Solution Code: 1EDF
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Question 1.
(a).
Most of the rules for decision making under uncertainty express a different degree of decision maker´s optimism.
We will describe the most used rules for the choice of the best alternative supposing that the decision criterion expresses the requirement of maximization
Themaximax ruleis appropriate for extreme optimists that expect the most favorable situation (they choose the alternative that could result in the maximum payoff). Under this rule, the decision maker will find the largest payoff in the decision matrix and select the alternative associated with it (the largest payoff is determined for each alternative and then the largest payoff from these values is selected; therefore “maximax”).
Themaximin rule(Wald criterion) represents a pessimistic approach when the worst decision results are expected. The decision maker determines the smallest payoff for each alternative and then chooses the alternative that has the best (maximum) of the worst (minimum) payoffs (therefore “maximin”).
TheHurwicz-criterionrepresents a compromise between the optimistic and the pessimistic approach to decision making under uncertainty. The measure of optimism and pessimism is expressed by anoptimism - pessimism index,<0,1> . The more this index is near to 1, the more the decision maker is optimist. By means of the index, a weighted average of the best payoff (its weight =)and the worst payoff (its weight =1-) is computed for each alternative and the alternative with the largest weighted average should be chosen
Theprinciple of insufficient reason(Laplace criterion) assumes that all states of nature are equally likely. Under this assumption, the decision maker can compute the average payoff for each row (the sum of the possible consequences of each alternative is divided by the number of states of nature) and, then, select the alternative that has the highest row average. (Orms.pef.czu.cz, 2015)
(b).
(i) If Joe is an optimist he will buy equipment A, because he will follow MAXIMAX criteria.
(ii) If Joe is a pessimist he will buy equipment C, because he will follow MAXIMIN criteria.
(iii) Joe would select Equipment A if he follows the criterion of regret.
(iv) If Joe believes that there is a 60% chance of a good economy which type of investment should he select?
For Equipment A = (80000*0.6) + (-20000*0.4) = 40000
Equipment B = (30000*0.6) + (20000*0.4) = 26000
Equipment C = (23000*0.6) + (23000*0.4) = 23000
Joe will buy Equipment A.
(C).
Marginal Profit= $20,000
Marginal Loss= $10,000
Minimum Probability using marginal analysis= ML/MP+ML
= 10,000/10,000+20,000
= 10,000/30,000
= 0.333%
Question 2.
Probability of S1=0.7 & S2=0.3
S1=0.7 | S2=0.3 | |
A1 | $6,000 | $10,000 |
A2 | -1,000 | 25,000 |
(a).Expected value of information a1= 6,000*0.7+10,000*0.3= 7,200*
Expected value of information a2= -1,000*0.7+ 25,000* 0.3= 6,800
(b). Expected value of perfect information
EVWPI= 6,000(0.7) + 25,000(0.3) = 11,700
STEP1:
P (Y1/S1) =0.8 P (Y1/S2) =0.7
P (Y2/S1) =0.2* P (Y2/S2) =0.3*
1 1
STEP 2: Revised Probabilities
P(Y/S1) = Probability that demand will be poor as predicted.
P(Y/S2) = Probability that demand will good as predicted.
Si | P(Si) | P(Y1/Si) | P(Si,Y1) | P(Si/Y1) |
S1 | 0.7 | 0.8 | 0.560 | 0.727 |
S2 | 0.3 | 0.7 | 0.210 | 0.273 |
1 | 0.770 | 1.000 |
If Y2 has been predicted
Si | P(Si) | P(Y2/Si) | P(Si,Y2) | P(Si/Y2) |
S1 | 0.7 | 0.2 | 0.140 | 0.609 |
S2 | 0.3 | 0.3 | 0.090 | 0.391 |
1 | 0.23 | 1.000 |
770[P (y1)] + 0.230 [P (y2)] =1.000
STEP 3: Now we should calculate using new probabilities:
S1=0.727 | S2=0.273 | |
A1 | $6,000 | $10,000 |
A2 | -1,000 | 25,000 |
E(U/a1,y1,n)= 6000(0.727)+ 10,000(0.273)= 7092*
E(U/a2,y1,n)= -1000(0.727)+ 25,000(0.273)=6098
E (U/a1, y2, n) = 6,000(0.609) +10,000(0.391) = 7564
E(U/a2,y2,n)= -1,000(0.609) + 25,000(0.391)= 9166*
Step 4: Expected Utility
E (U/n) = E (U/a*,yk,n) P(yk/n)
= (7092) p(y1) + (9166) p(y2)
= 7092(0.770) + 9166 (0.230)
=$7569.02
Step 5: The benefit of information
= Expected Utility- Expected value
= 7569.02- 7200
=369.02
(c). It is not worth paying consultant $3,000 as information has not increased profit much. The cost of opinion will be more, compare to profit we are garnering.
