WILFRID LAURIER UNIVERSITY
WATERLOO, ONTARIO
BUSINESS/ECONOMICS 275
FINAL EXAMINATION (SOLN), WINTER 2002
20-8-02
TIME:
3 hours (180 minutes)
PAGES:
16
INSTRUCTORS:
Dr. R. Craig
Dr. S. Isotupa
Prof. K. Li
Prof. A. Marshall
Prof. K Raftis
INSTRUCTIONS:
1. Total Mark Value: 110
Number of questions: 7
Number of pages: 16 + queueing formulas (separate handout)
2. 1 page ( 8 ½ x 11 inches, double sided) notes are permitted.
3. Standard Calculators only are permitted.
4. Budget your time carefully.
5. Answers are to be given in the space provided. However, should you require additional
space for a complete answer, use the blank page (page 16) attached for this purpose.
6. For all problems where calculation space is provided, show your reasoning and work.
Unsubstantiated answers will usually receive no marks.
7. You are to stop writing immediately upon being told that the exam is over.
TOTAL MARKS
Question 1
30
Question 2
15
Question 3
13
Question 4
12
Question 5
16
Question 6
12
Question 7
12
TOTAL
110
MARKS RECEIVED
Page 2
Question #1 ( 30 Marks)
Circle the correct (better/best) answer, or fill in the blanks.
TRUE/FALSE (1 mark each)
T F 1. For a transportation problem to be feasible the demand must be greater than the
supply.
T F 2. Reneging is the act of refusing to join a queue.
T F 3. A customer who arrives at a queuing location where the utilization factor is 0.5 will
never have to wait in line.
T F 4. Using the AHP approach, if I prefer A over C by a value of 6, then I prefer C over A
by a value of -6.
T F 5. In goal programming all goals with the same priority level must have the same
importance.
T F 6. Introducing a new constraint to a solved LP always requires that you re-solve the LP.
T F 7. Model verification identifies whether or not the model performs as intended.
T F 8. In a single-channel waiting line model, the system works most cost efficiently when
ρ=λ/μ=1
T F 9. Little’s flow equations only work for M/M/1 and M/M/k waiting line models.
T F 10. In waiting line situations, a decision maker usually wants to balance the service
quality and service cost.
MULTIPLE CHOICE (1 mark each)
1. Which of the following is not a principle of the AHP method?
(a)
Hierarchies
(b)
Different scoring scales possible
(c)
Logical Consistency
(d)
Priorities
2. When adding deviation variables to a ‘greater than’ constraint, the new objective is to:
(a)
minimize u > 0
(b)
minimize v > 0
(c)
minimize both u > 0 and v > 0
(d)
minimize u < 0
3. Which of the following is not part of the basic framework of a scoring model?
(a)
goal(s)
(b)
alternatives
(c)
criteria
(d)
predictions
4. In a simulation, the cost of a particular part follows a uniform continuous distribution,
between $43 and $47. For a random number of 0.2187, what is the part cost?
(a)
$46.13
(b)
$43.87
(c)
$41.91
(d)
$44.10
(e)
None of the above
Page 3
5. An approach to decision making that recommends the alternative that provides the best of the
worst possible payoffs, without using probabilities, is:
(a)
Optimistic
(b)
Expected Value
(c)
Pessimistic
(d)
Minimax Regret
6. Which of the following techniques does not deal with probabilistic inputs?
(a)
Utility and Decision Making
(b)
Simulation
(c)
Linear Programming
(d)
Markov Processes
7. The Excel function NORMDIST(1/2,1/2,1/3,1) equals
(a)
0.167
(b)
0.333
(c)
0.5
(d)
none of the above
8. The 100% rule in linear programming applies when:
(a)
Changing multiple objective function and RHS coefficients together.
(b)
Changing multiple objective function or RHS coefficients together.
(c)
Reduced costs are zero.
(d)
Dual prices are zero.
(e)
None of the above.
9. Rounding the solution to an LP Relaxation to the nearest integer value provides:
(a)
A feasible but not necessarily optimal integer solution.
(b)
An integer solution that is optimal.
(c)
An integer solution that may be neither feasible nor optimal.
(d)
An infeasible solution.
10. The solution to the LP Relaxation of a maximization integer linear program provides:
(a)
An upper bound for the value of the objective function.
(b)
A lower bound for the value of the objective function.
