ACE TEST

Question

exact ans of this question is 3.20% or 3.10 %

can i solve using this    1/1+2a??   

Comments

3.10%, you should go with 1 + 2a.

Answer 1

Bandwidth is given 2 Mbps,

L=512*8=4096 bits

RTT :2*32 = 64 msec

so in 1 second it can transfer :2x10⁶

in 64 msec it can transfer 128 x 10³

 so efficiency =useful packet / total packet =4096/128000 =0.032

in % 0.03*100=3.2

Comments

why we can't use 1/1 + 2a.

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@ Rahul_Rathod_

here 512bytes are taking not only 64msec but Tansmission time should also added with this.. 

by using 1+2a we can solve this

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you can't do with 1/1+2a because in question clearly mentioned that "WINDOWS" means it is using SLIDING WINDOW PROTOCOL not STOP & WAIT 

1/1+2a  is only for stop & wait because it sends only one packet in 1+2a time

Answer 2

The Transmission Delay in sending $512B \ = \ ( 4096)$ bits of data = $\Large\frac{4096}{2 \times 10^6}$ $=2.048$ msec.

One way delay of propagating data = $32$ msec 

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Total propagating delay $=$ delay in sending data $+$ delay in getting the acknowledgement .

Total propagating delay $=$ $64$ msec

Efficency $=$ $\Large \frac{Useful time}{Total time (Useful + Propagating)} =$ $\Large \frac{2.048}{2.048+64}$

Efficency$\%$ $= $ $0.03100 \times 100=\color{Red}{3.10078\%}$