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Calculate the effective throughput for transferring a 1000 KB file assuming TCP using
slow start congestion control technique. Given the round trip time 100 ms, and maximum
segment size is 1460 bytes. Assume there are no losses and both the bandwidth and the
receiver window size is infinite.
  (A) 5MBPS
  (B) 10Mbps
  (C) 1MBPS
  (D) 1Mbps
in Computer Networks by Active (1.2k points) | 98 views
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1 MBPS
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cwnd size at start=1 MSS = 1460 B

cwnd size approximately double on every RTT in case of slow start phase.

Number of MSS to send = 1000000 B/1460 B = 684.93 = 685

1 MSS -> 1 RTT

2 MSS -> 1 RTT

4 MSS -> 1 RTT

16 MSS-> 1 RTT

32 MSS-> 1 RTT

64 MSS-> 1 RTT

128 MSS-> 1 RTT

256 MSS-> 1 RTT (511 MSS sent till now)

173 MSS -> 1RTT

Total time to send 685 MSS =  9 * RTT = 900 ms

Total Data sent = 1000KB = 1000000B

Throughput = $\frac{\textrm{total data sent} }{\textrm{Total time}}$= $\frac{\textrm{1000000 B} }{900 * 10^{-3} sec}$= $\frac{10^{7}B}{9 sec}$ = 1 MBPS

1 Answer

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Here, maximum segment size(1 MSS)=1460B.,so in one RTT 1460 B is transferred.To send 1000KB we need 1000*1024/1460 MSS= 702 MSS (appx).

Using slow start algorithm,

                                          1 MSS |  2 MSS | 4 MSS | 8 MSS |  16 MSS | 32 MSS | 64 MSS | 128 MSS | 256 MSS |512 MSS | 1024MSS.So in total we need 10 RTT (10* 100 ms=1000 ms) to send 1000KB.

We know , throughput is the number of bytes sent per second.

So, to send 1000KB we need 1000 ms.So, in 1 sec, it will send 1000 Kb.

So,throughput =1000kbps=1 mbps(D)
by (357 points)

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