maximum size of the send and receive windows the number of b

maximum size of the send and receive windows, the number of bits in the sequence number field (m), and an appropriate time-out value for the timer.

Solution

Bit rate of link= 200 Mbps

Link length= 10,000 km= 10⁵ km= 10⁸ m

Propogation time= 10⁸ m/(2 * 10⁸ m/s) = 1/2 s

Round trip time = 2* Propagation time = 2* 1/2 s =1s

In the round trip time, given the bandwidth, one can send 200 * 10⁶ bits i.e 2 * 10⁸ bits.

Average packet size is 10⁵ bits.

Hence the number of packets sent in one RTT = 2 * 10⁸ / 10⁵ = 2000.

For 2000 packets we will need a 10-bit sequence number.

Window size is governed by the number of bits in sequence number. Since we are using GO- Back -N the maximum window size can be 2¹⁰-1 =1023.

An appropriate estimate for time-out value of a timer is equal to the RTT and hence is equal to 1s