1 3. Process Flow Measures Auto-Moto Financial Services- The Old Process Auto-Moto receives 1,000...

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3. Process Flow Measures

Auto-Moto Financial Services- The Old ProcessAuto-Moto receives 1,000 applications per month. In the old process, each application is handled in a single activity, with 20% of applications being approved. 500 were in the process at any time. Average flow time T = ?

The firm recently implemented a new loan application process. In the new process, applicants go through an initial review and are divided into three categories.Discussion: How operational power destroys the walls of poverty.

RT = IT = I/R = 500/1,000 months = 0.5 month or 15 days.

ProcessIp=500

1000/month200/month

800/month

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New Process: The Same R, But smaller I

Initial Review

Subprocess A Review

Subprocess B Review

Accepted

Rejected

1000/month

25%

50%

25%

70%

30%

90%

10%

800/month

200/month

R = 1000I = IIR + IA + IB = 200 + 25 + 150 = 375Inventory reduced to 375 from 500 in the old process. Since R is constant, therefore T has reduced.T = I/R = 375/1000 = 0.375 month or 0.375(30) = 11.25 daysThe new process has decreased the processing time from 15 days to 11.25 days.

IB = 150

IA = 25

IR = 200

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Compute average flow time.

Compute average flow time at Initial Review Process.

Compute average flow time at Subprocess A.

Compute average flow time at Subprocess B.

Compute average flow time of an Accepted application.

Compute average flow time of a Rejected application.

4


Flow Time at Each Sub-process (or activity) Average Flow Time for sub-process IR. Throughput RIR = 1,000 applications/month Average Inventory IIR = 200 applications TIR = 200/1,000 = 0.2 months = 6 days in the IR sub-

processAverage Flow Time for sub-process A. Throughput RA = 250 applications/monthAverage Inventory IA = 25 applicationsTA = 25/250 months = 0.1 months = 3 days in sub-

process A.Average Flow Time for sub-process B. Throughput RB = 250 applications/monthAverage Inventory IB = 150 applicationsTB = 150/250 months = 0.6 months = 18 days in sub-

proces B

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Routing, Flow Time, and Percentage of Each Flow units

One flow unit at very macro level: Application1000 flow units/month at very micro level: Each specific applicationTwo flow units: Accepted and rejected

Accepted-A: IR, AAccepted-B: IR, B

Five flow units: Accepted-A, Accepted-B Rejected-IR, Rejected-A, Rejected-B

TIR = 6 days

TA = 3 days

TB = 18 days

We also need percentages of each of the five flow units

Rejected-IR: IR

Rejected-A: IR, A

Rejected-B: IR, B

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New Process: Intermediate Probabilities

Initial ReviewT = 6

Subprocess A ReviewT = 3

Subprocess B ReviewT = 18

Accepted

Rejected

100%

25%

50%

25%

70%

30%

90%

10%

80%

20%

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New Process: Intermediate Probabilities

Initial ReviewT = 6

Subprocess A ReviewT = 3

Subprocess B ReviewT = 18

Accepted

Rejected

100%

25%

50%

25%

17.5%

7.5%

22.5%

2.5%

80%

20%

50%

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Flow Time of the Accepted Applications

Under the Original Process – the average time spent by an application in the process is 15 days (approved or rejected).

In the new process: 15 days reduced to 11.25 days.On average, how long does it take to approve an applicant? On average, how long does it take to reject an applicant?

Accepted-A: IR, A Accepted-A(T) = 6 + 3 = 9 Accepted-A = 17.5 %Accepted-B: IR, B Accepted-B(T) = 6 + 18 = 24 Accepted-B = 2.5 %

Average Flow time of an accepted application =

[0.175(9)+0.025(24)]

/ (0.175+.025)

= 10.875

The average flow time has reduced from 15 to 11.25. In addition, the flow time of accepted applications has reduced to 10.875. That is what the firm really cares about, the flow time of the accepted applications.

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Flow Time of Rejected Applications

Rejected-IR: IR Rejected-IR(T) = 6 Rejected-IR(%) = 50% Rejected-A: IR, A Rejected-A(T) = 6+3 = 9 Rejected-A(%) = 7.5%Rejected-B: IR, B Rejected-B(T) = 6+18 = 24 Rejected-B(%) = 22.5%

Average Flow time of a rejected application == 11.343

)24(8.0

225.0)9(

8.0

075.0)6(

8.0

5.0

Check our computations:Average flow time of an application0.8(11.343)+0.2(10.875) = 11.25

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A hospital emergency room (ER) is currently organized so that all patients register through an initial check-in process. Each patient is seen by a doctor and then exits the process, either with a prescription or with admission to the hospital. 55 patients per hour arrive at the ER, 10% are admitted to the hospital and the rest will leave with a simple prescription. On average, 7 people are waiting to be registered and 34 are registered and waiting to see a doctor. The registration process takes, on average, 2 minutes per patient. Among patients who receive prescriptions, average time spent with a doctor is 5 minutes. Among those admitted to the hospital, average time is 30 minutes. Assume the process to be stable; that is, average inflow rate equals average outflow rate.

