College of Computer and Information Sciences Department of Computer Engineering CEN444 Computer Networks Midterm 2 Exam Second Semester 1434/1435

Transcription

1 College of Computer and Information Sciences Department of Computer Engineering CEN444 Computer Networks Midterm 2 Exam Second Semester 1434/1435 Student Name ID Time Allowed: 2.0 Hours. Closed Book, Closed Notes. Problem Score/Points 6 6 Total Score 40 20

2 Problem 1 ( points) 1. (4 pts) Frames arrive randomly at a 20-Mbps channel for transmission. If the channel is busy when a frame arrives, it waits its turn in a queue. Frame length is exponentially distributed with a mean of 8,000 bits/frame. For the frame arrival rate 2100 frames/sec, give the delay experienced by the average frame, including both queueing time and transmission time. The delay is calculated using the formula: T = 1/(μC λ). Data rate C = = 2 10 bits/sec. Mean frame length 1 1 = 8000 bits/frame. Thus μ = = 1.25 μ frames/bit. Mean service rate μc = frames/bit 2 10 bits/sec = 2500 frames/sec Arrival rate λ = 2100 frames/sec. T = 1 = 1 = = 2.5 msec μc λ (3 pts) A group of N stations share a 36-kbps pure ALOHA channel. Each station outputs a 2880-bit frame on average once every 10 sec, even if the previous one has not yet been sent (e.g., the stations can buffer outgoing frames). What is the maximum value of N? In pure ALOHA, the maximum throughput is Maximum usable bandwidth = bits/sec = 6624 bits/sec. Each station requires So, N = bits 10 sec = 23 stations. = 288 bits/sec. Page 1 of 6

3 Problem 2 ( points) 1. (4 pts) A large population of ALOHA users manages to generate 60 requests/sec, including both originals and retransmissions. Time is slotted in units of 50 msec. What is the chance of success on the first attempt? a. G = number of transmission + retransmissions per frame (slot) time Slot time = 50 msec = 0.05 sec/slot. Number of slots per sec = 1 Frames transmissions = 60 frames/sec. G = frames/sec 20 slots/sec = 3 frames/slot. Chance of success on first attempt P 0 = e G = e 3 = = 20 slots/sec. 2. (3 pts) What is the length of a contention slot in CSMA/CD for a 10-km multimode fiber optic cable (speed is 65% speed in vacuum)? Signal propagation speed = = m/sec. Propagation time = τ = Distance Speed = m m/sec = sec. Length of contention slot = 2τ = = sec. Page 2 of 6

4 Problem 3 (6 pts) 1. In PPP state, peers exchange a series of LCP packets, to select the PPP options. (a) AUTHENTICATE (b) ESTABLISH (c) NETWORK (d) TERMINATE 2. Within AAL5 used in ADSL, PPP framing is not needed because: (a) ADSL uses different error correction than ATM. (b) ATM is a virtual circuit protocol. (c) ATM cell is too small. (d) ATM and AAL5 already provide framing. 3. The following is an example of static channel allocation method: (a) FDM. (b) Aloha. (c) Slotted Aloha. (d) HDLC. 4. In slotted Aloha with mean of G frames per frame time (old and new), the highest throughput occurs at G = (a) 2 (b) 1 (c) 0.5 (d) e 5. In non-persistent CSMA, if channel is busy, the station: (a) reports failure. (b) continuously senses the channel until it becomes idle. (c) waits a random time and repeats the algorithm. (d) waits 2τ and starts transmission. 6. In p-persistent CSMA with p = 0.3. If the station senses the channel and the channel is idle, the station will: (a) continuously sense the channel with probability 0.3. (b) wait for 0.3 seconds and repeats the algorithm. (c) transmit in the current slot with probability 0.. (d) transmit in the current slot with probability 0.3. Answer b d a b c d Page 3 of 6

5 Problem 4 (6 points) 1. How long does a station, s, have to wait in the worst case before it can start transmitting its frame over a LAN that uses the basic bit-map protocol? Explain your Answer. The worst case is where all stations want to send and s is the lowest-numbered station. Wait time N bit contention period + (N 1) d bit for transmission of frames. The total is N (N 1) d bit times. 2. Give an example to show that the RTS/CTS in the protocol is a little different than in the MACA protocol. In the figure above is an example of an exposed terminal problem that MACA can solve but does not. In this example, when B wants to transmit to A in MACA, it sends an RTS that both C and A can hear however when A sends the CTS, C can not hear it; so, C it is free to transmit. In , even if C hears just the RTS without the CTS, it is going to remain silent during the potential transmission time. Page 4 of 6

6 Problem 5 ( points) 1. Suppose the round-trip propagation delay for 10 Mbps Ethernet is 46.4 µs. This yields a minimum packet size of 512 bits (464 bits corresponding to propagation delay + 48 bits of jam signal). What happens to the minimum packet size if the delay time is held constant, and the signaling rate rises to 100 Mbps assuming same 48 bits of jam signal? Assuming 48 bits of jam signal was still used, the minimum packet size would be bits = 586 bytes 2. Let A and B be two stations attempting to transmit on an Ethernet. Each has frames ready to send all the time; A s frames will be numbered A1, A2, and so on, and B s similarly. Let T = 51.2 µs be the exponential backoff base unit. Suppose A and B simultaneously attempt to send frame 1, collide, and happen to choose backoff times of 0 T and 1 T, respectively, meaning A wins the race and transmits A1 while B waits. At the end of this transmission, B will attempt to retransmit B1 while A will attempt to transmit A2. These attempts will collide, but now A backs off for either 0 T or 1 T, while B backs off for time equal to one of 0 T,, 3 T. Give the probability that A wins this second backoff race after this collision; that is, A s choice of backoff time is less than B s. A can choose ka=0 or 1; B can choose kb=0,1,2,3. A wins outright if (ka, kb) is among (0,1), (0,2), (0,3), (1,2), (1,3); there is a 5/8 chance of this. Page 5 of 6

7 Problem 6 ( points) 1. Suppose that a 10-Mbps b LAN is transmitting 128-byte frames back-to-back over a radio channel with a bit error rate of How many frames per second will be damaged on average? A frame contains 1024 bits. The bit error rate is p = The probability of all 1024 of them surviving correctly is (1 p) 1024, which is about The fraction damaged is thus about The number of frames/sec is /1024 or about Multiplying these two numbers together, we get about damaged frame per second. 2. How can hidden terminals be detected in networks? Give an example and explain. In the figure above, A wants to send to B and C wants to send to D. Those two transmissions cannot happen in the same time as collision is going to occur at B. The solution is the RTS and CTS as in When A sends an RTS to B and B replied back by a CTS, this CTS will be heard by C and it will remain silent during the expected transmission time and this will solve the problem. The reason is called hidden problem is that C and A are not aware of each other transmissions as they are outside the radio range of each other or they are hidden from each other. Page 6 of 6