Given: Frames have constant length (L) = kRS Where; k – Constant value R – Transmission rate S – Length of the slot Consider that the propagation delay is less than the length of the slot. To find: Efficiency of the protocol for the fixed “N” (Nodes) and “p” (probability). Solution: Consider the random variable “A” that denotes the number of slots is P(A=m)=β(1-β) m-1 . Here, β is the probability of success. This is considered as the geometric distribution, which has the mean 1 β . The consecutive wasted slots is W = A - 1 . That is, x = E[W] Substitute the value of “W” in the above equation. We get, x = E[A-1] x = 1-β β β=Np(1-p) N-1 Formula to find the efficiency is: x= k k+x

Question

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Answer 1

Expert Solution & Answer

Chapter 6, Problem P20P

a.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency of the protocol for the fixed “N” (Nodes) and “p” (probability).

Solution:

Consider the random variable “A” that denotes the number of slots is $P(A=m)=β(1-β)m-1$ . Here, $β$ is the probability of success.

This is considered as the geometric distribution, which has the mean $1β$ . The consecutive wasted slots is $W = A - 1$ . That is,

$x = E[W]$

Substitute the value of “W” in the above equation. We get,

$x = E[A-1]x = 1-βββ=Np(1-p)N-1$

Formula to find the efficiency is: $x=kk+x$

b.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: To find the probability (p) that maxims the efficiency for the fixed number of nodes (N).

Solution:

By minimizing “x”, the efficiency can be maximized which is equivalent to maximize $β$

c.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency that approaches to infinity.

Solution:

We know that: $Efficiency = kk+1-Np(1-p)N-1Np(1-p)N-1$

Substitute $1e$ for $Np(1-p)N-1$

d.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency that approaches to 1...

Want to see the full answer?

Check out a sample textbook solution

See solution

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I need help with this problem and an step by step explanation of the solution from the image described below. (Introduction to Signals and Systems)

Goal: Understand how to build a concurrent server and deal with signals Description: Use TCP socket programming in C to implement a pair of client and server programs. The client simply asks the user to enter a string (null-terminated), and then displays a menu for the user to select an operation on that string. The string and the type of operation are sent to the child server which performs the required operation and returns the result back to the client. That is, the server (parent) creates a child for each request of the client (you need to print the port and IP address of the client on the server terminal). Then it waits for the child to finish processing client's request and prints the process ID of the terminated child. The connection between your client and server should stay open, so that the client can send another request. However, the exit will happen when the user sends the client program the SIGQUIT signal. The handler will print a notifying message (for example, "Good…

I need help with this problem and an step by step explanation of the solution from the image described below. (Introduction to Signals and Systems)

Answer 2

Expert Solution & Answer

Chapter 6, Problem P20P

a.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency of the protocol for the fixed “N” (Nodes) and “p” (probability).

Solution:

Consider the random variable “A” that denotes the number of slots is $P(A=m)=β(1-β)m-1$ . Here, $β$ is the probability of success.

This is considered as the geometric distribution, which has the mean $1β$ . The consecutive wasted slots is $W = A - 1$ . That is,

$x = E[W]$

Substitute the value of “W” in the above equation. We get,

$x = E[A-1]x = 1-βββ=Np(1-p)N-1$

Formula to find the efficiency is: $x=kk+x$

b.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: To find the probability (p) that maxims the efficiency for the fixed number of nodes (N).

Solution:

By minimizing “x”, the efficiency can be maximized which is equivalent to maximize $β$

c.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency that approaches to infinity.

Solution:

We know that: $Efficiency = kk+1-Np(1-p)N-1Np(1-p)N-1$

Substitute $1e$ for $Np(1-p)N-1$

d.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency that approaches to 1...

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Expert Solution & Answer

Chapter 6, Problem P20P

a.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency of the protocol for the fixed “N” (Nodes) and “p” (probability).

Solution:

Consider the random variable “A” that denotes the number of slots is $P(A=m)=β(1-β)m-1$ . Here, $β$ is the probability of success.

This is considered as the geometric distribution, which has the mean $1β$ . The consecutive wasted slots is $W = A - 1$ . That is,

$x = E[W]$

Substitute the value of “W” in the above equation. We get,

$x = E[A-1]x = 1-βββ=Np(1-p)N-1$

Formula to find the efficiency is: $x=kk+x$

b.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: To find the probability (p) that maxims the efficiency for the fixed number of nodes (N).

Solution:

By minimizing “x”, the efficiency can be maximized which is equivalent to maximize $β$

c.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency that approaches to infinity.

Solution:

We know that: $Efficiency = kk+1-Np(1-p)N-1Np(1-p)N-1$

Substitute $1e$ for $Np(1-p)N-1$

d.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency that approaches to 1...

