MGST391
BUSINESS ANALYTICS
Illustrative Exams & Solutions
Mid-Term & Final
2020-2021
MGST391
Business Analytics
Illustrative Mid-Term Exam Questions
This set of questions is illustrative of the type of questions to expect on the mid-term
exam. As such, this set of questions may not include all the aspects of the course which
are covered on the mid-term. This practice exam is meant to convey exam style and
structure rather than content.
Business Analytics and Excel Modeling
1.
Which of the following is FALSE about modelling?
(a)
(b)
(c)
(d)
2.
Which of the following is not part of the Business Analytics discipline?
(a)
(b)
(c)
(d)
3.
Raw data is of little use by itself; you need to model to make sense of it
The final quality of the analysis improves poor data.
Modelling can help us to understand the Business process
Counter-intuitive results can be very valuable.
Excel Solver
Simulation
Forecasting
Linear Programming
Which of the following Excel functions would be best to determine the distribution of
assigned grades in a course?
(a)
(b)
(c)
(d)
VLOOKUP Exact
VLOOKUP Range
COUNTIF
IF
MGST391 Illustrative Exams & Solutions
Page 1
4.
A feasible solution to a linear programming problem:
(a)
(b)
(c)
(d)
(e)
5.
must satisfy all of the problem's constraints simultaneously
need not satisfy all of the constraints, only some of them
must give the minimum possible cost
must give the maximum possible profit
None of the above is true
Infeasibility in a linear programming problem occurs when:
(a)
(b)
(c)
(d)
(e)
there is an infinite solution set
a constraint is not needed
more than one solution is optimal
there is no solution that satisfies all of the constraints
None of the above is true
6. Which of the following is TRUE?
(a)
(b)
(c)
(d)
Re-Solving an LP problem in Solver will yield the same change to the objective
function as using sensitivity analysis.
Defining integer constraints avoids non-linearity errors.
Iso-profit lines must be parallel to the constraint furthest from the origin.
Non-binding constraints should be removed because they are redundant.
7. The allowable increase and decrease for constraints:
(a)
(b)
(c)
(d)
8.
tells us the change of the right-hand side of a constraint that may be added or
subtracted without changing the profit (objective function value)
tells us the change that may be made in the right-hand side of a constraint before
the shadow price changes
is shown on the sensitivity output of integer programming problems
all of the above are correct
A shadow price is:
(a)
(b)
(c)
(d)
the coefficient of a changing cell
the value of one additional unit of a resource
the maximum change for a resource
always positive for maximization problems
MGST391 Illustrative Exams & Solutions
Page 2
USE THE FOLLOWING INFORMATION TO ANSWER THE NEXT FIVE INDEPENDENT QUESTIONS
9.
How many of the constraints are binding?
(a)
(b)
(c)
(d)
10.
0
1
2
3
Suppose we are required to increase our contract for A to 6 units. How will this impact
our profit?
(a)
(b)
(c)
(d)
No impact
Increases the profit by $10.00
Decrease the profit by $10.00
Cannot tell
MGST391 Illustrative Exams & Solutions
Page 3
11.
Suppose we were offered a 50 kg pallet of material for $350. You must take it all. Should
we take it?
(a)
(b)
(c)
12.
Suppose we could increase our available labour hours to 195 hours. What would our
new profit level be?
(a)
(b)
(c)
(d)
13.
$750 (stay the same)
$1073.75
at most $528, but can’t be determined further with the information given
at least $972, but can’t be determined further with the information given
Which of the following is TRUE about Solver for this problem?
(a)
(b)
(c)
(d)
14.
No, don’t take it
Yes, take it
Cannot tell
The reduced cost is 0 because one does not want any more of a non-binding
constraint.
This is an example of an integer problem.
The complete optimal solution is to make 5 of Product A, 10 of Product B, and 15
of Product C.
The model is linear.
Let X1, X2 and X3 represent binary decision variable where a value of variable Xi equal to
1 represents the performance of project i. Which formulation includes the constraint that
one can only perform project 1 or project 2 (but not both) when project 3 is performed?
(a)
(b)
(c)
(d)
(e)
X1 + X2 < X3
X1 + X2 < 2 X3
X1 < X3 and X2 < X3
Two of the above are correct
None of the above are correct
MGST391 Illustrative Exams & Solutions
Page 4
THE NEXT QUESTION SHOULD BE ANSWERED USING THE FOLLOWING
INFORMATION.
Farmer Sally is raising ostriches to sell at market. She wants to feed the ostriches a diet that
meets all nutritional requirements at minimum cost. The diet will consist of a mixture of the 3
types of feed available. The following table shows the nutrition in 1 kilogram of each type of
feed, along with their costs, and the minimum daily requirement of each nutritional ingredient.
