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Question

A 1-km-long, 10-Mbps CSMA/CD LAN (not 802.3) has a propagation speed of
200 m/μsec. Repeaters are not allowed in this system. Data frames are 256 bits long,
including 32 bits of header, checksum, and other overhead. The first bit slot after a
successful transmission is reserved for the receiver to capture the channel in order to
send a 32-bit acknowledgement frame. What is the effective data rate, excluding
overhead, assuming that there are no collisions?

ans=3.8 mbps

Answer 1

Actual data sent= 256-32= 224 bits

Contention slot for sender= 2*propagation delay= 10micro-sec.

Transmission time= 256/10Mbps=25.6 micro-sec.

Propagation time =5micro-sec.

Contention slot for receiver =10 micro-sec.

Transmission time for 32 bits(ack) =32/10Mbps = 3.2micro-sec.

Propag time= 5 micro-sec.

Total time to send 224 bits of data =10micro-sec+25.6 micro-sec+5micro-sec+10 micro-sec+3.2micro-sec+ 5 micro-sec

                                                      =58.8 micro-sec

Data rate = 224bits/58.8 micro-sec =3.809Mbps