Question

You need coffee and FAST! You hop out of your Uber at your favorite place, but with a history of a line. Usually you can expect there to be 17 people in line any time you come here (there is only one order taker). Service time averages 40 seconds. SHOW ALL WORK!

Remembering ρ=λ/μ, L=λW, W=1/(μ-λ), L_q=ρL=L-ρ, W_q=ρW=W-1/μ,P_n=(1-ρ) ρ^n

What is the utilization of the server? _____%

What is the expected time between people arriving to this system? _____ seconds

What is the expected wait until service starts? _____ seconds

What is the probability of 3 or fewer people in line upon your arrival? ____%

Answer 1

Average number of people in queue, Lq = 17

Average service time = 40 seconds

Average service rate, μ = 1/Average service time = 1/40 per second = 0.025

Using the queuing formulas, ρ=λ/μ, L=λW, W=1/(μ-λ), Lq=ρL=L-ρ, Wq=ρW=W-1/μ, Pn=(1-ρ) ρ^n

Lq = ρL = ρλW = ρλ/(μ-λ) = (λ^2/μ)/(μ-λ)

We are given Lq and μ, substituting these values in the above expression, we get,

17 = (λ^2/0.025)/(0.025-λ)

0.425-17λ = 40λ^2

or, 40λ^2 + 17λ - 0.425 = 0

Solving for λ, we get

λ = (-17+sqrt(17^2+4*40*0.425))/(2*40) = 0.023681

Utilization of the server, ρ=λ/μ = 0.023681/0.025 = 94.7 %

Expected time between people arriving to this system (interarrival time) = 1/λ = 1/0.023681 = 42.23 seconds

Expected wait until service starts, Wq = Lq/λ = 17/0.023681 = 718 seconds

Probability of 3 or fewer people in line upon arrival = Probability of 4 or fewer people in system (including one being served)

= P0+P1+P2+P3+P4

= (1-0.947)*0.947^0+(1-0.947)*0.947^1+(1-0.947)*0.947^2+(1-0.947)*0.947^3+(1-0.947)*0.947^4

= 23.75 %