(c)
An upper bound for the value of the decision variables.
(d)
A lower bound for the value of the decision variables.
11. In order to change a “ ≤ ” constraint to an equality, one can:
(a)
add a slack variable
(b)
add a surplus variable
(c)
subtract a slack variable
(d)
subtract a surplus variable
12. You learn decision analysis by:
(a)
memorizing questions and answers from past exams
(b)
doing decision analysis
(c)
cramming the night before the midterm and final
(d)
studying statistics
(e)
it is impossible to learn decision analysis
13. Which of the following LP special cases does not require reformulation of the problem in
order to obtain a solution?
(a)
alternate optimality
(b)
infeasibility
(c)
unboundedness
(d)
all of the above cases require a reformulation.
14. A negative shadow price for a constraint in a maximization problem means
(a)
as the right-hand side increases, the objective function value will increase.
(b)
the constraint is non-binding.
(c)
the constraint is of < = type
(d)
the constraint is of > = type.
Page 4
15. The constraint 5x1 + 3x2 < 150 is modified to become a goal equation, and priority one is to
avoid overutilization. Which of the following is appropriate?
(a)
5x1 + 3x2 + u = 150
(b)
5x1 + 3x2 - v = 150
(c)
Min u : 5x1 + 3x2 + u - v = 150
(d)
Min v: 5x1 + 3x2 + u - v = 150
16. The difference between the transportation and assignment problems is that
(a)
total supply must equal total demand in the transportation problem
(b)
the number of origins must equal the number of destinations in the transportation
problem
(c)
each supply and demand value is 1 in the assignment problem
(d)
there are many differences between the transportation and assignment problems
Fill in the Blanks: (2 marks each)
1. The probability that a customer will make a purchase this week depends only on whether or
not she purchased last week, and can be defined as a Markov process. If the customer
purchased last week, the probability she will purchase this week is 0.7. We also know that
the steady state probability of purchasing on any given week is 0.4. Determine the following
probabilities
(1.a) P(will not purchase next week | purchased this week):
0.3
(1.b) P(will purchase next week | no purchase this week):
0.2
Calculation Area:
p(SS purchase) = 0.4, so p(SS no purchase) = 1 - 0.4 = 0.6
The transition probability matrix is given by
Purchase
No purchase
Purchase
0.7
0.3
No purchase
x
1-x
(1 mark)
(1 mark)
p(SS purchase) (0.7) + p(SS no purchase) (x) = p(SS purchase)
(0.4)(0.7) + (0.6)x = 0.4
0.28 + 0.6 x = 0.4
0.6 x = 0.12
x = 0.2
And 1 - x = 0.8
2 marks each if all correct
Partial marks possible if work shown (see above)
Total #1: ________
Page 5
Question #2 (15 Marks)
The local supermarket, Zyhrs, is reviewing staffing for the Deli counter during the weekday
shift. Currently, Zyhrs employs one cashier during this shift. Analysis shows that customers
arriving at the Deli counter can be adequately modeled with a Poisson distribution having an
average rate of 1 person per 3 minutes. Service time follows an exponential distribution, taking
an average of 2 minutes to serve a customer.
FOR THE FOLLOWING QUESTIONS, PLEASE KEEP ALL CALCULATIONS CORRECT
TO THREE SIGNIFICANT DECIMAL PLACES. ALL QUESTIONS BELOW ASSUME THE
TIME PERIOD DURING THE WEEKDAY SHIFT. PUT YOUR ANSWERS INTO THE
BOXES BESIDE THE QUESTIONS.
Calculation Area (Unsupported answers will not be credited)
M/M/1 model with λ = 20 customers/hour and μ = 30 customers/hour; ρ = 2/3
25 ⋅ e −2
(a) λ = 2 customers/6 min, so P(x = 5) =
= 0.036 = POISSON(5, 2, 0)
5!
(b) Point probability for a continuous distribution is always 0.
(c) P(1min < x < 3min|μ = 30/hr)) = P(x < 3min) - P(x < 1min) = P(x < 3/60h)-P(x <
−30⋅ 3
−30⋅ 1
−1
−3
1/60h) = (1 − e 60 ) − (1 − e 60 ) = e 2 − e 2 = 0.383 = EXPONDIST(3, 1/2, 1) –
EXPONDIST(1, 1/2, 1); or EXPONDIST(1/20, 30, 1) – EXPONDIST(1/60, 30, 1)
(d) Lq =
(e) Wq =
(a)
λ2
202
4
=
= = 1.333 customers
μ ( μ − λ ) 30(30 − 20) 3
Lq
λ
=
20
=
1
hours (= 0.067 hours) = 4 min (or 4.02 min)
15
What is the probability that exactly 5 customers shop at the Deli during the next 6
minutes. Show the calculated value or the EXCEL expression.
