Answer: 7 bits
For Go Back to N:
- 'N' buffers are maintained on sender side and
- 1 buffer is maintained on receiver side
So, we need unique N + 1 sequence numbers
So the bits required would be Ceil $log_{2}(N + 1)$ bits to represent these numbers.
Now let us find what the N should be.
- Since the propagation delay is 6 μsec/km, which means for 1 Km it takes 6 μsec to travel from one end to another
- So, for 3000 Km it would take 3000*6 μsec to travel from one end to another end, which is 18000 μsecs of total propagation delay.
- So RTT would be $2*18000$
So, to achieve full capacity, sender must keep on transmitting for this time.
T1 trunk has a 1.544 Mbps transmission speed and is full duplex.
This means,
- In 1 sec, $1.544 * 10^{6}$ bits can be loaded on link
- Or $1,000,000$ μsec, $1.544 * 10^{6}$ bits can be loaded on link
- Or 1 μsec, $\frac{1.544 * 10^{6}}{1000000}$ bits can be loaded on link
- So, In $2*18,000$ μsec, $\frac{1.544 * 10^{6} * 2*18000}{1000000}$ bits can be loaded on link
- Which comes to $55,584$ bits or $6948$ bytes
Frame size is given as 64 bytes. So $\frac{6948}{64}$ packets or 108 packets can be sent
So, our N is 108 packets, so $N + 1$ is $109$ packets
Bits needed for Ceil $log_{2}109$ which is 7 bits