window size

Question

Answer 1

$T_{x}$ = (512 * 8 ) / (1024 * 10^ 3) = 4 ms

$T_{p}$  = 50 ms

Therefore within this time of transferring one packet in the link and receiving the acknowledgement the time gap is

2 * $T_{p}$ + 4 = 104 ms

within this time the number of packets we can send  is 104 /4 =26 in order to fully utilize the link .

Thats what am getting . Correct me if am getting wrong .

Comments

sir what is the problem by getting the solution with bw*dely /packet size

Answer 2

To find minimum window size we need to find no of frames which can be sent in one RTT

Window size=1024kbps*100ms/512*8

                      =25

Comments

It is sender's Window Size.. Ryt?

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:O

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what sir?

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she didnt take the transmission time in RTT. And yes, it is sender window size.

Answer 3

$T_{transmission} = \frac{512 \times 8\ bits}{1024 \ Kbits} = 4\ ms$

to get full rate throughput $\left( \eta \right)$ should be $1$.

$$\begin{align*} \eta &= \frac{W \times T_{transmission}}{T_{transmission} + 2\times T_{propagation}} \\ 1 &= \frac{W \times 4ms}{4ms + 2 \times 50ms} \\ \frac{104ms}{4ms} &= W \\ W &= 26 \end{align*}$$

Answer 4

BW*DELAY PRODUCT GIVES THE MAXIMUN CAPACITY UP TO WHICH WE CAN EXTEND

SO HERE BW*DELAY=1024Kb*100msec=1024*100 bits...

we have to use upto this capacity with 512B

=>(1024*100)/(512*8)=25