Question 3.
(a).
For the number demanded the previous match.
Normal view:
Formula view:
For 2500 units-
Normal view-
Formula view-
(b).
Production | Average profits | Increase in average profits |
2300 | 6900 | 0 |
2400 | 7085 | 185 |
2500 | 7155 | 70 |
2600 | 7080 | -75 |
2700 | 6980 | -100 |
Simulations were run on games by marginally adding 100 production units and the profits were calculated per game. The marginal calculation shows that the profit increases up to 2500 and then substantially started decreasing, thus 2500 is the optimum number of production.
(c)
Business report:
Introduction: Profit can occur only when there is an appropriate strategy is used. Manager will find the analysis and can decide what strategies are viable for the company and the number of programs they should print.
Recommendation: After considering these excel sheets we can figure it out that both the strategies are only giving the profits at every stage, but the key point is that the profit at each stage in 2500 programs strategy is more than the 2400 programs printed strategy. So Firstly I would recommend the manager for 2500 programs strategy for each game for getting more profits and then 2400 programs strategy because this strategy also giving profits and there is no chance of loss in both the strategies.
Conclusion: 2500 programs strategy is best for every game to get more advantages without suffering loss. Also, 2400 programs strategy is also good so far. As a Manager, we can apply this approach in every game without fear of loss.
Question 4.
(b).
(c).
Question 5.
(a).
Fixed cost = $40000
SP = $10
Cost = $5
CM = $10 - $5 = $5
CM ratio = ($10 - $5)/$10 = 0.5
(i) Breakeven point in units = fixed cost / Contribution margin = $40000 / $5 = 8000 units
(ii) Breakeven point in sales dollar = Fixed cost / CM ratio = $40000 / 0.5 = $80000
(iii) Let x be the number of pizzas to be sold.
Profit = sales – variable cost – fixed cost
$60000 = 10x – 5x - $40000
$100000 = 5x
X = $100000/5 = 20000 units
(b).
(i)
X | Y | |
Selling price | 19 | 15 |
Variable cost | 10 | 9 |
Contribution/unit | 9 | 6 |
Fixed cost = 11000
Average contribution margin = 1/3 (9) + 2/3 (6) = 3+4 = 7
Required volume = 11000 + 10000/7 = 3000 units
(ii)Ratio = 2:2
Average contribution margin = 2/4 (9) + 2/4 (6) = 7.5
Required volume = Fixed cost + ?/ contribution margin
12000 = 11000 + ?/7.5
? = $79000
(C).
(i) Expected Annual profit
? = (p – b) x-a
E ? = E (p – b) x-a
= (p – b) E(x) - a
= (4000 – 3500) 12000 – 2000000
= 6000000 – 2000000
= 40,00,000
(ii) -?X =(p - b) * ?X
= (4000 – 3500) * 4000
= 500 * 4000
= 20,00,000
(iii)Selling price = 4000
Variable cost = 3500
Fixed cost = 2000,000
Mean = 12000 units
Standard Deviation = 4000 units
Breakeven point = Fixed cost
Selling price – Variable Cost
= 2000000
4000 – 3500
= 2000000/500
= 4000 units
This area is under the curve and where breakeven point is less than mean.
