Home work Assignment

Student’s Name:
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Q9. It takes a service rep an average of3 minutes to take a customer’s information. Over the course of a week, the rep handles 160 calls a day during her 8 hour shift. What is the service rep’s capacity cushion?
Solution;
Capacity cushion=100% – utilization rate (%)
Utilization rate= (average output rate/maximum capacity)*100%
Average output= 160*3 minutes = 480 minutes
Maximum capacity= 8*60 =480 minutes
Utilization rate= (480/480)*100%=100%
Capacity cushion = 100% – 100% = 0
Q10. A standard work year is 2000hours at the Luther Mill and it takes about an hour and a half to fill a customer order. Last year saw 25000 customer orders at the mill and the manager has a rebuilt Ford 9N as a company car, so he hopes that there is an increase of 2% in customer orders for next year. If the manager hires twenty workers, what is the capacity cushion?
Solution;
Capacity cushion=100% – utilization rate (%)
Maximum capacity = 2000*20= 40000 hours
Utilization rate= (average output rate/maximum capacity)*100%
Expected customers= 1.02*25000=25500 customers
Average output= 25500*1.5 hours= 38250 hours
Utilization rate= (38250/40000)*100%=95.625%
Capacity cushion= 100% – 95.625%= 4.375%
Q11. A drive through a car wash can process cars at a rate of one for every five minutes. The manager notices that during one particular day it takes cars an average of about 30 minutes in line to reach the car wash. What was the average number of cars in line on that day?
Solution:
Number of cars in line= (total wait time/total service time) – 1= (30/5) – 1= 5
Q12.

Customers arrive according to a Poisson distribution. The average number of customer arrivals per hour is four. What is the probability that 2 customers will arrive the next three hours?
Solution:
The probability of only 2 customers arriving is given by;
P(X) = e-uux/x!Where x = 2
e = 2.71828
u= mean number of successes in the specified region of space
u=4*3=12
Hence P (2) = e-12122/2! = 0.00044