350.00	$  225.00	$  150.00	$    75.00	$  150.00	0.048394145
20	$  550.00	$  425.00	$  300.00	$  225.00	$  300.00	0.079788456
25	$  750.00	$  625.00	$  500.00	$  375.00	$  500.00	0.048394145
20	$  550.00	$  425.00	$  300.00	$  225.00	$  300.00	0.079788456
Expected Profit(EMV)	$  142.62	$  109.77	$    79.33	$    56.87	$    79.33

Table 7

(a) The prior probability of success is said to be 0.3 and that of failure is 0.7. The profit gained when success achieved is $1000000 and the loss when failure is faced is $600000. Therefore the expected profit is found to be   which equals -$120000, that is, a loss of $120000. Hence the product should not be marketed given the prior probability of success and failure.

(b) The expected value of perfect information (EVPI) is given by the difference in expected monetary value and the expected value given perfect information. The expected monetary value was computed to be $-120000 and the expected value given perfect information was computed as $300000. Thus the EVPI was computed as $420000.

2 Value of information

(c) The probability of favourable result given success was given as 0.7 and unfavourable result given failure was given as 0.8. Then,

P (favourable, success) = 0.7× P (success) = 0.21

P (unfavourable, failure) = 0.8× P (failure) = 0.56

Then P (favourable, failure) = P (failure) – P (unfavourable, failure) = 0.14

The P (favourable) = P (favourable, failure) + P (favourable, success) = 0.35 and

P (unfavourable) = 1- P (favourable) = 0.65.

(d) The posterior probability P(success | favourable) was computed using Bayes’ theorem,

P (unfavourable |success) = =  = (0.3-0.7×0.3)/0.3=0.09=0.3.

Thus P(favourable | success) = (0.7/(0.7+0.3)) = 0.7.

The following figure shows the formula used for the construction of the model as specified in the assignment. It is said that Heartbreak Hotel having experienced no-shows routinely during peak season which follow the distribution specified in the cells in the Excel sheet from A3 to C9, owing to which the hotel overbooks rooms to check the incidence of rooms ending up vacant. In case that they fail to provide a room to the guest owing to over booking they send the guest to a competing hotel while paying $125 on behalf of the guest. Hence in cases where the hotel has more than three no shows it ends up having to pay $50 as opportunity cost for each room. The following model represents the opportunity cost per vacant room, cost of overbooking and the average daily cost as per the number of rooms overbooked and vacant.

Formulae used in Model

Figure 1

Model Outputs:

Number of rooms overbooked= 3

Figure 2

The value of the average daily cost for 0 to 5 overbooked rooms are given in the following table. The minimum average daily cost was observed for 0 overbooked rooms and hence the management at Heartbreak hotel should try to ensure that there are no overbooked rooms.

Rooms Overbooked	Average daily cost
0	$130.00
1	$85.83
2	$76.67
3	$114.17
4	$180.83
5	$300.00

 Table 8

The following tables 9, 10 show the summary of regression of Price against Mileage. The model explains variation in mileage as -0.09553 change in price with unit change in Mileage. The adjusted R Squared which is the measure of goodness of fit of the model was found to be 0.6873.

Regression Statistics
Multiple R	0.849757
R Square	0.722087
Adjusted R Square	0.687348
Standard Error	1532.393
Observations	10

Table 9

	Coefficients	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 following tables 11 and 12 show the summary of the regression of price against age. The model explains variation in age as -839.658 change in price with unit change in age in the positive direction. The adjusted R squared which is the measure of goodness of fit of the model was found to be 0.6975.

Regression Statistics
Multiple R	0.855068
R Square	0.731142
Adjusted R Square	0.697534
Standard Error	1507.222
Observations	10

Table 11

	Coefficients	Standard Error	t Stat	P-value
Intercept	16226.39	971.1019	16.70926	1.66E-07
Age	-839.658	180.019	-4.66427	0.001614

Table 12

The following tables 13 and 14 show the summary of the regression of price against mileage and age. The model explains variation in age and mileage as -0.0393 change in price with unit change in mileage and -507.303 change in price due to unit change in age in the positive direction. The adjusted R squared which is the measure of goodness of fit of the model was found to be 0.66412.

Regression Statistics
Multiple R	0.859528
R Square	0.738789
Adjusted R Square	0.664157
Standard Error	1588.208
Observations	10

Table 13

	Coefficients	Standard Error	t Stat	P-value
Intercept	16699.52	1462.676	11.4171	8.88E-06
Mileage	-0.0393	0.086807	-0.45269	0.664466
Age	-507.303	758.2802	-0.66902	0.524926

Table 14

Among the two simple regressions, that is, price on mileage and price on age, the regression of price on age would be preferable since its value of adjusted R squared is greater than the former.

The coefficients of age and mileage of the multiple regression of price on age and mileage were found to be negative. This means unit increase in those predictors correspond to decrease in price equal to the magnitude of the coefficients of those predictors.

The adjusted R squared for the multiple regression was found to be 0.66 and thus show a fairly good fit. The correlation coefficient between the two independent variable was however seen to be 0.968 suggesting multi-collinearity and this however suggests that the model is not reliable.

Using the model shown in the following table 14, the number of units of Product A with selling price $15, variable cost per unit $7 and fixed cost $2400 and break-even was found to be 300. The computation was done using Goal Seek.

Known Parameters
Selling Price per unit	15
Fixed cost	2400
Variable cost per unit	7
Variables
Number of units X	300
Results
Total revenue	4500
Fixed cost	2400
Total variable costs	2100
Total costs	4500
Profit	0

Table 14

Using the same model, the number of units for a profit of $1600 was found to be 500 as shown in the following table 15.

Known Parameters
Selling Price per unit	15
Fixed cost	2400
Variable cost per unit	7
Variables
Number of units X	500
Results
Total revenue	7500
Fixed cost	2400
Total variable costs	3500
Total costs	5900
Profit	1600

Table 15

Let X be the number of units of product B and then number of units of A is 2X. The using the model in Table 16, with sales price of additional product B being $20, its variable cost being $10 and total fixed cost being $3600, X was found to be 1000 for the total profit of $20000. Thus number of units A required is 2000 and that of B is 1000.

X	1000
Product	A	B	Total
Sales Unit	2	1	3
Sales price	$15	$20	$35
Variable cost	$7	$10	$17
Total fixed costs	$2,400	$3,600	$6,000
Total Variable cost	$14,000	$10,000	$24,000
Total Costs	$16,400	$13,600	$30,000
Revenue	$30,000	$20,000	$50,000
Profit	$13,600	$6,400	$20,000
Units of A	2000
Units of B	1000

Table 16