Finally, she does not want more than 40% of the feed to be of type Feed X.
We can develop an algebraic representation of this problem and solve it as a linear program.
Let X represent the number of kilograms of Feed X, Y the number of kilograms of Feed Y, and Z
the number of kilograms of Feed Z that Sally will decide to use for her mixture.
Carbohydrates (g)
Protein (g)
Vitamin A (mg)
Fat (g)
Cost (in cents)
15.
Kilogram of
Feed X
50
15
10
14
Kilogram of
Feed Y
20
70
25
12
Kilogram of
Feed Z
30
40
5
7
21
42
36
Minimum daily
requirement
60
45
35
15 (min)
35 (max)
Formulate this problem as a Linear Programming model:
MGST391 Illustrative Exams & Solutions
Page 5
16.
Mugsy’s Muffins is trying to decide how many batches of their Plain Bran (P), Raisin
Bran (R), Banana Bran (B) and Date Bran (D) muffins to bake. They use bran, eggs,
flour, sugar, baking soda, raisins, bananas, and dates to make their different bran
muffins.
For each batch of muffins, they use:
0.8 kg of bran,
2 eggs,
0.6 kg of flour,
0.25 kg of sugar, and
3 mL of baking soda.
In addition,
Raisin Bran requires ½ kilogram of raisins for each batch,
Banana needs one banana for each batch, and
Date requires ¼ kilogram of dates for each batch.
In addition to the total ingredient limitations, the store will offer its selection of bran muffins with
the following requirements:
at least 4 batches of Banana Bran,
no more than 2 batches of Date Bran,
and no more than 30% of all the bran muffins can be with raisins or dates.
Each batch of bran muffins sells for $9.00 regardless of type. Mugsy wants to maximize his
profit.
Ingredient
Bran
Eggs
Flour
Sugar
Baking Soda
Raisins
Bananas
Dates
Cost
$2.00/kg
$0.10/egg
$1.50/kg
$1.00/kg
$0.01/mL
$1.80/kg
$0.15/banana
$2.40/kg
Supply
200 kg
4 dozen
160 kg
40 kg
500 mL
20 kg
15 bananas
12 kg
Note: although this question was used in a previous exam, the current question is much
shorter. But still just as difficult, if not more so.
MGST391 Illustrative Exams & Solutions
Page 6
Formulate (but do not solve) this problem in standard Linear Programming
format. Do not formulate this as an Excel Solver problem.
MGST391 Illustrative Exams & Solutions
Page 7
USE THE FOLLOWING INFORMATION TO ANSWER THE NEXT 4 QUESTIONS.
17.
Which of the following statements about the Solver solution output is FALSE?
(a)
(b)
(c)
(d)
18.
Produce 400 units of Product 1 at Plant A.
All of the product is produced.
There are 4 changing cells.
Product 2 is made at both plants.
Which of the plants has excess capacity?
(a)
(b)
(c)
Plant A
Plant B
neither
19.
What would occur in the cost minimization problem if both constraints were made to be < ?
(a)
no change
(b)
infeasible
(c)
changing cell values would become 0
(d)
unbounded
20.
What value does the objective function coefficient for Product 1 at Plant B need to be (at
least) in order to have this product scheduled at Plant B in the optimal solution?
(a)
(b)
(c)
10.00
13.00
16.00
MGST391 Illustrative Exams & Solutions
Page 8
21.
Solve the following Linear Programming model using graphical methods.
The Valley Wine Company produces two types of wines – Nectar and Red. The company has
48 tonnes of grapes this season. However, they only have 54 cubic meters of storage space for
ageing and 32 hours of processing time. Solve the LP model to determine the optimal number of
100-liter batches of Nectar and Red to produce to maximize profit.
MAXIMIZE
Subject to
4N
+
3R
(Profit, $000)
4N
6N
4N
+
+
+
6R
6R
2R
N, R
<
<
<
>
48
54
32
0
(grapes, tonnes)
(storage space, meters3)
(processing time, hours)
(non-negativity)
RED
Label the constraints and feasible region clearly. Show solution method (iso-profit for $24,000 or
corner point). Explicitly state the optimal solution.
NECTAR
MGST391 Illustrative Exams & Solutions
Page 9
USE THE FOLLOWING INFORMATION FOR THE NEXT FOUR QUESTIONS:
The Poppa’s Pizza Parlour is a small restaurant catering to patrons with a European taste for
pizza. One of its specialties is Poppa’s Prize pizza. The manager must forecast weekly demand
for these special pizzas so that he can order the special pizza shells weekly. Recent demand
has been as follows:
Week of
June 2
June 9
June 16
June 23
June 30
July 7
22.
