OIS 3660 Operations ManagementSpring 2012

Homework Set #1 SOLUTION

1 D

2 C

3 B

4 B

5 D

6Business Process Model

Chevrolet Volt Assembly Line

Sugar Refinery Process Flow

New Salt Lake City Airport Project

Eastern Gear Job Shop

7. Production rate is: 600 / (5*8) = 15 baskets per hourCycle time is 1/15 or 0.067 hours per basket or 4 minutes per basket

8. (a) Longest cycle time is at station 2 at 7 min/job. This is the bottleneck and cycle time of system. Convert to production rate of 1/7 or 0.143 jobs/min.

(b) Throughput (at WIP) = WIP * CT = 20 jobs * 7 min/job = 140 minutesTotal throughput of process is 5 + 140 + 7 = 152 minutes

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OIS 3660 Operations ManagementSpring 2012

Homework Set #1 SOLUTION

A(100)

B(200)

D(450)

E(500)

C(500) 3 parts

1 part

9. The flow chart shown above depicts a general manufacturing process. The capacity of each station in gal/hr is shown in each box as a number in parentheses.

(a) What is the system capacity?

Starting with Station A, we see that it can handle 100 gal/hr. B can process 200 gal/hr, but only 100 gal/hr can be passed from A.

Station D must receive product in the ratio of 1 part from Station B for every 3 parts from Station C. Thus if Station B can provide 100 gal/hr, Station C would need to provide 300 gal/hr (3*100), for a total of 400 gal/hr Station D can process up to 450 gal/hr, so all 400 gal/hr from B & C can pass through.

Station E can process 500 gal/hr, which is more than enough to handle the 400 gal/hr coming through from D, and this 400 gal/hr is the maximum output of the system.

System capacity is 400 gallons per hour

(b) What is the utilization at each station?

Utilization A: (400*1/4)/100 capacity = 100%Utilization B: (400*1/4)/200 capacity = 50%Utilization C: (400*3/4)/500 capacity = 60%Utilization D: 400/450 capacity = 89%Utilization E: 400/500 capacity = 80%

(c) Where is the bottleneck in this system?

Our bottleneck is station A (notice that we have 100% utilization at A)

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OIS 3660 Operations ManagementSpring 2012

Homework Set #1 SOLUTION

10.(a) Arrival rate is 1/10; service rate is 1/8

Utilization is arrival rate/service rate = 0.8Idle time is 1 – utilization = 0.2, so clerk is idle 20% of the time

(b) Wq = .1/(.125*(.125-.1)) = 32min

(c) Ws = 1/(.125-.1) = 40min

(d) Lq = .1^2/(.125*(.125-.1)) = 3.2

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