OPRE 6302 Sample mid mid id--term -term exam with detailed solutions. Question 1 Val d’Opim is a legendary village for winter sports in the French Alps. Given its enormous popularity among French, Swiss, German, Austrian, and Italian skiers, the 1,200 beds in the village are always booked in the winter season. On average, skiers stay in Val d’Opim for 10 days.
How many new skiers are arriving – on average – in Val d’Opim every day? a) 100 b) 120 c) 1000 d) 1200 e) None of the above The answer is b. Here, the inventory is 1200 and the flow time is 10 days. Using Little’s Law, we know the flow rate is Inventory 1200 Flowrate = = = 120. 10 Flowtime A study done by the French office of A. D. Little has shown that skiers spend on average $100 per person on the first day and $80 per person on each additional day in local restaurants. The study also forecasts that – due to increased hotel prices – the average length of stay for the 2001/2002 season will be reduced to 5 days. What will be the percentage change that skiers spend in local restaurants compared to the last year (when skiers still stayed for 10 days)? a) Spending would decrease by 10% b) Spending would remain unchanged c) Spending would increase by 10% d) Spending would increase by more than 20% e) None of the above
The answer is e. Last year, on average, skiers spend 100*1+80*9=$820 per person per visit. For a 5-day stay, skiers spend 100*1+80*4=$420. So spending would reduce (820-420)/820=49%.
Question 2 Metal frames for kick scooters are manufactured in two steps: Stamping and assembly. Each frame is made up of three pieces: one unit of part A and two units of part B. The parts are fabricated by a single stamping machine that requires a set up time of 90 minutes switching between two part types. Once the machine is set up, the activity time for parts, regardless of type, is 30 seconds each piece. Currently, the stamping machine rotates its production between one batch of 120 part A’s and 240 part B’s. Completed parts move only when the entire batch is produced. At assembly, parts are assembled manually to form the finished products. The three parts and a number of small purchased components are required for each unit of final product. Each product requires 30 minutes of labor time to assemble. There are 12 workers in assembly. There is sufficient demand to sell every scooter the system can make. At the current batch sizes, the bottleneck of the system is a. Stamping b. Assembly c. They both have the same capacity d. Cannot be determined
The answer is a. The capacity at stamping is 120/(90+120*0.5+90+240*0.5’*60=20 units per hour. The capacity at assembly is 1/30*12*60=24 units per hour. Therefore, stamping is the bottleneck.
At the current batch sizes, what is the process capacity in units per hour? Circle the answer below that is closest to the correct answer. A unit refers to a complete scooter frame (i.e. one part A and two parts B’. a. 1 units/hour b. 5 units/hour c. 10 units/hour
d. 20 units/hour e. 30 units/hour f. 40 units/hour
The answer is d. Since stamping is the bottleneck, its capacity is also the process capacity.
One way to increase process capacity is to a. increase the batch size at the Stamping step b. decrease the batch size at the Stamping step c. add more workers at Assembly d. none of the above The answer is a. At a batch-producing step, increasing the batch size increases the capacity at the step.
Which batch size for the stamping machine would minimize inventory without decreasing the current flow rate? Circle the answer below that is closest to the correct answer. a. 60 sets b. 120 sets c. 180 sets d. 240 sets b. 300 sets The answer is c. At a batch size of 180, the capacity at stamping is 180/(90+180*0.5+90+360*0.5)*60=24 units per hour.
Question 3 Cheapest Car Rental (CCR’ rents cars at the DFW airport. The car rental market consists of two segments: the short-term segment, which rents for an average of 0.5 week, and the long-term segment, which rents for an average of 2 weeks. CCR currently rents an average of 200 cars a week to the short-term segment and 100 cars a week to the long-term segment. Approximately 20% of the cars returned (evenly distributed across both segments’ are found to be defective and in need of repairs before they can be made available for rent again. On average, there are 100 cars waiting to be cleaned. The average cost of this operation is $5 per car. Cars needing repairs spend an average of 2 weeks in the repair shop and incur an average cost of $150 per car. Assume that cars are rented as soon as they are available for rent, that is, as soon as they have been cleaned or repaired. Short-term renters pay $200 per week, while long-term renters pay $120 per week. The process flow diagram is shown in the following figure.
