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4 years ago

Is bit stuffing necessary in the control or address field in theHDLC protocol? why?

Computers and Technology

1 answer:

steposvetlana [31]4 years ago

7 0

Answer:

Many network and communication protocols require bit stuffing for the following purposes: to avoid the interpretation of data as control information. For example, with six consecutive 1 bits, X.25 uses the bit stuffing,signal the beginning and end of a frame.

High-level Data link control (HDLC) is a bit-oriented protocol for point-to-point and multi-point communication.P/F bits are present in HDLC control field.

So,Yes bit stuffing is necessary in control field.

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What do you understand by technological depenence?

son4ous [18]

Answer:

It helps you understand stuff faster

Explanation:

7 0

3 years ago

Add the following 2's complement binary numbers. Also express the answer in decimal. a. 01+ 1011b. 11+ 01010101c. 0101+ 110d. 01

kenny6666 [7]

Answer:

(a) 01 + 1011

Convert the two numbers into their 4-bit representation.

=> 01 = 0001

=> 1011

Addition of the two numbers (in their 2's complement form) gives the following:

0 0 0 1 ------------------> + 1 (in decimal)

1 0 1 1 ------------------> - 5 (in decimal)

1 1 0 0 -------------------> - 4 (in decimal)

Explanation for (a)

01 or 0001

Since the most significant bit is 0 (leftmost bit), it shows that it is a positive number. Therefore, it can be directly converted to its decimal representation as follows:

0001 = 0 x $2^{3}$ + 0 x $2^{2}$ + 0 x $2^{1}$ + 1 x $2^{0}$ = 1

0001 is + 1 in decimal.

1011

Its most significant bit(MSB) is 1 showing that it is negative.

Flipping all its bits and adding 1 to the result, will convert it to it's positive counterpart.

i.e 1011 => 0100 + 1 = 0101.

Converting the result (0101) to decimal gives

0101 = 0 x $2^{3}$ + 1 x $2^{2}$ + 0 x $2^{1}$ + 1 x $2^{0}$ = 5

1011 is -5 in decimal

01 + 1011 or 0001+ 1011 = 1100

The result (1100), of the addition of the two numbers, is a negative number as the MSB is 1.

Convert it to it's positive counterpart.

i.e 1100 => 0011 + 1 = 0100.

Converting the result (0100) to decimal gives

0100 = 0 x $2^{3}$ + 1 x $2^{2}$ + 0 x $2^{1}$ + 0 x $2^{0}$ = 4

Therefore, the result, 1100 is -4 in decimal.

===================================================

(b) 11 + 01010101

Convert the two numbers into their 8-bit representation since one of them is in 8-bit.

PS: Taking 11, its MSB is 1 showing that it is a negative number. Also, it's 8-bit representation will mean pre-padding it with ones(1s) rather than zeros as follows. In other words, pre-padding a 2's complement number is done using its sign bit.

=> 11 = 11111111

=> 01010101

Addition of the two numbers (in their 2's complement form) gives the following:

1 1 1 1 1 1 1 1 ------------------> - 1 (in decimal)

0 1 0 1 0 1 0 1 ------------------> + 85 (in decimal)

1 0 1 0 1 0 1 0 0 -------------------> + 84 (in decimal)

Discard the carry-out bit (1) making the result 01010100

Explanation for (b)

11 or 11111111

Its MSB is 1 showing that it is negative.

Convert it to it's positive counterpart.

11111111 => 00000000 + 1 = 00000001

Converting the result (00000001) to decimal gives 1

11 or 11111111 is - 1 in decimal.

01010101

Its MSB is 0 showing that it is also positive. Now, convert to decimal

01010101 = 85

01010101 is + 85 in decimal.

11111111 + 01010101 = 01010100

The result (01010100), is also positive as the MSB is 0. Now, convert to decimal.

01010100 = 84

Therefore the result (01010100) is + 84 in decimal.

===================================================

Convert the two numbers into their 4-bit representations.

=> 0101 = 0101

=> 110 = 1110

Addition of the two numbers (in their 2's complement form) gives the following:

0 1 0 1 ------------------> + 5 (in decimal)

1 1 1 0 -----------------> - 2 (in decimal)

1 0 0 1 1 -------------------> + 3 (in decimal)

Dicard the carry-out bit (leftmost bit) to get 0011 as result.

Explanation for (c)

0101

Since the MSB is 0 (leftmost bit), it shows that it is a positive number. Now, convert to decimal.

0101 = 5

0101 is +5 in decimal.

110 or 1110

Its MSB is 1 showing that it is negative.

Convert it to it's positive counterpart.

i.e 1110 => 0001 + 1 = 0010.

Converting the result (0010) to decimal gives

0010 = 2

1110 is -2 in decimal

0101 + 110 = 0101+ 1110 = 0011

The result (0011), is a positive number as the MSB is 0.

Converting the result (0011) to decimal gives

0011 = 3

Therefore, the result, 0011 is 3 in decimal.

===================================================

(d) 01 + 10

Convert the two numbers into their 4-bit representation.

=> 01 = 0001

=> 10 = 1110

Addition of the two numbers (in their 2's complement form) gives the following:

0 0 0 1 ------------------> + 1 (in decimal)

1 1 1 0 ------------------> - 2 (in decimal)

1 1 1 1 -------------------> - 1 (in decimal)

Explanation for (d)

01 or 0001

Since the MSB is 0 (leftmost bit), it shows that it is a positive number. Now convert to decimal.

0001 = 1

0001 is + 1 in decimal.

1110

Its MSB is 1 showing that it is negative.

Convert it to it's positive counterpart.

i.e 1110 => 0001 + 1 = 0010.

Converting the result (0010) to decimal gives

0010 = 2

1110 is -2 in decimal

01 + 1011 or 0001+ 1110 = 1111

The result (1111), is a negative number as the MSB is 1.

Convert it to it's positive counterpart.

i.e 1111 => 0000 + 1 = 0001.

Converting the result (0001) to decimal gives

0001 = 1

Therefore, the result, 1111 is -1 in decimal.

7 0

4 years ago

Over a TCP connection, suppose host A sends two segments to host B, host B sends an acknowledgement for each segment, the first

Margaret [11]

Answer:

a. Host A will retransmit neither segments

Explanation:

The options are:

a. Host A will retransmit neither segments

b. Host A will retransmit first segment

c. Host A will retransmit the second segment

d. Host A will retransmit both segments

The first acknowledgment is lost but the second acknowledgment arrives before the timer of the first segment expires, and hence the host A gets confirmation that Host B has received both the segments. And hence Host A will retransmit neither segments.

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3 years ago

Why is musical notation important? What benefits do musicians and others receive from being able to write down and note aspects

madreJ [45]

Most importantly, musicians can share their works with others. Other people can see their musical ideas and can try and perform them too. Nuances such as tempo, dynamics (loud soft, sweet, "harsh", are just some examples) can be understood even if the composer is not present and there is no recording to listen to.
Financial benefits can be realized from the sale of sheet music, scoring the piece, arranging the piece for bands, orchestras, etc. Conductors can lead an entire musical ensemble through the piece.

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4 years ago

Computers are often referred to as _____.

GalinKa [24]

Smart machines I'm pretty sure

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3 years ago