Number of Poppa’s Prize Pizzas
50
65
53
56
55
60
Using a four−week moving average, what is the forecast for the week of July 14?
(a)
(b)
(c)
(d)
FJuly 14 < 55
55 < FJuly 14 < 60
60 < FJuly 14 < 65
FJuly 14 > 65
23.
If a four-week weighted moving average were used, what would the forecast for the
week of July 14 be? (The weights are 0.4, 0.3, 0.2, and 0.1, with the highest weight for
the most recent period).
(a)
FJuly 14 < 55
(b)
55 < FJuly 14 < 60
(c)
60 < FJuly 14 < 65
(d)
FJuly 14 > 65
24.
If the forecast of sales for July 7 is 64 and α = 0.2, what is the simple exponential
forecast for July 14?
(a)
FJuly 14 < 55
(b)
55 < FJuly 14 < 60
(c)
60 < FJuly 14 < 65
(d)
FJuly 14 > 65
MGST391 Illustrative Exams & Solutions
Page 10
25.
Assume that a forecast of 56 was used for the 6 weeks. Calculate the following (use
chart below):
(a)
Bias:
(b)
MAD:
(c)
Standard Error:
Yt
50
Ft
56
65
56
53
56
56
56
55
56
60
56
Error = Yt -Ft
|Error|
(Error)2
Total
26.
Which one of the following statements about forecasting is FALSE?
(a)
(b)
(c)
(d)
In order to achieve the objective of developing a useful forecast from the
information at hand, the forecaster must select the appropriate techniques. This
choice sometimes involves a trade-off between forecast accuracy and cost.
There are a number of general types of forecasting techniques used for demand
forecasting, including time series analysis, causal methods, and qualitative
techniques.
Time series express the relationship between the factor to be forecasted and
related factors, such as promotional campaigns, economic conditions, and
competitor's actions.
A time series is a list of repeated observations of a phenomenon, such as
demand, arranged in the order in which they actually occurred.
MGST391 Illustrative Exams & Solutions
Page 11
27.
A pet food retailer has experienced demands for wild birdseed over three years by
trimester as indicated in the table below (in kilograms).
Compute the forecasts for each trimester of Year 4. Determine the seasonal index (to 2
decimal places) and assume a growth rate of 4% of the last three years average for the
trend component.
Year
1
Jan-Apr
4010
May-Aug
2670
Sep-Dec
3070
2
4070
2590
3220
3
4050
2610
3410
Total
Total
Seasonal
Index
Forecast
for Year 4
28.
Which one of the following forecasting methods will generate accurate forecasts when
demands have a consistent trend pattern?
(a)
(b)
(c)
(d)
29.
Simple moving average method.
Weighted moving average method.
Single exponential smoothing method.
None of the above.
Which of the following statements about forecasting is FALSE?
(a)
(b)
(c)
(d)
The method for incorporating a trend in an exponentially smoothed forecast is
called single exponential smoothing.
The sum of absolute forecast errors is useful in measuring the MAD in a forecast.
The standard deviation and the mean absolute deviation measure the dispersion
of forecast errors.
A negative bias in a forecast indicates that the forecast is likely to be higher than
the actual.
MGST391 Illustrative Exams & Solutions
Page 12
30.
Which of the following is not usually a reason for incorrect forecasts?
(a)
(b)
(c)
(d)
(e)
Wrong model is chosen.
Past data is not collected properly.
Unexpected outside events.
The past is not a reflection of what will happen in the future.
All of the above are reasons for incorrect forecasts.
Use this Information to answer the next two questions:
Minnie has just come up with a new invention. Initially, she did not know what her demand
would be, but has tracked its progress over the last number of weeks. She has asked you for
some advice regarding the appropriateness of her forecast, shown in the spreadsheet output
below.
31.
What would be the best thing Minnie could do to improve her forecast?
(a)
(b)
(c)
(d)
32.
verify that the data she has collected is correct
change to a model that includes a seasonality component
change the value of the smoothing constant in her current model
change to a model that includes a trend component
Which of the following would improve the way for Minnie to judge how good her
forecasting model is for these data?
(a)
(b)
(c)
(d)
do a sensitivity analysis to determine how much additional data is needed
use a moving average model
either set the first time period’s forecast equal to the actual demand, just test it
with the last few time periods here, or test it against new data
make the forecast more accurate by increasing the number of decimals by one
MGST391 Illustrative Exams & Solutions
Page 13
MGST391 Illustrative Mid-Term Exam SOLUTIONS
1.