(1 mark)
0.036
(b)
4
3
POISSON(5, 2, 0)
What is the probability that a customer is served in exactly 2 minutes? Show the
calculated value or the EXCEL expression.
(1 mark)
0
Can’t calculate a point probability for a continuous rv
(c)
What portion of customers are served in between 1 and 3 minutes? Show the
calculated value or the EXCEL expression.
(1 mark)
EXPONDIST(3, 1/2, 1) – EXPONDIST(1, 1/2, 1); or
0.383
EXPONDIST(1/20, 30, 1) – EXPONDIST(1/60, 30, 1)
(d)
What is the average number of customers waiting to be served?
(1 mark)
(e)
What is the average time a customer must wait before he/she
can be served? (1 mark)
1.333
4 min or
1/15 hours
Page 6
In order to improve service, Zyhrs is considering employing one more cashier (at the same cost
rate as its current employee, $10/hour for wages and benefits). Assume the new cashier is as
experienced as the current one, and her service time follows the same distribution. In addition, in
order to improve operating efficiency, a single waiting line will be formed and the first customer
in the line will be served by the first available cashier. Management estimates customer waiting
costs (including serving time) at $8/hour/customer
Calculation Area (Unsupported answers will not be credited):
M/M/2 model with λ = 20 customers/hour and μ = 30 customers/hour and k = 2,
(f) P0 =
λ 2
=
μ 3
1
1
= = 0.5
( )
2 ⋅ 30
2
(1 + ) +
⋅
1!
2! 2 ⋅ 30 − 20
(g) Lq =
(h) Wq =
2 2
3
2
3
( 23 ) 2 × 20 × 30
1
⋅ P0 = = 0.083
2
(2 × 30 − 20)
12
Lq
λ
=
1
12
20
=
1
hours = 0.25 min
240
(i) For single-channel model L = Lq +
For two-channel model L = Lq +
λ 4 2
= + =2
μ 3 3
λ 1 2 9
= + = = 0.75
μ 12 3 12
Cost1 = 2*8 + 10 = 26
Cost2 = 0.75*8 + 2*10 = 26
(f)
What is the probability that neither of the cashiers are working
at any point in time? (2 marks)
0.5
(g)
What is the average number of customers waiting to checkout?
(2 marks)
0.083
(h)
What is the average time that a customer has to wait before
she/he can be served by either cashier? (2 marks)
1/240 hrs,
0.25 min
(i)
What is the net savings (or net loss) in hiring two cashiers
compared to hiring one cashier each weekday (between 9:00am
and 5:00pm)? (4 marks)
$0
Total #2: _______
Page 7
Question #3 (13 marks)
Two Guys and a Gyro is a new venture being contemplated by two young business students for
the upcoming summer. The idea is a street vendor, who sells Gyros and Souvlaki prepared fresh,
similar to the street vendors selling hotdogs and sausages. The young entrepreneurs, well versed
in the art of management science, sought the use of their quantitative skill to improve their
operations. At present they are struggling with how much preparation and precooking they
should perform. Precooking would drastically reduce service times, but may reduce quality and
distort their ‘fresh’ image. They initially estimated customers would arrive at their stand at an
average rate of one every two minutes during peak periods (11 am to 3 pm) and assumed that
inter-arrival times, as well as service times followed an exponential distribution. One student
would operate the stand at any given time.
(3.a)
What is the largest average service time they could contemplate if they hoped to have a
stable system – i.e., reach some sort of steady state condition? (1 mark)
< 2 min.
Okay: μ > 0.5, λ / μ < 1, utilization factor < 1
Incorrect if they confuse IAT and arrival rate
(3.b)
They are allowing for an average service time of 90 seconds.