ER1; Problem 3.4 MBPF

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Directions

o) Draw the flow process chartb) On average how long a patient spend in ER?c) On average how many patients are in ER?Hints:Compute flow time in buffer 1Compute average activity time of DoctorCompute the average flow time in this processCompute average flow time for a simple prescription

patient Compute average flow time for a potential

admission patient Compute number of patients in Doctor activityCompute the average number of patients in the

process.

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R I TBuffer 1 55 7Registeration 55 2Buffer 2 55 34Doctor 55

Re-Check

7.6

7.5

37.1

54.2TD= 0.9(5)+0.1(30) = 7.5

TSP= 7.6+2+37.1+5 = 51.7 TP= 0.1(76.7)+0.9(51.7) = 54.2

Flow Time

T1 = 5min T2 = 30min

I = 34 I = 7

Buffer 1 Registration Buffer 2 Doctor

Potential admits

Simple Prescription

R =55/hr

10%

90% T = 2 min

TPA= 7.6+2+37.1+30 = 76.7

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I = 34 I = 7


Potential admits

Simple Prescription

R =55/hr

10%

90% T = 2 min

R I TBuffer 1 55 7 7.6Registeration 55 2Buffer 2 55 34 37.1Doctor 55 7.5

54.26.9

Inventory

1.8

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I = 34 I = 7


Potential admits

Simple Prescription

R =55/hr

10%

90% T = 2 min

R I TBuffer 1 55 7 7.6Registeration 55 1.8 2Buffer 2 55 34 37.1Doctor 55 6.9 7.5

54.249.7

I = 49.7, R = 55. Is our T= 54.2 correct?

RT= I

R = 55/60 = 0.916667 minutes

RT = I T=I/R = 49.7/0.916667 = 54.2

Recheck

The Little’s Law

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A triage system has been proposed for the ER described in Exercise 3.4. As mentioned earlier, 55 patients per hour arrive at the ER. Patients will be registered as before and it takes an average of 2 minutes per patient. They will then be quickly examined by a nurse practitioner who will classify them as Simple Prescriptions or Potential Admits. Planners anticipate that the initial examination by triage nurses will take 3 minutes. They expect that, on average, 20 patients will be waiting to register and 5 will be waiting to be seen by the triage nurse. Planners expect eh Simple Prescriptions area to have, on average, 15 patients waiting to be seen. As before, once a patient’s turn come, each will take 5 minutes of a doctor’s time. The hospital anticipates that, on average, the emergency area will have only 1 patient waiting to be seen . As before, once that patient’s turn comes, he or she will take 30 minutes of a doctor’s time. Assume that, as before, 90% of all patients are Simple Prescriptions, assume, too, that the triage nurse is 100% accurate in making classifications.


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Directions

o) Draw the flow process charta) On average how many patients are in ER?b) On average, how long a patient spend in ER?c) Compute average flow rate in buffer 3 and buffer

4. d) Compute average flow time in all buffers.e) Compute average number of patients in all

activities. f) On average, how long does a simple prescription

patient spend in the ER? d) On average, how long does a Potential Admit

patient spend in the ER?

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Buffer 1 Registration Buffer 2 Triage Nurse

Buffer 3 Potential admits

Buffer 4 Simple Prescription

55/hr

5.5/hr

49.5/hr

R R/min I TBuffer 1 55 0.916667 20Registeration 55 0.916667 2Buffer 2 55 0.916667 5Triage Nurse 55 0.916667 3Buffer 3 5.5 0.091667 1Potential Admission 5.5 0.091667 30Buffer 4 49.5 0.825 15Simple Prescription 49.5 0.825 5

Process Flow and TR=I Table

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R R/min I TBuffer 1 55 0.916667 20 21.8Registeration 55 0.916667 1.8 2Buffer 2 55 0.916667 5 5.5Triage Nurse 55 0.916667 2.8 3Buffer 3 5.5 0.091667 1 10.9Potential Admission 5.5 0.091667 2.8 30Buffer 4 49.5 0.825 15 18.2Simple Prescription 49.5 0.825 4.1 5

IER= 52.5a) On average, how many patients are in ER?

Method 1: Macro Method, a single flow unit

RER= 55/60 flow units / min , IER= 52.5

b) On average, how long will a patient spend in ER?

