EBK COMPUTER NETWORKING
EBK COMPUTER NETWORKING
7th Edition
ISBN: 8220102955479
Author: Ross
Publisher: PEARSON
Expert Solution & Answer
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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...

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