T 305 A Digital Communications Problems Solving I

T 305 A: Digital Communications Problems Solving I Arab Open University Tutorial 1 2009 -2010 1

Question 1 What is the minimum frame size necessary to ensure the correct operation of the CSMA/CD protocol at a bit-rate of 1500 Mbit/s in a star topology with a maximum segment length of 200 m and two repeaters? Delay in the repeater is 0. 75 � �s, signal propagation in cable is: 1. 77 x 108 m/s. Solution: The worst-case cable distance between two stations is 2 x 200 m = 400 m. The propagation delay through the cable is: Delay= 400/ (1. 77 x 108 ) = 2. 26 �s. So the round-trip delay is 2 x 2. 26 = 4. 52 �s. The total repeater delay is 4 x 0. 75 = 3� �s. Therefore, the total round-trip delay (slot time) is: 4. 52 + 3 = 7. 52 �s. The number of bits that can be transmitted during the slot time is: # of bits =1500 x� 106 x� 7. 52 x 10 -6 = 11280 bits

Question 2 Two computers A and B are connected through an Ethernet link having a bit-rate of 10 Mbit/s. The Round-Trip Time (RTT) between the two computers is assumed to be 2 milliseconds. Computer A is sending messages containing 250 octets to computer B and waits for an ACK after each message. a- What is the percentage utilization of the channel capacity by computer A in this scenario? b- If a sliding window algorithm with a window width of seven messages is used, what would be then, the percentage utilization? Solution: a- Channel Capacity= bit-rate x RTT = (10 x 106 ) x (2 x 10 -3 ) = 2 x 104 bits A message of 250 octets transports 8 x 250 = 2000 bits. The percentage utilization of the channel = the ratio between the number of bits sent within an RTT period and the number of bits that could be sent (i. e. the channel capacity) x 100: Percentage utilization= (2000 / (2 x 104 )) x 100= 10 % b- Using a sliding window algorithm with a window width of 10 messages, the number of bits that is actually sent is given by: 7 x 2000 = 14000 bits The new Percentage utilization= (14000 / (2 x 104 )) x 100= 70 %

Question 3 Duration X (<5) Y(≥ 5) L (Local-calls) 11358 4365 O (Overseas calls) 3124 1153 The table below gives information on 20000 telephone calls. Calls are classified by distance and duration and the table gives the number of calls of each type. Category X represents calls lasting less than five minutes and Y represents calls lasting five minutes or longer. L represents local calls, and O represents overseas calls. a- What is the probability that an Overseas call lasts less than 5 minutes? b- What is the probability that a Local call lasts 5 minutes or more? c- What is the probability that a call would be a local call? Solution: a- The total number of overseas calls= 3124+1153= 4277 and 3124 of these lasted less than five minutes. So, the probability that an overseas call lasts less than five minutes= 3124/4277=0. 73. b- The total number of local calls=11358+4365=15723 and 4365 of these lasted 5 minutes or more. So, the probability that a local call lasts five minutes or more =4365/15723=0. 277. C- The total number of local calls=15723 and the total number of calls = 20000 calls. So, the probability that a call would be a local call = 15723/20000=0. 786.

Question 4 A network contains eight switches. Each has a probability of 0. 85 of being in operation at any given time. a- What is the probability of all the switches being in operation? b- What is the probability of all the switches being out of operation? c- What is the probability of five of the switches being in operation? Solution: a- The probability of one switch being in operation=0. 85. So, the probability of one switch being out of operation is: = 1 -0. 85=0. 15 Therefore, the probability of all switches being in operation is= (0. 85) 8 = 0. 272 b- The probability of all switches being out of operation is= (0. 15) 8 = 2. 56 x 10 -7 c-The probability of five of the eight switches being in operation is given by Formula of the binomial distribution: P(r) = Cnr pr (1 -p)n-r & Cnr = n! / (r! x (n-r)!) Where r=5; n=8; p=0. 85 P( r ) = C 58 (0. 85)5 (0. 15)8 -5; C 58 =8!/( 5!x 3!) = 56 P(r) = 56 x (0. 85)5 x (0. 15)3 = 56 x 0. 4437 x 0. 003375=0. 08386.

