Q1. (25 pts) This elementary problem begins to explore propagation delay and tra
ID: 3703560 • Letter: Q
Question
Q1. (25 pts) This elementary problem begins to explore propagation delay and transmission 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. Express the propagation delay, dprop, in terms of m and s. a. Determine the transmission time of the packet, drans, in terms of L and R b. Ignoring processing and queuing delays, obtain an expression for the end-to-end delay c. Suppose Host A begins to transmit the packet at time r-0. at time t-dirans where is the last bit of the packet? d. Suppose dprop is greater than dirans At time tedrans where is the first bit of the packet? e. Suppose dprop is less than drans At time tdrans, where is the first bit of the packet? f. Suppose s-2.5*10 8, L-100 bits, and R-28 kbps. Find the distance m so that dprop equals dransExplanation / Answer
Data tranfer rate between the hosts = R bps
Distance between the hosts = m meters
Propagation speed along the link = s meters/sec
Packet size = L bits
Propagation delay, dprop = Distance between the hosts / Propagation speed
= m / s sec
(a) Transmission time, dtrans = Packet size / Data Transfer Rate
= L / R sec
(b) End-to-end Delay = Transmission time (time to keep data over the link from Host A) + Propagation Delay (time taken by single bit to travel from Host A to Host B) + Frame Processing Time + Queuing Delay
Ignoring processing and queuing delays
End-to-end Delay = dtrans + dprop
(c) Transmission time or dtrans is the time taken by host to keep the entire packet over the link. If Host A started transmitting data at time t = 0, it will keep the last bit of packet onto the link at time t = dtrans. So last bit will just be put on the link at t = dtrans.
(d) dprop > dtrans, that is, time taken by a single bit to reach the destination host is greater than time taken to put the entire packet on the link. At time t = dtrans, last bit will be kept on the link, hence the entire packet is on the link. First bit will reach the destination at t = dprop, considering sender started transmitting at t = 0, therefore for 0 < t < dprop, first bit will be on link.
(e) First bit will reach the destination at t = dprop, considering sender started transmitting at t = 0. Since dprop < dtrans, first bit will reach the destination before the entire packet is put onto the link. Therefore, at time t = dtrans, first bit reaches the destination host.
(f) dprop = m / s sec
dtrans = L / R sec
Since, dprop = dtrans
Therefore, m / s = L / R
s = 2.5 * 10^8 meters/sec (Considering speed is given in meters/sec)
L = 100 bits
R = 28 kbps (kilo bits per sec) = 28 * 1000 bps (bits per sec)
m = (L * s) / R
= (100 * 2.5 * 10^8) / 28 * 1000
= 892,857.14 meters
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