P(r >0) = P (Z > 4000 – 12000/4000)
= P (Z > -2) = 1- F (-2)
= P (Z < 2)
After checking the Z table, we get the value of Z= 0.97725
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QUESTION 1 Probability
Classical probability technique deals with relative frequency of each and every event in the sample space. Let us consider the experiment of rolling a die twice and the possible outcomes are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. The frequency of occurrence is given in the table below
Outcome | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Frequency | 1 | 2 | 3 | 4 | 5 | 6 | 5 | 4 | 3 | 2 | 1 |
Relative Frequency | 0.028 | 0.056 | 0.083 | 0.111 | 0.139 | 0.167 | 0.139 | 0.111 | 0.083 | 0.056 | 0.028 |
Going through the above relative frequency distribution, we can find the most frequent occurrence of outcomes
When the random variable takes a finite number of values, then the distribution involved will be discrete probability distribution. For example, tossing two coins and counting the number of heads turned up is an example of discrete probability distribution
When the random variable falls between two specified values, then the distribution involved will be continuous probability distribution. For example, the weight of the students studying in class X will fall between 30 kgs and 36 kgs
Sales Unit | Number of Days | Probability |
8 | 10 | 10/50 = 0.2 |
9 | 12 | 12/50 = 0.24 |
10 | 14 | 14/50 = 0.28 |
11 | 8 | 8/50 = 0.16 |
12 | 6 | 6/50 = 0.12 |
Total | 50 |
d.The information in the given problem is represented with the following notations
Sample Space = n (s) = 40
Number of females and Australian citizens = n (F ? A) = 12
Number of males and Australian citizens = n (M ? A) = 14
Number of female and overseas citizens = n (F ? OS) = 10
Number of male and overseas citizens = n (M ? OS) = 4
American | Overseas | Total | |
Male | 14 | 10 | 24 |
Female | 12 | 4 | 16 |
Total | 26 | 14 | 40 |
2.(i) P (Male) = 24/40 = 0.6
(ii) P (Overseas student) = 14/40 = 0.35
(iii) P(Australian given a female) = 12/16 = 0.75
(iv) P (Overseas student given a male) = 10/24 = 0.417
e.The information in the given problem is represented with the following notations
Population Mean = µ = 50 Weeks
Population Standard deviation = ? = 10 Weeks
1.PX?54=Px-?<54-5010
=PZ<54-5010, where Z=x-?~N(0,1)
= P (Z < 0.40)
= 0.6554 (by referring normal distribution table)
Therefore, the probability that the project will be completed in 54 weeks or less is 0.6554
2.PX?62=Px-?<62-5010
=PZ<62-5010, where Z=x-?~N(0,1)
= P (Z < 1.2)
= 0.8849 (by referring normal distribution table)
Therefore, the probability that the project will be completed in 62 weeks or less is 0.8849
3.PX>70=1-Px-?<70-5010
=1-PZ<70-5010, where Z=x-?~N(0,1)
= 1-P (Z < 2)
= 1 – 0.97725 (by referring normal distribution table)
0.02275
Therefore, the probability that the project will take longer than 70 weeks is 0.02275
QUESTION 2 Research Question, Constructing data table and calculating probabilities
1.The latest figures of age and sex of the Australian population is taken form the link https://populationpyramid.net/australia/2016/
2.
Male | Female | Total | ||
Age in Years | 0-14 | 2333664 | 2212119 | 4545783 |
15-24 | 1628703 | 1531467 | 3160170 | |
25-54 | 5056272 | 5007654 | 10063926 | |
55-64 | 1385613 | 1409922 | 2795535 | |
65 and over | 1750248 | 1993338 | 3743586 | |
Total | 12154500 | 12154500 | 24309000 |
3.
QUESTION 3 Statistical Decision Making and Quality Control
The information in the given problem can be represented with the following notations
Population mean = µ = 20
Population Standard deviation = ? = 9
Sample size = 36
x-Z1-2*n,x+Z1-2*n
=20-1.96*936,20+1.96*936
= (20 – 2.94, 20 + 2.94)
= (17.06, 22.94)
x-t1-2*n,x+t1-2*n
=20-2.306*99,20+2.306*99
= (20 – 6.918, 20 + 6.918)
= (13.082, 26.918)
3.Sampling more frequently using 9 observations and setting confidence intervals of 80
The 80% confidence interval for the true mean labor time to produce a product is calculated by using the formula given below
x-t1-2*n,x+t1-2*n
=20-1.397*99,20+1.397*99
= (20 – 4.19, 20 + 4.19)
= (15.81, 24.19)
Maintaining 95% confidence intervals and increasing sample size to 64 observations
The 95% confidence interval for the true mean labor time to produce a product is calculated by using the formula given below
x-z1-2*n,x+z1-2*n
=20-1.96*964,20+1.96*964
= (20 – 2.205, 20 + 2.205)
= (17.795, 22.205)
Setting 95% confidence intervals and using sample sizes of 100 observations.