(i) How long would the average customer spend waiting for his/her Gyro? (1 mark)
λ = ½ (per min), μ = 1/1.5 = 2/3 (per min)
30 (per hr)
40 (per hr)
W = 1 / (μ − λ ) = 1 / (2/3 - 1/2 ) = 1 / (1/6) = 6 min
= 1 / (40 - 30) = 1 / 10 = 0.1 hr
Must have λ and μ correct
(ii) How many people should they expect to see at the stand at any given time? (1
mark)
L = λ * W = ½ * 6 = 3 people
or L = λ / (μ − λ) = ½ / (2/3 - ½) = 3 people
Okay if formula correct & wrong values carried forward from i)
(iii) What is the probability that an arriving customer would have to wait for someone
else to be served? (1 mark)
Pw = λ / μ = ½ / (2/3) = 3/4 = 0.75
or 1 - P0 = 1 - .........................
Okay if formula correct & wrong values carried forward from ii)
(3.c)
After being in operation for a few weeks they noticed that customers seemed to arrive at a
very uniform rate. They collected some data and found that the time between customer
arrivals was uniformly distributed over the range 0 to 2 minutes. They also observed that
their average service time was actually closer to 30 seconds versus 90 (still exponentially
distributed).
What is the new average arrival rate? (1 mark)
U ( 0, 2 ) -- mean is 1, so λ = 1 customer/min (or 60/hr); accept 1 min.
Page 8
(3.d) With this data in mind the Two Guys decided to simulate their operation rather than rely on
standard waiting line equations. The following table shows how they set up their
simulation model. Complete the table, calculating how long each of the first five
customers in the simulation waits for lunch. (4 marks)
TWO GUYS & A GYRO SIMULATION
All times in minutes
Customer Random
Number
1
2
3
4
5
0.60
0.06
0.86
0.95
Interarrival
Time
0
1.2
0.12
0.12
1.72
1.9
Arrival
time
Random
Number
Service
Time
Exit
System
0
1.2
1.32
3.04
4.94
0.03
0.69
0.39
0.86
0.62
1.75
0.19
0.47
0.08
0.24
1.75
1.94
2.41
3.12
5.18
Time to
get
Lunch
1.75
0.74
1.09
0.08
0.24
Calculation Area:
1 ½ marks for each correct column
- ½ each column error (to column max)
Must understand server can’t start new customer until current customer service completed
Explain how the Interarrival Times were calculated: (2 marks)
Limit your answer to 15 words:
Continuous Uniform distribution between 0 and 2; use formula 0 + 2*rand()
Explain how the service time was calculated: (2 marks)
Limit your answer to 15 words:
Exponential service times, with μ = 0.5 min; use Excel expression = − 0.5*ln(rand()); use
random numbers to generate random variates for an Exponential distribution with mean
0.5 min.
Total #3: _______
Page 9
Question #4 (12 marks)
A toy manufacturer is planning to produce new toys. The setup cost of the production facilities
and the unit profit for each toy are given below:
Toy Setup cost ($) Profit ($)
1
45000
12
2
76000
16
The company has two factories that are capable of producing these toys. In order to avoid
doubling the setup cost only one factory should be used for each toy. However one factory
could produce both toys.
The production rates of each toy are given below (in units/hour):
Toy 1 Toy 2
min/toy
Factory 1
52 38
1.154 1.579
Factory 2
42 23
1.429 2.609
hr/toy
.0192 .0263
.0238 .0435
Factories 1 and 2, respectively, have 480 and 720 hours of production time available for the
production of these toys. The manufacturer wants to know which of the new toys to produce,
where and how many of each (if any) should be produced so as to maximize the total profit.
Using appropriate decision variables (including 0-1 type) formulate the above problem as an
integer program.
Decision Variables & units:
We need to decide whether to setup a factory to produce a toy or not so let fij = 1 if factory i
(1 mark)
(i = 1,2) is setup to produce toys of type j (j = 1,2), fij = 0 otherwise
We need to decide how many of each toy should be produced in each factory so let xij be
the number of toys of type j (j = 1,2) produced in factory i (I = 1,2).
