#7 We want transmission rate to equal propagation delay. What is RTT for a 2km guided link? RTT/2 = 2km/2* 10^8 mbps = 10^-5 sec Suppose the packet size is 100 bytes. What should the transmission rate(Tx) be? 100 bytes/Tx bps = 10^-5 sec. 800 bits/Tx bps = 10^-5 sec. 800b /10^-5 sec = Tx Tx = 8 * 10^7 bps similarly, if the packet size is 512, Tx = 4.096 * 10^8 bps #5 A question on the transfer rate of a file using stop-and-wait, in which an ack must be returned after each packet is received before the next packet can be sent. The last packet is not acknowledged but we need a half of a RTT for it to get to the other side. Assume an initial 2RTT is required for a handshake and that RTT is 100 ms (alternatively you might have to compute it as above) Assume that a 1000KB file is broken in packets of 1 KB each and that the transmission rate is 1.5Mbps. The number of packets that must be sent is 1000. The transmission time of each packet is 1024 * 8 bits/1.5Mbps or about .00546 sec Then total transfer time is 2.5RTT + 1000 (.00546sec) + (1000 RTT - RTT)(the last one is not acknowledged) 1001.5 (.1) + 5.46 sec or about 105.61 seconds. What is the bit length in a 10 meter copper wire if the transmission speed is 5Gbps? (10 meters* 5* 10^9 bps) / 2*10^8 mps check your metrics; the meters and seconds cancel, leaving you with an answer in bits, which is what the question asked for 250 bits A packet of 250 bits will "fill" a 1-way pipe for the above numbers. Sample questions for midterm in computer networks 1) An example of digital information is _________ (computer data, the status of on/off switches)) 2) An example of analog information is __________ (a handwritten signature) 3) An advantage of analog signals is ___________ (only type of signal that can be used with wireless media and broadband coax; can be limited to a narrowband, allowing FDM and spread spectrum technologies) 4) An advantage of digital signals is ___________________ (computers use them internally) 5) An advantage of digital transmission is (repeaters can recover from small amounts of noise; errors over repeaters are not cumulative; VLSI of components allows for cost savings; enables the use of TDM & CSMA; integration of many different types of data) 6) An advantage of analog transmission is (it's in place in many local loops; attenuation is less) 7) An example of a media that uses analog signals is _______ ( broadband coax, wireless media) 8) A method for transforming voice into digital form by sampling and encoding analog signals is ______ (PCM) 9) A device for transforming voice into digital form is ______ (codec) 10) A device for transforming digital signals into analog signals ( modem) 11) A device for transforming analog signals into digital signals is ans: depends what the analog signals represent - can be codec - see question 9 can be demodulating part of modem - the latter if the underlying data has been digitized but modulated over an analog carrier Problems 1) If a sixteen-level signal is sent over a 5 kHz channel with signal-to-noise ratio of 1000 (30dB; 100 is 20dB), what is the maximum achievable data rate? DISCUSSION: sixteen-level suggests Nyquist's theorem - try Max Data Rate = 2 Hz log (base 2) (# of levels) = 10k (4) = 40kbps signal-to-noise ratio suggests Shannon Max Data Rate = Hz log (1+ S/N) = 5k log (1001) about 50kbps Since both of these are upper limits on Max Data Rate you must pick the smaller value, i.e., 40kbps 2) Using the generator polynomial x^4 + x + 1 and the frame message: 010001111100 a) What frame value will be transmitted? 10011 | 010001111100 0000 10011 10111 10011 10010 10011 10 000 10 011 0110 CRC frame transmitted 010001111100 0110 b) Give an example of an error that will not be detected by the CRC method. any frame that the divisor will divide with a 0 remainder - for example, if the entire frame was shifted one place to the left (a synchronization error) 3) Assume that the office SneakerNet carries an average of 3 floppies, each of which holds 1.4 Mbytes, an average of .25 km per trip. Assume that its velocity is 1 km/ hour. What is its average data rate in bps? 3(1.4Mbytes) (8bits/byte) = 33.6 *1.024 * 1.024 Mbits (Mbytes = 1024 * 1024 bits) 1 km/hour means than .25 km takes .25 hour or 900 sec. Our data rate is about 39 kbps 4) Consider an audio signal with spectral components in the range 300 to 3400 Hz. Assume that a sampling rate of 8 kHz is used to generate a PCM signal with 256 quantizing levels. What data rate is required for the digital information? What frequency