1. ORF 401 - Electronic Commerce Spring 2009 April 15, 2009 Week 10 Capacity Analysis Deterministic Model –Assume: each HTTP request takes T r (request time) –and requests arrive at a know known time sequence: A s –What is service time per request, T s No conflicts: N(t) t First request 2nd request T r = 1/2 As: 0, 1, 2, 3, … (no conflicts) T s = 1/2

2. ORF 401 - Electronic Commerce Spring 2009 April 15, 2009 Week 10 Sequenced with conflict: Round Robin: –Reqest 1 arrives at 0, works ‘till 0.25., waits ‘till.5 works ‘till.75 then ends –Reqest 2 arrives at.25, works ‘till 0.5., waits ‘till.75 works ‘till 1.0 then ends T s = (0.75 + 0.75)/2 = 6/8 First-come-first-serve T r = 1/2 A s : 0,.25, 1., 1.25,… (same # requests, sequenced differently) T s = (0.5 + 0.75)/2 = 5/8 First request starts N(t) t 2nd request starts Round Robin

3. ORF 401 - Electronic Commerce Spring 2009 April 15, 2009 Week 10 Observations If server is handling a single page, then deterministic service times is reasonable, otherwise NOT The long term behavior depends on the arrival sequence, A s, AND the service rate, T s F-C-F-S is a lower bound on the round robin

4. ORF 401 - Electronic Commerce Spring 2009 April 15, 2009 Week 10 Probabilistic Model (F-C-F-S) Let inter arrival time and the service times be : –independent, identically distributed (iid) random variables having an exponential distribution (denoted by M ) –a r.v. X has an exponential density iff: f(x) =  e -  x x > 0 = 0 x < 0 So P rob { X 0 P rob { X > x } = e -  x x > 0 E(X) = 1/  var(X) = 1/  

5. ORF 401 - Electronic Commerce Spring 2009 April 15, 2009 Week 10 Assume server has been running for a while and reached steady state In steady state (prob. of being in state j, (j waiting to be served)) : P j j = P j+1  j+1 or P j+1 = P j ( j /  j+1 ) (recursion relationship) Specifically: P 1 = ( 0 /  1 ) P 0 P 2 = ( 1 /  2 ) P 1 = ( 1 /  2 ) ( 0 /  1 ) P 0 … P n+1 = ( n /  n+1 ) P n = (( n n-1... 0 )/ (  n+1  n...  1 )) P 0 Let’s let: C n = ( n-1 n-2... 0 )/ (  n  n-1...  1 ), n = 1, 2, … Then the Steady State Probabilities are P n = C n P 0, n = 1, 2, … Arrival Rates Service rates States 12nn+10... n 1  22 11  n+1

6. ORF 401 - Electronic Commerce Spring 2009 April 15, 2009 Week 10 What about P 0 ? –Since  n   P n = 1, then  n   C n P 0 = 1 P 0 +  n   C n P 0 = 1 P 0 [1+  n   C n ] = 1  P 0 = 1 / [1+  n   C n ] Now, the number of requests in the system, L L = 1 P 1 +2 P 2 +3 P 3 +... =  n   (n P n ) Then the expected waiting time (likely time to be served),W = L /  average arrival time

7. ORF 401 - Electronic Commerce Spring 2009 April 15, 2009 Week 10 Simple case: M/M/1/FCFS/  /  Now suppose n =  n =  n = 1,2, 3, … then C n = ( n /  n ) = (  ) n  n = 1,2, 3, … let  thenP 0 = 1 / (1 +  n   n ) = 1 /  n   n ) Now recall  n  x n = (1- x n+1 ) / (1- x ) for any x and if |x| < 1  n  x n = 1/ (1- x ) We need to have  or the system will blow up! So  Thus P 0 = 1 / (1 / ( 1  and P n =  n P 0 =  n ( 1  n = 1,2, 3, …

8. ORF 401 - Electronic Commerce Spring 2009 April 15, 2009 Week 10 What about L (expected # of requests in queuing system) ? L =  n   n P n =  n   n (1  )  n = (1  )  n   n  n Now a little trick: Observe that n  n = n   n-1 So L = (1  )   n   n  n-1 Why bother? Because n  n-1 should be familiar. In particular d/d  (  n ) = n  n-1 Thus L = (1  )   n   d/d  (  n ) = (1  )  d/d  (  n   (  n ) ) = (1  )  d/d  (  n   (  n ) ) but d/d  (  n  inf (  n ) is a geometric series = (1  )  d/d  (1  ) -1, but d/d  (1  ) -1 = -1(1  ) -2 ( -1 ) SoL =   (1   ) -1 and we can show L =  (  ) -1 and expected time in system W = L /  =  (  ) -1 -1 = (  ) -1

9. ORF 401 - Electronic Commerce Spring 2009 April 15, 2009 Week 10 More of M/M/1/FCFS/  /  Queues L q = expected number of customers in line = 0 P 0 +  n  ( n  1) P n =  n  ( n P n ) -  n  ( P n ) = L  P 0 ) = L  = 2  (  )) L s = expected number of customers in service = 0 P 0 +  n  1 P n = 1 - P 0 = 1  ) =  =  W q = expected time of customer spends in line = L q /  =  (  )) W s = expected time of customer spends in service = L s /  = 

10. ORF 401 - Electronic Commerce Spring 2009 April 15, 2009 Week 10 What about multiple processors: C n = ( n -1 n-2... 0 )/ (  n  n-1...  1 ), n = 1, 2, … We again let n =  for every n Now assume that there are s processors. So when n < s we have  n = n  and when n < s we have  n = s  So C n = n  / (  n  n-1...  1 ) we has s terms of n  and n-s terms of s  Hence C n = n  / (n!  n ) n < s = n  / (s!)(s n-s )  n ) n > s If < s  as before we can show : P n = 1/ (  n  0, s -1 {(1 / n! ) ( /  ) n + (1 / s! ) ( /  ) s )(1 - ( / (s  )))}

11. ORF 401 - Electronic Commerce Spring 2009 April 15, 2009 Week 10 Finally: P n = P 0 (  n  (n!) n < s = P 0 (  n  ( (s!) s n-s ) n > s L = P 0 (  s (  s  )  ( s! ((1-  s  ) 2 )) +  W = P 0 (  s (  s  )  (s!) (1- (  s  ) 2  + ( 1 +  )  Where L= s = # processors  = arrival rates  = processing rates

12. ORF 401 - Electronic Commerce Spring 2009 April 15, 2009 Week 10 Realistic Server Performance Consider one server connected to the internet Arrivals Departures Initial processing “Server” BrowserInternet Model each as a single server queue

13. ORF 401 - Electronic Commerce Spring 2009 April 15, 2009 Week 10 Realistic Server Performance, cont. Assume: –File sizes are exponentially distributed –Service rate of browser and Internet are exponentially distributed Parameters: –Arrival rate –Mean file size (5275 bytes) –Initialization time (independent of size) –Bandwidth @ Server (T 1 = 1.5 Mbit/s, T 3 = 6Mbit/s ) –Bandwidth @ Client (700Kbit/s) Arrivals Departures Initial processing “Server” BrowserInternet p 1-p Response Time Arrival Rate 35/sec

14. ORF 401 - Electronic Commerce Spring 2009 April 15, 2009 Week 10 More Complicated Model Load balancing strategies (above not very smart) Arrivals Departures Initial processing “Server” BrowserInternet p 1-p “Server” q 1-q