This elementary problem begins to explore propagation delay and transmis- sion delay, two central concepts in data networking. Consider two hosts, A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meters, and suppose the propagation speed along the link is s meters/sec. Host A is to send a packet of size L bits to Host B a. Express the propagation delay, dprops in terms of m and s

b. Determine the transmission time of the packet, drans, in terms of L and R.

c. Ignoring processing and queuing delays, obtain an expression for the end-

d. Suppose Host A begins to transmit the packet at time t = 0. At time t = dtrans'

e. Suppose drop is greater than dran . At time t = d, ans, where is the first bit of

f. Suppose dprop is less than dtrans. At time t = dtrans, where is the first bit of

g. Suppose s = 2.5-108, L = 120 bits, and R = 56 kbps. Find the distance m so that dprop equals drans

A. dprops = m /s seconds.

B. drans = L / R seconds.

C.  delay(end −to−end) =  (m /s + L / R) seconds.

D. The bit hasjust been sent to Host B or just left Host A.

E. The first bitis in the link and has not reached Host B.

F. The first bit has reached Host B.

G. m = 535.714 km.

Explanation:

The transmission time or delay of packets in a network medium is the packet size L, divided by the bit rate R (in seconds). The propagation time or delay is the ratio of the distance or length of the transmission cable, m, and the propagation speed of the cable, S (in seconds).

The end-to-end delay or the Packet delivery time is the total delay in transmission, which is the sum of the propagation delay and the transmission delay.

To get the distance where the propagation delay is equal to the transmission delay;

distance (m) = L /R  

= (120/56 ×10^3) 2.5 ×10^8   = 535.714 km

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