band is required assuming a signal to noise ratio of 30dB? I am describing PCM above. We need 8 bits for the 256 levels, and 8000 of them per second, i.e, 64kbps. Using Shannon, we get 64kbps =(Frequency Band in Hz) (1 + S/N) = (Frequency band in Hz) log (1+1000) (about 10) 6.4k Hz = Frequency Band <== answer But I only have 3.1 kHz, by above. We can't seem to satisfy our requirements. One possibility that would work is to use Adaptive Differential Pulse Code Modulation, an encoding scheme that needs only 4 bits for the 256 levels since it encodes differentials rather than absolute values. Then place digital equipment in the end office to lower your S/N ratio. Then "steal" some of the upload capacity. You should be able to satisfy the requirements. 5) How many constellation points will a modem need to send at a data rate of 64kbp over a 4 kHz noiseless channel? Discussion: You must use Nyquist's theorem here. 64kbps = 2* 4k cycles/sec *log ( levels)/cycle 8 = log (levels) 2^8 = levels 256 levels needed for our signal (and modem points) to send 8 bits 6) what is the wavelength for signals sent at a frequency of 100 Hz. ans. wavelength = c/ f = 3*10^8 m/sec divided by 100 cycle/sec = 3 * 10,000,000 meters (per cycle implicit) That is pretty large and one reason why signals are not sent over low fequencies. The antennae would have to be pretty big. For the following questions, pick the BEST answer (At most one of these questions will be on the midterm; they are just examples of the type of question to expect.) 1. Advantages of computer networks include a) greater security b) faster computation c) faster communication d) simplicity of software (faster communication) 2. The highest transmission rates are achieved using a) fiber optics b) twisted pair type 3 c) coaxial cable d) radio waves (fiber optics) 3. The lowest error rates are achieved in a) fiber optics b) twisted pair c) coaxial cable d) radio waves (fiber optics) 4. The highest error rates are achieved in a) fiber optics b) twisted pair c) coaxial cable d) radio waves (radio waves) 5. Attenuation a) Does not change the strength of the signal b) Is a large factor in the use of fiber optics c) Distorts signals differently at different frequencies d) Is caused by intermodule noise (distorts signals differently at diff freq) 6. Given a wavelength of 4 meters for optical fiber, what is the frequency. a) 25 MHz b) 50 MHz c) 75 MHz d) 100 MHz (b) Use Lf = c, where c = 2 * 10^8 meters/second (guided media) f = 2* 10^8 mps / 4 meters per cycle f = 50MHz NOT COVERED IN CLASS Fill in the following blanks: 7) Advantages of fiber cables over copper cables is _______ and _____ I'd accept better lower error rates; better security; higher bandwidth 8) SONET frames include control bytes for _______________ Lots of possible answers, but I prefer frame synchronization. pointer to the first byte of user data (SPE) is also good. NOT COVERED IN CLASS From Midterm spring 2004 1) Assume that you have a 16kHz band, with a S/N of 1000 and a modem transmitting at 16 levels. What is the maximum capacity of the channel?. (10 pts) You must try Nyquist and Shannon's formulae and pick the smaller of the two numbers. 2) A modem constellation diagram has data points at (0, 1) and (0 ,3). What type of modulation scheme is used? If it is operating at 1200 baud, what is the bit rate? (5 pts) ans: AM; 1200 bps 3) For a terrestrial link of 15000 meters sending at 100Mbps with a propagation speed of 2 * 10^8 m/seconds, , and a frame size of 1000 bits, what is the maximum utilization of the link? How many bits should we use for a sequence number to fully utilize the channel? (10pts) Propagation delay (Tp) is 15000m/(2 * 10^8 m/sec) = .000075 sec transmission rate (Tx) is 1000b/ 100Mbps = .00001sec A bound on the max utilization = Tx/ (Tx+2Tp) = .00001 / .00016 We need 16 frames on the line to fill the pipe - i.e., 17 values in our window. This requires 5 bits in our sequence number. 