1 3. Process Flow Measures Problem1; Admission Flow 1. Marshall provides higher education to executives and receives about 1000 applications per month

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3. Process Flow Measures

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)

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?

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Admission Flow Chart

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

stageApplicants from group B receive an answer in 6

days on average

1000

50%

50%

200

100

Reject Process

Accept Process

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Admission Flow

b) How long do the applicants from group A spend in the advanced review stage?I =100 , R = 1000(0.5) = 500TR= I T=I/R T= 100/500 = 0.2Applicants 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.

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Admission Flow

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.

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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.

5

Reject

Accept 80%

20%

20 25

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

loan?c)What is the average processing time required for an

accepted loan (including the initial processing stage)?

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Problem 2

a) What is the average processing time required for a loan request?R = 20 loans/hourI = 5+25 = 30 loansAverage processing time = Throughput TimeRT= I T = I/R = 30/20 = 1.5 hours

b) What is the average processing time required for a rejected loan?R = 20 loans/hourI = 5 loansAverage processing time = Throughput Time RT= I T = I/R = 5/20 = 0.25 hours

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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

Additional Processing Stage:R = 20 × 0.8 = 16 loans/hour, I = 25 loansT = I/R = 25/16 = 1.5625

Average Total Processing Time = = 0.25 + 1.5625 = 1.8125 hours

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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?

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Problem 3

1000 Agents

50/month 50/month

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

R = 50 people/monthI = 1000 peopleAverage 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.

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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

?/month ?/month

2 years

1000 Agents

50/month 50/month

20 months

Current annual cost for hiring and training:Throughput Rate = 50 people/month = 600 people/yearAnnual hiring and training cost is 600 (1000) = $600,000

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Problem 3

New annual cost for hiring and training:Average working time = Throughput Time = 24 months = 2 yearsThroughput Rate R= I/T = 1000 people / 2 years = 500 people/year 41.7 per month

Annual hiring and training cost is 500 (1000) = $500,000Annual saving on hiring and training cost is $600,000-

$500,000 = $100,000c) How much the monthly salary of each agent can be

increased?Average # of employees = 1000Annual saving on hiring and training cost = $100,000Monthly saving = $8,333.338333.33/1,000 = $8.33 per employee/month

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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?

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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?

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Problem 4

240

car0.5

0.1

0.1

0.3

motorcycle

boat

house

I = 300 carsR = 0.5(240) =120

claims/wkTR = I T(120) = 300T = 300/120= 2.5

weeksc) How long, on average, does it take to process a house insurance claim? Average # of claims in process = 720720–300 car–114 motorcycle–90 boat = 216Average # of claims for house: I = 216House claims are 1- 0.5-0.1-0.1 = 0.3 of all claimsR = 0.3(240) = 72 TR = I T = I/R T = 216/72= 3 weeks

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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. 4

0 3 10.6(40) = 24

How long on average must customers that purchase fresh

orange juice wait?

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Problem 5

How long on average must customers that purchase fresh orange juice wait?

In the ordering processRT =I 40T = 3 T = 3/40 hours T = 60(3/40) = 4.5 minutesIn the juice squeezing processRT =I 24T = 1 T = 1/24 hours T = 60(1/24) = 2.5 minutesAverage waiting time = T = 4.5 + 2.5 = 7 minutes

40 3 1

24

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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 hourR = 30/60 = 0.5 per minuteT = 6 minutesRT = I I = 0.5(6) = 3 customers are waiting

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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/minuteI = 2Average time a customer will wait: T = I/R = 2/1 = 2 minutes

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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/minuteI = 2 (ordering station) and 1 (filling station) = 3RT = I 1.5T = 3 T = 2 minutes

360/4hrs 2 1

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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”

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Problem 6

At “UCLA-Slurps” we have I = 3 people and T = 15 minutesTR = I 15R = 3 R= 1/5 per minute R = 60(1/5) = 12

customers per hour or = 48 customers/dayBusiness at “CSUN-Stop” has decreased by 48

customers/dayThe new (lower) arrival rate to the “CSUN-Stop”?120-48 = 72 customers/daye) 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/minuteT = 6 minutesRT = I I = 0.3(6) = 1.8 customers

Documents

1 3. Process Flow Measures Problem1; Admission Flow 1. Marshall provides higher education to executives and receives about 1000 applications per month