1) A network uses byte stuffing on frames of 1000 bytes each. If frames have 20%

Question

1) A network uses byte stuffing on frames of 1000 bytes each. If frames have 20% of their data that contains FLAG or ESCP bytes in average, what is the average bandwidth efficiency of the framing mechanisms? Count the FLAG byte at the start and end of each frame when providing answer.

2) A network uses bit stuffing on frames of 1000 bytes each. If frames have 20 contiguous sequences of six 1's in average, what is the average bandwidth efficiency of the framing mechanisms? Count the six 1 bits at the start and end of each frame when providing answer.

3) We have the following code with 3 data bits (bold) and 2 redundant bits:

C0= 00010

C1= 00101

C2= 01010

C3= 01101

C4= 10010

C5= 10100

C6= 11001

C7= 11111

Find the Minimum Hamming Distance for this code?

What is the maximum number of bits in error that this code can detect?

What is the maximum number of bits in error that this code can correct?

Explanation / Answer

1.

Byte stuffing on frame of 1000 byte each.

Since all the frame is start and end with flag data. Thus, each frame will have 2-byte extra

data.

In the 1000-byte data, the 20% data is either flag or ESCP data. Thus, the flag or ESCP

byte is equal to the 200-byte data.

Therefore, the actual data in the frame = 800-byte.

Another escape byte or flag byte to be inserted to avoid the confusion and misinterpretation

of data = 200-byte.

Consider the bandwidth equals to frame size = 1000-byte/second.

The complete flag data which is stuffed = 200+200+2

                                                                = 402-byte

Thus, the total utilization of sending 1000-byte data = 60% approximately.

Therefore, the efficiency = (402*100)/1000

                                        = 60% approx.

2.

Network is using bit stuffing on 1000-byte. It means the 8000-bits data.

Since the frames have 20 contiguous sequences of six 1's in average, thus, 20 bit will be

added to the frame in bit stuffing.

Frame size is now 8002 bits.

Consider the bandwidth = size of the frame.

Therefore, the average efficiency of bandwidth is = ((8000 + 20 + 12) *100)/1600

                                                                              = 50.2 %

Where, 12 bit is the inserted bits at the end of the frame.

3.

Hamming distance,

[c0, c1] = 3

[c1, c2] = 4

[c2, c3] = 3

[c3, c4] = 5

[c4, c5] = 2

[c5, c6] = 3

[c6, c7] = 2

Minimum hamming distance = 2

Maximum number of bits in error that this code can detect = hamming distance – 1

                                                                                            = d – 1

                                                                                            = 2 – 1

                                                                                            = 1 bit

maximum number of bits in error that this code can correct = (distance – 1)/2

                                                                                               = (2 – 1)/2

                                                                                               = .5 bits’ error

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