Question

Assume a standard customer, Johnny Rose, at a shop will spend 20 mins in the shop. The shop has 4 customers on average and service time is random with known distribution. The shop has 4 customers on average and service time is random with known distribution. How many customers does the shop serve during a typical open hour?

Now let's say the shop would like to decrease the average number of customers at any point, but assume it cannot control/influence arrival patterns of customers or service rate. How can this be achieved? Support with calculations and clear principles.

Answer 1

using Little's law :
Ls = λWs
where Ls is the average number of customers in system
λ is the customer arrival rate
Ws is the waiting time in system

given, Ls = 4; Ws = 20min = 1/3 hrs

Therefore,  

λ = 12 customers per hour {i.e., 12 customers arrive per hour in the shop}

so,

Ls = ρ / (1 − ρ)

where ρ is utilization of the server = = λ/µ;  µ is the mean service rate (assuming single server)

Therefore, ρ = 0.8

& µ = 15 ; i.e., the shop can serve 15 customers per hour

Now, to decrease the average number of customers in shop, they would have to increase their service rates. This could be achieved by adding one more server.

Adding one more server would double their service rate,

(µ = 30 (assuming new server is as efficicent as the old one))

then, ρnew = 12/30 = 0.4

Lsnew = 0.4 / (1-0.4) = 2/3 or on an average less than one customer in the shop.

Even if we assume that the new server isn't as effecient and the service rate goes upto µ = 20 only;

then, ρnew = 12/20 = 0.6

Lsnew = 0.6 / (1-0.6) = 1.5; the number of customers in the shop still go down substantially

(please upvote)