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Suppose there are $4$ routers connected between two systems $S_1$ and $S_2$. $S_1$ wants to send a python file of size $8 \:KB$ to $S_2$. But the data rates at each routers are different. The data rates between the routers are $1 \: Mbps$ , $2 \: Mbps$, $1 \: Mbps$ and $512 \: Kbps$ respectively.  Find the maximum possible bandwidth for the file transfer in Kbps.  

(Assume that the time to transfer $1$ bit from $S_1$ to $S_2$ is $32$ sec and there is no traffic between these two systems)
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The data can be transmitted with minimum data rate throughout the path. (i.e) $512$ Kbps Given there is no traffic between these two systems. Propagation delay is $32$ sec. Length of the file is $64$ Kbits.
Transmission delay $= L/B = 64/512 =1/8$ sec
Propagation delay $= 32 $ sec
Maximum possible bandwidth means the throughput. (i.e) how much bandwidth can be used.
Throughput = utilization * bandwidth
$=\frac{1}{1+2(\frac{32}{1/8})} * 512$ Kbps
$\approx 1$ Kbps

2 Comments

Shouldn't we count the transmission time 5 times?
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wrong answer.

64 Kbits is 64*1024 bits and not 64*1000 bits which above question used

also transition times will need to count three times.

so formulae (3*Transmission times) / (3*Transmission times + Propagation time)

I am assuming Propagation time and not RTT because , it is not mentioned in question that it is using stop and wait.
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Answer:

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