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1 3. Process Flow Measures Auto-Moto Financial Services- The Old Process Auto-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 = I T = I/R = 500/1,000 months = 0.5 month or 15 days. Process I p =500 1000/month 200/month 800/month

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2 3. Process Flow Measures 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 = 1000 I = I IR + I A + I B = 200 + 25 + 150 = 375 Inventory 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 days The new process has decreased the processing time from 15 days to 11.25 days. I B = 150 I A = 25 I R = 200

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3 3. Process Flow Measures 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.

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4 3. Process Flow Measures Flow Time at Each Sub-process (or activity) Average Flow Time for sub-process IR. Throughput R IR = 1,000 applications/month Average Inventory I IR = 200 applications T IR = 200/1,000 = 0.2 months = 6 days in the IR sub-process Average Flow Time for sub-process A. Throughput R A = 250 applications/month Average Inventory I A = 25 applications T A = 25/250 months = 0.1 months = 3 days in sub-process A. Average Flow Time for sub-process B. Throughput R B = 250 applications/month Average Inventory I B = 150 applications T B = 150/250 months = 0.6 months = 18 days in sub-proces B

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5 3. Process Flow Measures Routing, Flow Time, and Percentage of Each Flow units One flow unit at very macro level: Application 1000 flow units/month at very micro level: Each specific application Two flow units: Accepted and rejected Accepted-A: IR, A Accepted-B: IR, B Five flow units: Accepted-A, Accepted-B Rejected-IR, Rejected-A, Rejected-B T IR = 6 days T A = 3 days T B = 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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6 3. Process Flow Measures New Process: Intermediate Probabilities Initial Review T = 6 Subprocess A Review T = 3 Subprocess B Review T = 18 Accepted Rejected 100% 25% 50% 25% 70% 30% 90% 10% 80% 20%

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7 3. Process Flow Measures New Process: Intermediate Probabilities Initial Review T = 6 Subprocess A Review T = 3 Subprocess B Review T = 18 Accepted Rejected 100% 25% 50% 25% 17.5% 7.5% 22.5% 2.5% 80% 20% 50%

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8 3. Process Flow Measures 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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9 3. Process Flow Measures 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 Check our computations: Average flow time of an application 0.8(11.343)+0.2(10.875) = 11.25

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10 3. Process Flow Measures 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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11 3. Process Flow Measures Directions o) Draw the flow process chart b) On average how long a patient spend in ER? c) On average how many patients are in ER? Hints: Compute flow time in buffer 1 Compute average activity time of Doctor Compute the average flow time in this process Compute average flow time for a simple prescription patient Compute average flow time for a potential admission patient Compute number of patients in Doctor activity Compute the average number of patients in the process.

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12 3. Process Flow Measures Re-Check 7.6 7.5 37.1 54.2 T D = 0.9(5)+0.1(30) = 7.5 T SP = 7.6+2+37.1+5 = 51.7 T P = 0.1(76.7)+0.9(51.7) = 54.2 Flow Time T PA = 7.6+2+37.1+30 = 76.7

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13 3. Process Flow Measures 6.9 Inventory 1.8

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14 3. Process Flow Measures 49.7 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 Littles Law

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15 3. Process Flow Measures 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 patients turn come, each will take 5 minutes of a doctors time. The hospital anticipates that, on average, the emergency area will have only 1 patient waiting to be seen. As before, once that patients turn comes, he or she will take 30 minutes of a doctors time. Assume that, as before, 90% of all patients are Simple Prescriptions, assume, too, that the triage nurse is 100% accurate in making classifications. ER2; Problem 3.5 MBPF

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16 3. Process Flow Measures Directions o) Draw the flow process chart a) 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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17 3. Process Flow Measures Process Flow and TR=I Table

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18 3. Process Flow Measures I ER = 52.5 a) On average, how many patients are in ER? Method 1: Macro Method, a single flow unit R ER = 55/60 flow units / min, I ER = 52.5 b) On average, how long will a patient spend in ER? T ER = 52.5/(55/60) = 57.2 minutes Inventory and Flow Time; Macro Method

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19 3. Process Flow Measures Method 2: Micro Method, two flow units, Potential Admission and Simple Prescription T PA = 21.8+2+5.5+3+10.9 +30 = 73.2 T SP = 21.8+2+5.5+3+18.2 +5 = 55.5 T ER =.1(73.2)+.9(55.5) = 57.3 c) On average, how long will a potential admission patient spend in ER? 73.2 Flow Time; Micro Method

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20 3. Process Flow Measures 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. ER3; Problem 3.6 MBPF

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21 3. Process Flow Measures Directions o) Draw the flow process chart a) 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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22 3. Process Flow Measures Problem 3.6 5.5 Process Flow and Throughputs

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23 3. Process Flow Measures 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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24 3. Process Flow Measures Micro Method Compute SP and PA first Common = 21.8 +2+5.5+3 = 32.3 T SP = 32.3+18+5 = 55.3 Flow Time; Simple Prescription

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25 3. Process Flow Measures Flow Time; Potential Admission and Overall 0.9 0.01 0.09

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26 3. Process Flow Measures Recheck T= 55.3 (0.9) + 73.2 (0.09) + 96.2 (0.01) T= 57.32 T PA1 = 32.3 +10.9+30 = 73.2 ……(4.95 PA patients out of 5.5 PA patient: 90%) T PA2 = 73.2+18+5 = 96.2 (0.55 PA patients out of 5.5 PA patient: 10%) T PA = 73.2(.9) + 96.2(.1) =75.5 Flow Time; Potential Admission and Overall Recheck Again T SP = 55.3 is 90% of flow units, and T PA =75.5 is for 10% of flow units T = 55.3 (.9) + 75.5(.1) = 57.32

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27 3. Process Flow Measures 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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28 3. Process Flow Measures Process Flow and Throughputs

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29 3. Process Flow Measures Process Flow and Throughputs

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30 3. Process Flow Measures Process Flow and Throughputs

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31 3. Process Flow Measures Process Flow and Throughputs

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32 3. Process Flow Measures Average flow time = T = I/R = 43.68/(80/60) = 32.76 Process Flow and Throughputs

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33 3. Process Flow Measures Micro Method Compute SP and PA first Common = 6.75+6+1.5+5 = 19.25 T SP = 19.25+10.95 = 30.2 Process Flow and Throughputs

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34 3. Process Flow Measures T PA1 = 19.25 +30.87 = 50.12 ……(7.2 PA patients out of 9.2 PA patients) T PA2 = 19.25 +10.95 +30.87 = 61.07 (2 PA patients out of 9.2 PA patients) T PA = 50.12(7.2/9.2) + 61.07(2/9.2) = T PA = 50.12(0.782609) + 61.07(0.217391) =52.5 Recheck T= 30.20(70.8/80) + 50.12 (7.2/80) + 61.02 (2/80) T = 32.76 Process Flow and Throughputs

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