search
Log In
4 votes
700 views
Consider that the stop and wait protocol is used on a link with bit rate of 128 Kbps and 40 msec as round trip time. Assume that the transmission time for the acknowledgment and processing time at nodes are negligible. If the frame size is 512 B then the link utilization is ________. (In % upto 2 decimal places)
in Computer Networks
edited by
700 views
0
$44.44  $ ?
0
yes it is 44.44 %

2 Answers

3 votes

Given Data :

 bit rate  =  128 Kbps 

 round trip time(2Tp) = 40 msec 

Therefore Tp =  20 msec

length of one frame = 512 Bytes = 512*8 bits

Transmission Time = length  / bandwidth  = 512*8 / (128kbps) = 32msec

efficiency  = (Tt / Tt + 2Tp) = 32 / (32 + 2*40)  = 0.44444444

Answer :

Efficiency percentage = 44.44 %

1
I used following formula and i got 80%

Link capacity: (Packet size) / (Channel capacity)

channel capacity : 2 * tp * Bandwidth = RTT * Bandwidth

Where im wrong?? tell me .. if we can't use this formula then tell me why??

and where we can use this formula??

Thanks ,
1

please refer this

http://nptel.ac.in/courses/106105080/pdf/M3L3.pdf

it will be helpful

0
Nice

Comment if u have more such links it would be helpful to all :p
1
Thank you so much..
0
Hey, were you able to understand, when and when not to use this formula?
1 vote
first find transmission time by using formula ( L / B )

find propagation time ( RTT / 2 )

to find efficiency use formula 1 / (1+2a) ; where a=(propagation time / transmission time)

multiply it by 100 to get answer in %

that's it
0
Thanks bro... @Rahul_Rathod_

Related questions

1 vote
2 answers
3
497 views
Consider a wireless link, where the probability of packet error is 0.6. To transfer data across the links, Stop and Wait protocol is used. The channel condition is assumed to be independent from transmission to transmission. The average number of transmission attempts required to transfer x packets is 500. The value of x is _______.
asked Dec 17, 2017 in Computer Networks Shubham Kumar Gupta 497 views
0 votes
1 answer
4
993 views
The distance from earth to a distant planet is approximately 9*(10^10) m. What is the channel utilization if a stop-and-wait protocol is used for frame transmission on a 64Mbps point-to-point link? Assume that the frame size is 32KB and the speed of light is 3*(10^8) m/s.
asked Jul 5, 2018 in Computer Networks Sid865 993 views
...