4) What is the frequency of a 5 meter wavelength using a microwave link ? (5 pts) NOT COVERED IN CLASS see #6 5a) Using an error checking polynomial of x^5 + x^3 + 1 and an input value of 01011010101110, find the value of the frame that will be sent (ignoring delimiters). (4 pts) ans: divide 101001 into 01011010101110 00000 (5 0's added with x5 poly) 101001 100010 101001 101011 101001 1010 00 1010 01 01000 (you need 5 digits for x^5) Frame sent: 01011010101110 01000 5b) Assume that a bit synchronization error occurred losing the first bit. Show me if this will be caught by the CRC computation. (1 pt) Ans: It won't be caught. Do the above manipulation without the leading bit. 6) For the input value of 100101101, find the check bits for the Hamming code that will allow you to correct a 1 bit value. Assume even parity (5 pts) _ _ 1_ 0 0 1 _ 0 1 1 0 1 the 1st check bit is the parity bit of bits 3, 5, 7, 9, 11 and 13 the 2nd check bit is the parity bit of bits 3, 6, 7, 10, 11 etc. 7) Given the following CDMA chip sequences for +1 (0 has the complementary chip sequence): A: (-1 -1 -1 +1 +1 -1 +1 +1) B: (-1 -1 +1 -1 +1 +1 +1 -1) C: (-1 +1 -1 +1 +1 +1 -1 -1) D: (-1 +1 -1 -1 -1 -1 +1 -1) Show the transmission and recovery at the base station when A and C send 1's, B sends a 0 and D does not send. (5 pts) NOT COVERED SUFFICIENTLY IN CLASS 8) Using stop and wait, assume that acknowledgments are NOT numbered and that a transmitted frame (numbered 0) is received correctly and acknowledged. The sender times out prematurely and transmits a duplicate frame. What action will the receiver take (accept the duplicate frame? send an ack?) ASSUMING THE RECEIVER RECEIVES THE DUPLICATE CORRECTLY, HE WILL DISCARD IT BUT SEND BACK AN ACK Now the sender receives the first acknowledgment and transmits frame number 1. Frame 1 is corrupted in transmission and the receiver discards it. The sender, meantime, receives the ack for duplicate frame 0 and accepts it as the ack for frame 1. What action will the sender take? THE SENDER WILL SEND THE NEXT (NEW) FRAME WITH SEQUENCE NUMBER 0. Is it possible that a frame is lost or that a duplicate frame is accepted with this scenario? (10 pts) YES, FRAME #1 WAS LOST. Fill in the following (3 pts each) 9) A function of the data link layer (besides providing services for the network layer) is ___________ possible ans: FRAME DELIMITING 10) An example of an HDLC control field is ___________ possible ans: SEQUENCE NUMBER 11) The function of the above field is ____________ possible ans: ERROR CONTROL, FLOW CONTROL 12) An advantage of standards is ____________ possible ans: ENABLE DIFFERENT MANUFACTURERS TO PROVIDE EQUIPMENT 13) An example of a standards maintaining organization is _____________ possible ans: ITU, IETF 14) The difference between baud rate and bit rate is ______________ possible ans: BAUD RATE IS SYMBOLS/SEC BIT RATE IS BITS/ SECOND IT IS POSSIBLE TO SEND MULTIPLE (OR FRACTIONAL) BITS/ BAUD 15) A protocol whose bit rate and baud rate are different is ______________ possible ans: ETHERNET 16) A function of the Physical Layer (besides providing services for the data link layer) is _________ possible ans: BIT ENCODING 17) An advantage of digital transmission over analog transmission is ______________ possible ans: ERRORS ARE NOT CUMULATIVE OVER REPEATERS 18) The result of bit stuffing 011111101111011111000 is _____________ ans: 01111101011110111110000 19) The media with the lowest error rate is __________________ ans: FIBER 20) A disadvantage of connectionless service is that _________________ possible ans: EACH DATA UNIT MUST INCLUDE COMPLETE SOURCE AND DESTINATION ADDRESSES 21) Traceroute returns _____________ possible ans: IP addresses of each router in the path that a packet might take between host and destination 22) CODECs may use _________ to digitize voice. possible ans: PCM 23) Match the protocol with its layer (5 pts) a) physical layer 1) http b) data link layer 2) SONET c) media access layer 3) PPP d) network layer 4) Ethernet e) transport layer 5) IP f) application layer 6) TCP ans: physical layer: SONET data link layer: PPP media access layer: ETHERNET network layer: IP transport layer: TCP application layer: HTTP Pick the best answer for the following (3 pts each) 24) A 100 km long cable runs at 1Gbps. The propagation speed is 2 * 10^8 m/sec. What is the bit length of the cable? a) 50 bit b) 5000 bits c) 50000 bits d) 500000 bits e) 5000000 bits ans: propagation delay of the cable is 100km/ 10^8 m/sec = 10^-3 sec. 1Gbps * 10^-3 sec = 10^6 bits 25) GSM uses a) analog signals and analog transmission b) analog signals and digital transmission c) digital signals and analog transmission d) digital signals and digital transmission ans: b) 26) SONET overhead includes a) a pointer to the beginning of the SPE (Synchronous Payload Envelope) b) a CLP (cell loss priority) bit to indicate that a frame can be dropped c) a CRC (cyclic redundancy check) for backward error detection d) a RNR (receive not ready) field to separate flow control and error control e) a Protocol field to identify the network protocol that is using SONET ans: a