(1 mark)
Objective Function:
Maximise total profit, z = 12(x11+ x21) + 16(x12+ x22) - 45000(f11 + f21) - 76000(f12 + f22)
(1 mark)
Constraints:
1
at each factory cannot exceed the production time available
x11/52 + x12/38 <= 480
or 52x11 + 38x12 <= 480
or 42x21 + 23x22 <= 720
x21/42 + x22/23 <= 720
480 hrs = 28,800 min; 720 hrs = 43,200 min
2. cannot produce a toy unless we are set up to do so
x11 <= 52(480)f11
or x11 <= 24,960f11
x12 <= 38(480)f12
or x12 <= 18,240f12
x21 <= 42(720)f21
or x21 <= 30,240f21
x22 <= 23(720)f22
or x22 <= 16,560f22
(1 mark)
(1 mark)
(1 mark)
(1 mark)
(1 mark)
(1 mark)
- 3marks if 4 constraints of type 52x11 <= 480f11
- 2 marks if 4 constraints of type x11 <= 52f11
3. only one factory should be used for each toy
only 1 mark if equality
f11 + f21 ≤ 1
f12 + f22 ≤ 1
xij >=0 and integer
fij binary (or 0/1)
(1 mark)
(1 mark)
(1 mark)
- if only 2 binary variables, max marks = 3
- no marks if no binary variables
Total #4: _______
Page 10
Question #5 (16 marks)
The manager for the pop star known as “BDA” (Bayesian Decision Analysis) must make a
decision as to the next video production. The outcome of the decision taken will be dependent
upon the acceptance of the very critical TV stations. This is very polar and is termed Low
Demand or High Demand.
The alternatives are as follows:
• Produce an elaborate, top of the line production
• Produce a medium quality music video, or
• Produce a video re-using same production set and materials as last video.
The following is the payoff table (profit in millions), for the three alternatives.
Produce Top of the Line (D1)
Produce Medium Quality (D2)
Produce Re-run Quality (D3)
Low Demand
8
9
15
High Demand
20
12
10
(a) Which option should BDA choose according to the following criteria: (4 marks)
Decision Criteria
Optimistic (Maximax)
Conservative (Maximin)
Minimax Regret
LaPlace
Decision
D1
D3
D1
D1
Calculation area:
PAYOFF TABLE:
D1
D2
D3
0.25
S1
8
9
15
0.75
S2
20
12
10
Opt
20
12
15
S1
7
6
0
S2
0
8
10
Max
7
8
10
Pess
8
9
10
Equal
28
21
25
EV
17
11.25
11.25
REGRET TABLE
D1
D2
D3
- if switch from profit to cost, only lose initial marks
(b) If BDA’s management believes the probability distribution of the economic climate is shown
in the table below, what decision should be chosen based upon expected value, and what is
its expected value? (1 mark)
Low Demand
.25
Probability
High Demand
.75
D1, with EV = 17 (see work above)
Cost basis: Tie of D2 & D3
(c) What is the expected value of perfect information? (1mark)
EVPI = __1.75_________
Cost basis:
11.25 – 9.5 = 1.75
Perfect Information:
Payoff
Decision
0.25
S1
15
D3
0.75
S2
20
D1
EPPI
18.75
EVPI
1.75
Page 11
(d) A leading market research firm has proposed to BDA to perform a market survey which
would yield an indication of the level of demand for the new video. Given the good track
record of past survey results, the probability of a favourable indicator given that the demand
was high is .90. Similarly, the probability of an unfavourable indicator given that the
demand was low is .75.
(i) What is this survey information worth to BDA? (8 marks)
Bayesian
Update:
Favourable Indicator (F)
E
P(E)
P(F| E)
P(FE)
PE | F)
0.25
0.25
0.0625 0.084746
S1-low
0.75
0.9
0.675 0.915254
S2-high
1
0.7375
1
=P(F)
(2)
Unfavourable Indicator (U)
E
P(E)
P(U| E)
P(UE)
PE | U)
0.25
0.75
0.1875 0.714286
S1-low
0.75
0.1
0.075 0.285714
S2-high
1
0.2625
1
=P(U)
(1)
For Favourable Survey (p = 0.7375)
0.084746 0.915254
S1
S2
EV
8
20
D1
18.98305
9
12
11.74576
D2
15
10
10.42373
D3
(2)
For Unfavourable Survey (p = 0.2625)
0.714286 0.285714
S1
S2
EV
8
20
11.42857
D1
9
12
9.857143
D2
15
10
D3
13.57143
(1)
EV wSI
EVwoSI
EVSI
17.5625
17
0.5625
(2)
- if P(F|E) wrong & rest right, 2/3 marks
(ii) What is the efficiency of this sample information? (2 marks)
EVSI =
EVPI =
Efficiency =
0.5625
1.75
0.321429
- watch use of EPSI vs EPPI
- with wrong posterior probabilities, can get Efficiency > 100%; 2 marks if they note
this is not correct, 1 mark if no comment
Total #5: _______
Page 12
Question #6 (12 Marks)
Heidelburg Meat Products (HMP) makes four types of dried beef jerky: Regular (R), Honey (H),
Teriyaki (T) and Spicy (S). The decision variables are number of 50-kg batches produced each
week. All versions of the product sell for the same amount, but profit per batch varies due to
differences in processing times and ingredient costs. HMP operates only one 40-hour shift per
week.There are three drying ovens, two seasoning rooms and two packaging lines. Marketing has
placed limits on the proportion of each product produced.