TER = 52.5/(55/60) = 57.2 minutes

Inventory and Flow Time; Macro Method

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R R/min I TBuffer 1 55 0.916667 20 21.8Registeration 55 0.916667 1.8 2.0Buffer 2 55 0.916667 5 5.5Triage Nurse 55 0.916667 2.8 3.0Buffer 3 5.5 0.091667 1 10.9Potential Admission 5.5 0.091667 2.8 30.0Buffer 4 49.5 0.825 15 18.2Simple Prescription 49.5 0.825 4.1 5.0

Method 2: Micro Method, two flow units, Potential Admission and Simple Prescription

TPA= 21.8+2+5.5+3+10.9 +30 = 73.2

TSP= 21.8+2+5.5+3+18.2 +5 = 55.5

TER= .1(73.2)+.9(55.5) = 57.3c) On average, how long will a potential admission patient spend in ER? 73.2

Flow Time; Micro Method

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Refer again to Exercise 3.5. Once the triage system is put in place, it performs quite close to expectations. All data conform to planners’ expectations except for one set-the classifications made by the nurse practitioner. Assume that the triage nurse has been sending 91% of all patients to the Simple Prescription area when in fact only 90% should have been so classified. The remaining 1% is discovered when transferred to the emergency area by a doctor. Assume all other information from Exercise 3.5 to be valid.


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Directions

o) Draw the flow process charta) On average how many patients are in ER?b) On average, how long a patient spend in ER?c) Compute average flow rate in buffer 3 and buffer 4. d) Compute average flow time in all buffers.e) Compute average number of patients in all

activities. f) On average, how long does a simple prescription

patient spend in the ER? d) On average, how long does a Potential Admit patient

spend in the ER?

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Buffer 1 Registration Buffer 2 Triage Nurse

Buffer 3 Potential admits

Buffer 4 Simple Prescription

55/hr

4.95/hr

50.05/hr

0.55/hr

R R/min I TBuffer 1 55 0.916667 20Registeration 55 0.916667 2Buffer 2 55 0.916667 5Triage Nurse 55 0.916667 3Buffer 3 5.5 0.091667 1Potential Admission 5.5 0.091667 30Buffer 4 50.05 0.834167 15Simple Prescription 50.05 0.834167 5

Problem 3.6

5.5

Process Flow and Throughputs

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Macro: Average number of patients in the system =

20+1.8+5+2.8+1+2.8+15+4.2 = 52.6

Average flow time = I/R = 52.6/(55/60) = 57.3

Inventory and Flow Time; Macro Method

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Micro Method Compute SP and PA firstCommon = 21.8 +2+5.5+3 = 32.3

TSP= 32.3+18+5 = 55.3

Flow Time; Simple Prescription

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R R/min I TBuffer 1 55 0.916667 20 21.8Registeration 55 0.916667 1.8 2Buffer 2 55 0.916667 5 5.5Triage Nurse 55 0.916667 2.8 3Buffer 3 5.5 0.091667 1 10.9Potential Admission 5.5 0.091667 2.8 30Buffer 4 50.05 0.834167 15 18Simple Prescription 50.05 0.834167 4.2 5

Flow Time; Potential Admission and Overall

Simple Prescripti

on23

Common32.3

Potential Admissio

n40.9

0.9

0.01

0.09

0.9 55.30.09 73.20.01 96.2

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Recheck

R R/min I TBuffer 1 55 0.916667 20 21.8Registeration 55 0.916667 1.8 2Buffer 2 55 0.916667 5 5.5Triage Nurse 55 0.916667 2.8 3Buffer 3 5.5 0.091667 1 10.9Potential Admission 5.5 0.091667 2.8 30Buffer 4 50.05 0.834167 15 18Simple Prescription 50.05 0.834167 4.2 5

T= 55.3 (0.9) + 73.2 (0.09) + 96.2 (0.01)

T= 57.32

TPA1= 32.3 +10.9+30 = 73.2 ……(4.95 PA patients out of 5.5 PA patient: 90%)

TPA2= 73.2+18+5 = 96.2 (0.55 PA patients out of 5.5 PA patient: 10%)

TPA = 73.2(.9) + 96.2(.1) =75.5

Flow Time; Potential Admission and Overall

Recheck Again

TSP= 55.3 is 90% of flow units, and TPA =75.5 is for 10% of flow units

T = 55.3 (.9) + 75.5(.1) = 57.32

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80 patients per hour arrive at a hospital emergency room (ER).All patients first register through an initial registration process.On average there are 9 patients waiting in the Rg-Buffer. The registration process takes 6 minutes.Patients are then examined by a triage nurse practitioner;m.On average there are 2 patients waiting in the Tr-Buffer in front of the triage process, and the triage classification process takes 6 minutes.On average, 91% of the patients are sent to the Simple-prescription process and the remainder to Hospital-admission.On average, there are 5 patients waiting in the Simple-prescription buffer (Sp-Buffer) in front of this process.A physician spends 6 minutes on each patient in the Simple-prescription process.In addition, on average, 2 patients per hour are sent to the Hospital-admission buffer buffer (Hs-Buffer) after being examined for 6 minutes in the Simple prescription process.On average, 0.9 patients are waiting in the Hospital-admission buffer (Hs-Buffer).A physician spends 25 minutes on each patient in the Hospital-admits process.0

