1.i) N=10MB=10240KB f=100samples/sec B=100KBPS T=10sec S=1KB E= Pi*(N/f - (N*S)/B - T) + Pa*((N*S)/B + T) = 10*(102.4 - 102.4 - 10) + 50*(102.4 + 10) = -100 + 5620 = 5520mW There are some doubts among students as the first-part has become negative which ideally it should not be. This is a problem with the question. I have checked just the sum before exam not each part. ii). E(Q1)=5nJ, E(Q2)=10nJ, E(Q3)=2nJ & E(Q4)=8nJ and the probabilities of individual predicates being true are P(Q1)=0.5, P(Q2)=0.3, P(Q3)=0.8 & P(Q4)=0.2. Moreover, the probability of (Q1 AND Q2) being true is 0.35 & the probability of (Q3 OR Q4) being true is 0.4. Q1 & Q2 NAC Q1= 5/0.5= 10 NAC Q2= 10/0.7=14.3 So, LR evaluation E(Q1&Q2)=0.5*15+0.5*5=10 Q3 OR Q4 NAC Q3=2/0.8=2.5 NAC Q4=8/0.2=40 So, LR evaluation E(Q3ORQ4)=0.8*2+0.2*(8+2)=1.6+2=3.6 NAC (Q1&Q2)=10/0.65=15.38 NAC (Q3|Q4)=3.6/0.6=6 So, RL evaluation E((Q1 AND Q2) OR (Q3 OR Q4))= 0.4*3.6 + 0.6*(3.6+10)=1.44+8.16=9.6nJ 2. i) Modp marker indices: 2 6 9 13 Maxp marker indices: 4 6 15 ii) MODP matching using Max match between markers: 2 [0,1,2,3,4] and 13 [11,12,13,14,15] iii) Maxp needs more memory because entire data-streams to be stored in server whereas for chunk-match we just need to store the hash-values. 3. i) All cells have max. strength at locations inside boundary. No cell can be out of the boundary according to the "boundary filtering" approach. ii) New trace after cell combining A -40 B -39 C -45 D -42 E -45 F -50 G -51 According to strongest RSSI predicted location B According to weighted Centroid X= 5*-40 + 8*-39 + 7*-45 + 11* -42 + 14*-45 + 19*-50 + 16*-51 /(-40-39-45-42-45-50-51) = 11.81 Y= 9.37 7. the difference in throughput between signal strength 99 and 60 at 60th percentile ==> Approx 1.3 At what percentile does the throughput for signal strength 80 reach the throughput attained by signal strength 99 in 30th percentile ===> Around 80-82 percentile