Question

In: Operations Management

During the last 10 hours a total of 23 patients arrived at an emergency clinic with...

During the last 10 hours a total of 23 patients arrived at an emergency clinic with one doctor. The average time for the doctor to examine a patient is 21.5 minutes.

What is the average number of patients waiting to be assisted?
What is the average time waiting to be assisted (in minutes)?
What is the probability there are no patients waiting?
What is the probability there are at most 2 patients waiting?

Please provide the answer as:

=x.x patients (for example 12.5 patients, 2.7 patients)
=x.x (for example 234.2 minutes, 0.6 minutes)
=x.x% (for example 67.4%, 34.1%)
=x.x%

Expert Solution

λ = Arrival rate( number of customers arriving per unit time ) = 23/10 = 2.3 patient per hour

Service Time = 21.5 minutes per patient

=> μ = Service Rate of the customers ( number of customers that can be served per unit time ) = 1/21.5 patient per minute = 60/21.5 = 2.79 patient per hour

p = λ/μ = average utilization = 2.3/2.79 = 0.82

L = Avg number of patients in the service system

L = 2.3/(0.49) = 4.69

and L(q) = p*L = Avg number of patients waiting in the line

1. What is the average number of patients waiting to be assisted?

= L(q) = 0.82*4.69 = 3.84

2. What is the average time waiting to be assisted (in minutes)?

= W(q) = p*W = 0.82*(1/(0.49)) = 1.67 minutes

3. What is the probability there are no patients waiting?

P(0) = 1 – λ / μ

i.e the probability that there are 0 customers in the system = 1- 2.3/2.79 = 1-0.82 = 0.18 = 18 %

4. What is the probability there are at most 2 patients waiting?

Pn = (λ / μ)n P0 = Probability of 'n' customers in the system

P(1) = (λ / μ)1 P0 = 0.82*0.18 = 0.14

P(2) = (λ / μ)2 P0 = 0.82*0.82*0.18 = 0.12

Probablity of atmost 2 persons waiting = P(0) + P(1) + P(2) = 0.18 + 0.14+0.12 = 0.44 = 44 %

keosha answered 1 year ago

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