Customers
Cleaning
Repairing
a. Identify throughput, inventory, and throughput time at each stage. b. What profit does CCR earn each week with the current system? Assume that each car loses $40 in value per week because of depreciation.
c. CCR is comparing two possible improvements: 1. Decrease time in repairs from 2 weeks to 1 week. 2. Decrease cost per repair from $150 per car to $120 per car while keeping throughput time in repairs in 2 weeks. Assume that the effort that is required in each case is the same. Which change do you think will be more effective? Why? Solutions: Part a.
Customers
Clean area Repair area
Throughput
300/week
240/week
60/week
Inventory
200×0.5 +
100 cars
120 cars
0.42
2 weeks
100×2 = 300 cars Flow Time
1 week
weeks
Part b:
Number of cars owned by Cheapest = 300 + 100 + 120 = 520 Number of cars on rent = 300 Revenue per week = (#of cars on rent for short term’ x $200/week + (#of cars on rent for LT’ x $120/week = I_ST x $200/wk
+ I_LT x $120/wk
Recall from [5a] that I_ST = R_ST x T_ST = 200/wk x 0.5 wks = 100 and I_LT = 100x2=200 = 100×$200
+ 200×$120 = $44,000/week
Cleaning cost per week = 240×$5 = $1,200/week Repair cost per week = 60×$150 = $9,000/week Depreciation per week = 520×$40 = $20,800/week Profit per week = 44,000 - (1,200 + 9,000 + 20,800’ = $13,000/week
Part Par t c:
Decreasing flow time in repairs by 1 week will lower the inventory in repairs from 120 to 60. This will reduce the number of cars required by 60 and thus weekly depreciation by $2,400. All other revenues and costs will be unchanged. Decreasing repair cost per car by $30 lowers weekly repair cost by 60×30 = $1,800. Decreasing the flow time in repairs is thus more effective since it results in larger savings.
Question 4 The state government has decided to open a clinic in downtown Houston for dispensing the flu vaccine to children from low-income households. The budget is tight and so the process must be as efficient as possible.
The program
administrator estimates that 450 children will receive vaccinations during each 10 hour day. A child, along with his or her guardian, enters the clinic and waits in the lobby until the receptionist can see them. There is only one receptionist and it takes him about 1 minute to fill out the required form for each child. The next step is to
wait for one of the two nurses to become available for drug allergy consultation. Each nurse can handle about 25 patients per hour. After talking with a nurse, the child and his or her guardian go to the vaccination waiting area where three technicians are administering vaccinations. It takes approximately 5 minutes per child to administer the shot. a. Draw the process flow diagram for the new clinic. b. How much waiting do you expect to find in front of the receptionist and the nurses, respectively? Why? c. Do you expect much waiting in the vaccination waiting area? Why? d. Assuming that patients arrive evenly during the day, draw the inventory buildup diagram for children waiting in the vaccination waiting area. How much overtime will the technicians have to work each day? e. On average, how many people will be waiting in the waiting area? What is the average waiting time per child? f. How many seats do you need in the seating area to ensure that everyone (child and guardian’ waiting can have a seat at all times during the day?
Solution: a. Fill out form
Drug allergy
(60/hr’
consultation (50/hr’
Take the shot (36/hr’
b. We do not expect to see much waiting in front of the receptionist and the two nurses because the arrival rate is less than their capacities. c. We do expect to see waiting in the vaccination waiting area because the arrival rate is greater than the vaccination capacity. d. The inventory buildup diagram is attached. Note that I made several assumptions to simplify the graph a little bit. In particular, I assume the vaccination stage starts giving shots as soon as children arrive. 90
3 0
10
12.5
The technicians have to work overtime for 2.5 hours each day. e. In order to compute the # of waiting children, we first compute the area of the up-triangle (the triangle above the horizontal line with the height 3’. The base of the triangle is (90-3’/90*12.5=12 and the height is (90-3’=87. So, the area is 12*87/2=522 children*hr. It follows that the average # of children waiting is 522/12.5=42. Note that the throughput rate is 35 children/hour. By Littles Law, the average waiting time (per child’ is 42/35=1.2hr. f. From the inventory buildup diagram, we see that the maximum inventory is 90. So the maximum waiting children is 90-3=87. To ensure all people have a seat, we will need to provide 87*2=174 chairs in the waiting room.
