Submitted by: Submitted by amu23
Views: 170
Words: 499
Pages: 2
Category: Business and Industry
Date Submitted: 10/20/2013 06:30 AM
Q8.8
a. Average capacity utilization is .625, So security officers available to satisfy a new request = .375*8 = 3
b. Average time from the moment a student calls for an escort to the moment the student arrives at his destination = 26.82 minutes
c. M cap unit and N-m = 0 sheet
Customer lost per minute = .07538, customer lost per hour = .07538*60 = 4.5 customers
d.
To satisfy around 80% of the caller, University should have 9 security officers
Q8.9
a. m cap units –approx.
Average utilization = 80%
b. Time it takes on an average from filling a patent to being launched into the market: 68.39261 months
c. Patent life left = 20*12 - 68.39261 = 171.6 months = 14.3 years
Q8.10 ( I am giving it a try, but am not sure if am going in the right direction)
a. Container of packages come every 3 hours i.e. 180 minutes and on average it contains 10 basic, 10 business and 5 oversized packages. To process all 3 type of packages Davis will take = 5*10+4*10+6*5 = 120 minutes
Capacity Utilization = 120/180 = 66.67%, so processing packages in any order is not going to make difference
b. average time a package spends in the contanier will change, based on the order of processing.
Chapter 9
Q9.1
a. m cap units and N-m = 0
Probability that all seven servers are utilized = .15280
b. Effective Flow Rate = .77725 * 60 = 46.635 units
c. Units lost every hour = .14018*60 = 8.41 units
Q9.2
a. 74%
b. .14804*60*24*30*500 = 8.8824 = $3197664
Q9.3
a.
1 - .62444 = .3755
b. .23472*24*5 = $28.17
c. .39028*24*1 = $9.3666
d. As below
Q9.4
2% of 1000/hour = 20
So average interrarival time = 60/20 = 3 minutes
a. Probability that all 6 spots are taken is .00563
b. Costumer served every hour = .33146 * 60 = 19.8876
Q9.5
a.
Scenario 1:
Scenario 2:
50% changes
Remaining 50% changes:
So average flow rate = .10101
Scenario 3:
50% probablity
Remaining 50%
Average flow...