Answer:
1) The sender will not be kept busy
Throughput = 4 Mbps
2) Minimum sender window size to achieve full utilization = 10
Number of bits = 4
Step-by-step explanation:
1) Bandwidth = 8 Mbps
Bandwidth = 8 * 10^6 bps
Packet size = 4000 bytes
Propagation delay = 20 msec
Sender window size (i.e. number of packets) = 5
Size of data sent in bits = (packet size) * ( number of packets) * ( Number of bits)
Size of data sent = 4000 * 5 * 8
Size of data sent = 160000 bits
Size of data that can be sent = 2 * ( Propagation delay) * (Bandwidth)
Size of data that can be sent = 2 * ( 20 * 10^-3 ) * ( 8 * 10^6)
Size of data that can be sent = 320000 bits
Efficiency = ( Size of data sent)/(Size of data that can be sent)
Efficiency = 160000/320000
Efficiency = 0.5
Throuput = Efficiency * Bandwidth
Throuput = 0.5 * 8 * 10^6
Throuput = 4 * 10^6
Throughput = 4 Mbps
Since there is a throughput of 4 Mbps, the sender will not be kept busy.
2) Minimum sender window size to achieve full utilization
Full utilization means that efficiency = 1
Efficiency = ( Size of data sent)/(Size of data that can be sent)
Size of data that can be sent = 320000 bits
Size of data sent in bits = (packet size) * ( number of packets) * ( Number of bits)
Size of data sent = 4000 * n * 8
Size of data sent = 32000n
1 = (32000n)/320000
320000 = 32000n
n = 10 packets
The minimum sender window size = 10
Number of bits needed for the sequence number field
2^3 = 8
3 bits will not be sufficient because 8 is not up to 19
4 bits will be sufficient because 2^4 =16 which is greater than 10
2^3 = 8
Therefore, the number of bits needed = 4