RSM 270 L0101/L0201/L0301 Midterm Examination 2016 Winter Term
1. Productivity [15 minutes] (16 points)
(a) A company that makes shopping carts for supermarkets and other stores recently purchased some new equipment that reduces the labor content of the jobs needed to produce the shopping carts. Prior to buying the new equipment, the company used four workers. Each worker produced an average of 80 carts per hour. Labor cost was $10 per worker per hour, and machine cost was $40 per hour. With the new equipment, it was possible to transfer one of the workers to another department. Machine cost increased by $10 per hour while output increased by four carts per worker per hour, and the labor costs stayed the same.
i. (2 points) Calculate labor productivity before and after the new equipment. Use carts per worker per hour as the measure of labor productivity.
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ii. (4 points) Calculate the multifactor productivity before and after the new equipment. Use carts per dollar cost (labor plus machine) as the measure.
iii. (3 points) Calculate and comment on the changes in productivity according to the two measures. Which one do you believe is more appropriate for this situation?
(b) A chocolate manufacturer who was preparing for Valentine’s Day had the fol- lowing output and input in the last three weeks before February 14:
Solution: Before: 80 carts per worker per hour. After: 84 carts per worker per hour.
Solution: Before: Costs ($10 × 4 =)$40 + $40 = $80; hence MFP = 80 × 4/$80 = 4 cart/$1 Cost.
After: ($10 × 3 =)$30 + $50 = $80; hence 84 × 3/$80 = 3.15 carts/$1 Cost.
Solution: Labour productivity increased by 4/80 = 5%.
Multifactor productivity decreased by 0.85/4 = 21.25%.
Multifactor productivity is more appropriate because there is not only a change in workers but also a change in machines and their cost.
Units produced:
No. of workers:
Hours per week per worker: Material (kg):
1,000 1,500 1,500 2 2 3 40 60 40 150 250 300
Average labor cost is $10 per hour during regular time (first 40 hours of a week) and $15 per hour during overtime (any hours in excess of 40 hours a week). Chocolate cost was $20 per kg.
i. (7 points) Calculate the multifactor productivity for each week.
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RSM 270 L0101/L0201/L0301 Midterm Examination 2016 Winter Term
Solution: Multifactor productivity material cost)
= units produced / (labor cost +
Units produced: Labor cost:
Material Cost:
Total Cost: Multifactor prod.: (units per $ input):
1,000 2(40)($10) = $800 150($20) = $3,000 $3,800 1,000/$3,800 = 0.26
2 1,500 $800 + 2(20)($15) = $1,400 250($20) = $5,000 $6,400 1,500/$6,400 = 0.23
3 1,500 3(40)($10) = $1,200 300($20) = $6,000 $7,200 1,500$7,200 = 0.21
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RSM 270 L0101/L0201/L0301 Midterm Examination 2016 Winter Term
2. Process Analysis [25 minutes] (22 points)
Consider a small manufacturing process with four consecutive stages A, B, C, and D with the following capacity rates: 120 units/hr, 150 units/hr, 110 units/hr, and 140 units/hr, respectively.
(a) (2 points) What is the capacity rate of this process and which station(s) is (or are) the bottleneck? Please explain!
(b) (2 points) What is the cycle time of this process?
(c) (2 points) What is the theoretical flow time to finish one unit (ignoring time spent waiting in buffers between the stages)?
Assume now for the rest of the question that the demand on the system is 130 units/hr.
(d) (5 points) What is the average utilization of all the resources?
Solution: throughput rate = min( capacity rate, input rate) = min(110, 130) = 110 units/hr.
average utilization = (110/120 + 110/150 + 110/110 + 110/140)/4 = 85.89%.
(e) (3 points) Which of the following three actions would yield the greatest increase in process capacity? Explain briefly!
1. increase the capacity of Operation A by 15 percent;
2. increase the capacity of Operation B by 10 percent; or 3. increase the capacity of Operation C by 10 percent.
(f) (4 points) Briefly explain where you would add buffers to the system? Hint: Do not add any unnecessary buffers.
Solution: 110 units/hr,
because C is the process with the lowest capacity rate.
Solution: The bottleneck determines the capacity rate of the whole system. With the previous part we have capacity rate = 110 units / hr. Cycle time = 1 / capacity rate = 1 / 110 = 32.72 secs between two finished units.
Solution: Sum of flow times: 1/120+1/150+1/110+1/140 hours = 0.03123 hours = 112.44 seconds.
Solution: Operation C by 10%, because operation C is the bottleneck.