The following is the Excel™ formulation of the HMP production problem:
R
H
T
S
0
0
0
0
ObFnCoef
500
450
450
475
Drying
Seasoning
Packaging
R-Max
R-Min
H-Min
T-Min
S-Min
6
3
3
0.6
0.7
-0.1
-0.15
-0.2
6
4
5
-0.4
-0.3
0.9
-0.15
-0.2
6
4
4
-0.4
-0.3
-0.1
0.85
-0.2
6
5
5
-0.4
-0.3
-0.1
-0.15
0.8
DecVar
0
LHS
0
0
0
0
0
0
0
0
<=
<=
<=
<=
>=
>=
>=
>=
Constraint
Amount
RHS
120
72
72
0
0
0
0
0
The Answer report follows:
Target Cell (Max)
Cell
Name
$F$5 ObFnCoef
Original Value
0
Final Value
8769.230769
Original Value
0
0
0
0
Final Value
7.384615385
1.846153846
5.538461538
3.692307692
Adjustable Cells
Cell
$B$3
$C$3
$D$3
$E$3
Name
DecVar R
DecVar H
DecVar T
DecVar S
Constraints
Cell
$F$7
$F$8
$F$9
$F$10
$F$11
$F$12
$F$13
$F$14
Name
Drying LHS
Seasoning LHS
Packaging LHS
R-Max LHS
R-Min LHS
H-Min LHS
T-Min LHS
S-Min LHS
Cell Value
Formula
110.7692308 $F$7<=$H$7
70.15384615 $F$8<=$H$8
72 $F$9<=$H$9
2.22045E-16 $F$10<=$H$10
1.846153846 $F$11>=$H$11
5.55112E-17 $F$12>=$H$12
2.769230769 $F$13>=$H$13
0 $F$14>=$H$14
Status
Not Binding
Not Binding
Binding
Binding
Not Binding
Binding
Not Binding
Binding
Slack
9.230769231
1.846153846
0
0
1.846153846
0
2.769230769
0
Page 13
The Sensitivity Report follows:
Adjustable Cells
Cell
$B$3
$C$3
$D$3
$E$3
Name
DecVar R
DecVar H
DecVar T
DecVar S
Final
Value
7.384615385
1.846153846
5.538461538
3.692307692
Reduced
Cost
Final
Value
110.7692308
70.15384615
72
2.22045E-16
1.846153846
5.55112E-17
2.769230769
0
Shadow
Price
Objective Allowable
Allowable
Coefficient Increase
Decrease
0
500
1E+30 155.8139535
0
450
125
3775
0
450 186.1111111 89.88095238
0
475 102.027027
2375
Constraints
Cell
$F$7
$F$8
$F$9
$F$10
$F$11
$F$12
$F$13
$F$14
Name
Drying LHS
Seasoning LHS
Packaging LHS
R-Max LHS
R-Min LHS
H-Min LHS
T-Min LHS
S-Min LHS
0
0
121.7948718
171.7948718
0
-121.7948718
0
-96.79487179
Constraint
R.H. Side
120
72
72
0
0
0
0
0
Allowable
Increase
1E+30
1E+30
1.894736842
2.88
1.846153846
2.666666667
2.769230769
2.666666667
Allowable
Decrease
9.230769231
1.846153846
72
1.8
1E+30
1.894736842
1E+30
3.891891892
Required:
(a) What is the optimal production plan and how much profit will be earned? (3 marks)
Decision Variables
Value
R
7.38
H
1.85
T
5.54
S
3.69
The operating profit earned is: $8769.23
at least 2 decimal places for answer values
(b) Could the R-Min and R-Max constraints both be binding in the same solution? (2 marks)
YES _____ or NO __X___
Explain (in 20 words or less):
One specifies the maximum proportion (40%) of regular style production, while
the other specifies the minimum proportion (30%).