ER4; Problem 3.6b MBPF

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RgBuff:9 Rg:5 TrBuff:2 Tr:6

HsBuff:0.9 Hs:25

SpBuff:6 Sp:6

80/hr

9%

91%

2/hr

R R/min I T

RgBuff 80 1.333333 9

RgProc 80 1.333333 6

TrBuff 80 1.333333 2

TrProc 80 1.333333 5

SpBuff 72.8 1.213333 6

SpProc 72.8 1.213333 6

HsBuff 7.2 0.12 0.9

HsProc 7.2 0.12 25


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R R/min I T

RgBuff 80 1.333333 9

RgProc 80 1.333333 6

TrBuff 80 1.333333 2

TrProc 80 1.333333 5

SpBuff 72.8 1.213333 6

SpProc 72.8 1.213333 6

HsBuff 9.2 0.153333 0.9

HsProc 9.2 0.153333 25



HsBuff:0.9 Hs:25

SpBuff:6 Sp:6

80/hr

9%

91%

2/hr

30


R R/min I T

RgBuff 80 1.333333 9 6.75

RgProc 80 1.333333 6.00

TrBuff 80 1.333333 2 1.50

TrProc 80 1.333333 5.00

SpBuff 72.8 1.213333 6 4.95

SpProc 72.8 1.213333 6.00

HsBuff 9.2 0.153333 0.9 5.87

HsProc 9.2 0.153333 25.00



HsBuff:0.9 Hs:25

SpBuff:6 Sp:6

80/hr

9%

91%

2/hr

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R R/min I T

RgBuff 80 1.333333 9 6.75

RgProc 80 1.333333 8.0 6.00

TrBuff 80 1.333333 2 1.50

TrProc 80 1.333333 6.7 5.00

SpBuff 72.8 1.213333 6 4.95

SpProc 72.8 1.213333 7.3 6.00

HsBuff 9.2 0.153333 0.9 5.87

HSProc 9.2 0.153333 3.8 25.00



HsBuff:0.9 Hs:25

SpBuff:6 Sp:6

80/hr

9%

91%

2/hr

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R R/min I T

RgBuff 80 1.333333 9 6.75

RgProc 80 1.333333 8.0 6

TrBuff 80 1.333333 2 1.5

TrProc 80 1.333333 6.7 5

SpBuff 72.8 1.213333 6 4.945055

SpProc 72.8 1.213333 7.3 6

HsBuff 9.2 0.153333 0.9 5.869565

HsProc 9.2 0.153333 3.8 25

43.68

Average flow time =

T = I/R =

43.68/(80/60) = 32.76



HsBuff:0.9 Hs:25

SpBuff:6 Sp:6

80/hr

9%

91%

2/hr

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Micro Method Compute SP and PA firstCommon = 6.75+6+1.5+5 = 19.25

TSP= 19.25+10.95 = 30.2

R R/min I T

RgBuff 80 1.333333 9.00 6.75

RgProc 80 1.333333 6.67 5.00

TrBuff 80 1.333333 2.00 1.50

TrProc 80 1.333333 8.00 6.00 19.25SpBuff 72.8 1.213333 6.00 4.95

SpProc 72.8 1.213333 7.28 6.00 10.95HsBuff 9.2 0.153333 0.90 5.87

HsProc 9.2 0.153333 3.83 25.00 30.87


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TPA1= 19.25 +30.87 = 50.12 ……(7.2 PA patients out of 9.2 PA patients)

TPA2= 19.25 +10.95 +30.87 = 61.07 (2 PA patients out of 9.2 PA patients)

TPA = 50.12(7.2/9.2) + 61.07(2/9.2) =

TPA = 50.12(0.782609) + 61.07(0.217391) =52.5Recheck

T= 30.20(70.8/80) + 50.12 (7.2/80) + 61.02 (2/80)

T = 32.76


R R/min I T

RgBuff 80 1.333333 9.00 6.75

RgProc 80 1.333333 8.00 6.00

TrBuff 80 1.333333 2.00 1.50

TrProc 80 1.333333 6.67 5.00 19.25SpBuff 72.8 1.213333 6.00 4.95

SpProc 72.8 1.213333 7.28 6.00 10.95HsBuff 9.2 0.153333 0.90 5.87

HsProc 9.2 0.153333 3.83 25.00 30.87

1 3. Process Flow Measures Auto-Moto Financial Services- The Old Process Auto-Moto receives 1,000...

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