Data Comm Midterm Study Guide
Twenty signals, each requiring 4000 Hz, are multiplexed onto a single channel using FDM. What is the minimum bandwidth required for the multiplexed channel? Assume that the guard bands are 200 Hz wide.
Minimum bandwidth = 4000 x 20 + 200 x 19 = 83,800 Hz
A 12-bit Hamming code whose hexadecimal value is 0xE4F arrives at a receiver. What was the original value in hexadecimal? Assume that not more than 1 bit is in error.
101001001111
When should a selective repeat protocol be used versus a go-back-n protocol.
Go-Back N retransmits all the frame after its lost, selective repeat only retransmits the lost frame.
A bit string, 011111101111101111110, needs to be transmitted at the data link layer. What is the string actually transmitted after bit stuffing? Underline the additional bits you insert.
Need to break up every 6 continuous ones with a zero. Given Data: 011111101111101111110 Transmitted Data is ----> 011111010111110011111010
If a binary signal is sent over a 3-kHz channel whose signal-to-noise ratio (SNR) is 40dB, what is the maximum achievable data rate?
Shannon theorem: B*log_2^(1+S/N) dB=10log_10^(S/N)40dB=10log_10^(S/N) ==> S/N=10^4=10,000 3k*log2^(1+S/N)=3log2^10,001kbps=39.87kbps. Maximum data rate = 6kbps according to Nyquist rate.
What SNR is needed to put a T1 carrier on a 50-kHz line?
Shannon's theorem = Blog_2(1+S/N) = 50kHz data rate is 1.544Mbps ==> S/N=2^(1.544M/50k)-1=1.976e9=92.9581dB
A 1024-bit message is sent that contains 992 data bits and 32 CRC bits. CRC is computedusing the IEEE 802 standardized, 32-degree CRC polynomial. For each of thefollowing, explain whether the errors during message transmission will be detected by the receiver: There were 47 isolated bit errors
The CRC can catch all the odd number of isolation errors. - Yes
Provides a simple parametric formula to quantify the size [in bytes] of a PPP frame.
X + 10; x is the bytes in payload and 10 is the max number of bytes in other fields.
Calculate the total time required to transfer a 20 MB file assuming an RTT of 100 ms, an initial 2 x RTT of "handshaking" before data is sent, and a bandwidth of 10 Mbps (assume data can be sent continuously).
(2 x 0.1 )+ (100 ms x 2^20 · 8/(10 x 2^20)) + (0.1/2) = 16.25 seconds
Imagine a sliding window protocol using so many bits for sequence numbers that wraparound never occurs. What relations must hold among the four window edges and the window size W, which is constant and the same for both the sender and the receiver? Use the following notation for the sender and receiver:Sl = Sender-lower and Su = Sender-upper to designate the lower and upper bounds of the sender's window. Likewise, use (Rl , Ru) for the receiver's window.
0 <= Su - S1 + 1 ≤ W Ru - R1 + 1 = W S1 <= R1 <= Su + 1
The following character encoding is used in a data link protocol:A: 01100111 B: 11110011 FLAG: 01111110 ESC: 11111100Show the bit sequence transmitted (in binary) for the four-character frame A B ESCFLAG when each of the following framing methods is used: Byte count
00000101 01000111 11100011 11100000 01111110
The following bit sequence arrives at the network adapter:011011101101011111011010111111101001111110Assuming an HDLC protocol, mark any stuffed bits, bits indicating the beginning or end of the frame, and bits indicating that an error has occurred.
011011101101011111011010111111101001111110
A regional telephone company has one million subscribers. Each of their telephones is connected to a central office by a copper twisted pair. The average length of these twisted pairs is 10 km. How much is the copper in the local loops worth? Assume that the cross section of each strand is a circle 1 mm in diameter, the density of copper is 9.0 grams/cm3, and that copper sells for $6 per kilogram.
Cross section of each strand = π/4 sq. mmmaterial volume = 2 x π/4 x 10 ^ -2 m^3 = 1.57 m^3 = 15708 cm^3Mass of loop with gravity is 141 kg.Weight of wire owned by company = 141 x 10^7 kg = 1.41 x 10^9 kgPrice of 1 loop = $6/kgPrice of all loops = 6 x 1.41 x 10 ^9 = 8.46 x 10 ^ 9Price of all loops = 8.46 billion $
Specify the fields that make up a Point-to-Point protocol [PPP] frame.
Flag (1 byte): Marks the beginning and the end of the frame. The bit pattern of the flag is 01111110. Address (1 byte): Provides the Ip address of the destination , all are 1's that means that its a broadcast address. Control (1 byte ): Is set to a constant value of 11000000 shows that frame doesn't contain any sequence number. Protocol (1 or 2 bytes): It define the type of data that will be present in the payload field. Payload: Carries the data from the network layer. The maximum length of the payload field is 1500 bytes. FCS: It is a 2 byte or 4 bytes frame check sequence for error detection.
The following character encoding is used in a data link protocol:A: 01100111 B: 11110011 FLAG: 01111110 ESC: 11111100Show the bit sequence transmitted (in binary) for the four-character frame A B ESCFLAG when each of the following framing methods is used: flag bit with byte stuffing
Flag: 01111110 A: 01000111 B: 11100011 Esc (byte Stuffed): 11100000 Esc: 11100000Esc (byte stuffed): 11100000Esc (byte stuffed): 01111110flag: 01111110
The following character encoding is used in a data link protocol:A: 01100111 B: 11110011 FLAG: 01111110 ESC: 11111100Show the bit sequence transmitted (in binary) for the four-character frame A B ESCFLAG when each of the following framing methods is used: Starting and ending flag bytes with bit stuffing.Provide you answers in hexadecimal format.