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RSM 270 L0101/L0201/L0301 Midterm Examination 2016 Winter Term
(g) (4 points) Assume that stage C had a variable capacity rate of 130±10 units/hr, instead of the original given 110 units/hr. Would you add or remove any buffer from the system compared to the previous part of the question. Please explain!
Solution: Between B and C, because C is the bottleneck and would there- fore block stage B.
Furthermore before A, because the current input rate is greater than A’s capacity rate.
No other buffers are useful.
Solution: A now becomes the bottleneck and the throughput rate of the system is 120 units/hr. C always has a capacity rate of at least 120 units/hr even though it shows variability. As B only receives 120 units/hr from A, we do not need a buffer in front of C anymore. We do not need to add a buffer between C and D either because first of all C’s capacity rate will never be greater than D’s capacity rate, and secondly and more importantly, the throughput rate of the system is 120 units/hr, so D will be able to keep up with the demand.
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RSM 270 L0101/L0201/L0301 Midterm Examination 2016 Winter Term
3. Inventory Build-up [15 minutes] (11 points)
A company manufactures toasters. It buys some of the components, but it makes the heating element, which it can produce at the rate of 800 per day. Toasters are assembled 250 days a year, at a rate of 300 per day. Because of the disparity between the production and usage rates, the heating elements are periodically produced in batches of 2,000 units. To meet the annual demand of 75,000 units, there are 37.5 batches per year.
(a) (4 points) If production of a batch begins when there is no inventory of heating elements on hand, how much inventory will be on hand (i.e., in inventory) two days later? Show all your calculations.
Solution: They are always producing batches of 2,000 units, this means it will take them 2.5 days to make them. The input rate for those 2.5 days is 800 units per day. Then they will stop production which means that the input rate will be 0 until the inventory is depleted.
The capacity rate of the process is constant at 300 units per day.
The number of units produced in two days = (2 days)(800 units/day) = 1,600 units
The number of units used in two days = (2 days) (300 units/day) = 600 units
Current inventory of the heating element = 0
Inventory build up after the first two days of production = 1, 600 − 600 = 1, 000 units
Total inventory after the first two days of production = 0 + 1, 000 = 1, 000 units.
(b) (4 points) Draw an inventory build-up diagram of heating elements for 20 con- secutive working days, assuming each production cycle begins when there is no inventory on hand. You may further assume that the production of heating elements and toasters is done uniformly over each day if that simplifies your calculations.
(c) (3 points) What will be the average level of inventory of the heating element, assuming each production cycle begins when there is no inventory on hand? Show all your calculations.
Solution: Draw an inventory build-up diagram based on the following data: From 0 the highest inventory level will be reached after 2.5 days at 1250 units. The inventory is depleted at time 6.67 and the next cycle starts.
The next cycle ends at time 13.33 and a third cycle end at day 20, so exactly three triangles where the buildup is steeper than the draw down.
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RSM 270 L0101/L0201/L0301 Midterm Examination 2016 Winter Term
Solution: Avg. inventory = 625 units. This can easily be seen from the diagram which shows three identical triangles with height 1250.
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RSM 270 L0101/L0201/L0301 Midterm Examination 2016 Winter Term
4. Inventory Build-up and Little’s Law [15 minutes] (14 points)
Consider the following regional distribution centre which is typically placing orders to its suppliers on Mondays. The following data only refer to the unloading process of arriving trucks delivering the products ordered. During the five business days of the week the number of warehouse workers is set to six workers per day but experience has shown that on Mondays usually only four of them will show up for work. During the weekend the warehouse is staffed with only three workers, both on Saturday and on Sunday. The workers have to work 10 hours per day and we assume that one worker can unload one truck per hour. Because most orders are placed on Mondays, trucks arriving on Mondays bring in goods that were supposed to be delivered last week but did not arrive on time. Thus, only 20 trucks arrive on Mondays. Most of the orders placed on Mondays arrive on the next day: 90 trucks. 50 trucks arrive on Wednesdays and most of the other orders arrive before the weekend, leading to an arrival rate of 70 trucks per day on Thursdays and Fridays. Only very few deliveries arrive on Saturdays and Sundays so that only 10 trucks per day arrive during the weekend. Trucks have to wait at the distribution centre until they are unloaded.
(a) (4 points) Draw an inventory build-up diagram of trucks for one week. Start the week with Monday and assume that no trucks are waiting from the day before.
(b) (3 points) What is the average inventory?
(c) (4 points) Calculate the system throughput rate for each day, and the average system throughput rate.
