Delay - Computer Networks

Question

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.

Answer 1

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

Comments

I am not getting from where you get 0.13 ms ?

You are doing everything right.But a small mistake at the end.

That is,

Ans : (0.88 + 99* 0.22) = 22.66 ms

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i agree with your solution except one thing

in this equation 0.88 + 0.13*99 . It should be 0.12 instead of 0.13. because propagation time is overlapped for every packet after 1st and hence we just require frame transmission time.

Answer 2

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

Comments

1500 bytes are given and 100Mbps is band width. bytes must be converted into bits by 1500*8.

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Al lot of faults in that solution.

Start with transmission time, as prev comment says translate byte to bit first of all. 1500 bytes = 1500*8 bit = 12000 bit.

Then: transmission time t_trans= packet length / bandwidth: t = 12000 / 100*10^6 = 12*10^(-5) = 0,12 ms (per packet)

Then we get to the propagation time. You have the wrong propagation speed. It should be 2*10^8 not 20*10^8. Thus, propagation time t_prop = link length /propagation speed = 20 km / 2*10^8 = 2*10^4/2*10^8 = 1*10^(-4) = 10 ms

And the number of hops should be four, not five as your sketch suggests.

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@sofiage, I agree with your solution but shouldn't the t_prop=0.1 ms rather than 10ms?

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Yes, your right.

Just want to add that my comment is not a solution, just a comment.

Answer 3

given solution is wrong answer is 175 milli seconds