Operational Management

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...