Question

Problem 7-20 L. Winston Martin (an allergist) has an excellent system for handling his regular patients who come in just for allergy injections. Patients arrive for an injection and fill out a name slip, which is then placed in an open slot that passes into another room staffed by one or two nurses. The specific injections for a patient are prepared, and the patient is called through a speaker system into the room to receive the injection. At certain times during the day, patient load drops and only one nurse is needed to administer the injections. Use Exhibit 7.12 Let's focus on the simpler case of the two-namely, when there is one nurse. Also, assume that patients arrive in a Poisson fashion and the service rate of the nurse is exponentially distributed. During this slower period, patients arrive with an interarrival time of approximately 4 minutes. It takes the nurse an average of 2.80 minutes to prepare the patients' serum and administer the injection.

Answer 1

(c)

L = average arrival rate = 1 in 4 min = 15 per hour
M = average service rate = 1 in 2.8 minutes = 21.429 per hour

P_n>2 = Probability of 3 or more patients in the premises = (L/M)²⁺¹ = (2.8/4)³ = 0.343

(d)

Utilization = L/M = 2.8 / 4 = 0.7 or 70%

(e)

M = 1 in 4 minutes = 15 per hour, so, L/ M = 1.0

Lq = average number of patients waiting can be found using the table for L/M = 1.0

So, Lq = 0.0454

Using Little's Law, Wq = average time patient in the queue = Lq / L = 0.0454 / 15 hour = 4 x 0.0454 minute = 0.1816 minutes

Ws = average time patient in the system = 0.1816 + 4.0 = 4.18 minutes

[Note: this answer may change a bit in rounding off as I have used my table which is not that is given with this question which you haven't posted]