(d) (3 points) Determine the average waiting time for trucks. Namely, the average time spent in line before being unloaded.
Solution: Draw an inventory build-up diagram based on the following data: Time Mon Tues Wed Thurs Fri Sat Sun
Inventory 0 30 20 30 40 20 0
Solution: correct formula: (0 + 30 + 20 + 30 + 40 + 20 + 0)/7 Average inventory = 20 trucks.
Time Mon Tues Wed Thurs Fri Sat Sun
20 60 60 60 60 30 30 Avg. throughput rate = (20+60+60+60+60+30+30)/7 = 45.71 trucks/day.
Throughput give throughput for each day
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RSM 270 L0101/L0201/L0301 Midterm Examination 2016 Winter Term
Solution: Little’s Law:
Avg. waiting time = (avg. inventory)/(avg. throughput rate) = 0.438 days.
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RSM 270 L0101/L0201/L0301 Midterm Examination 2016 Winter Term
5. Queuing Analysis [15 minutes] (16 points)
(a) Tim’s Second Buck, a new coffee shop franchise, has had problems with its drive-through window operations. Customers arrive at an average rate of one every 3 minutes. Current service time has averaged 2.5 minutes with a standard deviation of 2 minutes. A suggested process change, when tested, results in an average service time of 2.5 minutes with a standard deviation of 1 minute.
i. (2 points) As a result of implementing this change, will the average wait time in queue increase, decrease, or remain unchanged? Briefly justify your answer.
ii. (2 points) As a result of implementing this change, will the average server utilization increase, decrease, or remain unchanged? Briefly justify your answer.
iii. (3 points) Explain why Tim’s Second Buck cannot operate at full utiliza- tion, even if the average input rate was equal to the average capacity rate.
(b) Answer the following questions.
i. (2 points) Give a real life example for an M/D/c queuing system. You may
choose the number of servers, c.
ii. (3 points) Why do (most) big supermarkets not use the single line to mul- tiple server system (Type 2)?
Solution: Cs decreases from 2.0/2.5 = 0.8 to 1.0/2.5 = 0.4.
All other parameters held constant this decreases variability will decrease the average number of customers in the queue, and therefore, by Little’s law decrease the wait time.
Solution: Utilization ρ = λ/μ. Input rate λ and service rate μ do not change and therefore the average utilization does not change.
Solution: (Answers should include: ) This is due to variability. If there is no variability, then full utilization may be possible. Near full server utilization means very long waiting lines, long waiting times and therefore unhappy customers.
Solution: (Answers can include but are not limited to: ) It is sufficient if the service time is close to deterministic.
Vending machine, massage/physiotherapy, exercise/yoga classes, lectures at UofT, car wash. . . .
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RSM 270 L0101/L0201/L0301 Midterm Examination 2016 Winter Term
iii. (4 points) Mention 4 hints from the Psycology of Queueing that can reduce the perception of the waiting time.
Solution: (Answers can include but are not limited to: ) There is only restricted room for shopping carts to line up without blocking the aisles in the shop. Cashiers are far apart from each other, so that it would take a long time for customers to reach the cashier.
Solution: Unoccupied time feels longer than occupied time Process waits feel longer than in process waits
Anxiety makes waits seem longer
Uncertain waits seem longer than known, finite waits Unexplained waits are longer than explained
Unfair waits are longer than equitable waits
The more valuable the service, the longer the customer will wait Solo waits feel longer than group waits
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RSM 270 L0101/L0201/L0301 Midterm Examination 2016 Winter Term
6. Queuing Models [25 minutes] (21 points)
If you need to make any assumptions, please state them clearly!
The successful coffee shop chain Moonbucks is interested in evaluating a new project, developed by the consulting company Analytics Edge. It consists of a cell phone app that would allow their customers to use it to place an order, pay, and pick it up at their preferred time in one of the 150 locations in Toronto without waiting in line!
In order to evaluate this project, Moonbuck considers a representative store in Toronto. All the average measures that are valid for this representative store can be multiplied by 150 to get a good estimate of the impact that a full implementation in the city of Toronto would have. At the representative store, the inter-arrival times for customers are 30 seconds on average, with a standard deviation of 1 minute; while the service times are of 1 minute and 20 seconds on average, with a standard deviation of 2 minutes.
Analytics Edge has presented convincing evidence, collected from small scale exper- iments, that suggests that the app would reduce variability in both the customers’ arrival process and in the service process. Moreover, they claim that the implemen- tation of the app could reduce the standard deviation of the inter-arrival times by 50%, while at the same time reducing the standard deviation of the service times by one third.
