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GI/GI/1

GI/GI/1. GI/GI/1. Interarrival times are independent and generally distributed {t_n} is an i.i.d. s equence of interarrival times t _n is distributed according to density f Arrival instants: T_{n+1} = T_n + t_n Service times are independent and generally distributed

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GI/GI/1

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  1. GI/GI/1

  2. GI/GI/1 • Interarrival times are independent and generally distributed • {t_n} is an i.i.d. sequence of interarrival times • t_n is distributed according to density f • Arrival instants: T_{n+1} = T_n + t_n • Service times are independent and generally distributed • {s_n} is i.i.d. sequence of service times • s_n is distributed according to density g

  3. Waiting times / Lindley recursion • The first customer in a busy period experiences no waiting time. • The next has waiting time is s1 > tn w2 = (s1-t1)= (s1-t1)+=(w1 + s1-t1)+ • 2 subsequent customers arrive at Tn and Tn+1 respectively • Completion time of the first customer is Cn = Tn + wn + sn • The 2nd customer waits from Tn+1 to Cn, i.e. from Tn + tn to Tn + wn + sn • Thus he waits for wn+1 =(Cn - Tn+1 )+ =(wn + sn - tn)+ • If wn = 0 the busy period has ended

  4. A closed form solution • wn+1 =(wn + sn - tn)+ • Sn = i=1n (si-ti) • Sn+1 = Sn + sn – tn • wn = max{Sn ,Sn-S1 ,Sn-S2 ,.., Sn-Sn} • Proof wn+1¸ wn + sn – tn wn+1 - wn¸ sn – tn ¸Sn+1 –Sn • So wn – wn-k¸ Sn – Sn-k forall n ¸ k ¸ 0

  5. Closed form solution • Proof (cont’d) • wn – wn-k¸ Sn – Sn-k forall n ¸ k ¸ 0 • wn¸ Sn – Sn-k + wn-k ¸ Sn – Sn-k forall n ¸ k ¸ 0 => wn¸ max{Sn ,Sn-S1 ,Sn-S2 ,.., Sn-Sn} • Conversely if n is in a busy period shared with k all waiting times in between are positive and wn = i=kn (si-ti) = Sn – Sn-k • If n initiates a busy period wn = 0 = Sn-Sn • q.e.d.

  6. A comparative sequence • Mn = maxi <= n Sn • Let Zn =Sn and Zn-k = Sn –Sk for n ¸ k ¸ 0 • We have Zn-k+1= Zn-k + (sk-tk), i.e. {Zn-k} is a random walk of increments from {sk-tk} • So {Sn, Sn-1, .., S1, 0} has the same distribution as {Sn ,Sn-S1 ,Sn-S2 ,.., Sn-Sn} • Thus wn is distributed like Mn • We may study {Mn} instead of wn

  7. Limit properties • Let E(sk)=1/¹ and E(tk)=1/¸ • Let ½ = ¸ / ¹ • Assume ½ <1 => E(sk)-E(tk)<0 • From the LLN (Strong low of large numbers) Sn /n -> E(sk)-E(tk)<0 for n -> 1 (w.P.1) • So Sn -> -1 (w.P.1) • So for some n, Mn = M= sup n Sn (w.P.1) • Wn =DMn ->D M

  8. Convergence in distribution • P({9 n Mn =M})=1 =P([n {Mn=M}) =lim L -> 1P([nL{Mn=M}) = lim L -> 1P( {ML=M}) = 1 lim L -> 1P({ML M}) = 0 A={ML2 A} B={M 2 A} A [ B =A [ (B/A) = B [ (A/B) P(A) + P(B/A) = P(B) + P(A/B) P(A)=P(B)+P(A/B)-P(B/A) -> P(B) ?? A/B µ{ML M}, B/A µ{ML M} limL -> 1 P(A/B) = limL -> 1 P(B/A) ·lim L -> 1P({ML M}) = 0

  9. Limit properties • Now let s and t be generic service and interarrival times respectively • Consider (s-t+M)+ = max {0,s-t, s-t +(s1 –t1), s-t +(s1–t1) +(s2 –t2),..} = max {0,s0-t0, s0-t0 +(s1 –t1), s0–t0 +(s1 –t1) +(s2 –t2),..} = (by renumbering) max {0,s1-t1, s1-t1 +(s2 –t2), s1–t1 +(s2 –t2) +(s3 –t3),..} =D M

  10. Limit properties – Lindley equation • (s-t+M)+ =D M (ii) • wn ->D M • Let H be the distribution of M • Then from (ii) • H(m) = s H(m-x) fs(x) dx • Where fs is the density of s-t • Solution of (ii) is somewhat complicated !! • However since we know {wn} converges in distribution for ½ < 1, we may just remove indexes in the recursion for w_n, i.e w = (w+s-t)+