(c) St. Jacobs Herbs and Spices wants to buy some drying oven time from HMP. Based on the
solution, how much oven drying time would HMP be willing to sell? (2 marks)
Cannot determine, must resolve _____
Would be willing to sell __9.23___ hours per week
Explain (in 15 words or less):
HMP has 9.23 hours of slack/spare oven drying time per week
Page 14
(d) Currently the Seasoning and Packaging lines have 10% downtime for cleaning between
batches. If HMP offered a $50 per hour bonus for reducing this downtime by 4 hours, how
could this affect the solution? (4 marks)
Offer incentive for Seasoning Line:
Yes _____ or No __X___ or Must Resolve _____
(1 mark)
Offer incentive for Packaging Line:
Yes _____ or No ______ or Must Resolve __X___
(1 mark)
Explanation (25 words or less):
- no reason to offer this incentive for the seasoning line, since there is spare
capacity
- packaging line is at capacity and $50 incentive is less than shadow price of
$121.79, but the 4 hr increase is not within the allowable increase of 1.89 hr
(2 marks)
(e) Interpret the S-Min constraint in the Formulation (15 words or less): (1 mark)
Explanation (20 words or less):
This constraint sets the minimum proportion of Spicy style to 20%.
Total #6: _______
Page 15
Question #7 (12 marks)
Carmen Quantjock is doing a BBA combined with a math major. Carmen is also trying to
maintain a scholarship, which requires the maintenance of a 9.5 GPA. After getting back a
disappointing midterm, Carmen analyzed her typical week and came up with the following:
Activity
Sleep, personal hygene
Classes
Studying
Eating, Breaks, Light Social
Evening Socializing, Pubs, Parties
Part-time jobs
Ideal
63
20
60
20
12
20
195
Total
Minimum
50
15
40
10
6
8
129
Priority
2
1
1
3
3
2
Clearly, with only 168 hours in a week, the ideal is not achievable!
Required: Formulate Carmen’s problem using Goal Programming.
--------------------------------------------------------------Decision Variables & Units:
SPH, C, S, EBLS, ES, PTJ (hours/activity)
(3 marks total, ½ mark each)
Constraints:
Hard Constraints: (3 marks total, ½ mark each)
SPH >= 50
-1 if missed ‘hard’ & treated all as soft
C >= 15
some set dev. var. as hard; ok if full formulation right
S >= 40
EBLS >= 10
ES >= 6
PTJ >= 8
Sum <=168; Non-negativity
Soft Constraints, written as goals:
(1) SPH +u1 – v1 = 63
(2) C + u2 – v2 = 20
(3) S + u3 – v3 = 60
(4) EBLS + u4 – v4 = 20
(5) ES + u5 - v5 = 12
(6) PTJ + u6 – v6 = 20
(3 marks total, ½ mark each)
P2
P1
P1
P3
P3
P2
Objective Function:
Min P1(u2 + u3) +P2(u1 + u6) + P3(u4 + u5)
(3 marks total, 1/Priority)
OR
LP #1: Min u2 + u3
s.t. hard constraints
soft constraints (2) and (3)
LP #2: Min u1 + u6
s.t. hard constraints
LP #1 soln
Soft constraints (1) and (6)
LP #3: Min u4 + u5
s.t. hard constraints
LP #1 soln
LP #2 soln
Soft constraints (4) and (5)
Could also formulate for ‘as close to
ideal’ as possible; both u & v enter
objective function; -1 for any -v
Total #7: _______
Page 16
THIS PAGE INTENTIONALLY LEFT BLANK
--- USE IF NEEDED FOR ADDITIONAL CALCULATIONS ---
Page 17
QUESTION #1 MARKING KEY
TRUE / FALSE
MULTIPLE CHOICE
FILL-IN-BLANKS
1. F
1. b
1. 0.3
2. F
2. a
2. 0.2
3. F
3. d
1 mark for stating
P(SS no purchase) = 0.6
4. F
4. b
5. T
5. c
6. F
6. c
7. T
7. c
8. F
8. b
9. F
9. c
10. T
10. a
11. a
12. b
13. a
14. d
15. c
16. c
1 mark for correct
transition probability
matrix
0.7 0.3
x 1-x