Given, propagation time =25 ms.
Time required to transmit 1 frame = 1000 bits (frame size)/ 106 bps(bandwidth) = 1 ms
To fully utilize the channel, we need to transmit frames for propagation time(25 ms), 25 frames.
therefore, minimum no. of bits required for numbering 25 frames is = log 2(25) = 5 bits.
We need to calculate minimum waiting time after sending 2 I frames (32 frames) to initiate transmission of next frame, for that we need to understand for what time sender is in busy mode and for what time sender is in waiting mode.
Sender is in busy mode during transmission of all the 32 frames, that means, for 32 ms.
after transmission sender is in waiting mode which is
= ( time required to get the ack for 1st frame ) - ( time for which sender is busy)
= (1+25+1+25) - (32) = 20 ms.
1+25+1+25 = transmission time for 1st frame + propagation time for frame + transmission time for ack as it is piggy backed + propagation time for ack