Aspire Faculty ID #18748 · Topic: UGC NET Computer Science Sep 2022 (Paper II) · Just now
UGC NET Computer Science Sep 2022 (Paper II)

A $3000\ km$ long trunk operates at $1.536\ mbps$ and is used to transmit $64$ bytes frames and uses sliding window protocol. The propagation speed is $6\mu sec/km$

The sender window size to get the maximum efficiency is

Solution

$T_t=333.33\mu sec$

$T_p=18000\mu sec$

Now,

$a=\frac{T_p}{T_t}$

$a=\frac{18000}{333.33}$

$a=54$

For maximum efficiency in sliding window protocol, sender window size should be:

$W=1+2a$

$W=1+2(54)$

$W=109$

So, sender window size required for maximum efficiency is $109$.

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