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Another: an error correction coding method Hamming code did not elaborate, refer to [cat _fish] of [this blog] ~
1. The use of bit stuffing bit notation, after 0,111,110,001,111,110 framing bit string corresponding to a number?
Bit stuffing flag bit method, to "01111110" as the frame marker, i.e. the start and end of a frame of the previous frame. In order to avoid frame mark frame content appearing in each of five "1" back up "zeros." Therefore, the corresponding bit string is "011111 0 00011111 0 10" .
2. authentication modes in the Point to Point Protocol PPP, PAP is not only simple, but also solve the problem of denial of service attacks and transmitted in the clear.
Incorrect.
3. If the data link layer of the transmitting window size W = 4, after transmitting the No. 3 and No. 2 acknowledgment frame is received, the sender number of consecutive frames may also be transmitted? (A cumulative acknowledgment)
Since the cumulative acknowledgment is received acknowledgment frame No. 2, and No. 2 and the frame described previously No. 2 have been successfully received. No. 2 from the back frame 4 retransmission. Now that a number has been issued and 3 which did not receive the acknowledgment frame, it can also be continuously transmitted = 4-1 . 3 frames.
4. The main role is piggybacking: confirmation fitted into the outgoing data frame, without the need to send a separate frame, the improved channel utilization.
correct.
5. using a sliding window, each window 10 packets, the RTT is 100 ms, assuming there are 1250-byte packets, the maximum throughput is the number? (Note that 1250 byte packets is 10000. To find the maximum throughput, assuming that network capacity is not a limiting factor, ignoring packet loss)
I here only a default window ... the
amount of data transmitted by RTT is a 10 × 1250Bytes × 8bits / Byte = 100000bits
Maximum throughput = amount of data transmitted per unit time 100000bits = / 100ms = 000bits. 1 000 / S = 1Mbps .
6. The system uses a frame boundary with bit stuffing notation, a recipient string received from the network is 11101111100. So, what the original string sent by the sender is?
Method of bit stuffing or zero bit stuffing method or process flag bit stuffing bits to "01111110" as the frame marker, i.e. the start and end of a frame of the previous frame. In order to avoid frame mark frame content appearing in each of five "1" back up "zeros." The data for each recipient of five "1" immediately following a "0" removed, it can restore the original data.
111 011 111 0 0 = 1,110,111,110 .
7. codewords are used in systems 0000,0011, 1100, 1111 (corresponding to the code to be transmitted, respectively 00, 01, 10 and 11). Please use the Hamming distance determination: The system can detect the number of bit errors?
The minimum code distance is equal for all non-zero codewords minimum weight code. Here three non-zero code words corresponding to the code are 2,2,4 weight, the minimum code from the set of codewords is 2. E bit errors to be detected, require a minimum code distance d≥e + 1. Since 1 + 1 = 2, it is possible to detect a bit error.
8. The system uses a Hamming code error correction of the data to be transmitted is: 10101111, collection of even parity check codeword which is encoded under a?
Hamming code to correct a wrong, we need four 8-bit data parity bit.
The check bits inserted to afford 12 is 1 . 4 010 . 8 1111. (underlined with a validation bit at position 1, 4, 8 and so on)
3,5,7,9,11 bits of 3 "1", parity 1 up "a."
3,6,7,10,11 total of four bits "1", even parity 2 complement "0."
5,6,7,12 total of two bits, "1", even parity 4 complement "0."
9,10,11,12 total of four bits "1", even parity 8 complement "0."
Code word encoded 101,001,001,111 .
9. The system uses an error correcting Hamming code, the total length of the codeword is 11 (n = 8). Assume eight yards word represented by ABCDEFGHIJK, ask these bits, which is the location of the data bits?
This ah n = 8 did not know what is meant, but the Hamming code according to the coding rule, the power of the parity bit 2 position, the other bits are data bits. So ABDH on 1,2,4,8 is the check digit, and the rest by the data bits, the location is CEFGIJK .
10. The following information about stop - and - wait ARQ protocol, the description is correct?
Sent ACK only if the current frame falls * * (transmit window), the sender transmits the next frame.