Analytics Edge has offered to lead the successful implementation of this project, in- cluding the development and implementation of the app, training the staff in the new processes required to serve the customers using the app, and an advertising campaign in the subway, street cars, buses and the radio. For this whole project, Analytics Edge would charge Moonbucks CAD$1.5 million, and they claim that Moonbuck would al- ready be making a profit out of the project by the end of the first year after starting the implementation. Moonbuck has hired you to verify claim.
Further information that can be useful to evaluate Analytics Edge’s claim is that Moonbucks estimates that each minute that their customers wait in line on average in each store costs them CAD$10,000 annually, in terms of loss of goodwill. For this reason, the stores operate with the appropriate staff that ensures that the average wait time at the queue in their stores is no more than 5 minutes. On the other hand, each employee costs Moonbuck CAD$40,000 annually, therefore they keep the number of servers at each store at the minimum level such that this goal is met.
(a) (2 points) Explain briefly why it makes sense that the app developed by Ana- lytics Edge reduces the variability in the customers’ arrival process.
(b) (6 points) What are the average system utilization, individual server utilization, queue lenght, and waiting time at the representative store before the app is
Solution: If the customers let Moonbucks know when they want to pick up their order in advance using the app, then their arrival time is known a priori and not random any more.
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RSM 270 L0101/L0201/L0301 Midterm Examination 2016 Winter Term introduced? How many servers are working at the store?
Solution: λ = 1 = 2 customers/minute 1 .5
μ = 4/3 = .75 customers/minute Ca = 1 = 2
.5 Cs=2 =1.5
c = 4 servers
τ = λ/4μ = 2/3 is both the system and individual server utilization Iq = τ √2(c+1) × Ca2 +Cs2 = 2.6 customers
Tq = Iq/λ = 1.3 minutes from Little’s Law
Note that c=3 servers gives Tq = 10.08 > 5 minutes, which is not acceptable to Moonbucks.
(c) (6 points) What are the average system utilization, individual server utiliza- tion, queue length, and waiting time at the representative store after the app is introduced? How many servers are working at the store?
Solution: λ = 2 customers/minute μ = .75 customers/minute
Ca =50%·2=1
Cs = 2/3 · 1.5 = 1
c = 3 servers
τ = λ/3μ = 8/9 is both the system and individual server utilization
Iq = τ √2(c+1) × Ca2 +Cs2 = 6.45 customers 1−τ 2
Tq = Iq/λ = 3.23 minutes from Little’s Law
Note that c=3 servers is acceptable to Moonbucks now.
(d) (6 points) make money after one year of implementation if they go ahead with this project?
Solution: The gains are saving one employee per store (on average), for yearly total of 40, 000 · 150 = $6 million .
The costs are the project implementation, for CAD$1.5 million, and the extra waiting time of the customers, for a yearly total 10, 000 · (3.23 − 1.3) · 150 = $2, 895, 000 .
Hence, the net value of the project after the first year of implementation is 6 − 1.5 − 2.895 = $1.605 million. The claim by Analytics Edge is true.
(e) (1 point) What is your recommendation for Moonbucks regarding the imple- mentation of this project?
Solution: They should go ahead with it. They will make a profit in the first year, and continue to make a profit from it afterwards.
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RSM 270 L0101/L0201/L0301 Midterm Examination 2016 Winter Term 7. Balancing a Simple Line [only if you have time left] (4 points (bonus))
This question is a bonus/challenge question. You are not required to solve this question, but you will receive a bonus towards the examination mark for solving it.
A model kit packing line has three separate stages. It starts with operators assembling the boxes from cardboard outers and polystyrene filling pieces. The normal time for this operation is 0.2 minutes. The assemblers place each box on a conveyor that runs along behind a series of benches occupied by the packers. A packer picks a box from the conveyer and fills it with the appropriate mix of parts from trays within reach of the bench. Normal time for filling is 0.45 minutes after which the box is returned to another conveyor. This feeds the last process, an automated wrapping machine that covers each box with cellophane with a cycle time of 0.04 minutes. This stage cannot be performed by hand. The figure below shows the layout.
Your goal is to maximize the capacity rate of the process. You are allowed to hire as many workers as you want to for the assembly and packing station, but you should not hire more than necessary to achieve your goal.
Packing Section
Wrapping Machine
Box Assembly
(a) (1 point (bonus)) What is the maximum possible capacity rate of the whole packing line given that you hired a sufficient number of wor
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