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2 years ago

Which mechanisms do we use to transport binary data and memory addresses?

Computers and Technology

1 answer:

Sav [38]2 years ago

6 0

Answer:

The External Data Bus is used to transport binary data and Address bus is used to transport memory addresses.

Explanation:

External Data Bus is a combination of data bus and external bus. Data bus is used to carry data and instructions between two or more components in the system e.g. CPU and all other computer components.
External bus is also called expansion bus. It is used to connect external components to the computer. Its a communication medium between CPU and other components. These components can be peripheral device like USB or flash memory.
So External Data Bus is used to transport data between CPU and external components. It is a primary communication pathway for data in a computer. The external data components are connected to this bus and the instruction or data on this bus is available to all external components. But the data communication is slower as compared to that of internal bus.
Address Bus carries physical location or address of data and transports memory addresses. Processor uses address bus when it wants to read data from memory or write data to the memory by sending a read/write signal by placing the read/write address of the specific memory location on the address bus.

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Your computer has gradually slowed down. What's the most likely reason?

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2 years ago

Consider a short, 10-meter link, over which a sender can transmit at a rate of 150 bits/sec in both directions. Suppose that pac

Katarina [22]

Answer:

The Tp value 0.03 micro seconds as calculated in the explanation below is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP.

Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.

Explanation:

Given details are below:

Length of the link = 10 meters

Bandwidth = 150 bits/sec

Size of a data packet = 100,000 bits

Size of a control packet = 200 bits

Size of the downloaded object = 100Kbits

No. of referenced objects = 10

Ler Tp to be the propagation delay between the client and the server, dp be the propagation delay and dt be the transmission delay.

The formula below is used to calculate the total time delay for sending and receiving packets :

d = dp (propagation delay) + dt (transmission delay)

For Parallel downloads through parallel instances of non-persistent HTTP :

Bandwidth = 150 bits/sec

No. of referenced objects = 10

For each parallel download, the bandwith = 150/10

= 15 bits/sec

10 independent connections are established, during parallel downloads, and the objects are downloaded simultaneously on these networks. First, a request for the object was sent by a client . Then, the request was processed by the server and once the connection is set, the server sends the object in response.

Therefore, for parallel downloads, the total time required is calculated as:

(200/150 + Tp + 200/150 + Tp + 200/150 + Tp + 100,000/150 + Tp) + (200/15 + Tp + 200/15 + Tp + 200/150 + Tp + 100,000/15 + Tp)

= ((200+200+200+100,00)/150 + 4Tp) + ((200+200+200+100,00)/15 + 4Tp)

= ((100,600)/150 + 4Tp) + ((100,600)/15 + 4Tp)

= (670 + 4Tp) + (6706 + 4Tp)

= 7377 + 8 Tp seconds

Thus, parallel instances of non-persistent HTTP makes sense in this case.

Let the speed of propogation of the medium be 300*106 m/sec.

Then, Tp = 10/(300*106)

= 0.03 micro seconds

The Tp value 0.03 micro seconds as calculated above is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP. Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.

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2 years ago

In a fantasy world, your character must face hordes of demons. Each demon is vulnerable to a type of magical spell. This weaknes

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2 years ago

Read 2 more answers

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yan [13]

Answer:

import java.io.*;

public class Main {

public static void main(String[] args) throws IOException {

BufferedReader brObject = new BufferedReader(new InputStreamReader(System.in));

String str;

while ((str = brObject.readLine()) != null) {

int number = Integer.parseInt(str);

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}

Explanation:

Inside the main method, create an object of BufferedReader class to read lines from standard input.
Declare a string and run a while loop until it reaches the end of the input.
Inside the while loop convert the string into an integer data type.
Finally display the output by squaring the number.

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2 years ago

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sergij07 [2.7K]

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<u>Explanation:</u>

Microsoft PowerPoint provides an easy way for customizing the shapes according to the needs of the user.

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2 years ago