Question

Queueing theory M/M/K exercise

A supermarket currently has three cash machines. The manager wishes to improve the service either hiring a new checker or by installing bar detectors in existing cash machines. Historical facts indicate that customers arrive according to a poisson process at an arrival rate of 45 per hour and the time necessary to serve a customer follows an exponential distribution with service rate of 20 customers per hour. The manager estimates that modernizing cash machines with a barcode equipment would increase service rate in 20%. The accounting department suggests that the cost of a customer in the queue should be valued at $32 per hour and the cashier's cost per hour should increased to $18, including benefits. However, to cover the cost of the barcode equipment, the cashier's cost must increased to $ 23. Analyze the current situation and determine if the company should hire one more checker or installing a barcode equipment on existing cash machines. (Assume a single queue).

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Answer 1

Present condition

Arrival rate, λ = 45 per hr.
Service rate μ = 20 per hr.
No. of servers, s = 3

λ/μ = 2.25

Compute the idle server probability (P₀) and the average queue length (L_q) using the following two formulae:

So,

P₀ = 1 / ((1+(2.25) + (1/2)*(2.25)^2) + (1/6)*(2.25)^3*(3*20) / (3*20-45)) = 0.0748

L_q = 45*20*(2.25)^3*0.0748/(2*(3*20-45)^2) = 1.703

Cost of waiting = L_q * $32 = 1.703*32 = 54.5 per hour
Cost of the servers = s * $18 = 3*18 = 54 per hour

So, total cost pr hour = 54.5+54 = $108.5 per hour --------------1

Proposed-1: Hire a new server

Arrival rate, λ = 45 per hr.
Service rate μ = 20 per hr.
No. of servers, s = 4

λ/μ = 2.25

P₀ = 1 / ((1+(2.25) + (1/2)*(2.25)^2 + (1/6)*(2.25)^3) + (1/24)*(2.25)^4*(4*20) / (4*20-45)) = 0.0988

L_q = 45*20*(2.25)^4*0.0988/(6*(4*20-45)^2) = 0.31

Cost of waiting = L_q * $32 = 0.31*32 = 9.92 per hour
Cost of the servers = s * $18 = 4*18 = 72 per hour

So, total cost pr hour = 9.92+72 = $81.92 per hour-------------2

Proposed-2: Install barcode equipment

Arrival rate, λ = 45 per hr.
Service rate μ = 20 per hr. * (1+20%) = 24 per hr.
No. of servers, s = 3

λ/μ = 1.875

P₀ = 1 / ((1+(1.875) + (1/2)*(1.875)^2) + (1/6)*(1.875)^3*(3*24) / (3*24-45)) = 0.1322

L_q = 45*24*(1.875)^3*0.1322/(2*(3*24-45)^2) = 0.646

Cost of waiting = L_q * $32 = 0.646*32 = 20.67 per hour
Cost of the servers = s * $23 = 3*23 = 69 per hour

So, total cost pr hour = 20.67+69 = $89.67 per hour-------------3

So, after comparing 1, 2, and 3, it seems that hiring a new server will be more economical.