  11. Limit properties – Lindley equation • wn+1 = (wn+s-t)+ =(wn+s-t)+ (wn+s-t)+ - (wn+s-t) • (x)+ - x = max{0,x}-x = max{-x,0} = -min{x,0} • So (wn+s-t)+ =(wn+s-t) – min{wn+s-t,0} =(wn+s-t) –X • Now if wn+ sn- tn < 0 then it equals –In • In is the idle period until the (n+1)st customer

  12. Limit properties – Lindley equation • If I is a generic stationary idle period then • Now -E(X)=E(– min{w_n+s-t,0})= P(idle)E(I|idle) • So E(w)=E(w+s-t)-P(idle)E(I|idle) or • E(s-t)=P(idle)E(I|idle)=E(X) • E2(s-t)=E2(X) ·E(X2)

  13. An estimate (in red) • wn+1 = (wn+s-t)+ =(wn+s-t)+X • wn+1-X = (wn+s-t) • Squaring and taking expectations yields • E(wn+12)+E(X2)-2E(wn+1 X) =E(wn 2)+E((s-t)2)+2E(wn(s-t)) (ii) • Now X=min{wn+s-t,0} and wn+1 = max{0,wn+s-t} • So if X<0, wn+1=0 and if wn+1>0, X<0 which cancels the last l.h. term in (ii) • Also wnand (s-t) are independent • Altogether • Since {wn} converges in distribution to w we may ultimately write • E(X2) =E((s-t)2)+2E(w)E(s-t) or • E(w)= (E((s-t)2) - E(X2))/2/E(t-s) = (V(s)+V(t)+E((s-t)2) - E(X2))/2/E(t-s) ·(V(s)+V(t)) )/2/E(t-s)

  14. Waiting time probabilities • P(wn¸ W) for W > 0 ?? • wn+1 = (wn+sn-tn) + • wn+1¸ W  wn+sn-tn¸ W so • P(wn+1¸ W) = P(wn + sn -tn¸ W) = P(wn¸ W-(sn-tn)) =sP(wn¸ W-(sn- tn)|sn-tn=y) fs(y) dy =sP(wn¸ W-y) fs(y) dy (since wn, sn, tn independent)

  15. Waiting time probabilities • P(wn+1¸ W) = sP(wn¸ W-y) fs(y) dy = s-1W P(wn¸ W-y) fs(y) dy + sW1 fs(y) dy = s-1W P(wn¸ W-y) fs(y) dy + 1-Fs(W) • Consider a function f:R -> [0,1] • Assume • f(W) ·s-1W f(W-y) fs(y) dy + 1-Fs(W) • Then P(wn¸ W) · f(W) for all W>0  P(wn+1¸ W) · f(W) • Since w1=0, P(wn¸ W) = 0 · f(W) for all W>0 • So P(wn¸ W) ·f(W) for all W>0 and n>0

  16. Finding f • Let Á(S) = E(exp(S(s-t))) (Laplace transform of the distribution of (s-t)) • Á(0)=1 • Á’(0)=E(s)-E(t)=1/¹ – 1/¸ < 0 (for ¸ < ¹, ½ <1) • So if Á is analytic (!!!) in 0 there is S>0 so that Á(S)<1 • Choose f(W)=exp(-SW)

  17. Checking f • f(W) ·s-1W f(W-y) fs(y) dy + 1-Fs(W) • s-1W exp(-S(W-y)) fs(y) dy + 1-Fs(W) = s-11 exp(-S(W-y)) fs(y) dy -sW1 exp(-S(W-y)) fs(y) dy + 1-Fs(W) = exp(-SW) s-11 exp(Sy)) fs(y) dy -sW1 exp(-S(W-y)) fs(y) dy + 1-Fs(W) = exp(-SW) s-11 exp(Sy)) fs(y) dy -sW1 (1-exp(-S(W-y))) fs(y) dy = f(W) Á(S) -sW1 (1-exp(-S(W-y))) fs(y) dy · f(W)

  18. Result • Let S be the largest real so that Á(S)<1 • Then P(wn¸ W) ·exp(-SW) (Taking the largest real S gives the highest decay rate)

  19. Equivalent BW • Service time distribution may be scaled by bandwidth, i.e. P(BW s · S) • For a given model each BW gives rise to particular queueing statistics, i.e. EBW(w) or PBW(w ¸ W) • EBW (Equivalent Bandwidth) • EBW = arg minBWEBW(w) · W • EBW = arg minBWPBW(w ¸ W) ·P

  20. Miniproject • Consider the M/G/1 queue you previously simulated • Compute average queue length estimates based on the GI/GI/1 waiting time estimate (Hint: Little) • Compare with the PK result • Compute the Laplace transform Á(S) • Find the largest real S to that Á(S)<1 • Setup appropriate equivalent bandwidth conditions and show results for a representative set of input data

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