Question

The students are trying to access videos where each video is of size 850,000 bits and the average request rate from the students to Blackboard servers is 16 requests/second. Assume that the RTT to any of the Blackboard servers from the Internet router is on average 3 seconds and that the total average response time is the sum of the average access delay and the average Internet delay. Also assume that for the average access delay, the average time required to send an object over the access link (α) and the arrival rate of objects to the access link (β) follow the relation α/ (1−αβ) .

(a) Find the total average response time.

(b) Find the total response time when a cache is installed in the institutional LAN with miss rate of 0.4.

Answer 1

a)As we know that:

L(Size of the video)=850000bits

Internet Delay = 3 sec

As we understands the time to transmit an object of size L over a link i.e Rate L/R.

I am assuming Internet speed is 1.5Mbps(Mega bits per sec)

1.5Mbps=15,000,000 bps(bits/sec)

therefore he average time is the average size of the object divided by R=(850000bits)/(15,000,000bits/sec)=0.0567 sec (after round off)

The traffic intensity on the link is avg request rate * average time = (16 requests/sec)*(0.0567 sec/request) = 0.9072.

As we know that average access delay = average time/(1-traffic intensity)

therefore average access delay = (0.0567)/(1-0.9072)=0.61sec

Total Response Time = Average access delay + Internet Delay

so 0.61 sec + 3 sec = 3.61sec

b)We know that the miss rate is 0.4 that means 60% of the request are satisfied by the cache

so the average access delay is (0.0567 sec)/[1 – (0.4)(0.9072)] = 0.089 seconds.

The response time is approximately zero if the request is satisfied by the cache (60% of the time i.e for probability 0.6). Therefore the average response time is 0.089 sec+ 3 sec = 3.089 sec[Total Response Time = Average access delay + Internet Delay] for cache misses (40% of the time).

So the average response time is (0.6)*(0 sec) + (0.4)(3.089 sec) = 1.24 seconds which means it reduced from 3.61sec to 1.24 sec.