OPM Exam 3 review
CPU-on-Demand (CPUD) offers real-time high-performance computing services. CPUD owns one supercomputer that can be accessed through the Internet. Its customers send jobs that arrive, on average, every five hours. The standard deviation of the interarrival times is five hours. The average processing time of executing each job is 3 hours and the standard deviation of the processing time is 6 hours. 1. What is the utilization (as a percent) of CPUD's supercomputer? 2. . How long does a customer have to wait to have a job completed? 3. How many customers are waiting in the queue on average?
1. average interarrival time: a=5 hours average processing time: p= 3 hours Utilization= p/a=60% 2. a=5, CVa= 5/5=1, CVp=6/3=2, utilization= 0.6 3*(0.6/1-0.6)*(1^2+2^2/)= 11.25 T=p+tq = 3+11.25=14.25 3. Iq=Tq/a 11.25/5=2.25 customers
Find a Doctor is a small startup that helps people find a physician who best meets their needs (location, insurance accepted, etc.). During a "slow" time, on average, one call arrives every six minutes (with a standard deviation of six). Each staff member spends 20 minutes with each customer (on average, with a standard deviation of 20). 1. What is minimal number of staff members needed to ensure a stable queue? 2. If it has seven staff members taking calls from customers, what is the utilization?
1. interarrival time: a=6min, processing time: p=20min minimal number of servers>p/a = 20/6 = 3.33, so minimal number of staff members = 4 2. a=6min, p=20min, m=7 staff members utilization= p/a*m = 20/6*7=0.476
The Shady Farm Milk Company can process milk at a fixed rate of 7500 gallons/hour. The company's clients request 120,000 gallons of milk over the course of one day. This demand is spread out uniformly from 8 a.m. to 8 p.m. The company starts producing at 8 a.m. and continues to work until all of the demand has been satisfied. On average, how long does a client wait (in hours) to receive its product?
Average time to serve a unit =1/2 * 𝑇 * (demand/capacity -1) There are total 12 hours from 8 a.m. - 8 p.m, so demand rate = 120,000 𝑔𝑎𝑙𝑙𝑜𝑛𝑠 12 ℎ𝑜𝑢𝑟𝑠 = 10,000 gallons/hour. There are total 12 hours from 8 a.m. - 8 p.m, so demand rate = 120,000 𝑔𝑎𝑙𝑙𝑜𝑛𝑠 12 ℎ𝑜𝑢𝑟𝑠 = 10,000 gallons/hour. Average time to serve a client = 1/2 * T * (demand/capacity - 1) = 1/2 * 12 * (10000/7500 -1) = 2 hours