(b)
***The final quality of the analysis improves poor data. – it does not. Because
poor data will result in incorrect or poor decisions even with analytics.
2.
(a)
***Excel Solver - it is a modelling tool
3.
(a)
(b)
(c)
(d)
VLOOKUP Exact (for matching IDs)
VLOOKUP Range (for assigning grades)
*** COUNTIF (if the grades are already assigned, want to count the occurrence
of each letter grade)
IF (could be done but would be cumbersome)
4.
(a)
(b)
(c)
(d)
(e)
*** Correct answer.
The optimal solution must satisfy all the constraints.
Not true in a profit maximization problem.
Not true in a cost minimization problem.
Wrong answer.
5.
(a)
(b)
(c)
(d)
(e)
This means that the solution is unbounded.
This does not affect infeasibility.
This means that there are multiple solutions not infeasible ones.
*** Correct answer.
Wrong answer.
6.
(a)
*** Re-Solving an LP problem in Solver will yield the same change to the
objective function as using sensitivity analysis – TRUE
Defining integer constraints avoids non-linearity errors – non-sensical statement.
Iso-profit lines must be parallel to the constraint furthest from the origin – usually
not parallel to constraints.
Non-binding constraints should be removed from consideration because they are
redundant. – False since non-binding constraints can become binding if a
resource or objective function changes.
(b)
(c)
(d)
7.
(a)
(b)
(c)
(d)
is incorrect as RHS changes in scarce resources will change the objective
function value.
*** correct answer
there is no sensitivity output for an integer problem
wrong answer
MGST391 Illustrative Exams & Solutions
Page 14
8.
(a)
(b)
(c)
(d)
the objective function value (coefficient)
*** correct answer
the range of values for which a shadow price is valid
is false as the sign (+ or -) of the shadow price depends on the direction of the
equation (< or >).
9.
(d)
The solution is 3 since all of the constraints are binding (final value = Constraint
R. H. Side or shadow price is not 0)
10.
(c)
We do not want to make product A. For every item we make above the
minimum, the impact on the profit is the shadow price of –$10.00.
11.
(a)
We must compare the cost and benefit. 50 kg is more than we want (allowable
increase), but we would have to take it all. However, we will only use 40 kg with
a profit of $8.00 for every extra kg of product so the additional profit would be
$320. This is less than the $350 cost. So, the conclusion is not to take it.
12.
(d)
The shadow price of 9.25 is valid to an increase of 24 units ($222) for an
increase of profit to $972 (750 + 222). We cannot tell what the impact is on the
profit beyond the allowable increase.
13.
(a)
(b)
Definition is for shadow price, not reduced cost.
The solutions are integer but not defined as a constraint. If it was an integer
problem, at least one of the changing cells would be identified as integer in the
Solver Parameter constraint box. As well, there would not be a Sensitivity report.
The optimal solution must also include the optimal profit (value of the target cell).
*** correct answer
(c)
(d)
14.
Let X1, X2 and X3 represent binary decision variable where a value of variable Xi equal to
1 represents the performance of project i. Which formulation includes the constraint that
one can only perform project 1 or project 2 (but not both) when project 3 is performed?
(a)
(b)
(c)
(d)
(e)
X1 + X2 < X3 Correct Project 1 and 2 can never be performed at the same time
X1 + X2 < 2 X3 No because X3 = 1 when X1 = X2 = 1
X1 < X3 and X2 < X3 No because X3 = 1 when X1 = X2 = 1
Two of the above are correct
None of the above are correct
X1
0
0
1
1
X2
0
1
0
1
X3
0/1
1
1
-
MGST391 Illustrative Exams & Solutions
Page 15
15.
Decision Variables:
X = amount of feed X to feed (in kg)
Y = amount of feed Y to feed (in kg)
Z = amount of feed Z to feed (in kg)
MINIMIZE
21 X + 42 Y + 36 Z
subject to
Carbohydrates (g)
50 X + 20 Y + 30 Z > 60
Protein (g)
15 X + 70 Y + 45 Z > 45
Vitamin A (mg)
10 X + 25 Y + 5 Z > 35
Fat minimum (g)
14 X + 12 Y + 7 Z > 15
Fat maximum (g)
14 X + 12 Y + 7 Z < 35
Max Feed X (kg)
Non-negativity
X
< 0.40 (X+Y+Z)
X, Y, Z > 0
Note that this question is easier than the actual question on the exam.
