Spring 2006 May 11 2006 CMPE 150 NAME Mid Term Exam 1 10 pts Signals sent over long distances must be amplified to restore them to the proper intensity level Why is it better to send digital encoding of analog signals e g audio vs sending the analog signal Ans Digitally encoded signals when successively reconstructed after effects of noise and attenuation but while still perfectly decodable can be regenerated noise free Analog signals after attenuation and noise when amplified have signal strength restored but noise in signal bandwidth is also amplified 2 5 pts a Give at least two reasons why fiber is preferable to UTP for a data link 1 fiber has less noise susceptibility 2 fiber has less attenuation per unit length 3 so a long fiber link needs fewer repeaters 4 optical signals in fiber are not easily tapped as are electric signals so are more secure 5 pts b Give at least two reasons why most LANs still use UTP 1 UTP is cheaper and easier to install than fiber 2 many installations already have UTP installed 3 Network Interface Cards NIC cards for workstations for UTP are cheaper than for fiber LANs 3 10 pts Suppose your computing infrastructure has a mail server file server and a print server all attached to the server LAN and your machine and others attached to departmental LANs There is a bridge connecting the server LAN and the departmental LANs Suppose you change departments and move your machine from port 2 of LAN A to port 7 of LAN D What happens now to frames sent for example by the mail server to your machine How when does the bridge fix its table Ans They continue to go to port 2 of LAN A until the table entry for your port in the bridge times out After that when your machine is not in the table of the bridge it floods any new traffic directed to your MAC address and thus it gets to you When you acknowledge receipt of a frame at your new location or if you take some action from your new location such as trying to use the file server or reading your mail the bridge will then see traffic coming from your new location find you are not in its table and make an entry for you at port 7 of LAN D 4 10 pts We have seen that the efficiency of an Ethernet due to its random access scheme is far below 100 A proposal for a different protocol is as follows Suppose there are N stations using the LAN Time on the network is divided equally among each of the stations and they take turns on the LAN according to their allocated time slot time division multiplexing It is claimed that this protocol will achieve 100 efficiency Is this true Is this a good protocol for a LAN Explain Ans This protocol time division multiplexing will achieve 100 efficiency only if all stations have traffic to send all the time Otherwise for any station without traffic to send at a time slot that slot is just wasted Ethernet traffic is usually bursty and this protocol is a bad idea for such a LAN and will probably be significantly worse than the CSMA CD of Ethernet 5 5 pts a Why did 802 11 require a different protocol than 802 3 Ethernet for controlling access to the channel Ans Ethernet 802 3 uses CSMA CD Wireless can t do CD as it can t listen while sending Also hidden terminals keep if from hearing all the traffic of all the stations on the LAN 5 pts b If 802 11 used the same frames as Ethernet frames what would have been the consequences Ans The 802 11 wireless environment suffers from significant noise unlike a good LAN using UTP so long frames as in 802 3 would have a much higher probability of being damaged by noise than would short ones and thus would require more retransmissions and consequently less throughput 6 The POTS telephone sends voice after the local loop as digital data The analog signal is limited to approximately a 4 KHz bandwidth and is sampled at 8 KHz a 3 pts Why is it sampled at 8 KHz Ans To meet the Nyquist criteria 2x max frequency in the signal b 3 pts If the samples are quantized to 256 different levels what is the bit rate required for sending this digital signal Ans log2 256 8 bits of encoding sample 8 x 103 samples sec x 8 bits sample 64 x 103 bits second c 3 pts What is the bandwidth required for sending this digital signal when encoded in two levels Ans Nyquist data rate 64 x 103 bits sec 2H log2 V 2H as V 2 so H 32 K bps d 3 pts If a new VoIP Voice over IP phone system wanted to give better fidelity to voice traffic and chose to deliver up to 8 KHz of analog bandwidth how many bits per second would it use for this voice channel again using 256 quantization levels Ans 2 x 8 x 103 samples sec x 8 bits sample 128 bits second e 3 pts How might it achieve this still using the same digital channel used for the POTS as found in c Ans use more levels From Nyquist we need to have 128 x 103 bits sec 2 x 32 x 103 log2 V or log2 V 2 so V 4 levels needed 7 Rank the link layer protocols Stop and Wait S Go back N G and Selective Repeat R for a link with noise not error free and of non negligible length according to a 5 pts utilization of link bandwidth Ans R G S b buffering required of the i 5 pts sender Ans R G S ii 5 pts receiver Ans R G no buffering used for S 8 20 pts For a 10 Mbps Ethernet LAN the 802 3 specification allows multiple cable segments and multiple repeaters but no two transceivers may be more than 2 5 km apart and no path between any two transceivers may traverse more than four repeaters text pg 274 Later text pg 277 we find 500 bits is the smallest frame that is guaranteed to work this number was rounded up to 512 bits or 64 bytes a Show how this frame size is determined Ans For collision detection the number of bits must fill the segment and back so that a sender is still sending when the first bit gets back to the sender at the maximum distance But the maximum cable segment is 2 5 km so it is 5 km round trip distance at 2 x 108 meters second takes 5 2 x 108 2 5 x 10 5 seconds or 25 sec At 107 bits sec the sender would in 25 sec send 250 bits So the standard to allow for the time delay in repeaters doubled this to 500 bits and then rounded it up to 64 bytes b In specifying this frame size how …
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