Flag: 7E A: 47 B: 1A3 Esc: 1C0 Flag: FA Flag: 7E
What is the minimum bandwidth needed for communication link to transfer a video/tv stream of 4Kx4K pixels resolution, in real time / broadcast mode, assuming 64-bit encoding of colors per pixel? Can the bandwidth you calculated be achieved using all three communication mediums, namely: electrical, optical, and the wireless spectrum. Justify your answer. As motivation for the question: Consider the needs to support site synchronous connectivity for classrooms between the Statesboro and Armstrong campuses.
Formula: (n bits x (streamxstream)mil x (60)fps)/1024
A 100-km-long cable runs at the T1 data rate. The propagation speed in the cable is 2/3 the speed of light in vacuum. How many bits fit in the cable?
Propagation Delay = Length / speed Propagation speed = 2/3 * 3 *10^8 m/sec = 2*10^8 m/sec = 2*10^5 km/sec Propagation delay is 100/ 2*10^5= 0.5 *10 ^ - 3 RTT= Round trip time --->2* Propagation Delay RTT= 2*0.5 * 10^-3 = 1/1000 sec Total data carry by the cable = RTT * T1 Total data carry by cable= 1/1000 * T1 * 8 bits
A 1024-bit message is sent that contains 992 data bits and 32 CRC bits. CRC is computedusing the IEEE 802 standardized, 32-degree CRC polynomial. For each of thefollowing, explain whether the errors during message transmission will be detected by the receiver: There was a 24-bit long burst error
The CRC can catch errors with a burst length less than 32. -Yes
A 1024-bit message is sent that contains 992 data bits and 32 CRC bits. CRC is computedusing the IEEE 802 standardized, 32-degree CRC polynomial. For each of thefollowing, explain whether the errors during message transmission will be detected by the receiver: There was a 35-bit long burst error
The CRC cannot catch errors with a burst length greater than 32. - No
A 1024-bit message is sent that contains 992 data bits and 32 CRC bits. CRC is computedusing the IEEE 802 standardized, 32-degree CRC polynomial. For each of thefollowing, explain whether the errors during message transmission will be detected by the receiver: There were two isolated bit errors
The CRC catches all errors for any long message. - Yes
A 1024-bit message is sent that contains 992 data bits and 32 CRC bits. CRC is computedusing the IEEE 802 standardized, 32-degree CRC polynomial. For each of thefollowing, explain whether the errors during message transmission will be detected by the receiver: There was a single-bit error
The CRC catches all the errors - Yes
A 1024-bit message is sent that contains 992 data bits and 32 CRC bits. CRC is computedusing the IEEE 802 standardized, 32-degree CRC polynomial. For each of thefollowing, explain whether the errors during message transmission will be detected by the receiver: There were 18 isolated bit errors
The CRC might not catch all isolated errors of even number. - No
An image is 1600 × 1200 pixels with 3 bytes/pixel. Assume the image is uncompressed. How long does it take to transmit it over a 56-kbps modem channel? Over a 1-Mbps cable modem? Over a 10-Mbps Ethernet? Over 100-Mbps Ethernet? Over gigabit Ethernet?
The image is 1600*1200*3 bytes = 5,760,000 bytes or 46,080,000 bits. At 56,000 bits/seconds it takes 822.857 seconds. At 1,000,000 bits/sec it takes 46.08 seconds.
Ten signals, each requiring 4000 Hz, are multiplexed on to a single channel using FDM. How much minimum bandwidth is required for the multiplexed channel? Assume that the guard bands are 400 Hz wide.
There are ten 4000 Hz signals. We need nine guard bands to avoid any interference. The minimum bandwidth required is 4000*10+400*9=43,600 Hz.
The distance from earth to a distant planet is approximately 9 × 1010 m. What is the channel utilization if a stop-and-wait protocol is used for frame transmission on a 64 Mbps point-to-point link? Assume that the frame size is 16 KB and the speed of light is 3 × 10^8 m/s.
Tx = 32KB/64Mbps = 0.004 seconds (assuming base 10 for kilo and mega) RTT = 2 * (9*10^10m) / (3*10^8m/s) = 600 seconds Utilization = 0.004 / (600 + 0.004) = 6.667x10^-6 = 6.667x10^-4 %
Suppose a sliding window protocol is used instead. For what send window size will the link utilization be 100%? You may ignore the protocol processing times at the sender and the receiver.
period = 600.004 seconds 1 Tx = 0.004 seconds
What is the minimum bandwidth needed to achieve a data rate of B bits/sec if the signal is transmitted using Manchestor encoding?
the bits are either high '1' or low '0' voltage. Cycles are completed every bit. Therefore a bandwidth "B Hz" is needed.
What is the minimum bandwidth needed to achieve a data rate of B bits/sec if the signal is transmitted using NRZ encoding?
the cycles between 1s and 0s. so it completes the cycle every 2 bits. Therefore, the bandwidth B/2 Hz is needed to achieve data rate "B bits/sec
