GATE-CS-2006 [closed]

Question

Station A uses 32 byte packets to transmit messages to Station B using a sliding window protocol. The round trip delay between A and B is 80 milliseconds and the bottleneck bandwidth on the path between A and B is 128 kbps. What is the optimal window size that A should use?

My Approach:
RTT= 2*Tp + Tt (Is this wrong) - It is taken as RTT = 2*Tp in some solutions
RTT = 80ms (Given)
Tt=L/B = 2ms

so for 2ms - 32 Bytes are transmitted and
so for 1 ms - 16 bytes are transmitted, so for 80ms - 80*16 bytes are transmitted

Since Each packet is 32 bytes - (80*16)/32 - Number of packets
Then optimal Window size is 40
If the RTT=2*Tp then we get Ws = 41, what is the correct approach ?