11. a use of an error correcting Hamming code system, the transmission of data bits 7, redundant bits 4, now the receiver receives a code word as follows: 00,111,000,100, which of the following statements is correct?
The default here is even parity le.
Bit 1 Parity: 1,3,5,7,9,11 total of three bits "1" is not an even number, wrong.
Bit 2 Parity: 2,3,6,7,10,11 each bit and a "1" is not an even number, wrong.
4 parity bit: 4,5,6,7 total of two bits, "1" is an even number, right.
8 bits Parity: 8,9,10,11 a total of "1" is not an even number, wrong.
Error occurs: (1 + 2 + 8) × 1 + 4 × 0 = 11 bits. Modified correct codeword 00111000101 .
(If odd parity, then a parity of correctly 1,2,8, 4 bit parity error, an error in the code word (1 + 2 + 8) × 0 + 4 × 1 = 4 bits after the correct code is modified 00101000100)
12. a system uses a cyclic redundancy check error detection code. If the generator polynomial is G (x) = x ^ 3 + x ^ 2 + 1, the original codeword 1111 are to be transmitted, calculate using the CRC code word encoded is the number?
1101,4 bit generating polynomial is, we make 4-1 = 0 3 behind the original code word, to give 1111000. It then 1111000 ÷ 1101 = 1,011,111, the remainder is added to the end of the original code word 111, to obtain encoded codeword, is 1,111,111 .
13. a system uses a cyclic redundancy check error detection code. If the generator polynomial is G (x) = x ^ 3 + x ^ 2 + 1, the receiver received codeword is 1100101, on this codeword, which of the following statements is true?
1101 is a generator polynomial, due 1100101 ÷ 1101 = 1001, exactly divisible no remainder, it is correct . And we know that the generator polynomial is four, so there are three redundant bits. CRC code word is encoded data bits + redundant bits, after 3 we can get rid of data bits, the data bits is "1100" .
14. Hamming distance of a code word in the system is 5, the 4-bit error can be checked.
correct.
Hamming distance of a code word in the system is the smallest 5 = 5 yards from it? I do not know. But here I = 5 in accordance with the minimum code distance is calculated, and the e bit errors to be detected, require a minimum code distance d≥e + 1. Because 4 + 1 = 5, the 4 bit errors can be detected.
15. Hamming distance of a codeword system 5 is capable of correcting four errors.
Incorrect.
To correct t bit errors require a minimum code distance d≥2t + 1. Since 5 = 2 × 2 + 1, we can correct up to two errors.
16. There is a string 1001011, j using odd parity for error detection. What Calculate the check digit should be appended to the string is?
String has four "1", odd parity, so to make a 1, make up an odd number of "1." So the parity bit is " 1 ."
17. In the system using error detection code, the receiving end if an error is found, usually Which of the following measures?
The receiving end do not know what made, so to regenerate a look at the raw data does not fly ah.
The system uses an error detection code, note that not error correction code Oh, so there is no error correction capability, automatic error correction can not.
Terrible mistake their own way, the upper layer protocol to trouble you doing it ...
Finally is the feedback retransmission compare actual friends.
18. In the selective retransmission protocol, when the frame number is 4 bits, and transmit and receive windows for the same dimensions, the maximum transmission window size?
Selective retransmission protocol, the size of transmitting window size of the reception window can not be played, and the receive window size range of 0 to (the maximum frame number +1) / 2. 4-bit frame number, and that the maximum frame number = 1111 (2) = 15, the maximum receive window size = (15 + 1) / 2 = 8 .
19. The data transmission using the backward N frame protocol (a GBN) between the host A and the host B, the transmission window size of 1000 A, the data frame length of 1000 bytes, the channel is 100Mbps, each B immediately receives a frame using a short frame (ignored transmission delay) for confirmation. If the one-way propagation delay between A and B is 50ms, A can reach a maximum average transfer rate which is about?
The maximum average rate is within a RTT data can just all the windows are finished.
Data of a total of all windows: 1000 (window) × 1000 (bytes / window) × 8 (bits / byte) = 8 000 000bits.
A RTT = 50ms × 2 = 100ms.
The maximum average transfer rate 8000000bits = / = 80 000 000bps 100ms = 80Mbps .
