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# Problem1; Admission Flow - PowerPoint PPT Presentation

Problem1; Admission Flow . 1. Marshall provides higher education to executives and receives about 1000 applications per month. The evaluation starts with a preliminary classification: Group A: Applicants with desired recommendations, working experience, etc. ( 50% of the applicants )

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• 1. Marshall provides higher education to executives and receives about 1000 applications per month. The evaluation starts with a preliminary classification:
• Group A: Applicants with desired recommendations, working experience, etc. (50% of the applicants)
• Group B: Other applicants. (50% of the applicants)
• Applicants in group A will be further considered through an advanced review. Applicants in group B will be rejected.
• On average there were are 200 applications in the preliminary review stage, and 100 applications in the advanced review stage
• a) How long does group A spend in the application process?
• b) How long does group B spend in the application process?
• c) How long is the average process time?

Accept Process

100

50%

1000

200

Reject Process

50%

a) How long do the applicants spend in the preliminary stage?

RT=I  T = I/R=200/1000 = 0.2 month

Applicants spend 0.2(30) = 6 days in the first stage

Applicants from group B receive an answer in 6 days on average

• b) How long do the applicants from group A spend in the advanced review stage?

I =100 , R = 1000(0.5) = 500

TR= I  T=I/R  T= 100/500 = 0.2

Applicants from group A spend 0.2(30) = 6 days on average in the advanced review stage.

Applicants from group A receive answer in 12 days (6 + 6) on average.

• c) What is the average processing time?
• 0.5(6)+0.5(12) = 9 days
• Is there an alternative way to calculate the average waiting time? Do it------

Little’s Law holds for complicated systems.

Problem 2
• 2. Bank XYZ receives 20 loan requests per hour on average. Each loan request goes through an initial processing stage, after which an “accept or reject” decision is made. Approximately 80% of the loans are accepted, and these require additional processing. Rejected loans require no additional processing. Suppose that on average 5 loans are in the initial processing stage, and 25 (accepted) loans are in the additional processing stage.

Accept

80%

• What is the average processing time required for a loan
• What is the average processing time required for a rejected loan?
• What is the average processing time required for an accepted loan (including the initial processing stage)?

25

20

5

Reject

20%

Problem 2

a) What is the average processing time required for a loan request?

R = 20 loans/hour

I = 5+25 = 30 loans

Average processing time = Throughput Time

RT= I  T = I/R = 30/20 = 1.5 hours

b) What is the average processing time required for a rejected loan?

R = 20 loans/hour

I = 5 loans

Average processing time = Throughput Time

RT= I  T = I/R = 5/20 = 0.25 hours

Problem 2

c) What is the average processing time required for an accepted loan (including the initial processing stage)?

Initial Stage: As computed for rejected applications the time for the initial process is

T = I/R = 5/20 = 0.25 hours

R = 20 × 0.8 = 16 loans/hour, I = 25 loans

T = I/R = 25/16 = 1.5625

Average Total Processing Time =

= 0.25 + 1.5625 = 1.8125 hours

Problem 3
• 3. A call center employs 1000 agents. Every month 50 employees leave the company and 50 new employees are hired.
• a) How long on average does an agent work for this call center?
•  Now suppose the cost of hiring and training a new agent is \$1000. The manager of this call center believes that increasing agents’ salary would keep them working longer term in the company. The manager wants to increase the average time that an agent works for the call center to 24 months, or 2 years.
• b) If an agent works for the call center for 24 months on average, how much can the company save on hiring and training costs over a year? Hint: first determine the current annual cost for hiring and training, then determine the new annual cost for hiring and training.
• c) How much the monthly salary of each agent can be increased?

1000

Agents

50/month

50/month

Problem 3

a) How long on average does an agent work for this call center?

R = 50 people/month

I = 1000 people

Average working time = Throughput Time

= I/R = 1000/50 = 20 months or 20/12 = 1.67 years

Suppose the cost of hiring and training a new agent is \$1000. The manager of this call center believes that increasing agents’ salary would keep them working longer term in the company. The manager wants to increase the average time that an agent works for the call center to 24 months, or 2 years.

Problem 3

b) If an agent works for the call center for 24 months on average, how much can the company save on hiring and training costs over a year? Hint: first determine the current annual cost for hiring and training, then determine the new annual cost for hiring and training.

1000

Agents

1000

Agents

?/month

50/month

?/month

50/month

Current annual cost for hiring and training:

Throughput Rate = 50 people/month

= 600 people/year

Annual hiring and training cost is 600 (1000) = \$600,000

20 months

2 years

Problem 3

New annual cost for hiring and training:

Average working time = Throughput Time = 24 months = 2 years

Throughput Rate  R= I/T = 1000 people / 2 years = 500 people/year  41.7 per month

Annual hiring and training cost is 500 (1000) = \$500,000

Annual saving on hiring and training cost is \$600,000-\$500,000 = \$100,000

c) How much the monthly salary of each agent can be increased?

Average # of employees = 1000

Annual saving on hiring and training cost = \$100,000

Monthly saving = \$8,333.33

8333.33/1,000 = \$8.33 per employee/month

Problem 4

4. Time Travelers Insurance Company (TTIS) processes 12,000 claims per year. The average processing time is 3 weeks. Assume 50 weeks per year.

