Workers come to a tool store room to receive special tools (required by them) for accomplishing a particular project assigned to them. The average time between two arrivals is 60 seconds and the arrivals are assumed to be in Poisson distribution. The average service time (of the tool room attendant) is 40 seconds. Determine:

i. Average queue length

ii. Average length of non-empty queues

iii. Mean waiting time of an arrival

iv. Average waiting time of an arrival (worker) who waits

Mean arrival rate: λ = 1/60 s; Mean service rate: $μ = 1/40 s$

Traffic intensity (or utilization factor): $ρ = \dfracλ μ = \dfrac{1/60}{1/40} = 23$

1. Average queue length:

   $Lq = \dfrac{ρ^2}{1-ρ} = \dfrac{(2/3)^2}{1-2/3} = 1.33$

2. Average length of non-empty queues:

   $L = \dfrac{1}{1-ρ} = \dfrac{1}{1-2/3} = 3$

3. Mean waiting time of an arrival:

• Waiting time in the system (including service time):

  $Ws = \dfrac{1}{μ - λ} = Error! = 120s$

• Waiting time in the queue (excluding service time):

  $Wq = \dfrac{ρ}{(μ - λ)} = \dfrac23 × 120 = 80s$

1. Average waiting time of an arrival who waits:

   $W = \dfrac{1}{μ(1-ρ)} = \dfrac{40}{1-2/3} = 120s$