MOS 3330 – Exam
Operations Management
1. The welding department supplies parts to the final assembly line.
Management decides to implement a kanban system and has collected
the following data.
The daily demand is 2,200 units.
The production lead time is 2 days (this includes processing time,
transport time, and waiting time).
Management has decided to have 1 day(s) of safety stock
One container fits 475 units
Answer：
We first determine the amount of demand during the production lead time.
Demand during lead time = Lead time × Daily demand = 2 days × 2,200 units / day = 4,400 units
We translate the 1 day safety stock into a quantity, which is 1 day(s) × 2,200 units / day = 2,200
units.
With 6,600 units and a container size of 475 units / container, we need:
Number of kanban containers = (Demand during lead time + Safety stock) / Container size =
(4,400 + 2,200) / 475 = 6, = 14
So, we need 14 containers.
2. A copy machine is available 24 hours a day. On a typical day, the
machine produces 113 jobs. Each job takes about 4 minutes on the
machine, 2 minutes of which is processing time and 2 minute is setup
time (logging in, defining the job). About 16 Percent of the jobs need to
be reworked, in which case the setup time and the processing time have
to be repeated. The remainder of the time, the equipment is idle. What is
the OEE of the equipment?
Answer：
The copy machine is available 60 × 24 = 1440 minutes a day. The value add time is 226 minutes.
So, the OEE is:
,OEE = Value add time / Time available = 226 / (24 × 60) = 15.7.
3. In the welding operations of a bicycle manufacturer, a bike frame has a
flow time of about 13.75 hours. The time in the welding operation is
spent as follows: 3 hours waiting in front of the cutting machine for the
batch to start, 1 hours waiting for the setup of the machine, 2 hour
waiting for the other pieces of the batch to go through cutting, 2 minute
at the cutting machine, and 3 hours waiting for the transfer to the
welding machine. Then, at the welding machine, the unit spends 1 hour
waiting in front of the welding machine for the batch to start, 2 hour
waiting for the setup of the welding machine, 0.75 hour waiting for the
other pieces of the batch to go through welding, 0.4 minute at the
welding machine, and 1 hour waiting for the transfer to the next
department.
Answer：
The exact flow time is 827.40 minutes. Of this, only 2.4 minutes are value-added time. Thus, the
percentage of value-added time is:
Value-added time of the flow unit / Flow time = 2.4 minutes / 827.40 minutes = 0.002901 =
0.2901%
4. A process makes two products, A and B and operates 5 days per
week, 8 hours per day. Demand for the two products for the next
week is as follow:
Day 1: A: 20 B: 26
Day 2: A: 10 B: 15
Day 3: A: 20 B: 19
Day 4: A: 18 B: 20
Day 5: A: 12 B: 20
Answer：
The total demand across both products is 180 units that we have to process over the
course of the 40 hour work week. So, the takt time would be:
, Takt time = Available time / Demand = 40 hours / 180 units = 0.22222 hours / unit
A level production schedule would almost alternate between A and B as follows:
ABABABABB ABABABABB… This puts the flow-units in a relationship of 8:10, as is the
overall demand mix.
5. Consider again that the company making tires for bikes is concerned
about the exact width of its cyclocross tires. The company has a lower
specification limit of 22.5 mm and an upper specification limit of 23.1
mm. The standard deviation is 0.20 mm and the mean is 22.80 mm.
What is the probability that a tire will be too narrow?
What is the probability that a tire will be too wide?
What is the probability that a tire will be defective?
Answer：
a..
Probability{too narrow} = NORM.DIST(22.5, 22.8, 0.2, TRUE) = 0.066807
b.
Probability{too wide} = 1 − NORM.DIST(23.1, 22.8, 0.2, TRUE) = 0.066807
c.
Probability{part defective} = 0.066807 + 0.066807 = 0.133614
Operations Management
1. The welding department supplies parts to the final assembly line.
Management decides to implement a kanban system and has collected
the following data.
The daily demand is 2,200 units.
The production lead time is 2 days (this includes processing time,
transport time, and waiting time).
Management has decided to have 1 day(s) of safety stock
One container fits 475 units
Answer：
We first determine the amount of demand during the production lead time.
Demand during lead time = Lead time × Daily demand = 2 days × 2,200 units / day = 4,400 units
We translate the 1 day safety stock into a quantity, which is 1 day(s) × 2,200 units / day = 2,200
units.
With 6,600 units and a container size of 475 units / container, we need:
Number of kanban containers = (Demand during lead time + Safety stock) / Container size =
(4,400 + 2,200) / 475 = 6, = 14
So, we need 14 containers.
2. A copy machine is available 24 hours a day. On a typical day, the
machine produces 113 jobs. Each job takes about 4 minutes on the
machine, 2 minutes of which is processing time and 2 minute is setup
time (logging in, defining the job). About 16 Percent of the jobs need to
be reworked, in which case the setup time and the processing time have
to be repeated. The remainder of the time, the equipment is idle. What is
the OEE of the equipment?
Answer：
The copy machine is available 60 × 24 = 1440 minutes a day. The value add time is 226 minutes.
So, the OEE is:
,OEE = Value add time / Time available = 226 / (24 × 60) = 15.7.
3. In the welding operations of a bicycle manufacturer, a bike frame has a
flow time of about 13.75 hours. The time in the welding operation is
spent as follows: 3 hours waiting in front of the cutting machine for the
batch to start, 1 hours waiting for the setup of the machine, 2 hour
waiting for the other pieces of the batch to go through cutting, 2 minute
at the cutting machine, and 3 hours waiting for the transfer to the
welding machine. Then, at the welding machine, the unit spends 1 hour
waiting in front of the welding machine for the batch to start, 2 hour
waiting for the setup of the welding machine, 0.75 hour waiting for the
other pieces of the batch to go through welding, 0.4 minute at the
welding machine, and 1 hour waiting for the transfer to the next
department.
Answer：
The exact flow time is 827.40 minutes. Of this, only 2.4 minutes are value-added time. Thus, the
percentage of value-added time is:
Value-added time of the flow unit / Flow time = 2.4 minutes / 827.40 minutes = 0.002901 =
0.2901%
4. A process makes two products, A and B and operates 5 days per
week, 8 hours per day. Demand for the two products for the next
week is as follow:
Day 1: A: 20 B: 26
Day 2: A: 10 B: 15
Day 3: A: 20 B: 19
Day 4: A: 18 B: 20
Day 5: A: 12 B: 20
Answer：
The total demand across both products is 180 units that we have to process over the
course of the 40 hour work week. So, the takt time would be:
, Takt time = Available time / Demand = 40 hours / 180 units = 0.22222 hours / unit
A level production schedule would almost alternate between A and B as follows:
ABABABABB ABABABABB… This puts the flow-units in a relationship of 8:10, as is the
overall demand mix.
5. Consider again that the company making tires for bikes is concerned
about the exact width of its cyclocross tires. The company has a lower
specification limit of 22.5 mm and an upper specification limit of 23.1
mm. The standard deviation is 0.20 mm and the mean is 22.80 mm.
What is the probability that a tire will be too narrow?
What is the probability that a tire will be too wide?
What is the probability that a tire will be defective?
Answer：
a..
Probability{too narrow} = NORM.DIST(22.5, 22.8, 0.2, TRUE) = 0.066807
b.
Probability{too wide} = 1 − NORM.DIST(23.1, 22.8, 0.2, TRUE) = 0.066807
c.
Probability{part defective} = 0.066807 + 0.066807 = 0.133614