0. 6 Question 5 0. 7 0. 6 0. 8 A subsystem of a communication system has 0. 6 the reliability diagram shown in the figure above that shows the individual components reliability. 0. 7 0. 6 0. 8 Calculate the overall reliability of the system to three significant figures. Solution: There are six components shown in the diagram above three of which are connected in parallel. R 1=0. 7; R 2=R 3=R 4=0. 6; R 5=0. 8; R 6=0. 8; The reliability for the three components in parallel = [1 -(1 -0. 6) 3 ] = 0. 936 Therefore, Rp= 0. 936 The reliability for the overall system is given by: RT= R 1 x Rp x. R 5 x. R 6= 0. 7 x 0. 936 x 0. 8=0. 419.

Day Question 6 Survivors at the start of day Failures during day 1 600 100 2 500 83 3 4 In tests of a certain type of component, it is found that 5 approximately one-sixth of the surviving components fail each day. The table shown below predicted results for the first two days of a 5 -day test of 600 of these components. Complete the missing fields in the above table (round your answers to the nearest integer). Calculate the value of the reliability R (t) at the end of day 4. Calculate the value of the failure function Q (t) at the end of the test period.

Day Survivors at the start of day Solution: 600 The completed table is shown below with taking 1 into consideration the rounding to the nearest integer. 2 where Ns(t) 500 is b- The reliability function R(t) at a time t = Ns(t)/No the number of components surviving at time 3 t & No 417 is the number of components under test. 4 347 No = 600; the number of surviving components at the end of day 4= Ns (t) = 289. 5 289 Therefore, reliability R(t) at the end of day 4 = 289/600=0. 481 c- The failure function Q(t) at a time t = Nf(t)/No where Nf(t) is the number of components which have failed by time t & No is the number of components under test From the table, the number of components that have survived by the end of day 5 is: 289 -48= 241. Then, the total number of components which have failed by the end of day 5 = Nf(t)= 600241=359. Therefore, the failure function Q(t) at the end of the test period= 359/600=0. 598 Arab Open University-Lebanon 2009 Failures during day 100 83 70 57 48 8

Question 7 The components in a new communication system are known to have a constant hazard rate. It is a requirement of the system that exactly 4% of the components should fail in the first 2000 hours of operation. What is the MTTF that corresponds to such a system? Solution: We have: 4% of the components failing during the first 2000 hour. This means that after, 2000 hours R(t)= 0. 96. Because components have a constant hazard rate, h, the reliability function R(t) is given by: R(t)= exp(-ht). So, R(2000)=exp(hx 2000)=0. 96. To find the value of h, take natural logarithms of both sides, this gives: -hx 2000=ln(0. 96). This implies: h=-ln (0. 96)/ 2000 = 0. 0000204. The MTTF for components with a constant hazard rate h is 1/h. so, the corresponding MTTF = 1/h=1/0. 0000204= 48993. 2 hours.

Question 8 An STM-4 signal is passed through a buffer of 3 bits width, for how long is the signal delayed? b) A BIP-8 calculation is carried out on the following block of data: 110010111 10111100 01100011 10111100 10011100. Determine the value of the BIP bits, assuming even parity? Solution: a) Time to transmit one bit is = 1/ (4*155520000) s= 1. 6075 ns 3 bits need 3*1. 6075 ns=4. 8225 ns So the delay will be 4. 8225 ns b) 110010111 10111100 01100011 10111100 10011100 -----10100011

