computer network

Question

A TCP machine is sending windows of 65,535 bytes over 512 Mbps channel  that  has  20  millisec  one  way  delay.  The  line  efficiency  is ________ (upto two decimal places).
 

Ans given 2.56 .............i got 0.0249(i.e 2.49%)

Answer 1

RTT = 2 * one way delay = 2 * 20 = 40 ms

using 512 Mbps channel, in 40 ms we can trasmit at max,

Bandwidth delay product = $40 \times 10^{-3} \times 512 \times 10^{6} = 20.48 \times 10^6$ bits

$\therefore$ line efficiency = $\frac{65535 \times 8}{20.48 \times 10^6} \times 100= 2.55996 = 2.56 \%$

Comments

@Lokesh can you please clear my doubt here

when we calculate efficiany sometimes we only consider only RTT but sometimes we also consider transmission time.here if we consider transmission time then efficiany will be 2.48%.There are so many questions sometimes they only consider Tp and sometimes Tt and Tp both.Please share your view about this.will be a great help.

---

To calculate efficiency we need to find total data we could have transmitted in that time. So, the problem lies what is which time??

$1$ RTT includes the time when packet is sent to when ack. is received. So, it covers all the time [$T_t + T_p + T_t \; ack\; + T_p$

Whereas one way delay is $T_t + T_p$ . means till it reaches other end.

So, $T_t$ is always included, but when RTT or one-way-delay is given $T_t$ is already included in these.

---

Ty @mcjoshi

@HRK , see we add transmission time for one packet
but here they have given whole window size and there no way to find the packet size from given data
so you can not include transmission time

and as Manish said if RTT or one-way-delay is given assume it is include
and if propagation delay is given then add transmission time

Answer 2

ya answer is 2.49%

Comments

please explain the steps