MGST391 Illustrative Exams & Solutions
Page 16
16.
Solution for Mugsy’s Muffins:
Note that this question is longer than the actual question on the exam.
17.
(a)
(b)
(c)
(d)
True – Changing cell C9 has a final value of 400.
True – The production used = production available for all three products
False – There are 6 changing cells (but 2 are zero in the optimal solution)
True – Product 2 has 200 units made at Plant A and 150 units made at Plant B.
18.
(b)
B has 500 capacity used compared to 650 capacity available
19.
(c)
The values would go to 0. It is feasible since all of the values are less than or
equal to the availability. It would also minimize cost at $0. However, this is not
the information that you would be looking for.
MGST391 Illustrative Exams & Solutions
Page 17
20.
(a)
21.
MAXIMIZE
Subject to
calculated from Objective Coefficient – Reduced Cost = 13.00 – 3.00 = 10.00
4N
+
3R
(Profit, $000)
4N
6N
4N
+
+
+
6R
6R
2R
N, R
<
<
<
>
48
54
32
0
(grapes, tonnes)
(storage space, meters3)
(processing time, hours)
(non-negativity)
Label the constraints and feasible region clearly. Show solution method (iso-profit for $24,000 or
corner point). Explicitly state the optimal solution.
Co-ordinates
(N,R)
(N= 0 , R= 0)
(N= 0, R=8 )
(N = 3, R = 6)
Profit
(N= 7, R = 2)
34
0
24
30
RED
(N=8, R=0)
32
The optimal solution is to make
7 – 100 litre batches of nectar
and 2 – 100 litre batches of red
for a maximum profit of $34,000.
Optimal solution
NECTAR
MGST391 Illustrative Exams & Solutions
Page 18
22.
(b)
Forecast = 60 + 55 + 56 + 53 = 56
4
23.
(b)
Forecast = 60(0.4) + 55(0.3) + 56(0.2) + 53(0.1) = 57
24.
(c)
ForecastJuly 7 = 64
Ft+1 = α At + (1 – α) Ft where α = 0.2
ForecastJuly 14 = 0.2(60) + 0.8(64) = 63.2
25.
Yt
50
65
53
56
55
60
Total
Ft
56
56
56
56
56
56
Error = Yt -Ft
-6
9
-3
0
-1
4
3
|Error|
6
9
3
0
1
4
23
(Error)2
36
81
9
0
1
16
143
Total
9750
9880
10070
29700 = 3300
9
Growth:
4%*3300
= 132
Bias = sum et = 3 = 0.5
n
6
MAD = sum | et | = 23 = 3.8
n
6
SE = SQRT [sum (et )2 / n ]= SQRT (143/6) = 4.9
26.
(c)
is referring to causal forecasting.
27.
Year
1
2
3
Total
Jan-Apr
4010
4070
4050
12130
May-Aug
2670
2590
2610
7870
Sep-Dec
3070
3220
3410
9700
Seasonal Index
4043.3
3300
= 1.23
(3300+132)*
1.23 =
4221
2623.3
3300
= 0.79
3300+2(132)*
0.79 =
2816
3233.3
3300
= 0.98
3300+ 3(132)*
0.98 =
3622
Forecast for
Year 4
MGST391 Illustrative Exams & Solutions
Page 19
28.
(d)
a consistent trend pattern implies a steady upward or downward state in the form
of a line; the three averaging methods listed will lag behind the trend.
29.
(a)
single (or simple) exponential smoothing does not involve trend
30.
(e)
*** all of the above are reasons for incorrect forecasts
31.
(b)
***change to a model that includes a seasonality component
32.
(c)
***either set the first time period’s forecast equal to the actual demand, just test it
with the last few time periods here, or test it against new data
MGST391 Illustrative Exams & Solutions
Page 20
MGST391
Business Analytics
Illustrative Final Exam Questions
This set of questions is illustrative of the type of questions to expect on the final exam.
As such, this set of questions may not include all the aspects of the course which are
covered on the final exam. This practice exam is meant to convey exam style and
structure rather than content.
1.
Draw a decision tree for Jack’s problem below. At each decision node, clearly label each
alternative. Include probabilities at all chance nodes. DO NOT try to SOLVE this decision
tree – just draw the tree and label it appropriately.
Jack has built a new cottage near the lake and needs water. He is trying to decide if he
should connect his cottage to the nearby town’s water supply or drill a well. If he drills
(without any test), he expects to be successful 75% of the time.
Jack can also hire a firm to do further testing on his property. This would cost him $800.
