RSM 270 L0101/L0201/L0301 2016 Fall Term
1. Warm-up questions[15 minutes](15 points)
(a) (2 points)For a coffeeshop opens at 7AM, we consider a 5 minute interval between
8-8:05AM. Is it possible to have the throughput rate between 8-8:05 to be higher
than the input rate between 8-8:05? If yes explain via an example, and if no
provide the justification.
Solution:
Yes. It is possible. If the inventory is not empty at 8:00 and the input rate is
less than the capacity, then the throughput rate can be higher than the input
rate between 8-8:05.
(b) (2 points)For an emergency room, explain the downsides of efficiency view and
the downsides of effectiveness view.
Solution:
The downside of efficiency is having long waiting times, and the downside of
effectiveness is having low utilization and overstaffing.
(c) (2 points) Clearly explain the information-capacity tradeoff in the OM triangle
using an example.
Solution: A family doctor who makes fixed appointments and who limits
the time he spends with each patient reduces the variability significantly, i.e.,
has more more information than a walk-in clinic where patients arrive purely
randomly. If the walk-in clinic wants to offer the same short wait times as
the family doctor above, they need to have several doctors on duty, i.e., much
higher capacity to meet the random demand.
(d) A candle making company operates with 5 workers, each working 8hours per day,
and getting paid $10 per hour, and each making 100 candles per day.
i. (1 point) What is the labor Productivity (use candle per worker per hour)?
Solution:
Labor productivity = 100/8 candles per worker per hour.
ii. (3 points) The company recently purchased a new machine, which helps them
operate with only 4 workers, each working 8 hours as before and getting paid
$10, however, they can now make 150 candles per day, and the machine costs
$100 per day. Calculate the new labor productivity (use candle per worker
per hour), as well the Multifactor productivity before and after adding the
new machine.
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RSM 270 L0101/L0201/L0301 2016 Fall Term
Solution: Labor productivity= 150/8 candles per worker per hour. Mul-
tifactor before: 100×55×10×8 =
5
4 candles per dollar. Multifactor Productivity
after: 150×4100+4×10×8 =
600
420 =
10
7 .
iii. (3 points) Which productivity measure shall we use to compare the two situ-
ations? Do you think adding the new machine was wise? Explain.
Solution:
Yes. We should multifactor productivity since we added a new resource,
so two types of costs involved.
(e) (2 point) What is the coefficient of variation of interarrival time, and the coefficient
of variation of service time in M/D/1 queues?
Solution: Ca = 1, Cs = 0.
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RSM 270 L0101/L0201/L0301 2016 Fall Term
2. [Process Analysis](15 minutes)[15 points] CleanCar is a car wash offering car cleaning
services. In the table below the steps taken to clean a car as well as the time needed
for each step has been explained:
Service Description Processing Time Resourced used
A.Wash Exterior washing 10 min 1 automated machine
B.Wax Exterior waxing 10 min 1 automated machine
C.Wheel Cleaning Cleaning of all wheels 7 min 1 employee
D. Interior Cleaning inside cleaning 20 min 1 employee
The store operates 12 hours a day, and has 1 automated machine for exterior washing,
one automated machine for exterior waxing, 1 employee for cleaning of all wheels,
and another employee for interior cleaning. In the first part of the question, assume
customers are going through all steps(stations) in the process in the car wash.
(a) (2 points) What is the maximum number of cars that can be served per day?
Which station is the bottleneck?
Solution:
Stage D is the bottleneck. 36 cars can be served per day.
(b) (2 points) What is the cycle time of the process assuming there is enough demand
and the process is capacity constrained.
Solution:
Cycle time = 20 minutes.
(c) (2 points) what is the theoretical flow time for one car?
Solution:
FTime = 10 + 10 + 7 + 20 = 47 minutes.
(d) (2 points) Assuming customers are visiting the station at the constant rate of 30
customers per day, what is the average utilization.
Solution:
Average utilization =
30
72
+ 30
72
+ 210
720
+ 30
36
4 .
Next consider the company decides to offer the following packages to their customers:
• Package 1: include only exterior wash(service A)
• package 2: include exterior wash and waxing (service A and B)
• package 3: include exterior wash, waxing and wheel cleaning (service A,B,C).
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RSM 270 L0101/L0201/L0301 2016 Fall Term
• package 4: include all 4 services (A,B,C,D).
