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melisa1 [442]
3 years ago
12

1 Explain the process of how the memory works with theprocessor​

Computers and Technology
2 answers:
butalik [34]3 years ago
5 0

Answer:

As every processor works on same stages i.e. fetch , decode , execute.

Most of the computers executes ONE instruction at a time. But speed is very fast .

So let's move to memory

The fastest memory is cache (registers)

Then RAM , then hard disk.

Before an instruction can be executed, program instructions and data must be placed into memory from an input device or a secondary storage device .

Then

1)The control unit fetches (gets) the instruction from memory.

2)The control unit decodes the instruction (decides what it means) and directs that the necessary data be moved from memory to the arithmetic/logic unit. These first two steps together are called instruction time, or I-time.

3)The arithmetic/logic unit executes the arithmetic or logical instruction. That is, the ALU is given control and performs the actual operation on the data.

4)Thc arithmetic/logic unit stores the result of this operation in memory or in a register. Steps 3 and 4 together are called execution time, or E-time.

Explanation:

NISA [10]3 years ago
3 0
As every processor works on same stages i.e. fetch , decode , execute. The fastest memory is cache (registers)

Then RAM , then hard disk.

Before an instruction can be executed, program instructions and data must be placed into memory from an input device or a secondary storage device .
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Computer Networks - Queues
lyudmila [28]

Answer:

the average arrival rate \lambda in units of packets/second is 15.24 kbps

the average number of packets w waiting to be serviced in the buffer is 762 bits

Explanation:

Given that:

A single channel with a capacity of 64 kbps.

Average packet waiting time T_w in the buffer = 0.05 second

Average number of packets in residence = 1 packet

Average packet length r = 1000 bits

What are the average arrival rate \lambda in units of packets/second and the average number of packets w waiting to be serviced in the buffer?

The Conservation of Time and Messages ;

E(R) = E(W) + ρ

r = w + ρ

Using Little law ;

r = λ × T_r

w =  λ × T_w

r /  λ = w / λ  +  ρ / λ

T_r =T_w + 1 / μ

T_r = T_w +T_s

where ;

ρ = utilisation fraction of time facility

r = mean number of item in the system waiting to be served

w = mean number of packet waiting to be served

λ = mean number of arrival per second

T_r =mean time an item spent in the system

T_w = mean waiting time

μ = traffic intensity

T_s = mean service time for each arrival

the average arrival rate \lambda in units of packets/second; we have the following.

First let's determine the serving time T_s

the serving time T_s  = \dfrac{1000}{64*1000}

= 0.015625

now; the mean time an item spent in the system T_r = T_w +T_s

where;

T_w = 0.05    (i.e the average packet waiting time)

T_s = 0.015625

T_r =  0.05 + 0.015625

T_r =  0.065625

However; the  mean number of arrival per second λ is;

r = λ × T_r

λ = r /  T_r

λ = 1000 / 0.065625

λ = 15238.09524 bps

λ ≅ 15.24 kbps

Thus;  the average arrival rate \lambda in units of packets/second is 15.24 kbps

b) Determine the average number of packets w waiting to be serviced in the buffer.

mean number of packets  w waiting to be served is calculated using the formula

w =  λ × T_w

where;

T_w = 0.05

w = 15238.09524 × 0.05

w = 761.904762

w ≅ 762 bits

Thus; the average number of packets w waiting to be serviced in the buffer is 762 bits

4 0
3 years ago
A database can be used to manage which of the following for a website?
Oliga [24]
Images would probably be the best choice here
7 0
3 years ago
What does 32mb SVGA do
Nadya [2.5K]

Answer:

hope it helps PLZ MARK ME BRAINLIEST,,!

Explanation:

They stabilize your grpahic image.

5 0
3 years ago
Write a Python program that takes a file name as input and generates the following output: File size in bytes and KBs Time the f
Airida [17]

Answer:

See explaination

Explanation:

import os

import time

from stat import *

file = input("Enter file name: ")

details = os.stat(file)#stores statistics about a file in a directory

print("File size: ", details[ST_SIZE])#retreiving size

print("File last accessed time: ", time.asctime(time.localtime(details[ST_ATIME])))#access time

print("File last modified time: ", time.asctime(time.localtime(details[ST_MTIME])))#modified time

print("File creation time: ", time.asctime(time.localtime(details[ST_CTIME])))#creation time

print("First character: ", file[0])

print("Middle character: ", file[len(file) // 2])

print("Last character: ", file[len(file) - 1])

num_words = 0

num_lines = 0

f=open(file, "r")

lines = f.readlines()

for line in lines:

num_lines = num_lines + 1

num_words = num_words + len(line.split())#add no. of words in that line

print("Number of lines: ", num_lines)

print("Number of words: ", num_words)

8 0
3 years ago
While recording a voice, if the narration is large,it is better to make a :a)a linked object b)an embedded object c)none d)does
goldenfox [79]

Answer:

c)none

Explanation:

Automatic updates can be a great problem in the case of the linked object and an embedded object. Hence, "a" and "b" are not the correct options, and since there is an effect, the d. the option is also not correct, as it does affect. And hence none of these options are correct. And the correct option is c) none.

4 0
3 years ago
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