1. GI/GI/1

2. 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 s 1 > t n w 2 = (s 1 -t 1 )= (s 1 -t 1 ) + =(w 1 + s 1 -t 1 ) + 2 subsequent customers arrive at T n and T n+1 respectively Completion time of the first customer is C n = T n + w n + s n The 2nd customer waits from T n+1 to C n, i.e. from T n + t n to T n + w n + s n Thus he waits for w n+1 = (C n - T n+1 ) + =(w n + s n - t n ) + If w n = 0 the busy period has ended

4. A closed form solution w n+1 =(w n + s n - t n ) + S n =   n (s i -t i ) S n+1 = S n + s n – t n w n = max{S n,S n -S 1,S n -S 2,.., S n -S n } Proof w n+1 ¸ w n + s n – t n w n+1 - w n ¸ s n – t n ¸ S n+1 –S n So w n – w n-k ¸ S n – S n-k forall n ¸ k ¸ 0

5. Closed form solution Proof (cont’d) w n – w n-k ¸ S n – S n-k forall n ¸ k ¸ 0 w n ¸ S n – S n-k + w n-k ¸ S n – S n-k forall n ¸ k ¸ 0 => w n ¸ max{S n,S n -S 1,S n -S 2,.., S n -S n } Conversely if n is in a busy period shared with k all waiting times in between are positive and w n =   n (s i -t i ) = S n – S n-k If n initiates a busy period w n = 0 = S n -S n q.e.d.

6. A comparative sequence M n = max i <= n S n Let Z n =S n and Z n-k = S n –S k for n ¸ k ¸ 0 We have Z n-k+1 = Z n-k + (s k - t k ), i.e. {Z n-k } is a random walk of increments from {s k -t k } So {S n, S n-1,.., S 1, 0} has the same distribution as {S n,S n -S 1,S n -S 2,.., S n -S n } Thus w n is distributed like M n We may study {M n } instead of w n

7. Limit properties Let E(s k )=1/ ¹ and E(t k )=1/ ¸ Let ½ = ¸ / ¹ Assume ½ E(s k )-E(t k )<0 From the LLN (Strong low of large numbers) S n /n -> E(s k )-E(t k ) 1 (w.P.1) So S n -> - 1 (w.P.1) So for some n, M n = M= sup n S n (w.P.1) W n = D M n -> D M

8. Convergence in distribution P({ 9 n M n =M})=1 =P( [ n {M n =M}) =lim L -> 1 P( [ n L {M n =M}) = lim L -> 1 P( {M L =M}) = 1 lim L -> 1 P({M L  M}) = 0 A={M L 2 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 {M L  M}, B/A µ {M L  M} lim L -> 1 P(A/B) = lim L -> 1 P(B/A) · lim L -> 1 P({M L  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 +(s 1 –t 1 ), s-t +(s 1 –t 1 ) +(s 2 –t 2 ),..} = max {0,s 0 -t 0, s 0 -t 0 +(s 1 –t 1 ), s 0 –t 0 +(s 1 –t 1 ) +(s 2 – t 2 ),..} = (by renumbering) max {0,s 1 -t 1, s 1 -t 1 +(s 2 –t 2 ), s 1 –t 1 +(s 2 –t 2 ) +(s 3 – t 3 ),..} = D M

10. Limit properties – Lindley equation (s-t+M) + = D M (ii) w n -> 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 {w n } 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 w n+1 = (w n +s-t) + =(w n +s-t)+ (w n +s-t) + - (w n +s-t) (x) + - x = max{0,x}-x = max{-x,0} = -min{x,0} So (w n +s-t) + =(w n +s-t) – min{w n +s-t,0} =(w n +s-t) –X Now if w n + s n - t n < 0 then it equals –I n I n 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) E 2 (s-t)=E 2 (X) · E(X 2 )

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

14. Waiting time probabilities P(w n ¸ W) for W > 0 ?? w n+1 = (w n +s n -t n ) + w n+1 ¸ W  w n +s n -t n ¸ W so P(w n+1 ¸ W) = P(w n + s n -t n ¸ W) = P(w n ¸ W-(s n -t n )) = s P(w n ¸ W-(s n - t n )|s n -t n =y) fs(y) dy = s P(w n ¸ W-y) fs(y) dy (since w n, s n, t n independent)

15. Waiting time probabilities P(w n+1 ¸ W) = s P(w n ¸ W-y) fs(y) dy = s - 1 W P(w n ¸ W-y) fs(y) dy + s W 1 fs(y) dy = s - 1 W P(w n ¸ W-y) fs(y) dy + 1-Fs(W) Consider a function f:R -> [0,1] Assume f(W) · s - 1 W f(W-y) fs(y) dy + 1-Fs(W) Then P(w n ¸ W) · f(W) for all W>0  P(w n+1 ¸ W) · f(W) Since w 1 =0, P(w n ¸ W) = 0 · f(W) for all W>0 So P(w n ¸ 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 - 1 W f(W-y) fs(y) dy + 1-Fs(W) s - 1 W exp(-S(W-y)) fs(y) dy + 1-Fs(W) = s - 1 1 exp(-S(W-y)) fs(y) dy - s W 1 exp(-S(W-y)) fs(y) dy + 1-Fs(W) = exp(-SW) s - 1 1 exp(Sy)) fs(y) dy - s W 1 exp(-S(W-y)) fs(y) dy + 1-Fs(W) = exp(-SW) s - 1 1 exp(Sy)) fs(y) dy - s W 1 (1-exp(-S(W-y))) fs(y) dy = f(W) Á (S) - s W 1 (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(w n ¸ 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. E BW (w) or P BW (w ¸ W) EBW (Equivalent Bandwidth) EBW = arg min BW E BW (w) · W EBW = arg min BW P BW (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