for question number 1 1. Node A wants to send data to node B, with an average ar

Question

for question number 1

1. Node A wants to send data to node B, with an average arrival rate being 1000 packets per second. There are two design options (a) We build a single link between A and B with the service rate 4000 packets per second (b) We build two parallel links i.e., two servers) with 2000 packets per second for each link. Each arriving packetwl be transmitted randomly via either link (i.e., each link only needs to transmit half of the packets) Assume the arrival is a Poisson random variable and the service time for each link is exponential random variable. What is the average time a packet spend at node A for each case? Which design choice is better? Any intuitive explanation? 2. What is the average time a packet spends in each node i? What is the average number of packets in each nodei? node 1 node 2 Poisson(a) 1/2T Exp H /2- Exp(2) node 3 node 4 1/2 Poisson ( Exp (u3) Exp(3) 3. Flows 1 and 2 use the network, with the link rate of A B and B-C being 1 packet and 0.5 packet, respectively, per time unit. Flow 1 uses links A - B and B - C. Flow 2 uses only link A - B. Let i and r2 be the maximum throughput of the flows 1 and 2, respectively. Suppose both flow use the full link rate of A-B. i.e., x1 +T2 = 1 . The utilities of both flow are ul(x)-2x and u2(x) = x. Find the rate allocation x1,x2 for the following cases (a) What is the rate allocation for maximizing the social welfare?

Explanation / Answer

Node A wants to send data to node B with an average arrival rate being 1000 packets per second.

For a system where arrivals are determined by poisson process and service rates ahve exponential process,

mean delay = 1/(u-v) where u is average servce rate and v is average arrival rate.

Now, for the two design options, we calculate the mean delay between any two consecutive packets sent from A to B. "Lesser the mean delay better is the design".

a)
single link between A and B with service rate 4000 packets per second.

average time a packet spend at node A or
mean delay = 1/(link_rate - service_rate)

           = 1/(4000-1000)
           = 1/3000
           = 0.333 ms

mean delay for design patter (a) is 0.33 ms

b)
Two parallel links with 2000 packets per second for each link.

In this case each link has service rate of 2000 packets/sec and sees an arrival rate of 500 packets/sec

for each link,
mean delay = 1/(link_rate - service_rate)

           = 1/(2000-500)
           = 1/1500
           = 0.66 ms

mean delay for design patter (b) is 0.66 ms

The first design pattern, single link between A and B, is better since by doubling the arrival rate and also the service rate, the average waiting time reduces by a factor of 1/2.

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