Question

Assume that you have a printer that can print an average file in two minutes. Every two and a half minutes a user sends another file to the printer. Assuming both inter-arrival and service time follow the exponential distribution, in steady state condition, (a) As an average, how long does it take before a user can get their output? (10 points) (b) To speed things up you can buy two similar printers that is exactly the same as the one you have. In steady state condition, how long will it take for a user to get their files printed if you had three identical printers (configured parallel to each other having one queue)? (10 points) (c) Another solution is to replace the existing printer with one that can print a file in an average of one minute following of exponential distribution. How long does it take for a user to get their output with the faster printer? (10 points) (d) For part b, (three parallel printers’ configuration) assume there is queue capacity of 12 jobs, calculate again and compare your results. (10 points)

Answer 1

Average arrival rate, λ = 1 in 2.5 minutes = 24 per hour
Average service rate, μ = 1 in 2 minutes = 30 per hour

(a)

Total time required, on an average, in the system (i.e. to get the jobs completed), W_s = 1 / (μ - λ) = 1 / (30 - 24) hours = 10 minutes.

(b)

Now the number of servers, s = 3. Use the following formulae to compute the idle server probability (P₀) and average time in the system (W_s).

Input λ = 24, μ = 30, and s = 3 in the above formulae to get, P₀ = 0.447 and W_s = 0.0341 hours = 2.05 minutes

(c)

Average service rate, μ = 1 in 1 minute = 60 per hour

Total time required, on an average, in the system (i.e. to get the jobs completed), W_s = 1 / (μ - λ) = 1 / (60 - 24) hours = 1.667 minutes.

(d)

λ = 24
μ = 30
s = 3
Queue capacity, N = 12

Utilization, ρ = λ /(s.μ) = 24/(3*30) = 0.267

The following formulae will be used:

Calculations will give us, P₀ = 0.447, P_N will be very close to zero. So, we can take λ_e = λ = 24, L_q will be 0.0189

So, Wq = Lq / λ_e = 0.0189 / 24 = 0.000788

Ws = Wq + 1/μ = 0.000788 + (1/30) = 0.034 hours = 2.05 minutes (which is very close to part-b answer). This is because the N=12 is quite large almost equaling to the infinite queue system.