He expects the test results to be favourable about 60% of the time. In either case, he
would still need to decide to drill a well or connect to the town. If the test results are
favourable, he expects his drilling to be successful 80% of the time. If the test results are
unfavourable, he expects the drilling results to be unsuccessful about 90% of the time.
MGST391 Illustrative Exams & Solutions
Page 21
2.
A retailer is trying to determine whether to order 1, 2, or 3 magazines. The payoff table
for profit is:
Demand
(# of magazines)
Ordering
Decision
1
2
3
1
2
-2
-3
2
-1
4
0
3
-3.5
2
6
The probabilities of a demand of 1, 2, or 3 magazines is 0.4, 0.25 and 0.35 respectively.
What is the retailer’s best decision, based on the MAXIMAX, MAXIMIN, MINIMAX
Regret, and EMV criterion?
MGST391 Illustrative Exams & Solutions
Page 22
3.
Decision trees are particularly useful when:
(a)
(b)
(c)
(d)
4.
The expected value of sample information (EVSI) can be used to
(a)
(b)
(c)
(d)
5.
establish a maximum amount to spend on getting information
calculate conditional probabilities
establish risk avoidance
none of the above
An insurance company plans to market a particular type of term life insurance. The
premium that will be paid for this depends on the expected amount that the insurance
company will have to pay out on each policy plus any cost and profits. Assume the
company wishes to make $50 profit on each policy and there is $40 overhead cost. The
probability a person in a certain age group would die in the coming year is 0.001. How
much would the company charge for a $100,000 life insurance policy?
(a)
(b)
(c)
(d)
(e)
6.
perfect information is available
formulating a conditional values table
a sequence of decisions must be made
none of the above
90
100
190
999
none of the above
A company is trying to determine if they should open a new café or expand their current
café. Their situation is shown in the decision tree below.
What is the EVPI?
(a)
(b)
(c)
(d)
12,900
23,000
120,000
180,000
MGST391 Illustrative Exams & Solutions
Page 23
Complete the Bayesian probability tree for this situation (10 values) and
answer the four questions that follow.
The marketing manager of a toy manufacturing company is considering marketing a new toy. In
the past, 35% of the toys introduced by this company have been successful and 65% have
been a failure. Before the toy is marketed, market research is conducted and a report, either
favourable or unfavourable, is compiled. In the past, 80% of the successful toys received a
favourable market research report, and 30% of the failed toys received a favourable market
research report.
13.
What is the probability of a toy being successful or unsuccessful?
a.
b.
c.
d.
e.
14.
0.00
0.20
0.40
0.60
0.80
<
<
<
<
<
p
p
p
p
p
< 0.20
< 0.40
< 0.60
< 0.80
< 1.00
What is the probability that the report is unfavourable given that the toy was
unsuccessful?
a.
b.
c.
d.
e.
0.00
0.20
0.40
0.60
0.80
<
<
<
<
<
p
p
p
p
p
<
<
<
<
<
0.20
0.40
0.60
0.80
1.00
MGST391 Illustrative Exams & Solutions
Page 26
15.
What is the probability that a toy will be successful and it receives a favourable report?
a.
b.
c.
d.
e.
16.
p
p
p
p
p
< 0.20
< 0.40
< 0.60
< 0.80
< 1.00
0.00
0.20
0.40
0.60
0.80
<
<
<
<
<
p
p
p
p
p
<
<
<
<
<
0.20
0.40
0.60
0.80
1.00
Which of the following is true?
(a)
(b)
(c)
(d)
18.
<
<
<
<
<
What is the probability that a toy will be unsuccessful and it receives an unfavourable
report?
a.
b.
c.
d.
e.
17.
0.00
0.20
0.40
0.60
0.80
In simulation, we always assume that the arrival rates are normally distributed.
For randomly generated numbers like service time in a line-up simulation, it is not
possible for a randomly chosen value to immediately follow itself. For example, a
"5" minute service time could not follow another "5" minute service time.
The goal of simulation is to create an environment in which it is possible to obtain
information about alternative actions via experimentation.
Using simulation is typically less work than using Queuing Theory
When simulating demand using @RISK, the average simulated demand over a large
number of iterations would be an estimate of the:
(a)
(b)
(c)
(d)
expected demand
maximum possible demand
sampled demand
daily demand
MGST391 Illustrative Exams & Solutions
Page 27
19.