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Expert Solution & Answer

Chapter 6, Problem P20P

a.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency of the protocol for the fixed “N” (Nodes) and “p” (probability).

Solution:

Consider the random variable “A” that denotes the number of slots is $P(A=m)=β(1-β)m-1$ . Here, $β$ is the probability of success.

This is considered as the geometric distribution, which has the mean $1β$ . The consecutive wasted slots is $W = A - 1$ . That is,

$x = E[W]$

Substitute the value of “W” in the above equation. We get,

$x = E[A-1]x = 1-βββ=Np(1-p)N-1$

Formula to find the efficiency is: $x=kk+x$

b.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: To find the probability (p) that maxims the efficiency for the fixed number of nodes (N).

Solution:

By minimizing “x”, the efficiency can be maximized which is equivalent to maximize $β$

c.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency that approaches to infinity.

Solution:

We know that: $Efficiency = kk+1-Np(1-p)N-1Np(1-p)N-1$

Substitute $1e$ for $Np(1-p)N-1$

d.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency that approaches to 1...

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Chapter 6, Problem P20P

a.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency of the protocol for the fixed “N” (Nodes) and “p” (probability).

Solution:

Consider the random variable “A” that denotes the number of slots is $P(A=m)=β(1-β)m-1$ . Here, $β$ is the probability of success.

This is considered as the geometric distribution, which has the mean $1β$ . The consecutive wasted slots is $W = A - 1$ . That is,

$x = E[W]$

Substitute the value of “W” in the above equation. We get,

$x = E[A-1]x = 1-βββ=Np(1-p)N-1$

Formula to find the efficiency is: $x=kk+x$

b.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: To find the probability (p) that maxims the efficiency for the fixed number of nodes (N).

Solution:

By minimizing “x”, the efficiency can be maximized which is equivalent to maximize $β$

c.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency that approaches to infinity.

Solution:

We know that: $Efficiency = kk+1-Np(1-p)N-1Np(1-p)N-1$

Substitute $1e$ for $Np(1-p)N-1$

d.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency that approaches to 1...

Answer 8

Chapter 6, Problem P20P

a.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency of the protocol for the fixed “N” (Nodes) and “p” (probability).

Solution:

Consider the random variable “A” that denotes the number of slots is $P(A=m)=β(1-β)m-1$ . Here, $β$ is the probability of success.

This is considered as the geometric distribution, which has the mean $1β$ . The consecutive wasted slots is $W = A - 1$ . That is,

$x = E[W]$

Substitute the value of “W” in the above equation. We get,

$x = E[A-1]x = 1-βββ=Np(1-p)N-1$

Formula to find the efficiency is: $x=kk+x$

Answer 9

Chapter 6, Problem P20P

a.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency of the protocol for the fixed “N” (Nodes) and “p” (probability).

Solution:

Consider the random variable “A” that denotes the number of slots is $P(A=m)=β(1-β)m-1$ . Here, $β$ is the probability of success.

This is considered as the geometric distribution, which has the mean $1β$ . The consecutive wasted slots is $W = A - 1$ . That is,

$x = E[W]$

Substitute the value of “W” in the above equation. We get,

$x = E[A-1]x = 1-βββ=Np(1-p)N-1$

Formula to find the efficiency is: $x=kk+x$

Answer 10

b.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: To find the probability (p) that maxims the efficiency for the fixed number of nodes (N).

Solution:

By minimizing “x”, the efficiency can be maximized which is equivalent to maximize $β$

Answer 11

b.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: To find the probability (p) that maxims the efficiency for the fixed number of nodes (N).

Solution:

By minimizing “x”, the efficiency can be maximized which is equivalent to maximize $β$

Answer 12

c.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency that approaches to infinity.

Solution:

We know that: $Efficiency = kk+1-Np(1-p)N-1Np(1-p)N-1$

Substitute $1e$ for $Np(1-p)N-1$

Answer 13

c.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency that approaches to infinity.

Solution:

We know that: $Efficiency = kk+1-Np(1-p)N-1Np(1-p)N-1$

Substitute $1e$ for $Np(1-p)N-1$

Answer 14

d.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency that approaches to 1...

Answer 15

d.

Explanation of Solution

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency that approaches to 1...

Answer 16

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 17

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Explanation of Solution

Explanation of Solution

Explanation of Solution

Explanation of Solution

Given: Frames have constant length (L) = kRS Where; k – Constant value R – Transmission rate S – Length of the slot Consider that the propagation delay is less than the length of the slot. To find: Efficiency that approaches to 1...

Want to see the full answer?

Chapter 6 Solutions

Given:

Frames have constant length (L) = kRS

Where;

k – Constant value

R – Transmission rate

S – Length of the slot

Consider that the propagation delay is less than the length of the slot.

To find: Efficiency that approaches to 1...