EL Wireless and Mobile Networking Spring 2002 Mid-Term Exam Solution - March 6, 2002

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1 Instructions: EL Wireless and Mobile Networking Spring 2002 Mid-Term Exam Solution - March 6, 2002 Be sure to write your name on your submission. This is an open book test. Use your class notes, the EL604 textbook. Be neat. Unreadable solutions will not be graded. Show all the steps in your solution. Points will be awarded for correct steps. Number of problems: 5 Total number of points: 35 Problem 1 (5 points): Consider a hybrid TDMA/FDMA wireless system that uses TDD. Assume that the total spectrum available is 20Mhz and that we divide this spectrum into carriers, each with a 1Mhz bandwidth. Assume that a 1Mbps signal can be carried per carrier. The framing structure used on each frequency consists of 20 time slots per frequency. For each communication session, a station requires 100Kbps in the reverse direction and 1Mbps in the forward direction. Also, 10% of the total wireless link bandwidth needs to be set aside for control messages. 1. What is the maximum number of stations that can be supported? 2. How many time slots are assigned in each direction for each communication session? 3. What is the minimum number of frequencies needed by each station per communication session? 4. If it is an FDD system instead of TDD, what is the maximum number of stations that can be supported? 5. For the FDD case, how should the spectrum be partitioned between upstream and downstream needs? Answer (1 point each): 1. Total wireless link bandwidth is 20Mhz. Of this 10% is used for control; therefore 18Mhz is available for user traffic. A data rate of 18Mbps can be supported on this bandwidth. Each call needs 1.1Mbps. Therefore a maximum of = 16 stations 1.1 can be supported. 2. Each time slot is kb/s or 50kbps. For the upstream direction, we need 2 time slots 20 and we need 20 time slots in the downstream direction. 3. At least two frequencies are needed because each frequency can only support 1Mbps but each station needs 1.1Mbps. 1 of 5

2 4. If the system is FDD, given we have a total of 20-2 (for control) = 18 carriers. This can be divided in the following manner: Upstream allocation (in carriers) Downstream allocation (in carriers) Number of hosts limited by upstream allocation Number of hosts constrained by downstream allocation etc. Therefore the maximum number of stations that can be supported occurs when we allocate 2 carriers for upstream and 16 for downstream. This maximum is 16 hosts. 5. The allocation is 2 upstream and 16 downstream, and 2 for control (which can be allocated between upstream and downstream in any way 1+1 or 1+ some time slots for one direction or the other). Problem 2 (10 points): Consider a 7-cell system covering an area of 3100km 2. The average number of subscribers in the seven cells is as follows: Number of subscribers Each user generates an average of 0.03 Erlangs of traffic, with a mean holding time of 120s. The system is designed for a grade of service of Determine the average number of calls per hour per subscriber. Since traffic intensity A per subscriber is 0.03 and A = λh, where λ is call arrival rate per hour and h is the holding time expressed in hours, the average number of calls made per subscriber is 0.03 ( ) = 0.9 calls/hour (1 point) 2. Determine the number of channels required in each cell. The traffic generated per cell is as follows (determined by multiplying number of subscribers with the number of Erlangs per subscriber) (2 points) Traffic (Erlangs) of 5

3 If P is and the erlangs is 14.21, then we need 24 channels (consulting the table presented in the book Table 10.3 or the table I gave in class in erlang.ppt.). (2 points) Number of channels Determine the average number of subscribers per channel across the system. (2 points) Total number of subscribers is Total number of channels is 132. Therefore the average is Determine the subscriber density per km 2. (1 point) 2416/3100 = What is the radius of a cell (assume hexagonal cells)? (2 points) The area of a cell is A = 1.5 3R 2. With 7 cells covering an area of 3100 km 2, A = = Therefore R = = 13 km Problem 3 (7 points): Compare a pull vs. push scheme for an Internet web site access. Assume that a service provider uses the current NA-TDMA system (IS136) for the pull scheme. Each user uses a full-rate channel for a data download of. Assume that there are users scattered uniformly in a coverage area of 105 cells (assume a reuse factor of 7 and 21 control channels as described in the class lecture). The alternative is to use a push scheme in which the web site information is pushed on a TV channel at a data rate of 19.2Mbps. The channel has a reach (coverage area) of all 105 cells. Ignore propagation delays and errors. Compare the best-case total times taken for all users to obtain this file in the two schemes. Answer: In the push scheme, data can be delivered to all users with one data transmission = 4.17s 19.2Mb s (2 points) Pull scheme: if users are scattered in 105 cells, there are at least 190 users per cell. In NA TDMA system, in each cell there are = channels. The multiplicative factor 7 of 3 explains the three full-rate channels per carrier frequency. Therefore in one parallel time range (the time for one transmission of the file), only 175 users can receive the information in pull mode. Either way there will be the remaining or , i.e., 15 or 16 users who will need to wait and get their files on the next round. Plus, the data rate is much lower than in the push scheme. Therefore, the total time required is 3 of 5

4 = 9876 seconds. (1) 16.2kb s Since the NA-TDMA full-rate channel uses 13kbps payload and 3.2kbps for control information, the correct answer is = seconds. (2) 13kb s Problem 4 (10 points): Consider the network shown below. There are 10 IEEE b 11Mbps Access Point (AP) 1 Ethernet R Internet MAC 1... MAC Access Point (AP) 10 Server access points on the Ethernet LAN and 10 hosts within the coverage area of each access point. Thus, there is a total of 100 hosts (not counting the server shown in the figure). If there are no ARP caches at any of the nodes in the network, what is the maximum number of IP packets/sec that the server can send per host if packets are sent uniformly to all hosts. Assume that the Ethernet packet payload is 1500 bytes. Ignore RTS/CTS and interframe spacings. Assume that the server only knows IP addresses of the hosts and that the Ethernet link is not a bottleneck. Answer: (1 point) ARP request length: 28 bytes (also ARP reply length) (1 point) MAC header: 28 bytes (34 is acceptable but address 4 should be omitted) (1 point) Payload length: 1500 bytes (1 point) ACK length: 14 bytes (1 point) For every packet sent, an ARP request/reply needs to occur, and data packet is ACK ed. (1 point) ARP request will use up bandwidth at all 10 APs (1 point) ARP reply will use bandwidth on only 1 AP (1 point) ARP reply is ACK ed because it is a unicast packet (1 point) ARP request is not ACK ed because it is a broadcast packet (1 point) final answer PHY header: bonus The ARP request is sent by all 10APs and hence the multiplier is 100 for ARP requests, but the ARP reply is unicast and hence only impacts one AP. If n is the number of packets/sec sent by the server per host, the ( 100n( ) + 10n( ) )8 = (3) 4 of 5

5 The first term (28+28) is for the ARP request + MAC header. The second term has 1500 for data payload, 28 for MAC header for the data packet, 14 for ACK, for ARP reply + MAC header + 14 for ACK for the ARP reply. Therefore n = 85 packets/sec. Problem 5 (3 points): Suppose that the ALOHA protocol is used to share a 11Mbps wireless LAN channel. If packets are 2300 bytes long, what is the maximum throughput of the system in packets/sec? The maximum throughput in ALOHA is 18% (1 points), i.e., S = S is the number of packets transmitted per X sec, where X = = 1.7ms. Therefore the maximum throughput in packets/sec is packets/sec. (2 points for the answer) 5 of 5