Question

A company has started a phone service that uses overseas doctors to provide emergency medical consultations. The responding doctors are based in a country with low wages but with a highly skilled pool of physicians. Responding to each call takes on average 15 minutes. At any given moment in time, there are 4 doctors overseas on duty. Calls arrive every 5 minutes on average and standard deviation of the inter-arrival time is 5 minutes. If no doctor is available overseas, the call is rerouted to the U.S. where a local physician answers the question. A local physician is always available to take a call. In this case, the firm pays $50 to the local physician.​

Now assume that the demand has increased to 20 calls per hour for the overseas doctors. Responding to each call still takes on average 15 minutes. At any given moment in time, there are 4 doctors overseas on duty. The coefficient of variation of customers’ inter-arrival time is 1. If no doctor is available overseas, the call is rerouted to the U.S. where a local physician answers the question.

Q18. What is the number of calls directed to the doctors in the US every hour?

Q19. The company has set a new goal to direct no more than 2 calls per hour to the doctors in the US. How many doctors do they need overseas?

Answer 1

Q 18.

The problem is that of Multiple-server Queuing Model.

Number of servers, M = 4.

Arrival rate, = 12 calls per hour.

Service rate, = 1 / (15 / 60) = 4 calls per hour.

Probability of zero customers in the system:

Probability of zero customers in the system, P₀ = 1 / {[(1/0!) x (3/1!) + (9/2!) x (27/3!)¹] + [(81/4!) x 4]} = 0.04

Let average number of calls waiting in line for service = L_q.

L_q =[..( / )^M / {(M – 1)! (M. – )²}] x P₀ = [{12 x 4 x (12/4)⁴} / {(4 – 1)! x ((4 x 4) – 12)²}] x 0.04 = 1.62

Number of calls directed to the doctors in US every hour = 1.62

Q 19.

Let number of servers M = 6.

Arrival rate, = 20 calls per hour.

Service rate, = 1 / (15 / 60) = 4 calls per hour.

Probability of zero customers in the system:

Probability of zero customers in the system, P₀ = 1 / {[(1/0!) x (5/1!) + (25/2!) x (125/3!) + (625/4!) (3125/5!)] + [(15625/6!) x 6]} = 0.005

Let average number of calls waiting in line for service = L_q.

L_q =[..( / )^M / {(M – 1)! (M. – )²}] x P₀ = [{20 x 4 x (20/4)⁶} / {(6 – 1)! x ((6 x 4) – 20)²}] x 0.005 = 3.26 (which is greater than 2).

Let number of servers M = 7.

Probability of zero customers in the system:

Probability of zero customers in the system, P₀ = 1 / {[(1/0!) x (5/1!) + (25/2!) x (125/3!) + (625/4!) (3125/5!) + (15625/6!)] + [(78125/7!) x (7/2)]} = 0.006

Let average number of calls waiting in line for service = L_q.

L_q =[..( / )^M / {(M – 1)! (M. – )²}] x P₀ = [{20 x 4 x (20/4)⁷} / {(7 – 1)! x ((7 x 4) – 20)²}] x 0.006 = 0.81 (which is less than 2).

Number of doctors needed overseas so as not to direct more than 2 calls per hour = 7.