Ted Glickman, the administrator at D. C. General Hospital emergency room, faces
ID: 445459 • Letter: T
Question
Ted Glickman, the administrator at D. C. General Hospital emergency room, faces the problem of providing treatment for patients who arrive at different rates during the day. There are 4 doctors available to treat patients when needed. If not needed, they can be assigned other responsibilities (such as doing lab tests, reports, X- ray diagnoses) or else rescheduled to work at other hours.
It is important to provide quick and responsive treatment, and Ted thinks that, on the average, patients should not have to sit in the waiting area for more than 6 minutes before being seen by a doctor. Patients are treated on a first- come, first- served basis and see the first available doctor after waiting in the queue. The arrival pattern for a typical day is as follows:
Time ............ Arrival Rate
9 a. m.- 3 p. m. ..... 8 patients/ hour
3 p. m.-8 p. m. ... 3 patients/ hour
8 p. m. -midnight ... 13 patients/ hour
Arrivals follow a Poisson distribution, and treatment times, 12 minutes on the average, follow the exponential pattern.
a) How many doctors should be on duty during each period to maintain the level of patient care expected?
b) What condition would exist if only one doctor were on duty between 8 p.m and midnight?
Explanation / Answer
a)
9 A.M. -3 P.M.
Lambda= 8 patients/hour
Mu= 6 patients/hour
Utilization factor (uf) = lambda/ mu
= 8/6=1.33
According to hall’s thumb rule
s>= max[1 , uf+sqrt(uf)]
max[1,1.33+sqrt(1.33)]
max(1,2.48)
= 2.48
The least no. of doctor required between 9 A.M. to 3 P.M. is 3
3 P.M. -8 P.M.
Lambda= 3 patients/hour
Mu= 6 patients/hour
Utilization factor (uf) = lambda/ mu
= 3/6=0.5
According to hall’s thumb rule
s>= max[1 , uf+sqrt(uf)]
max[1,0.5+sqrt(0.5)]
max(1,1.2)
= 1.2
The least no. of doctor required between 3 A.M. to 8 P.M. is 2
8 P.M. - Midnight
Lambda= 13 patients/hour
Mu= 6 patients/hour
Utilization factor (uf) = lambda/ mu
= 13/6=2.16
According to hall’s thumb rule
s>= max[1 , uf+sqrt(uf)]
max[1,2.16+sqrt(2.16)]
max(1,3.63)
= 3.63
The least no. of doctor required between 8 P.M. to Midnight is 4
b)
Lambda= 13 patients/hour
Mu= 6 patients/hour
Lambda> mu therefore, line would back up indefinitely
1 doctor can treat 4*6 =24 but no. of patients 4*13 = 52.therfore untreated patients will go untreated.
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