Dave’s Candies is a small family-owned business that offers gourmet chocolates and ice
cream fountain service. For special occasions, such as Valentine’s Day, the store must
place orders for special packaging boxes of chocolates several weeks in advance from
their supplier. One product, the Valentine’s Day Chocolate Delight, is bought for $7.50
per box and sells for $12.00. Any boxes not sold by February 14 are discounted by 50
percent and can always be sold easily. Historically, Dave’s Candies sold has followed a
normal distribution of 65 boxes with a standard deviation of 10 boxes each year. Dave’s
dilemma is deciding how many boxes to order for Valentine’s Day customers. If demand
exceeds the purchase quantity, then Dave loses profit opportunity. On the other hand, if
too many boxes are purchased, he will lose money by discounting them below cost.
Fill in the chart above.
Identify the following cells: Distribution: ________ Parameter: ______ Output: ______
20.
What are the main purpose of the Fixed Seed Value in @RISK?
___________________________________________________________________
21.
What is the purpose of confidence intervals?
(a)
(b)
(c)
(d)
To make you do more work (!)
To increase your confidence in the answer
To give you a range in which the population mean lies
To assign a confidence value to a result
MGST391 Illustrative Exams & Solutions
Page 28
Some simulation output is shown below. Use it to answer the next 2 questions.
22.
What is the 95% confidence interval for the Cost for the above example to 4 decimal
places? Note that this data is in 00s.
Lower 95% C.I. ______________
23.
Upper 95% C.I. _______________
What is the 90% confidence interval for the probability that the Cost will be less than or
equal to $13 (in % to 2 decimal places)?
Lower 90% C.I. ______________%
MGST391 Illustrative Exams & Solutions
Upper 90% C.I. ______________ %
Page 29
24.
Which of the following is TRUE regarding empirical distributions?
(a)
(b)
(c)
(d)
They are ‘named’ distributions (e.g., normal, exponential)
They systematically overestimate real variability
They convert observed data into a probability distribution
None of the above are true
25.
Which of the following is a disadvantage of using simulation?
(a)
they allow for the study of "what-if" questions
(b)
each simulation requires a unique model
(c)
allows the study of interaction of components or variables to determine which are
important
26.
Which queuing model best represents getting service from a cafeteria?
(a)
(b)
(c)
(a)
27.
What is the fundamental reason line-ups occur?
(a)
(b)
(c)
(d)
28.
single-server, single-phase
single-server, multi-phase
multi-server, single-phase
multi-server, multi-phase
people are not distracted
too few skilled servers
temporary mismatch between supply and demand
because people don’t complain
Jones, the mechanic at Golden Muffler Shop, is able to install new mufflers at an
average rate of three per hour (or about one every 20 minutes), according to a negative
exponential distribution. Customers seeking this service arrive at the shop on the
average of two per hour, following a Poisson distribution. The customers are served on a
first-in, first-out bases and come from a very large population of possible buyers.
The owner of Golden Muffler Shop, estimates that the cost of customer waiting time, in
terms of customer dissatisfaction and lost goodwill, is $10 per hour spent waiting in line.
a) What are the values of λ and µ per hour?
λ =
µ =
b) On average, how many cars are in the system?
MGST391 Illustrative Exams & Solutions
Page 30
c) What is the average waiting time in the system?
d) On average, how many cars are waiting in line?
e) What is the average waiting time per car?
f)
What is the utilization rate of Jones?
g) What is the probability that there at no cars at the shop?
h) How many cars are serviced in a typical 8 hour working day?
i)
What is the average waiting-time cost for Golden Muffler Shop?
MGST391 Illustrative Exams & Solutions
Page 31
29.
Consider a single-server line-up in a store. A customer will not wait for more than 10
minutes in the line-up. If the service start time is more than 10 minutes after they arrive,
cross them off and have the next person in line be served. Assume no customers are in
the store at 9:00. Complete the chart below.
Customer
Interarrival
Time
(minutes)
Arrival
Time
(hh:mm)
Leave?
(Yes/No)
Service
Time
(minutes)
Service
Start Time
(hh:mm)
Annelies
4
9:04
No
8
9:04
Benson
11
9:15
Carolyn
2
9
David
5
14
Erin
8
11
Fay
7
17
MGST391 Illustrative Exams & Solutions
Service
End Time
(hh:mm)
Waiting
Time (if
they do
not leave)
16
Page 32
30. Solve a DecisionTree
Kim Fleming has a race horse named Cocoa Futures; she is trying to decide whether to
have the horse race for another season or not. If the horse does not race, it will be sold
immediately. Another choice for Kim is to send Cocoa Futures to a new trainer who
would evaluate the horse at a cost of $5,000 and then make a recommendation as to
whether or not to continue racing. (Note that this cost IS included on the end nodes of
the decision tree.) Solve the decision tree below. Ensure to prune the
tree and to highlight the optimal strategy.
a) What is the proper course of action for Kim to take?
b) What is the EMV of the optimal solution?
c) What decision should Kim make if the horse was not evaluated by a trainer?
d) What is the expected profit if the trainer evaluation is not available?
e) If the horse was checked by a trainer and the trainer recommended ‘do not continue’
but Kim decided to race for the season, what is the expected profit?
f) What is the expected value of sample information (EVSI)?
g) What is the expected value of perfect information (EVPI)?
