AceTestSeries 2018: Computer Networks - Network Switching

Question

Consider a packet switched network the propagation delay is 20 msec per hop and the packet size is 1000 bits and the

data rate is 52 kbps. assuming  all packet follow the same 4 hop path,  then the delay in sending  an 100000 bit  message

is________

Answer 1

total time = 160 + 99*20 = 160+1980=2140ms

Comments

I think u missed transmisson of 99 packets

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@royal how 99th packet will be following 98th packet, 98th pacet will be following 97th packet and so on..

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We have to total 100 packets in which 1st packet will take his full time and remaining will use pipelining concept so according to this total time 160+99(20+20) , @manu u r considering the propogation delay of all 99 packet. I have u will understand.

Answer 2

4040 ms should be ans

Comments

iam getting 4037.6 as answer

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U r close to answer

Answer 3

Total packet = 100000/1000 = 100.

Transmission time for 1 packet in all mode = 

(1000/52 kbps) = 19.23 msec.

so , for 100th packet transmission delay = 19.23 * 100 

.                                                                       = 1923 msec.

now, as it is a 4 hop path, means, 3 switches are there..

So, total transmission delay in all nodes 

= 19.23 * 3

= 57.69 msec.

Now, add 4 hop's propagation delay = 4 * 20 = 80 msec.

 

so, total duration = 1923 + 57.69 + 80 = 2060.69 msec 

.                                               

Answer 4

Formula is,

H = Number of hops, P = number of packets , T = transmission time , S = Propagation time

Transmission time = 1000bit * 1000ms/52000bit = 19.23ms

Delay = ((P-1) + H)*T + H*S = (99+4)*19.23ms + 4*20ms

                                              = 103*19.23ms + 80ms = 1980.76ms+80ms = 2060.76