Question

The 911 number of the city of Turtle Creek receives emergency calls for a life-support vehicle (LSV) at a mean rate of 15 calls per hour. The interarrival time between these calls has an exponential probability distribution. The time that elapses from the dispatch of an LSV in response to a call until the LSV is available to respond to another call has an exponential probability distribution with a mean of 48 minutes. Turtle Creek defines the average response time as the average time between the receipt of a call and the dispatch of an LSV to attend to this call. Calls are processed on a first-come first-served basis. Turtle Creek wants an LSV fleet of sufficient size to keep average response time to less than 2 minutes.

1. Turtle Creek is considering two different sizes for its fleet of LSVs: 15 LSVs or 20 LSVs. For each of these fleet sizes, compute the average response time (how long, on average, a call has to wait in the queue until an LSV is dispatched to it).
2. Does either of the fleet sizes meet Turtle Creek’s goal of an average response time of 2 minutes.

Answer 1

(a)

Average arrival rate, ra = 15 per hour = (15/60) per minute = 0.25 per minute
Average service time, te = 48 minutes

We have to compute the average waiting time (CTq) in the queue for the number of servers, s = 15 and 20. This is also the average response time as it is defined.

Since the number of servers is very large, we will take the Kingman's approximation approach [otherwise, using software such as Excel QM will also give exact results]

Note that since the interarrivals and service times are Poisson and Exponentially distributed the coefficient of variation 'ca' and 'ce' will be 1.

Utilization, u = te * ra / m

For m = 15, u = 48*0.25/15 = 0.80

CTq = ((1^2+1^2)/2)*(0.8^(SQRT(2*(15+1))-1))*48 / (15*(1-0.8)) = 5.66 minutes

For m = 20, u = 48*0.25/20 = 0.60

CTq = ((1^2+1^2)/2)*(0.6^(SQRT(2*(20+1))-1))*48 / (20*(1-0.6)) = 0.36 minutes

Using software for M/M/s queue, we will get exact values as 5.107 minutes and 0.145 minutes for these above two approximations.

(b)

So, as we can see when the number of servers=20, the CTq value is < 2 minutes and the goal is met.