Computer Networks

Question

Assume a 10Mbps Ethernet has two nodes, A and B, connected by a 360m cable with three repeaters in between, and they each have one frame of 1,024 bits to send to each other. Further assume that the signal propagation speed across the cable is 2*10^8 m/sec, CSMA/CD uses back-off intervals of multiples of 512 bits, and each repeater will insert a store-and-forward delay equivalent to 20-bit transmission time. At time t=0, both A and B attempt to transmit. After the first collision, A draws K = 0 and B draws K = 1 in the exponential back-off protocol after sending the 48 bits jam signal.

What is the one-way propagation delay (including all repeater delays) between A and B in seconds? At what time is A's packet completely delivered at B?

Now suppose that only A has a packet to send and that the repeaters are replaced with switches. Suppose that each switch has an 8-bit processing delay in addition to a store-and-forward delay. At what time, in seconds, is A's packet delivered at B

Answer 1

Propagation delay  $T_p = \frac{360}{2  *  10^8} = 1.8 \mu s$

Store and Forward Delay $S_d = \frac{20}{10*10^6}=2\mu s$

Transmission time for a frame $T_t=102.4\mu s$

$(i)$ Total one way propagation delay $= T_p +3*S_d=7.8\mu s$

$(ii)$ both A and B transmit at $t = 0$ and at $t = 7.8\mu s$ both will know about the collision and after 48-bit jamming signal $i.e. 4.8\mu s$ both nodes will participate in back off algorithm.

A transmits immediately as $K=0$

B waits for $51.2 \mu s$ as $K=1$

Time taken for A's frame to get delivered at B $= T_t + 7.8\mu s  + 4.8\mu s = 115\mu s$

$(iii)$ Processing delay $P_d = 0.8\mu s$

Total delay at switches $=3(P_d+S_d)=8.4\mu s$

Time taken for A's frame to get delivered at B $=T_t + T_p+8.4\mu s=112.6\mu s$

Comments

You specify that the “time taken for A's frame to get delivered at B =Tt+7.8μs+4.8μs=115μs.”  Should this interval also include the propagation time to send the jamming signal? IE.  the total should include an additional 7.8 microseconds.