The 95% confidence interval for the true mean labor time to produce a product is calculated by using the formula given below
x-z1-2*n,x+z1-2*n
=20-1.96*9100,20+1.96*9100
= (20 – 1.764, 20 + 1.764)
= (18.236, 21.764)
On comparing the three confidence intervals, we see that the width of the 95% confidence intervals and using sample sizes of 100 observations is less and the confidence interval is narrower. This is mainly due to larger sample size. Increase in sample size will decrease the width of the confidence interval and hence the interval becomes narrower
In order to determine whether the mean time until delivery differ significantly from 120 days, we perform single mean z test. The null and alternate hypotheses are given below
Null Hypothesis: H0: µ = 120
That is, the mean time until deliver do not differ significantly from 120 days
Alternate Hypothesis: Ha: µ ? 120
That is, the mean time until deliver differ significantly from 120 days
Level of Significance
Let the level of significance be ? = 0.05
Test Statistic
The z test statistic is
Z=x-??/n=118.5-12012/100=-1.25
The p – value of z test statistic is
P (Z > |-1.25|) = 0.2113 (by referring normal distribution table)
Since the p – value of z test statistic is greater than 0.05, there is no sufficient evidence to reject the null hypothesis at 5% level of significance. Therefore, there is no sufficient evidence to support the claim that the mean time is not 120 days
Group Research project
Here, we wish to start building the first child care center in Australia. Currently, the Australian population is 24309000 people. Out of which 10063926 people aged between 25 years and 54 years live in Australia. The Australian Pyramid map showing the population shape in Australia is given below
It is also seen that, overall in Australia, there are 802197 male children and 753579 female children aged less than 5 years
In addition, we found that the states Victoria (Population = 5,938,100) and New South Wales (Population = 7,618,200) records the second and third largest states in Australia in terms of population size. Therefore, we can select these two states to build a new first child care center in Australia
Assessment item 3
QUESTION 1 Decision Analysis
a.Marginal analysis is considered as one of the most important decision-making tool and now days, it is widely used in many business situations. It allows business operators to calculate the additional benefits on comparing its production activity and cost function. When we are planning to decide whether the project is moving on the right track, we need to use marginal analysis to break down all available options. The main purpose of marginal analysis is to allocating their rare resources to increase their benefits of their products. The formula used to compute the marginal analysis is given below
Marginal Benefit = Increase in Total Benefits per unit of control variable
?TR / ?Qcv = MR
For example, let us consider the situation of doing exercise daily for five days. Then before performing the exercise on day 6, we must do marginal analysis on how much calories burnt in doing exercises and also measure the effort ratio and output ratio to determine whether we are doing the exercise correctly and the work is going in right track or not
Advantages
Disadvantages
b.(i)Maximum (Stock Market) = $160000
Maximum (Bonds) = $60000
Maximum (Term Deposit) = $46000
From the above table, it can be clearly seen that the maximum of the maximum payoff = $160000, which is corresponding to the Alternative Stock Market. Thus, an optimist would make a decision to choose Stock Market as their decision alternative
(ii)Minimum (Stock Market) = - $40000
Minimum (Bonds) = $40000
Minimum (Term Deposit) = $46000