Hours form start of the test Question 9 Total number of failure 1400 110 1600 138 2800 150 A batch of 4000 components is tested. Table 1 shows an extract from the test results. Estimate the value of the reliability function R (t) and the failure density function f(t) at time t= 2800 hours after the start of the test. Solution: R(t)=(4000 – 150) / 4000=0. 9625 at time t = 2800 hours Q(t 1) = 138/4000 (at time t 1 = 1600 hours) Q(t 2) = 150/4000 (at time t 2 = 2800 hours) Now we can deduce the value of f(t) at time t = 2800: f(t) = d. Q(t) / dt = (Q(t 2) – Q(t 1)) / (2800 – 1600) = (12/4000)/1200=0. 003/1200=0. 0000025

Question 10 In table 1 below you can depict the average daily loading of a wireless Ethernet LAN recorded over a period of 300 days. The LAN’s performance degrades significantly at loading above 40%. What is the probability that the LAN loading will exceed this value on a given day? Loading ≤ 10% 84 >10% and ≤ 20% 78 >20% and ≤ 30% 63 >30% and ≤ 40% 37 >40% and ≤ 50% 21 > 50% 17 Solution: The number of days on which the LAN loading exceeds 40% can be found by totaling the figures in the last two rows of the table: 21+17, giving 38. The total number of days on which the loading was measured is 300 days, so probability of a loading above 40% is 38/300, which is 0. 127. n Number of days

Question 11 The components in a new communication system are known to have a constant hazard rate. It is a requirement of the system that exactly 6% of the components should fail in the first 3000 hours of operation. What is the MTTF that corresponds to such a system? Solution: We have: 6% of the components failing during the first 3000 hour. This means that after, 3000 hours R(t)= 0. 94. Because components have a constant hazard rate, the reliability function R(t) is given by: R(t)= exp(-ht). So, R(3000)=exp(-hx 3000)=0. 94. To find the value of h, take natural logarithms of both sides, this gives: -hx 3000=ln(0. 94). This implies: h=-ln (0. 94)/ 3000 = 0. 0000206. The MTTF for components with a constant hazard rate h is 1/h. so, the corresponding MTTF = 1/h=1/ 0. 0000206 = 48484. 53214 hours.

Q 132. Probability n n A trial of internet providers has tested the availability of the dial-in lines over a period of weeks. It found that the best service providers gave access on 99% of the occasions. What is the probability of successful access using this service provider? For the worst service provider, the dial-in line was found to be engaged or unavailable for 22% of the time. What is the probability of successful access this service provider? Answer: probability is 0. 99 probability is 0. 78

Q 13. Probability n n n A digital transmission system uses sequences of 4 bits. It is known that all bit pattern are equally probable. What is the probability of receiving a sequence containing exactly three 1 s? Answer: There are 16(2^4) possible 4 bit sequences. So the probability that a sequence should contain three 1 s therefore 4/16=0. 25

Q 14. Probability n n A communication system uses frames consisting of 8 data bits followed by a checksum bit. It can be assumed that if an error occurs in a frame, only one bit is in error and it is equally likely to be any bit. If an errored frame is received, what is the probability that the errored bit is the checksum bit itself, rather than one of the data bits? How many data bits would a frame need to contain in order to reduce this probability to 0. 01? Answer: Probability that the checksum is the errored bit is 1/9=0. 11 A probability of 0. 01 nthat is the checksum digit which is errored corresponds to a 1 in 100 chance. Therefore there must be 100 bits in a frame, including checksum.

