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Operations Research
Multi Channel Queuing Model: M / M / C: (∞ / Fcfs)
Introduction: Customers are generated by limited pool of potential customers i.e. finite population. The total customer’s population is M and n represents the number of customers already in the system (waiting line), any arrival must come from M - n number that is not yet in the system.
Multi-channel queuing model:
- Here the length of waiting line depends on the number of channels engaged.
- In case the number of customers in the system is less than the number of channels i.e. n < c, then there will be no problem of waiting and the rate of servicing will be nμ as only n channels are busy, each servicing at the rate m. In case n = c, all the channels will be working and when n ≥ c, then n .
- c elements will be in the waiting line and the rate of service will be cμ as all the c channels are busy. Various formulae we have to use in this type of models are:
- Average number of units in waiting line of the system = E (n) = [ρ pc /(1− ρ)2 ] = {[ λ μ ( λ /μ ) ]/[( c-1)!( c μ- λ)2 ]} p0 ( λ /μ )
- Average number in the queue = E (L) = [ pcρ/(1− ρ)2 ] C.μ ={[λ.μ.(λ /μ)c ]/[(c −1)! (cμ − λ)2 ]} p
- Average queue length = Average number of units in waiting line number of units in service
- Average waiting time of an arrival = E (w) = (Average number of units in waiting line) / λ =
- (Average number of items in the queue) / λ = [( pcρ) / λ (1− ρ)] (C.μ / λ) μ× λ μ c c − cμ − λ 2 × p μ
- Probability that all the channels are occupied = p (n ≥ c) = [1/(1− ρ)] pc = [μ× (λ /μ)c ]p0 /[(c −1)!(cμ − λ)]
- Probability that some units has to wait = p (n ≥ c 1) = [ρ pc /(1− ρ)] = 1− p(n ≥ c) =1−[μ×(λ /μ)c ] p0 /[(c − i)!(cμ − λ)]
- The average number of units which actually wait in the system =
- Average number of idle channels = c . Average number of items served Efficiency of M/M/c model: = (Average number of items served) / (Total number of channels) Utilization factor = ρ = (λ / cμ)