Computers and Technology, 24.01.2020 23:31 breannaasmith1122

Suppose two hosts, a and b, are separated by 20,000 kilometers and are connected by a direct link of r = 2 mbps. suppose the propagation speed over the link is 2.5 x 10^8 meters/sec. a. calculate the bandwidth-delay product, r _ dprop. b. consider sending a file of 800,000 bits from host a to host b. suppose the file is sent continuously as one large message. what is the maximum number of bits that will be in the link at any given time? c. provide an interpretation of the bandwidth-delay product. d. what is the width (in meters) of a bit in the link? is it longer than a football field? e. derive a general expression for the width of a bit in terms of the propagation speed s, the transmission rate r, and the length of the link m.

Answers: 3

Show answers

Answers

Answer from: Tyrant4life

a ) The bandwidth-delay is 160,000 bits.

b)The maximum number of bits at a given time will be 160,000 bits.

c ) The maximum number of bits on the transmission line is equal to the product of he bandwidth and the delay

d) The length is 125 meters which is longer than a football field

e) The general expression is given as $\frac{S}{R}$

Explanation:

Step-by-Step Explanation

(a) We are old from he question that A and B are connected with a direct link.

The length between A and B is given as ( d ) = 20,000 Km = 2×10⁷ meters given that (1 Km = 10³ )

The rate of transmission on the direct connection between A and B

Denoted as (R) = 2 Mbps = 2 × 10⁶bps given that (1Mbps = 10⁶bps)

Speed of propagation of the link between A and B

Denoted as (S) = 2.5 × 10⁸ meters/sec

Formula to calculate propagation delay denoted as dₐ = $\frac{D}{S}$

dₐ = $\frac{D}{S}$

= $\frac{2 * 10 ^7}{2.5*10^8}$

= 0.08 sec

We multiply the rate of transmission with the propagation delay to get the bandwidth delay

i.e. R × dₐ = (2 × 10⁶) × 0.08

= 16 × 10⁴ bits

Hence the bandwidth delay is given as 160,000 bits.

Step 2 :

b) From the question we are given that file is been transmitted from host A to host B consistently as one large message.

Given that the size of the file o be transmitted is =800,000 bits

= 8 × 10⁵bits

The rate of transmission of the direct connection between A and B (R)

= 2 Mbps

= 2 × 10⁶bps (Given that 1Mbps = 10⁶ bps)

Note : That the maximum number of bits the connection can carry is not dependent on the file size

but on the bandwidth -delay i.e (R * da). It is also equal in value hence

At a given time the maximum number of bits is equal to the bandwidth product delay = 160,000 bits

Step 3

c) The highest number of bits on the link between A and B is gotten from the product of bandwidth and delay

Step 4:

Given from the question that the propagation speed of the connect A and B (S) = 2.5 × 10⁸ meters/sec

The rate of transmission through the direct connection between A and B (R) = 2 Mbps

= 2 × 10⁶ bps (Given that 1Mbps = 10⁶ bps)

We then have that the length of 1 bit on the link can be calculated with the formula $\frac{S}{R}$

Length of 1 bit = $\frac{S}{R}$

= $\frac{2.5 * 10^8}{2*10^6}$

= 125 m/bit

We can see from the answer that the length of 1 bit is longer than a football field

Step 5:

We can deduce the width of a bit by looking at the unites of the given parameters

Unit for the length of the link = m meters

Unit of propagation speed = S m/s

the transmission rate is = R bps

Note :Width of a bit is not dependent on the length but on the propagation speed and the rate of transmission and it is the ratio these parameter

Hence the width of a bit is given as = $\frac{S}{R}$

Answer from: Quest

uiy

explanation:

Answer from: Quest

Answerthe systems development lifecycle;

Another question on Computers and Technology

Computers and Technology, 22.06.2019 13:00

We as humans write math expression in infix notation, e.g. 5 + 2 (the operators are written in-between the operands). in a computer’s language, however, it is preferred to have the operators on the right side of the operands, i.e. 5 2 +. for more complex expressions that include parenthesis and multiple operators, a compiler has to convert the expression into postfix first and then evaluate the resulting postfix.write a program that takes an “infix” expression as input, uses stacks to convert it into postfix expression, and finally evaluates it. it must support the following operations: + - / * ^ % (example infix expression: (7 - 3) / (2 + 2)postfix expression: 7 3 - 2 2 + /result: 1guidelines: 1. you will need to use stacks in three placesa. one for the parenthesis check [char stack]b. one during infix to postfix [char stack]c. one during evaluation [int stack]for a and b above, you can use same array and same push, pop method as both ofthem are char. but for evaluation you have int stack and you might consider to createanother push pop method to handle it. maybe push_int, pop_int, etc. or find otherstrategy to utilize existing push pop method2. you can create a function for obtaining operator priority. that function should take anoperator as input and return its priority as an integer. this function will you a lot andreduce repeated code3. during evaluation you will need to convert char into integer. example for single digit: char c = '5'; int x = c - '0';

Answers: 2

Answer

Computers and Technology, 22.06.2019 23:30

Select all that apply. which of the following are proofreading options included in microsoft word? spell check find replace grammar check formatting check

Answers: 1

Answer

Computers and Technology, 23.06.2019 12:30

How is the brightness of oled of the diaplay is controled

Answers: 1

Answer

Computers and Technology, 23.06.2019 13:30

Select the correct answer from each drop-down menu. which types of computer networks are bigger as well as smaller than a man? a man is a network of computers that covers an area bigger than a , but smaller than a .

Answers: 1

Answer

You know the right answer?

Suppose two hosts, a and b, are separated by 20,000 kilometers and are connected by a direct link of...

Questions