Brian Chang
U71839200
CS350 Hw4
2/24/2007

1.
Utilization of the Firewall = (L1+L2) * 0.005 >= 1
Utilization of the Web Server = (L1+ 0.05 * L2) * 0.05 >=1
Web Site will buckle if either become greater than one.

2. N packets with mean of 30 packets per second processing time of 30msec.
lambda = 30 Ts = 0.03

2a.	p = lambda * Ts =
	30 * .03 = 0.9
	W = p2 / 1(1 – p) = .92 / .1 = 8.1 packets

2b.	Tw = W/lambda
	8.1/30 = 0.27 secs

2c.	Infinite buffer size = zero blips
	plop is when nothing is processing = 1 – p = 0.1
	0.1 / 0.9 = 0.111 in msec
	111plops 

2d.	w = ?
	w = q – p
	w = q – Ts * lambda’
	Ts = .03
	q = 	(p / (1 – p)) – 
((N+1)*pN+1) / (1 - pN+1) =
.9 / .1 –
6*.96 / 1-.96 =
2.195
	lambda’ =	lambda( 1 – Pr(x))
			Pr(x) = ((1- p)* pN)/ (1 - pN+1)
			Pr(x) = ((1- .9)* .95)/ (1 - .95+1) = .126023
	30 (1 - .126) = 26.2193
	plug back into w = q – Ts * lambda’
	w = 2.195 - .03 * 26.2193 = 1.40842 requests

2e. Tw = w / lambda’= .053717

2f.	numBlips = Pr(x) = ((1- p)* pN)/ (1 - pN+1) * 1000
	numBlips = (from 2d) = .126023 * 1000 = 127
	1000 – 127 = 873
	873* (1 – p / p) = ?
	p = p’ =lambda’* Ts = 26.2193 * 0.03 = .7866
	plug back into into p = 873* (1 – p / p) = 873 * (1 - .7866) / .7866 = 236.84

4. lambda = 20. Ts = 0.08

For 2 MM1 queues, lambda is split so lambda=10
p = lambda * Ts = 0.8
q = p / (1 – p) = .8/.2 = 4
Tq = p/lambda = 0.4s

For MM2 queue
p = (lambda * Ts)/ N = 20*.08u / 2 = 0.8u

































 
1.	
2.	   
3.	
4.	5solutionsAs we discussed in class, N independent queues tend to result in worse performance (measured in terms of average response time, for example) when compared to a system of one queue being served by N servers. To enable multiple servers to dequeue requests from a single pool often requires that these servers follow a synchronization protocol to ensure that concurrent accesses to the queue (e.g., two servers attempting to simultaneously dequeue a request) do not result in unintended results (e.g., a request being served twice). This "synchronization" protocol (examples of which we will encounter in following later parts of the course) may impose overhead, which would have the net result of slowing down the servers. In this problem, your job will be to asses the impact of such a slowdown. 
Consider the comparison we made between 2 M/M/1 systems and an M/M/2 system. Let us assume that the "price" of synchronizing access to the shared queue is to prolong the average service time by a constant factor U>1. In other words, if Ts is the average service time for a vanilla M/M/1 system, then U*Ts is the average service time for a request served by an M/M/2 system. For example if U=1.1, then service time in an M/M/2 will be 10% longer than that for a M/M/1 system. What values of U would make it not worth the trouble to have two servers share the queue? In other words, what vaules of U would make the M/M/2 system similar to 2 M/M/1 systems running in parallel (with incoming load being distributed equally)? 
You may use the following settings: 
o	Lambda = 20 
o	Ts = 0.08 
  
5.	The path between two hosts goes through three identical routers. Each router can be modeled as an M/M/1/K queue with a maximum queue size of 10 packets (i.e. K=10). The service time for each router is 1 msec. Assuming that the rate of packet arrivals to the first router is 1000 packets per second, answer the following questions: 
b.	What is the average latency through the network? 
c.	How many packets per second are dropped on the average from all three routers? 
Note: You may assume that the output of an M/M/1/K queue is Poisson.