retagged by
2,263 views
3 votes
3 votes

We wish to send a message of size $150,000$ bytes over the network. There are four hops, each of length $20km$ and running at $100 Mb/s$. Before sending we split the message into $1500$ byte packets. What is the end-to-end delay of the message? 

Make following assumptions :

# The switches "store and forward" packets along the path.

# For each hop, propagation speed = speed of light in copper $= c = 2 * 10^8 m/s$.

# The delay over each of the links is identical.

# Net queueing delay is zero.

# Net nodal processing delay is zero.

3 Answers

1 votes
1 votes
4 hops means

S->r1->r2->r3->D

No. of packets is =15*10^4 / 1500 = 100 packets
T.D=> 15*10^2*8 / 100* 10^6 (this is for 1 packet) = 0.12ms
P.D=> 20*10^3 / 2* 10^8 = 0.1ms
Total is =>0.1+0.12 = 0.22ms
4 hops takes => 0.22*4=0.88ms
0.88ms is for 1st packet only
other 99 packets reaches r3 by the time 1st packet reaches Destination

so answer is => 0.88 + 0.13*99=>13.75ms

Do comment is found Wrong
0 votes
0 votes

Message is divided into 100 packet each of size 1500 bytes

Transmission time = 1500/100 =15 microsecond

Propogation delay over a link = 20km/20*10^8=200 microsec

S----R1------R2------R3--------R4--------D

Since Router/switches are store and forward device we wont consider ack time 

below table shows packet1 and packet 2 reaches different hops

S R1 R2 R3 R4 D
0 215 430 645 860 1075
15 230 445 660 875 1090

Now to reach 100th packet at D  we need 1075+99*15 =2560microsecond

Related questions

0 votes
0 votes
2 answers
2
Sanjay Sharma asked Jul 11, 2016
3,729 views
A Multicomputer with 256 CPUs is organized as 16$\times$16 grid . What is the worst case delay(in hops) that a message might have to take16153230
0 votes
0 votes
1 answer
3
atulcse asked Jan 22, 2022
5,371 views
Consider the queuing delay in a router buffer (preceding an outbound link). Suppose packet length is L bits and the link rate is R bps, and that N packets arrive to...