From the above table, it can be clearly seen that the maximum of the minimum payoff = $46000, which is corresponding to the Alternative Term Deposit. Thus, a pessimistic would make a decision to choose Term Deposit as their decision alternative
(iii)In this procedure we first find the maximum of each economy structure (states of nature) and subtract it from the other elements. The next step is to find the maximum for each alternative and then we need to find the minimum of the maximum alternatives
Decision Alternative | Good Economy | Poor Economy |
Stock market | 0 | $86,000 |
Bonds | $100,000 | $6,000 |
Term deposit | $114,000 | $0 |
Maximum (Stock Market) = $86000
Maximum (Bonds) = $100000
Maximum (Term Deposit) = $114000
Thus the Minimax Regret criteria is to select the Decision Alternative Stock Market
(iv)EMV (Stock Market) = 0.7 * $160,000 + 0.3 * (-$40,000) = $ 100000
EMV (Bond) = 0.7 * $60,000 + 0.3 * $40,000 = $ 540000
EMV (Term Deposit) = 0.7 * $46,000 + 0.3 * $46,000 = $ 46000
Form the above workings, we need to select the Decision Alternative Stock Market
(v)In general, the expectation value of perfect information = EVPI = EV|PI –EOL
.where:
EMV = Expected monetary value = i = 1, 2, 3
EV|PI = Expected value given perfect information=
EV|PI = 0.7 * 160000 + 0.3 * 46000 = $125800
EVPI = EV|PI – EOL = 125800 – 100000 = $ 25800
QUESTION 2 Value of information
a.EMV (a1) = 0.4 * $24,000 + 0.6 * $15,000 = $ 18600
EMV (a2) = 0.4 * $4,000 + 0.6 * $40,000 = $ 22400
Form the above workings, we need to select the a2
b.Expected value of perfect information
EVWPI= 24000 (0.4) + 40000 (0.6) = 33600
It is given that
P (Y1/S1) =0.8 and P (Y2/S1) =0.2
P (Y1/S2) =0.7 and P (Y2/S2) =0.3
Then, we have
Si | P(Si) | P(Y1/Si) | P(Si,Y1) | P(Si/Y1) |
S1 | 0.7 | 0.8 | 0.560 | 0.727 |
S2 | 0.3 | 0.7 | 0.210 | 0.273 |
Suppose, if we predict Y2
Si | P(Si) | P(Y2/Si) | P(Si,Y2) | P(Si/Y2) |
S1 | 0.7 | 0.2 | 0.140 | 0.609 |
S2 | 0.3 | 0.3 | 0.090 | 0.391 |
c.Now, the posterior probability is
E(U/a1,y1,n)= 24000(0.727)+ 15000(0.273)= 21543
E(U/a2,y1,n)= -4000 (0.727)+ 40,000(0.273)=8012
d.Here, the alternative a1 is the optimal when we use first strategy Y1
E(U/a1,y1,n)= 24000(0.609)+ 15000(0.391)= 18519
E(U/a2,y1,n)= -4000 (0.609)+ 40,000(0.391)=22796
Here, the alternative a2 is the optimal when we use first strategy Y2
EV|PI = 0.77 * 21543 + 0.23 * 22796 = 21831.9
EVPI = 21831.9 – 22400 = -568.1
QUESTION 3 Simulation
a.The simulation is performed to calculate the profit and is given below
Arrival Time Probability | Service Time Probability | Arrival Time | Service Time | Sales Revenue |
0.338333 | 0.526782 | 4 | 5 | 25 |
0.759587 | 0.750612 | 6 | 7 | 25 |
0.842755 | 0.121937 | 6 | 7 | 25 |
0.229497 | 0.326476 | 4 | 6 | 25 |
0.489786 | 0.290529 | 5 | 8 | 25 |
0.309932 | 0.309861 | 4 | 6 | 25 |
0.319198 | 0.335065 | 4 | 7 | 25 |
0.575895 | 0.109284 | 6 | 7 | 25 |
0.150147 | 0.010189 | 3 | 6 | 25 |
0.264109 | 0.153145 | 4 | 6 | 25 |
0.638102 | 0.993949 | 6 | 6 | 25 |
0.896226 | 0.779099 | 6 | 6 | 25 |
0.835057 | 0.526628 | 6 | 5 | 25 |
0.601559 | 0.393617 | 6 | 6 | 25 |
0.06525 | 0.417294 | 3 | 6 | 25 |
0.097007 | 0.862628 | 3 | 7 | 25 |
0.763174 | 0.139116 | 6 | 6 | 25 |
0.124239 | 0.997282 | 3 | 7 | 25 |
0.92512 | 0.940369 | 7 | 8 | 25 |
0.402377 | 0.016425 | 5 | 6 | 25 |
Total | 97 | 128 | 500 |
b.Formula
QUESTION 4 Regression Analysis and Cost Estimation
To determine the slope using high value method, we first need to find the minimum and maximum value of the variables taken into consideration
For variable Administrative cost, the minimum value is y1 = 4100 and the maximum value y2 = 16100
For variable patient load, the minimum value is x1 = 300 and the maximum value x2 = 1500
Variable Cost per Unit=y2-y1x2-x1=16100-41001500-300=10
