So here is the question, Source S wishes to send a packet to destination D, and

Question

So here is the question,

Source S wishes to send a packet to destination D, and there are two alternatives for possible paths (see figure below). The first path is along nodes S-A-B-D, say Path 1, and the second path is along nodes S-X-Y-Z-D, say, Path 2. There are two possible choices of spectrum chosen at each node. The default channel is the licensed 600MHz channel, and there is a backup 2.4GHz unlicensed ISM band channel that can be used when the licensed channel is occupied by a primary user (PU). The values Pi = 0.4, p2 = 0.3, P3 = 0.2 indicate the probability with which the PUs cause channel switches into the ISM band at the node locations, as indicated in the figure. Whenever a channel switch occurs, there is switching time penalty of Tsw = 1ms. The transmission time of a packet in the 600MHz and ISM band are T600 = 3ms and TISM = 5ms, respectively. Note that transmission is always first attempted on the licensed channel (i,e., the licensed channel is the default choice) at each node. What are the best case and worst-case end-to-end latencies for each of the two paths? Which path would be chosen, considering the average end-to-end latency for the two paths?

Explanation / Answer

a) the best case latency for SABD is 9ms and worst case is 1*3 + 5*3 = 18ms

the best case latency for SXYZD is 12ms and worst case is 1*4 + 5*4 = 24ms

b) probability of best case in SABD = (1-0.4)*(1-0.3) = 0.42

probability of worst case in SABD = (0.4)*(0.3) = 0.12

avg time in SABD = ((0.42*9 )+ (0.12*18))/(0.42+0.12) = 11 ms

probability of best case in SXYZD = (1-0.2)*(1-0.2)*(1-0.2) = 0.512

probability of worst case in SXYZD= (0.2)^3 = 0.008

avg time in SXYZD = ((0.512*12 )+ (0.008*24))/(0.512+0.008) = 12.1846153846 ms

THUS PATH SABD MUST BE SELECTED.

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.