Consider an error-free $64$ kbps satellite channel used to send $512$-byte data frames in one direction with very short acknowledgments coming back the other way. What is the maximum throughput for a window size of $15$?

Given,

Channel bandwidth $=64$ kbps

Frame size $=512$ bytes

Frame size in bits:

$512\times 8=4096$ bits

Transmission time:

$T_t=\frac{4096}{64000}$

$T_t=0.064$ sec

$T_t=64$ ms

For a satellite channel, one-way propagation delay is generally taken as approximately $270$ ms.

So,

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

$a=\frac{270}{64}$

$a=4.21875$

For sliding window protocol, maximum efficiency is achieved when:

$W\ge 1+2a$

$1+2a=1+2(4.21875)$

$1+2a=9.4375$

Given window size is:

$W=15$

Since,

$15>9.4375$

So, the channel can be fully utilized.

Therefore, maximum throughput is equal to the channel bandwidth.

Maximum throughput $=64$ kbps