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@jatin khachane 1 yes

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ya sorry it was 10th frame only.

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It should be 9...if it's 10 then according to your formula efficiency will be more than 1

Answer 1

Given, 

Bandwidth(B)=64 kbps=64 x $10^3$ bps

$Tp = 270msec$

Length of packet (L)=$512 Byte= 512 \ast   8  bits $

$\therefore Tt= L/B=512 \ast 8/64 \ast 10^3 = 64 msec$

For maximum utilization window size should be,

$Ws= 1+2a$

      $= 1+ 2 Tp/Tt$

      $= 1+ 2 * 270/64$

      $ = 1+ 8.44 $

      $=9.44 \cong 10$