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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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...