After introducing the new packages, it is anticipated that the demand will raise to 80
customers per day. Of these customers 40 percent will be interested in package 1, 15
percent will be interested in package 2, 15 percent will be interested in package 3, and
30 percent will be interested in package 4. The mix does not change over the course of
the day.
(e) (6 points) Calculate the implied utilization of the resources.
Solution:
potential input going through stage A is 80, stage B is 48, stage C is 36
and stage D is 24. The implied utilizations are then IUA =
80
72 , IUB =
48
72 ,
IUC =
36
720
7
, IUD =
24
36 .
(f) (1 point) To meet the demand, the capacity of which stations (resources) should
be increased. Please explain your answer.
Solution: Only stage A needs capacity expansion.
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RSM 270 L0101/L0201/L0301 2016 Fall Term
3. [Forecasting](20 minutes)[15 points]
(a) (1 point) What is a major advantage of the Delphi Method compared to the Panel
Consensus?
Solution:
Delphi method is anonymous and as a result the people will not get intimidated
by their peers.
(b) The table below presents the actual demand and the forecasts using various ap-
proaches.
Week actual demand 3-week simple moving average 2-week weighted moving average exponential smoothing
week 1
week 2
week 3 80 95
week 4 96 90
week 5 92 92
i. (4 points) Calculate the actual demand of week1 and week2 (left blank in the
table).
Solution:
A2 = 92× 3− 96− 80 = 100, A1 = 90× 3− 80− 100 = 90.
ii. ( 3 points) What were the relative weights used to calculate the weighted
moving average?
Solution: 92 = 96×w1 + 80×w2, and w1 +w2 = 1. therefore w1 = .75,
and w2 = .25.
iii. (3 points) Calculate the exponential smoothing parameter α.
Solution:
F3 = F2 + α(A2 − F2) = 90 + α(10) = 95, therefore α = .5.
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RSM 270 L0101/L0201/L0301 2016 Fall Term
4. [Little’s Law](20 minutes)[15 points] CCfirm is a consulting firm that divides his
consultants into three classes: associates, managers, and partners. The firm has been
stable for the last 20 years and is expected to have 160 juniors, 60 managers and 20
partners.
In general each junior, After 4 years of working in this level, either leaves the company
or being levelled up; that is becomes a manager or is dismissed from the company.
Similarly after five years, a manager either becomes a partner or is dismissed. The
company recruits commerce students at the junior level. No hires are made at the
manager or partner level. A partner stays with the company for another 10 years (a
total of 19 years with the company).
(a) (5 points)How many new commerce graduates CCfirm does have to hire every
year? Hint: Calculate of the throughput rate of the juniors per year, i.e., how
many juniors are moving out of the process per year (i.e., becoming a manager or
are dismissed from the firm each year).
Solution:
Ij = Tj × Rj = 160 = 4 × Rj , therefore Rj = 40 juniors/year. To keep the
number of juniors fixed, they should hire 40 per year.
(b) (5 points) How many of the juniors are dismissed each year? how many of juniors
are becoming managers each year?
Solution:
Im = Tm × Rm = 60 = 5 × Rm, therefore Rm = 12 managers/year. To keep
the number of managers fixed, they should hire 12 managers per year from
juniors, therefore 40− 12 = 28 juniors will be dismissed each year.
(c) (5 points) What are the chances that a new hire at the junior level in CCfirm
will become a partner eventually (as opposed to leaving the company either after
4years or 9 years)?
Solution:
The probability is equal to 12/40× 212 = 120 .
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RSM 270 L0101/L0201/L0301 2016 Fall Term
5. [Inventory build up and Little’s Law](25 minutes)[20 points] Consider a single
security screening line at the Pearson International Airport, which is comprised of an
X-ray scanner operated by officer A, and a metal detector operated by officer B.
Office A (assigned to the X-ray scanner) is in charge of screening the bags as they pass
through the scanner.
Officer B, is in charge of screening the passengers as they walk through the metal
detector.
Arriving passengers either queue up or, if there is no queue on arrival, directly put their
bags on the scanner. A typical customer has 1.5 bags (one carry-on and one backpack
which is counted as half a bag). The X-ray scanner cannot process more than 18 bags
per minute. Each passenger takes 4 seconds to walk through the metal detector.
(a) (1 points) How would you define a “flow unit” through this process?
Solution: Passenger, with his/her 1.5 bags.
(b) (3 points) What is the capacity rate of this process? Please explain.
Solution:
12 passengers per minute, because each passenger has 1.5 bags and only 18
bags per minute can be scanned. This is less than going through the metal
detector (15 passengers/min).