MGST391 Illustrative Exams & Solutions
Page 33
$1,245,000
MGST391 Illustrative Exams & Solutions
Page 34
MGST391 Illustrative Final Exam SOLUTIONS
1.
MGST391 Illustrative Exams & Solutions
Page 35
2.
The Payoff Tables:
MGST391 Illustrative Exams & Solutions
Page 36
3.
(c)
a sequence of decisions must be made
4.
(a)
comparing the value to the cost of information
5.
(c)
The expected payout per policy is 0.001x100000 = $100. The overhead plus
profit is $90. So they need to charge $190.
6.
(a)
MGST391 Illustrative Exams & Solutions
Page 37
Solved Decision Tree for Questions 7 to 11:
7.
(a)
EMV for Large = -100, Small = -35, None = -20
8.
(b)
EMV for Large = 30.4, Small = 26.4, None = 0
9.
(a)
EMV = 40.6
10.
(e)
The course of action has to specify entire sequence. a), b) and c) don’t do that.
d) is wrong because if the study is negative, doing nothing (EMV = -20) is better
than building a large plant (EMV = -100) or small plant (EMV = -35). If the study
is positive the best option is a large plant with an EMV = 101.6
11.
(a)
EMV = -20
12.
(d)
EVSI assumes sample information is free, i.e., we have to add back the survey
cost to EV with SI. So EVSI = $63,000 – 50,000 = $13,000
MGST391 Illustrative Exams & Solutions
Page 38
See diagram at bottom of page.
13.
1.00 (all of the possibilities)
14.
0.70
15.
0.28
16.
0.455
17.
(c)
The goal of simulation is to create an environment in which it is possible to obtain
information about alternative actions via experimentation.
18.
(a)
expected demand [mean or average value]
0.8 * 0.35 = 0.28
0.2 * 0.35 = 0.07
MGST391 Illustrative Exams & Solutions
Page 39
19.
Distribution Cell: B3 Parameter Cell: B6 Output Cell: C11
20.
The purposes of the Seed value in @RISK is: 1) Specifying a non-zero RNG seed
@RISK causes repeat runs to use the same random numbers, which makes the results
of multiple model runs more comparable.
21.
(c)
22.
CI for mean: sample mean + z-value * mean standard error where MSE = sample
standard deviation / square root of the number of observation
Lower 95% CI: 11.6998 – 1.96 * (1.5494 / sqrt (1000)) = 11.6038
Upper 95% C.I.: 11.6998 + 1.96 * (1.5494 / sqrt (1000)) = 11.7958
23.
CI for proportion is p-bar + z * sqrt(p-bar * (1- p-bar)/n
Lower 90% C.I.: 0.778 - 1.645 * SQRT (0.778(0.222)/1000) = 0.7564 = 75.64%
Upper 90% C.I.: 0.778 + 1.645 * SQRT (0.778(0.222)/1000) = 0.7996 = 79.96%
24.
(a)
(b)
(c)
Should be theoretical
Should be underestimate
*** Correct answer
25.
(b)
*** Correct answer
26.
(b)
one server per station, many stations
27.
(c)
variability in arrivals and service rates
To give you a range in which the population mean lies *** within a certain
confidence level
MGST391 Illustrative Exams & Solutions
Page 40
28.
MGST391 Illustrative Exams & Solutions
Page 41
29.
Customer
Interarrival
Time
(minutes)
Arrival
Time
(hh:mm)
Leave?
(Yes/No)
Service
Time
(minutes)
Service
Start Time
(hh:mm)
Service
End Time
(hh:mm)
Waiting
Time (if
they do
not leave)
Annelies
4
9:04
No
8
9:04
9:12
0
Benson
11
9:15
No
16
9:15
9:31
0
Carolyn
2
9:17
Yes
9
David
5
9:22
No
14
9:31
9:45
9
Erin
8
9:30
Yes
11
Fay
7
9:37
No
17
9:45
10:02
8
MGST391 Illustrative Exams & Solutions
Page 42
$930,000
MGST391 Illustrative Exams & Solutions
Page 43
$1,245,000
MGST391 Illustrative Exams & Solutions
Page 44