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Chapter 4Simulation and analytical modelling pFrom this figure we can appreciate that the highest value for the variance is when ρ =0.5, which results in σ 2= 0.000000036 ≈ 0 . By substituting σ 2 = 0 in Equation 4.10,Xthe mean waiting time in queue is then given by:λ ⋅ X 2W =(4.32)2 ⋅ (1 − ρ)By substituting Equations 4.29 and 4.32 in Equation 4.4, the derivation of the end-toendpacket delay is complete and is given by:λλλ ⋅ XD ete = ( DSl_ Tx+ Dprop) ⋅ (1.5 −2⋅µ) + (0.5 + MCI+ Pkmci+2⋅µ) ⋅ MCIt+ (4.33)λ2 ⋅ (1 − )X2µ4.4.2 Throughput formulation for a single node scenarioHaving formulated the total end-to-end packet access delay, the throughput, S, is givenby the number of packets serviced per second (L x ) multiplied by the packet size, thus:S = L x⋅ Pk size(4.34)In order to obtain L x , we need to calculate first the mean number of packet serviced perbusy cycle,L bc . Then, the number of packets serviced per second can be easilyobtained by dividingL bc between the average time of a busy cycle, T bc, thus:LbcLx= (4.35)T bcFrom analysing Figure 4.17 a busy cycle ( Tbc) can be defined as the sum of a busyLσ 2 X (x10 -9 )a rriv a ld e p a rtu re5 04 03 02 01 0T bp 1 T ip1 0 T b p2T ip 2T bp 3TT bp n ip30 0 .1 0 .2 0 .3 0 .4 0 .5 0 .6 0 .7 0 .8 0 .9 1T b c 1T b c 2T b c 3T b c nρ = λ µFigure 4.17 – Time of busy cycle.Figure 4.16 – Variance of the mean service time.T ip ntim e4-28