Assume for simplicity that exactly three flights, each carrying 200 passengers, are
scheduled for departure each hour. For each flight with 200 passengers, we have the
following approximate arrival pattern: 75 passengers arrive 80 to 50 minutes early
(arriving at a constant pace over this 30 minutes), 100 arrive 50 to 30 minutes early
(arriving at a constant pace over this 20 minutes), and the remaining 25 arrive between
30 to 20 minutes before scheduled departure (arriving at a constant pace over this 10
minutes).
For the following questions consider 3 flights leaving at 8pm, and ignore flights leaving
at 7pm and 9pm whose passenger arrivals or queues might overlap with the 8pm flights.
(You may assume they are served by a different security screening line, which we will
not consider here.)
(c) (4 points) Draw an inventory build-up diagram (inventory of passengers waiting
in line) until the last passenger for the flights at 8pm has passed the security
check. (Use the graph next page.) Assume that the queue is empty when the first
passenger arrives for the 8pm flights.
Solution: The first passengers arrive at 6:40pm,but the arrival rate(7.5/min)
¡ capacity rate. Therefore, no inventory will build up during that first 30
min. The arrival rate increases to 15/min, and thus an inventory of 60 units
will build up until 7:30pm. Until 7:40pm the arrival rate is 7.5/min so we can
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RSM 270 L0101/L0201/L0301 2016 Fall Term
draw down inventory to a level of 15. We need 1.25 min to serve the remaining
passengers because no further customers arrive.
(d) (2 points) Precisely at what time does the last passenger pass the security check?
Solution:
7:41’15”.
(e) (3 points) What is the average inventory (from the time when the first passenger
arrives until the last passenger passed the security check)?
Solution:
Calculate the area under the curve and divide by the time span. Note: 7:30pm
- 7:41’15” is not one triangle.
(f) (3 points) What is the average throughput rate?
Solution:
Total passengers divided by time that is 600/61.25 passenger per minute.
(g) (2 points) Determine the average waiting time for passengers.
Solution:
T = I/R therefore average waiting time is part (e) divided by part (f).
(h) (2 points) What is A’s utilization (on average) during the 1.5-hour time interval
from 6:30pm-8pm?
Solution
He can process 12× 90 passengers in 90 minutes, and total of 600 passengers
are passing through. Therefore his utilization is 600/(12× 90) = 5/9.
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RSM 270 L0101/L0201/L0301 2016 Fall Term
6. Queuing Models(25 minutes)[20 points] The airport branch of “Awesome” car rental
company maintains a fleet of 50 cars. The interarrival time between request for a
rental car is 2.4 hours, on average with the standard deviation of 2.4 hours. Since it is
an airport branch, the company is open 24 hours a day and there is no indication of
systematic interarrival pattern over the course of the day. Assume that if all cars are
rented, customers are willing to wait until there is a car available. A car is rented on
average for 3 days with a standard deviation of 1 day.
(a) (2 point) Calculate the arrival rate of customers to the company.
Solution:
λ = 12.4 = .416 customers per hour.
(b) (2 point) Calculate the coefficient of variation of interarrival time.
Solution:
Ca =
2.4
2.4 = 1.
(c) (6 point) Calculate the utilization of each car (you can think of each car as one
server(resource)). Calculate the average number of cars being parked in the park-
ing lot.
Solution:
τ = λcµ =
3×24
2.4×50 = .6. The number of cars in use is Is =
λ
µ =
3×24
2.4 = 30.
Therefore the average number of cars parked in the lot is 50− 30 = 20.
(d) (4 points) What is the average time a customers has to wait to rent a car?
Solution:
Iq =
τ
√
2(c+1)
1−τ × C
2
a+C
2
s
2 =
.6
√
102
1−.6 ×
1+ 1
9
2 = .00798.
Tq =
Iq
λ = .00798× 2.4 = .01915hours = 1.14minutes.
(e) ( 6 points) How would the waiting time change if the company decides to limit
all cars rentals to exactly 4 days? Assume that if such a restriction is imposed,
the average interarrival time will increase to 3 hours, with standard deviation
changing to 3 hours.
Solution:
Ca =
3
3 = 1, Cs = 0. τ =
λ
cµ =
4×24
3×50 = .64. Therefore, Tq =
Iq
λ = 3× .64
√
102
1−.64 ×
0+1
2 = .045hourshours = 2.75 minutes.
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