• a) What is the average number of claims that are in process?
• b) 50% of all the claims that TTIS receives are car insurance claims, 10% motorcycle, 10% boat, and the remaining are house insurance claims. On average, there are, 300 car, 114 motorcycle, and 90 boat claims in process.
• c) How long, on average, does it take to process a house insurance claim?
Problem 4
• a) What is the average number of claims that are in process?

R per week = 12,000/50 = 240 claims/week

RT = I

I = 240(3) = 720 claims waiting

50% of all the claims that TTIS receives are car insurance claims, 10% motorcycle, 10% boat, and the remaining are house insurance claims. On average, there are, 300 car, 114 motorcycle, and 90 boat claims in process.

• b) How long, on average, does it take to process a car insurance claim?
Problem 4

I = 300 cars

R = 0.5(240) =120 claims/wk

TR = I  T(120) = 300

T = 300/120= 2.5 weeks

c) How long, on average, does it take to process a house insurance claim?

car

0.5

0.1

240

motorcycle

Average # of claims in process = 720

720–300 car–114 motorcycle–90 boat = 216

Average # of claims for house: I = 216

House claims are 1- 0.5-0.1-0.1 = 0.3 of all claims

R = 0.3(240) = 72

TR = I  T = I/R  T = 216/72= 3 weeks

0.1

boat

0.3

house

Problem 5
• 5. Consider a roadside stand that sells fresh oranges, and fresh orange juice. Every hour, 40 customers arrive to the stand, and 60% purchase orange juice, while the remaining purchase oranges. Customers first purchase their items. Customers that purchased oranges leave immediately after purchasing their oranges. Any customer that ordered orange juice must wait while the juice is squeezed. There are 3 customers on average waiting to purchase either oranges or orange juice, and 1 customer on average waiting for orange juice to be squeezed.

40

0.6(40) = 24

3

1

How long on average must customers that purchase fresh

orange juice wait?

Problem 5
• How long on average must customers that purchase fresh orange juice wait?
• In the ordering process
• RT =I  40T = 3 
• T = 3/40 hours  T = 60(3/40) = 4.5 minutes
• In the juice squeezing process
• RT =I  24T = 1 
• T = 1/24 hours  T = 60(1/24) = 2.5 minutes
• Average waiting time = T = 4.5 + 2.5 = 7 minutes

40

24

3

1

Problem 6
• 6. A recent CSUN graduate has opened up a cold beverage stand “CSUN-Stop” in Venice Beach. She takes life easy and does a lot of surfing. It sounds crazy, but she only opens her store for 4 hours a day. She observes that on average there are 120 customers visiting the stand every day. She also observes that on average a customer stays about 6 minutes at the stand.
• a) How many customers on average are waiting at “CSUN-Stop”?
• R = 120 in 4 hours  R = 120/4 = 30 per hour
• R = 30/60 = 0.5 per minute
• T = 6 minutes
• RT = I  I = 0.5(6) = 3 customers are waiting
Problem 6
• She is thinking about running a marketing campaign to boost the number of customers per day. She expects that the number of customers will increase to 240 per day after the campaign. She wants to keep the line short at the stand and hopes to have only 2 people waiting on the average. Thus, she decides to hire an assistant.
• b) What is the average time a customer will wait in the system after all these changes?
• R = 240/(4hrs*60min) = 1 person/minute
• I = 2
• Average time a customer will wait:
• T = I/R = 2/1 = 2 minutes
Problem 6
• The business got a lot better after the marketing campaign and she ended up having about 360 customers visiting the stand every day. So, she decided to change the processes. She is now taking the orders and her assistant is filling the orders. They observe that there are about 2 people at the ordering station of the stand and 1 person at the filling station.
• c) How long does a customer stay at the stand?
• R = 360/4 hours = 1.5 people/minute
• I = 2 (ordering station) and 1 (filling station) = 3
• RT = I  1.5T = 3  T = 2 minutes

360/4hrs

2

1

Problem 6
• A recent UCLA graduate has opened up a competing cold beverage stand “UCLA-Slurps”. The UCLA grad is not as efficient as the CSUN grad, so customers must stay an average of 15 minutes at “UCLA-Slurp”, as opposed to 6 minutes at the “CSUN-Stop”. Suppose there is an average of 3 customers at “UCLA-Slurps”. The total number of customers remains at 120, as it was before the marketing campaign. But now the 120 is divided between the “CSUN-Stop” and “UCLA-Slurps”.
• d) By how much has business at the “CSUN-Stop” decreased?
• First find how many customers “CSUN-Stop” is losing to “UCLA-Slurps”
Problem 6
• At “UCLA-Slurps” we have
• I = 3 people and T = 15 minutes
• TR = I  15R = 3  R= 1/5 per minute R = 60(1/5) = 12 customers per hour or = 48 customers/day
• Business at “CSUN-Stop” has decreased by 48 customers/day
• The new (lower) arrival rate to the “CSUN-Stop”?
• 120-48 = 72 customers/day
• e) What is now the average number of customers waiting at the “CSUN-Stop” if the flow time at CSUN-Stop remains 6 mins?
• R = 72/day = 72/(60min×4hrs) = 0.3/minute
• T = 6 minutes
• RT = I  I = 0.3(6) = 1.8 customers