Total Fixed Cost = y2 ? bx2 = 16100 – 10 * 1500 = 1100
Therefore, the regression model is
Administrative Cost = 1100 + 10 * Patient Load
b.Here, we perform three regression models with dependent variable as Administrative Costs
Model 1 (Dependent Variable – Administrative Costs and Independent Variable – Patient Load)
SUMMARY OUTPUT | ||||||
Regression Statistics | ||||||
Multiple R | 0.957909 | |||||
R Square | 0.91759 | |||||
Adjusted R Square | 0.909349 | |||||
Standard Error | 995.7963 | |||||
Observations | 12 | |||||
ANOVA | ||||||
df | SS | MS | F | Significance F | ||
Regression | 1 | 1.1E+08 | 1.1E+08 | 111.3447 | 9.69E-07 | |
Residual | 10 | 9916103 | 991610.3 | |||
Total | 11 | 1.2E+08 | ||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | |
Intercept | 2670.705 | 731.3395 | 3.651799 | 0.004449 | 1041.179 | 4300.231 |
Patient Load (Number) | 7.812068 | 0.74034 | 10.552 | 9.69E-07 | 6.162488 | 9.461649 |
Administrative Cost = 2670.705 + 7.812 * Patient Load
Going through the ANOVA table, we see that the value of F test statistic is 111.3447 and its corresponding p – value is 0.0000 < 0.05, indicating that the estimated regression model is good fit in predicting the dependent variable Administrative Cost
Model 2 (Dependent Variable – Administrative Costs and Independent Variable – Emergency Procedure)
SUMMARY OUTPUT | ||||||
Regression Statistics | ||||||
Multiple R | 0.685904 | |||||
R Square | 0.470464 | |||||
Adjusted R Square | 0.417511 | |||||
Standard Error | 2524.228 | |||||
Observations | 12 | |||||
ANOVA | ||||||
df | SS | MS | F | Significance F | ||
Regression | 1 | 56609415 | 56609415 | 8.884472 | 0.013792 | |
Residual | 10 | 63717252 | 6371725 | |||
Total | 11 | 1.2E+08 | ||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | |
Intercept | 2115.091 | 2668.472 | 0.792623 | 0.446394 | -3830.64 | 8060.818 |
Emergency Procedures (Number) | 734.5512 | 246.4372 | 2.980683 | 0.013792 | 185.4549 | 1283.648 |
Administrative Cost = 2115.091 + 734.5512 * Emergency Procedure
Going through the ANOVA table, we see that the value of F test statistic is 8.884472 and its corresponding p – value is 0.0138 < 0.05, indicating that the estimated regression model is good fit in predicting the dependent variable Administrative Cost
Model 3 (Dependent Variable – Administrative Costs and Independent Variables – Patient Load and Emergency Procedure)
SUMMARY OUTPUT | ||||||
Regression Statistics | ||||||
Multiple R | 0.963963 | |||||
R Square | 0.929225 | |||||
Adjusted R Square | 0.913497 | |||||
Standard Error | 972.7451 | |||||
Observations | 12 | |||||
ANOVA | ||||||
df | SS | MS | F | Significance F | ||
Regression | 2 | 1.12E+08 | 55905285 | 59.08194 | 6.68E-06 | |
Residual | 9 | 8516098 | 946233.1 | |||
Total | 11 | 1.2E+08 | ||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | |
Intercept | 1769.331 | 1029.328 | 1.718919 | 0.119748 | -559.17 | 4097.832 |
Patient Load (Number) | 7.101339 | 0.929748 | 7.637918 | 3.2E-05 | 4.998103 | 9.204576 |
Emergency Procedures (Number) | 148.5074 | 122.0906 | 1.21637 | 0.254781 | -127.681 | 424.6956 |
Administrative Cost = 1769.331 + 7.10134 * Patient Load + 148.51 * Emergency Procedures
Going through the ANOVA table, we see that the value of F test statistic is 59.08194 and its corresponding p – value is 0.00000686 < 0.05, indicating that the estimated regression model is good fit in predicting the dependent variable Administrative Cost
On comparing the coefficient of determination (R2) for three models, we see that the R2 value for Model 3 is high when compared with other two regression models . Therefore, we can say that the regression equation predicting Administrative cost using Patient Load and Emergency Procedures as independent variables as the best model
