Problem Set Errors 1
.txt
School
University College Dublin**We aren't endorsed by this school
Course
COMPUTER S COMP20320
Subject
Electrical Engineering
Date
Dec 16, 2024
Pages
2
Uploaded by CountIbexPerson1229
Problem Set Errors 1-----------------------------------------------------------Solutions1. Suppose there is a frame of size 12,000 bits and theprobability of bit error is one in a million. What is theresulting probability of frame error.Solution: approx 0.012, exact p = 1 - (1-pb)^L =2. Suppose there is a frame of size 6 kB and the probability of bit error is one in a million. What is the resulting probability of frame error.Solution: approx 0.048, probably too big, exact p = 1 - (1-pb)^L 3. The probability of bit error is one in a billion and wewant to have a probability of frame error less than 1 in amillion. What is the largest frame size in Bytes we can use.Solution: approx 10^-6 > Frame size x 10^-9approximate frame size < 1000 bits or 125 Bytes 4. The probability of bit error is one in a million and wewant to have a probability of frame error less than 1 in 10. What is the largest frame size in Bytes we can use.Solution: approx 0.1 > Frame size x 10^-6approximate frame size < 100,000 bits or 12,500 Bytesexact p = 1 - (1-pb)^L, 0.9 = (.999999)^LLog of these, -.045749 = -4.34294 10^-7 LL = 105,360 = 13,170 Bytes5. We have a k-hop path with each link having a BER of pb. Ifthe packet size is L bytes, then what is the probability thatthe packet gets to the other side correctly in terms of L, k and pb.Solution p (probability of error on a hop) 1 - (1-pb)^8*LThe probability of getting the packet correctly means thateach hop is correct or (1-pb)^L for each one. If there arek hops then it is given by ((1-pb)^8*L)^k6. We have a 5-hop path with each link having a BER of 10^-5. The packet size is 1200 bytes. What is the probability thatthe packet is in error over the whole path.Solution Correct is ((1-pb)^8*L)^k = (1-10^-5)^9,600^5= .90846^5 = .6188, so correct 38.12%7. For the example in Q6 if we want to have a probability ofbeing in error to be lower than 1 in a hundred, what is themaximum size of the packet in bytes.Solution: Correct is 99% = .99 = ((.99999)^8*L)^5-0.004365 = 5 * 8 * L * -0.434297 x 10^-71005.07 = 40 * L, L = 25 bytes
8. We have a path with 8 intermediate nodes and all the linkshave the same probability of bit error. The packet size isgiven as 1250B. What is the highest BER to achieve 99.9%of the packets to be recieved correctly.Solution: Correct is 99.9% = .999 9 hops as 8 intermediate nodes, so .999 = ((1-pb)^8*L)^9-4.34512 x 10^-4 = 8 * 9* 1250 * Log (1-pb)Log(1-pb) = -4.8279 x 10^-91-pb = 0.99999998888, or pb = 1.112 x 10^-8