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3.5 MBPF. 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. Under the.

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Presentation on theme: "3.5 MBPF. 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. Under the."— Presentation transcript:

1 3.5 MBPF. 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. Under the proposed triage plan, entering patients will classify them as Simple Prescriptions or Potential Admits. While Simple Prescriptions will move on to an area staffed for regular care, Potential Admits will be taken to the emergency area. Planners anticipate that the initial examination 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. Recall that registration takes an average of 2 minutes per patient. 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. Under the proposed plan, how long on average, will a patient spend in the ER? On average, how long will a Potential Admit spend in the ER? On average, how many patients will be in the ER? Assume the process to be stable; that is, average inflow rate equals average outflow rate.

2 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? Hints: Average flow rate is still 55 per hour (50 is typo) Planners estimate that initial examination takes 3 minutes. It is the flow time in triage activity. It is 3 it is not 1. One, as we will see in the future chapters is an estimate for the theoretical flow time. Compute average flow rate in buffer 3 and buffer 4. Compute average flow time in all buffers. Compute average number of patients in all activities. 

3 Problem 3.5

4 IER= 52.5 a) On average, how many patients are in ER?
b) On average, how long will a patient spend in ER? Method 1: Macro Method, a single flow unit RER= 55/60 flow units per minute IER= 52.5 TER = 52.5/(55/60) = 57.2 minutes

5 b) On average, how long will a patient spend in ER?
Method 2: Micro Method, two flow units, Potential Admission and Simple Prescription TPA= = 73.2 TSP= = 55.5 TER= .1(73.2)+.9(55.5) = 57.3 c) On average, how long will a potental admission patient spend in ER? 73.2

6 3. 6 MBPF. Refer again to Exercise 3. 5
3.6 MBPF. 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. On average, how long does a patient spend in the ER? On average, how long does a Potential Admit spend in the ER? On average, how many patients are in the ER? Assume the process to be stable; that is, average inflow rate equals average outflow rate.

7 Directions o) Draw the flow process chart
b) On average how many patients are in this process? c) On average, how long a patient spend in this process? Hints: Compute flow rates at Buffer 3 Compute flow rates at Buffer 4

8 Problem 3.6 5.5

9 Problem 3.6

10 Macro: Average number of patients in the system = 20+1. 8+5+2. 8+1+2
Average flow time = 52.6/(55/60) = 57.3

11 Re-Check Micro Method Common = = 32.3 TSP= = 55.3 TPA1= = 73.2 ……(4.95 PA patients out of 5.5 PA patient: 90%) TPA2= =96.2 (0.55 PA patients out of 5.5 PA patient: 10%) TPA = 73.2(.9) (.1) =75.5 TSP= is 90% of flow units, and TPA =75.5 is for 10% of flow units T = 55.3 (.9) (.1) = 57.32 Re-Check again T= 55.3 (49.5/55) (4.95/55) (.55/55) T= (re-check)


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