(c)The regression model chosen is
Administrative Cost = 2670.705 + 7.812 * Patient Load
When there were 1,000 Patients, we have
Administrative Cost = 2670.705 + 7.812 * 1000 = 10482.77292
QUESTION 5 CVP Analysis
a.The unit contribution margin for each product is given below
Unit Contribution Margin= Unit Specific Revenue-Unit Specific CostUnit Specific Revenue
=500-300500=0.40
(b) The break even workings is given below
Road Bikes | Sale Price Per Unit | Variable Cost | Fixed Cost | Total Cost | Revenue |
150 | 500 | 300 | 65000 | 110000 | 75000 |
175 | 500 | 300 | 65000 | 117500 | 87500 |
200 | 500 | 300 | 65000 | 125000 | 100000 |
225 | 500 | 300 | 65000 | 132500 | 112500 |
250 | 500 | 300 | 65000 | 140000 | 125000 |
275 | 500 | 300 | 65000 | 147500 | 137500 |
300 | 500 | 300 | 65000 | 155000 | 150000 |
325 | 500 | 300 | 65000 | 162500 | 162500 |
350 | 500 | 300 | 65000 | 170000 | 175000 |
375 | 500 | 300 | 65000 | 177500 | 187500 |
400 | 500 | 300 | 65000 | 185000 | 200000 |
425 | 500 | 300 | 65000 | 192500 | 212500 |
450 | 500 | 300 | 65000 | 200000 | 225000 |
475 | 500 | 300 | 65000 | 207500 | 237500 |
500 | 500 | 300 | 65000 | 215000 | 250000 |
525 | 500 | 300 | 65000 | 222500 | 262500 |
550 | 500 | 300 | 65000 | 230000 | 275000 |
Sales Price = 500 * 325 (Road bikes) = $162500
Cost = 65000 +300 * 325 (Road bikes) = $162500
Therefore, the manufacturer should sell at least 325 Road Bikes
(c) Sales Price = 500 * 325 (Road bikes) = $162500
The breakeven sales volume for the year in sales dollars is $162,500
(d) The table given below shows the revenue and profit workings
Road Bikes | Sale Price Per Unit | Variable Cost | Fixed Cost | Total Cost | Revenue | Profit |
500 | 500 | 300 | 65000 | 215000 | 250000 | 35000 |
525 | 500 | 300 | 65000 | 222500 | 262500 | 40000 |
550 | 500 | 300 | 65000 | 230000 | 275000 | 45000 |
575 | 500 | 300 | 65000 | 237500 | 287500 | 50000 |
600 | 500 | 300 | 65000 | 245000 | 300000 | 55000 |
625 | 500 | 300 | 65000 | 252500 | 312500 | 60000 |
650 | 500 | 300 | 65000 | 260000 | 325000 | 65000 |
675 | 500 | 300 | 65000 | 267500 | 337500 | 70000 |
700 | 500 | 300 | 65000 | 275000 | 350000 | 75000 |
725 | 500 | 300 | 65000 | 282500 | 362500 | 80000 |
750 | 500 | 300 | 65000 | 290000 | 375000 | 85000 |
765 | 500 | 300 | 65000 | 294500 | 382500 | 88000 |
790 | 500 | 300 | 65000 | 302000 | 395000 | 93000 |
815 | 500 | 300 | 65000 | 309500 | 407500 | 98000 |
840 | 500 | 300 | 65000 | 317000 | 420000 | 103000 |
865 | 500 | 300 | 65000 | 324500 | 432500 | 108000 |
890 | 500 | 300 | 65000 | 332000 | 445000 | 113000 |
Profit earned = $382500 - $294500 = $88000
(ii) The table given below shows the workings of profit calculated after tax
Road Bikes | Total Cost | Revenue | Profit | Profit after tax |
700 | 275000 | 350000 | 75000 | 52500 |
725 | 282500 | 362500 | 80000 | 56000 |
750 | 290000 | 375000 | 85000 | 59500 |
775 | 297500 | 387500 | 90000 | 63000 |
800 | 305000 | 400000 | 95000 | 66500 |
825 | 312500 | 412500 | 100000 | 70000 |
850 | 320000 | 425000 | 105000 | 73500 |
875 | 327500 | 437500 | 110000 | 77000 |
900 | 335000 | 450000 | 115000 | 80500 |
925 | 342500 | 462500 | 120000 | 84000 |
950 | 350000 | 475000 | 125000 | 87500 |
975 | 357500 | 487500 | 130000 | 91000 |
1000 | 365000 | 500000 | 135000 | 94500 |
1025 | 372500 | 512500 | 140000 | 98000 |
1050 | 380000 | 525000 | 145000 | 101500 |
1075 | 387500 | 537500 | 150000 | 105000 |
1100 | 395000 | 550000 | 155000 | 108500 |
Sales Price = 500 * 1075 (Road bikes) = $537500
Cost = 65000 +300 * 1075 (Road bikes) = $387500
Profit earned = $537500 - $387500 = $150000
Tax = 150000 * 0.3 = 45000
Profit after tax = $150000 - $45000 = $105000
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