Sequence numbers inside the sender’s window: Given: Window size = Size = 4 Range = 1024 Case 1: Let’s consider the receiver is “K” and acknowledged all the “k-1” data packets. The sender side window’s range will be of [ k , k+ Size - 1 ] sequence numbers

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Answer 1

Expert Solution & Answer

Chapter 3, Problem P22P

a.

Explanation of Solution

Sequence numbers inside the sender’s window:

Given:

Window size = Size = 4

Range = 1024

Case 1:

Let’s consider the receiver is “K” and acknowledged all the “k-1” data packets. The sender side window’s range will be of [ k , k+ Size - 1 ] sequence numbers

b.

Explanation of Solution

Values of ACK field:

Suppose the receiver is waiting for the packet “k”, then it has been received and acknowledged “k-1” packet and the “Size-1” packets before that. If NACKs of those packets have not been received by the sender, then the values of ACK field is [ k - size , k - 1 ] may propagate back.

Because the sender has sent packets k - 1, it states that the sender had already received an acknowledgement for k - Size -1

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- | العنوان I need a detailed drawing with explanation so A 4 شكا +x-pu +965 Taylor Series Approximation Example- H.W More terms used implies better approximation f(x)+ f(x) Zero order First order 1.0 0.5 x-0 Second order True f(x) f(x) • flx;+ 1) = f(x) + fƒ '\x;}h √(x,+ 1) = f(x) + f (x)h + "(x) 2 f(x1) X+1-1 f(x) 0.1x 0.15x³-0.5x²-0.25x+1.2 Taylor Series Approximation H.w: Smaller step size implies smaller error Errors 100+ F(x) Zero order First order Second order 0.5 Reduced step size 0 x, 0 f(x+1)f(x,) + ƒ\x,}h 51 f(x + 1) − f(x) + (x)+2 ((x1) +1-1 Using Taylor Series Expansion estimate f(1.35) with x0 =0.75 with 5 iterations (or & s= 5%) for f(x) 0.1x 0.15x³-0.5x2- 0.25x+1.2 マ 52

Home Work Use Taylor's series expansion to Compute the true and approximate percent relative errors & and εa for x = π/6, use 6 iterations for the series. sin(x) == x-

Taylor Series Approximation Example- H.W More terms used implies better approximation f(x) 4 f(x) Zero order f(x + 1) = f(x;) First order f(x; + 1) = f(x;) + f'(x;)h 1.0 Second order 0.5 True f(x + 1) = f(x) + f'(x)h + ƒ"(x;) h2 2! f(x+1) 0 x; = 0 x+1 = 1 x h f(x)=0.1x4-0.15x³- 0.5x2 -0.25x + 1.2 51 Taylor Series Approximation H.w: Smaller step size implies smaller error Errors f(x) + f(x,) Zero order f(x,+ 1) = f(x) First order 1.0 0.5 Reduced step size Second order True f(x + 1) = f(x) + f'(x)h f(x; + 1) = f(x) + f'(x)h + "(xi) h2 f(x,+1) O x₁ = 0 x+1=1 Using Taylor Series Expansion estimate f(1.35) with x0 =0.75 with 5 iterations (or & s= 5%) for f(x)=0.1x 0.15x³-0.5x²- 0.25x + 1.2 52

Answer 2

Expert Solution & Answer

Chapter 3, Problem P22P

a.

Explanation of Solution

Sequence numbers inside the sender’s window:

Given:

Window size = Size = 4

Range = 1024

Case 1:

Let’s consider the receiver is “K” and acknowledged all the “k-1” data packets. The sender side window’s range will be of [ k , k+ Size - 1 ] sequence numbers

b.

Explanation of Solution

Values of ACK field:

Suppose the receiver is waiting for the packet “k”, then it has been received and acknowledged “k-1” packet and the “Size-1” packets before that. If NACKs of those packets have not been received by the sender, then the values of ACK field is [ k - size , k - 1 ] may propagate back.

Because the sender has sent packets k - 1, it states that the sender had already received an acknowledgement for k - Size -1

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Check out a sample textbook solution

See solution

Answer 3

Expert Solution & Answer

Chapter 3, Problem P22P

a.

Explanation of Solution

Sequence numbers inside the sender’s window:

Given:

Window size = Size = 4

Range = 1024

Case 1:

Let’s consider the receiver is “K” and acknowledged all the “k-1” data packets. The sender side window’s range will be of [ k , k+ Size - 1 ] sequence numbers

b.

