Computer Network Homework 2

Computer Network

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< Data and computer communication >
4.1; 6.1; 6.8, 6.14, 6.15; 7.2, 7.5

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4.1

Suppose that data are stored on 1.4-Mbyte floppy diskettes that weigh 30 g each. Sup- pose that an airliner carries 104 kg of these floppies at a speed of 1000 km/h over a distance of 5000 km. What is the data transmission rate in bits per second of this sys- tem?

$$total: \frac{104 * 1000}{30} * 1.4 * 1024 * 8 = 39758506.666666664 bit$$
$$time: 5 hous = 5 * 60 * 60s = 18000 s$$
$$therefore, speed = \frac{total}{time}=2208.81 bit/s$$

6.1

Suppose a file of 10,000 bytes is to be sent over a line at 2400 bps.
a. Calculate the overhead in bits and time in using asynchronous communication. Assume one start bit and a stop element of length one bit, and 8 bits to send the byte itself for each character. The 8-bit character consists of all data bits, with no
parity bit.
b. Calculate the overhead in bits and time using synchronous communication.
Assume that the data are sent in frames. Each frame consists of 1000 characters
8000 bits and an overhead of 48 control bits per frame.
c. What would the answers to parts (a) and (b) be for a file of 100,000 characters?
d. What would the answers to parts (a) and (b) be for the original file of 10,000 characters except at a data rate of 9600 bps?

a.

$$overhead: \frac{10000 * 8}{8} * 2 = 20000bit$$
$$time: \frac{\frac{10000 * 8}{8} * (8 + 2)}{2400} = 41.67s$$

b.

$$overhead: \frac{10000}{1000} * 48 = 480bit$$
$$\frac{\frac{10000}{8000} * (8000 + 48)}{2400} = 4.19s$$

c.

• a
  

  $$overhead: \frac{100000 * 8}{8} * 2 = 200000bit$$
  $$time: \frac{\frac{100000 * 8}{8} * (8 + 2)}{2400} = 416.7s$$

• b
  

  $$overhead: \frac{100000}{1000} * 48 = 4800bit$$
  $$\frac{\frac{100000}{8000} * (8000 + 48)}{2400} = 41.9s$$

d.

• a
  

  $$overhead: \frac{10000 * 8}{8} * 2 = 20000bit$$
  $$time: \frac{\frac{10000 * 8}{8} * (8 + 2)}{9600} = 10.42s$$

• b
  

  $$overhead: \frac{10000}{1000} * 48 = 480bit$$
  $$\frac{\frac{10000}{8000} * (8000 + 48)}{9600} = 1.05s$$

6.8

Two communicating devices are using a single-bit even parity check for error detection. The transmitter sends the byte 10101010 and, because of channel noise, the receiver gets the byte 10011010. Will the receiver detect the error? Why or why not?

10101010

10011010

No.We can see that highlighted words is changed, and the number of changed word is 2.The single-bit even parity check just work well when the value of changed word is odd.

6.14

A CRC is constructed to generate a 4-bit FCS for an 11-bit message. The generator polynomial is X4 + X3 + 1.

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a. Draw the shift register circuit that would perform this task (see Figure 6.6).

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b. Encode the data bit sequence 10011011100 (leftmost bit is the least significant)
using the generator polynomial and give the codeword.

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c. Now assume that bit 7 (counting from the LSB) in the codeword is in error and

6.15

A CRC is constructed to generate a 4-bit FCS for an 11-bit message. The generator polynomial is

$$X^4 + X^3 + 1$$

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a. 10010011011

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b. Suppose the channel introduces an error pattern 100010000000000 (i.e., a flip
from 1 to 0 or from 0 to 1 in position 1 and 5). What is received? Can the error be
detected?
received: 00011011011

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c. Repeat part (b) with error pattern 100110000000000.

7.2

The number of bits on a transmission line that are in the process of actively being trans- mitted (i.e., the number of bits that have been transmitted but have not yet been received) is referred to as the bit length of the line. Plot the line distance versus the transmission speed for a bit length of 1000 bits. Assume a propagation velocity of 2 * 10^8 m/s.

$$\frac{l}{2 * 10 ^ 8 m/s} = \frac{1000 bit}{v}$$
$$l * v = 2 * 10^8 km/s$$

7.5

In Figure 7.10 frames are generated at node A and sent to node C through node B.
Determine the minimum data rate required between nodes B and C so that the buffers of node B are not flooded, based on the following:
• The data rate between A and B is 100 kbps.
• The propagation delay is 5 ms/km for both lines.
• There are full duplex lines between the nodes.
• All data frames are 1000 bits long; ACK frames are separate frames of negligible
length.
• Between A and B, a sliding-window protocol with a window size of 3 is used.
• Between B and C, stop-and-wait is used.
• There are no errors.
Hint: In order not to flood the buffers of B, the average number of frames enter- ing and leaving B must be the same over a long interval.