Q 15. Joint Probability n n n A telephone exchange uses two identical processors. A and B. The exchange can provide adequate service as long as one of them is working. Maintenance records show that probability, P (A or B) failing in any hour period is 0. 0008. P(AB) is 0. 0002. what is the probability that individual processor should fail. It can be assumed that this probability is the same for both processors. Answer: P(A or B)= P(A)+P(B)-P(AB) P(A)+P(B)=P(A or B)+ P(AB)=0. 0008+0. 0002=0. 001 P(A)=P(B)=0. 001/2=0. 0005

Q 16. Conditional Probability n n n Two nodes in a network are connected by two parallel links, A and B. Each of these links is found to be out of operation for 1% of the time. A). What is the probability that both links will be out of operation at any given time? State any assumptions you are making in your. answer: Assuming that the failure of both links A and B are independent events. P(AB)=P(A)*P(B)=0. 01*0. 01=0. 0001

Q 17. Binomial distribution n n A network contains 5 routers, each has a probability of 0. 01 of being out of operation at any given time. What is the probability of all routers being out of operation? What is the probability of 4 of them being out of operation? Answer: P, probability of one router being out of operation is 0. 01, the probability that 5 routers out of operation is (0. 01)^5. The probability of 4 of them being out of operation can be calculated using the binomial distribution: 5 C p 4(1 -p)(5 -4)=4. 95*10 -8 4

Q 18. Binomial distribution n n What is the probability that there will be three errors in a code word 16 binary digits long if the probability of a single errors is 10^-4 and errors occur independently. Answer: The probability of a single error is p=10^-4 and the probability of three errors out of 16 digits can be calculated using the binomial distribution: 16 C p 3(1 -p) (16 -3)=5. 6*10 -10 3

Q 19. Reliability n The table show the result s of a test of 1000 components over a period of 50 hours. The second column is the table gives the number of components failing in each 10 -hour period from the start of the test. What is the probability that a given component will have failed by 30 hours into the test? What is the probability that a given component will still be functioning at the end of the test period? Test period Number of failures <=10 hours 100 >10 and <=20 hours 90 >20 and <=30 hours 81 >30 and <=40 hours 72 >40 and <=50 hours 66

Answer : after 30 hours testing, the number of components which have failed is 100+90+81=271. The total number of components under test is 1000. So the probability that a given component will have failed by 30 hours into the test is 271/1000=0. 271 At the end of the test the number of component who have failed is 100+90+81+72+66=409. Still functioning are 1000 -409=591. so 591/1000=0. 591 is the probability that a given component is still functioning at the end of the test period. Arab Open University-Lebanon Tutorial 1 2008 -2009 22

Q 20. Failure density function n n Hours from Total a batch of 1000 components is tested. Table shows the start of number of an extract from the test results. Estimate the value test failures of the reliability function R(t) and the failure density function f(t) at time t=2000 hours after the start of 1900 the test. Answer: 2000 After 2000 hours there were 1000 -130=870 2100 surviving components. R(2000)-870/1000=0. 87. F (t) can be calculated from the number of failures per unit time divided by the total number of components. In each 100 -hour periods before and after 2000 hours there were(135 -130)=5 failures, so 0. 05 failure per hour f(t)=0. 05/1000=5*10^-5 per hour. 125 130 135

Q 21. Hazard rate n n n n A batch of components, which has already been through burn-in, is found to have a constant hazard rate of 0. 01 per day. If the size of the batch is 1000, how many components are likely to be in operation (a) after the first day of the test, (b) after the second day, © after 10 days? Answer: a hazard rate of 0. 01 means failing 1/100 every day. A) So in the first day 10 of 1000 will fail, leaving 990 survivors B) 10 (aprox) in the second day, leaving 980. C). having a constant hazrad rate R(t)=exp(-ht) So after 10 days, R(t)=exp(-0. 01*10)=0. 905

Q 22. Availability. A router is found to fail, on average, four times a year. A failure can be detected immediately, but it takes an average of 5 hours for the router to be brought back into service. What is the probability that, at any given time, the router will be operational. Answer: The MTTR is 5 hours. The routers fails, on average, four time a year, and for each failure spends on average 5 hours being repaired. Its MTBF is therfore (365 X 24/4)5 houtrs=2185 hours. The probability that the router will be in operation at any given time is the avaialibility A, which si: A/MTBF/(MTBF+MTTR)=2185/(2185+5)=0. 998