Explanation of Solution

Values of ACK field:

Suppose the receiver is waiting for the packet “k”, then it has been received and acknowledged “k-1” packet and the “Size-1” packets before that. If NACKs of those packets have not been received by the sender, then the values of ACK field is [ k - size , k - 1 ] may propagate back.

Because the sender has sent packets k - 1, it states that the sender had already received an acknowledgement for k - Size -1

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 4

Expert Solution & Answer

Chapter 3, Problem P22P

a.

Explanation of Solution

Sequence numbers inside the sender’s window:

Given:

Window size = Size = 4

Range = 1024

Case 1:

Let’s consider the receiver is “K” and acknowledged all the “k-1” data packets. The sender side window’s range will be of [ k , k+ Size - 1 ] sequence numbers

b.

Explanation of Solution

Values of ACK field:

Suppose the receiver is waiting for the packet “k”, then it has been received and acknowledged “k-1” packet and the “Size-1” packets before that. If NACKs of those packets have not been received by the sender, then the values of ACK field is [ k - size , k - 1 ] may propagate back.

Because the sender has sent packets k - 1, it states that the sender had already received an acknowledgement for k - Size -1

Expert Solution & Answer

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See solution

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Chapter 3, Problem P22P

a.

Explanation of Solution

Sequence numbers inside the sender’s window:

Given:

Window size = Size = 4

Range = 1024

Case 1:

Let’s consider the receiver is “K” and acknowledged all the “k-1” data packets. The sender side window’s range will be of [ k , k+ Size - 1 ] sequence numbers

b.

Explanation of Solution

Values of ACK field:

Suppose the receiver is waiting for the packet “k”, then it has been received and acknowledged “k-1” packet and the “Size-1” packets before that. If NACKs of those packets have not been received by the sender, then the values of ACK field is [ k - size , k - 1 ] may propagate back.

Because the sender has sent packets k - 1, it states that the sender had already received an acknowledgement for k - Size -1

Answer 8

Chapter 3, Problem P22P

a.

Explanation of Solution

Sequence numbers inside the sender’s window:

Given:

Window size = Size = 4

Range = 1024

Case 1:

Let’s consider the receiver is “K” and acknowledged all the “k-1” data packets. The sender side window’s range will be of [ k , k+ Size - 1 ] sequence numbers

Answer 9

Chapter 3, Problem P22P

a.

Explanation of Solution

Sequence numbers inside the sender’s window:

Given:

Window size = Size = 4

Range = 1024

Case 1:

Let’s consider the receiver is “K” and acknowledged all the “k-1” data packets. The sender side window’s range will be of [ k , k+ Size - 1 ] sequence numbers

Answer 10

b.

Explanation of Solution

Values of ACK field:

Suppose the receiver is waiting for the packet “k”, then it has been received and acknowledged “k-1” packet and the “Size-1” packets before that. If NACKs of those packets have not been received by the sender, then the values of ACK field is [ k - size , k - 1 ] may propagate back.

Because the sender has sent packets k - 1, it states that the sender had already received an acknowledgement for k - Size -1

Answer 11

b.

Explanation of Solution

Values of ACK field:

Suppose the receiver is waiting for the packet “k”, then it has been received and acknowledged “k-1” packet and the “Size-1” packets before that. If NACKs of those packets have not been received by the sender, then the values of ACK field is [ k - size , k - 1 ] may propagate back.

Because the sender has sent packets k - 1, it states that the sender had already received an acknowledgement for k - Size -1

Answer 12

Expert Solution & Answer

Answer 13

Expert Solution & Answer

Answer 14

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Answer 15

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Sequence numbers inside the sender’s window: Given: Window size = Size = 4 Range = 1024 Case 1: Let’s consider the receiver is “K” and acknowledged all the “k-1” data packets. The sender side window’s range will be of [ k , k+ Size - 1 ] sequence numbers

Solution Summary: The author explains the range of sequence numbers inside the sender's window.

Explanation of Solution

Sequence numbers inside the sender’s window:

Given:

Window size = Size = 4

Range = 1024

Case 1:

Let’s consider the receiver is “K” and acknowledged all the “k-1” data packets. The sender side window’s range will be of [ k , k+ Size - 1 ] sequence numbers

Explanation of Solution

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Chapter 3 Solutions