Test by Bikram | Computer Networks | Test 1 | Question: 19

Question

A $20$ kbps satellite link has propagation delay of $300$ ms , transmitter use Go Back $10$ ARQ scheme, if each frame is $50$ bytes long then the maximum data rate is ________  kbps

Comments

is this asking for throghput ?

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Yes, it is asking for Throughput.

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@Bikram i have given 6.45  got wrong

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Can you explain the procedure @Gabbar

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@rajesh170293  

efficiency * bandwidth will give the effective bandwidth ....

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please explain.

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@ rajesh170293   and @kavita_joshi  

Go back N used where n value is 10

Transmission time  = 50 * 8 bits /  20 kbps =  20 ms

Propagation Time = 300 ms

Efficiency = Window Size*Transmission Time/(Transmission Time + 2*Propagation Time)

= (10 * 20 ) / ( 20 + 2 * 300 )

= 200 / 620

= 0.322580

so Maximum Data Rate  or throughput is ( efficiency * bandwidth )

 =( 0.322580 * 20 ) kbps  [ bandwidth 20 kbps given ]

= 6.45 kbps

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thanks boss

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i have given answer as 6.451 asuming till 3 decimal places and got it as a wrong answer.how to know whether to give till 2 or 3 decimal places as it is not mentioned in the question?

Answer 1

Go back N used where n value is 10

Transmission time  = 50 * 8 bits /  20 kbps =  20 ms

Propagation Time = 300 ms
Efficiency = Window Size*Transmission Time/(Transmission Time + 2*Propagation Time)

= (10 * 20 ) / ( 20 + 2 * 300 )

= 200 / 620

= 0.322580

Maximum Data Rate = 0.322580 * 20 kbps = 6.45

Comments

Tt is 20ms and RTT is 60ms
you can't send more than 3 packets in 1 RTT though here the size of windos is 10.
hence throughout should be:

in 60ms we are sending 3packets or 3*50 = 150Bytes or 150*8 = 1200 bits
1 ms = 1200/60 = 20bits bits
1 sec =  20,000 bits = 20kbps  
Throughput = 20Kbps should be answer

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@manu00x

No,

maximum data rate or throughput is asked, throughput is efficiency * bandwidth .

Bandwidth 20 kbps is given .

you can see above comments ..they also got 6.45 as throughput.

see this for reference  https://gateoverflow.in/15759/throughput-calculation

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Please Update the range in GO Tests, or at least mention in question (up to two decimal places.)

Answer 2

Given Bandwidth = 20 kbps, T_p = 300 ms, Frame Size = 50 Bytes = 50 * 8 bits,

N = 10

Now, T_t = $\frac{Frame Size}{Bandwidth}$ = $\frac{50 * 8 b}{20 * 10^{3} bps}$ = 20 ms

We know Throughput = Efficiency * Bandwidth = $\frac{N}{1+2*\frac{Tp}{Tt}}* Bandwidth$

                                =$\frac{10}{1+2*\frac{300}{20}} * 20 kbps$ = 6.45 kbps