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ANTONII [103]
3 years ago
6

Which of the following is productivity strategy for collaboration?

Computers and Technology
1 answer:
mixer [17]3 years ago
8 0

Which of the following is a productivity strategy for collaboration?

  • A. Saving focused work for a time of day when you feel most creative.
  • B. Employing the 80/20 rule to prioritize tasks.
  • C. Posting files to a web-based shared site.
  • D. Using white noise to block distractions in the office.

<u>Answer:</u>

Posting files to a web-based shared site is a productivity strategy for collaboration.

  • C. Posting files to a web-based shared site.

<u>Explanation:</u>

Collaborative software or groupware is application programming intended to help individuals taking a shot at a typical errand to achieve their objectives. This permits individual to impart thoughts and their abilities to different individuals with the goal that the assignment can be done both proficiently and adequately.

Joint effort stages ordinarily incorporate an email customer, Web conferencing, internet based life sharing, video capacities, report sharing abilities, texting and that's just the beginning. Endeavor joint effort stages are intended to be introduced on-premises or conveyed by means of the Web as cloud-based administrations.

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There will be 10 numbers stored contiguously in the computer at location x 7000 . Write a complete LC-3 program, starting at loc
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Answer:

The LC-3 (Little Computer 3) is an ISA definition for a 16-bit computer. Its architecture includes physical memory mapped I/O via a keyboard and display; TRAPs to the operating system for handling service calls; conditional branches on N, Z, and P condition codes; a subroutine call/return mechanism; a minimal set of operation instructions (ADD, AND, and NOT); and various addressing modes for loads and stores (direct, indirect, Base+offset, PC-relative, and an immediate mode for loading effective addresses). Programs written in LC-3 assembler execute out of a 65536 word memory space. All references to memory, from loading instructions to loading and storing register values, pass through the get Mem Adr() function. The hardware/software function of Project 5 is to translate virtual addresses to physical addresses in a restricted memory space. The following is the default, pass-through, MMU code for all memory references by the LC-3 simulator.

unsigned short int get Mem Adr(int va, int rwFlg)

{

unsigned short int pa;

// Warning: Use of system calls that can cause context switches may result in address translation failure

// You should only need to use gittid() once which has already been called for you below. No other syscalls

// are necessary.

TCB* tcb = get TCB();

int task RPT = tcb [gettid()].RPT;

pa = va;

// turn off virtual addressing for system RAM

if (va < 0x3000) return &memory[va];

return &memory[pa];

} // end get MemAdr

Simple OS, Tasks, and the LC-3 Simulator

We introduce into our simple-os a new task that is an lc3 Task. An lc3 Task is a running LC-3 simulator that executes an LC-3 program loaded into the LC-3 memory. The memory for the LC-3 simulator, however, is a single global array. This single global array for memory means that alllc3 Tasks created by the shell use the same memory for their programs. As all LC-3 programs start at address 0x3000 in LC-3, each task overwrites another tasks LC-3 program when the scheduler swaps task. The LC-3 simulator (lc3 Task) invokes the SWAP command every several LC-3 instruction cycles. This swap invocation means the scheduler is going to be swapping LC-3 tasks before the tasks actually complete execution so over writing another LC-3 task's memory in the LC-3 simulator is not a good thing.

You are going to implement virtual memory for the LC-3 simulator so up to 32 LC-3 tasks can be active in the LC-3 simulator memory without corrupting each others data. To implement the virtual memory, we have routed all accesses to LC-3 memory through a get Mem Adr function that is the MMU for the LC-3 simulator. In essence, we now have a single LC-3 simulator with a single unified global memory array yet we provide multi-tasking in the simulator for up to 32 LC-3 programs running in their own private address space using virtual memory.

We are implementing a two level page table for the virtual memory in this programming task. A two level table relies on referring to two page tables both indexed by separate page numbers to complete an address translation from a virtual to a physical address. The first table is referred to as the root page table or RPT for short. The root page table is a fixed static table that always resides in memory. There is exactly one RPT per LC-3 task. Always.

The memory layout for the LC=3 simulator including the system (kernel) area that is always resident and non-paged (i.e., no virtual address translation).

The two figures try to illustrate the situation. The lower figure below demonstrates the use of the two level page table. The RPT resident in non-virtual memory is first referenced to get the address of the second level user page table or (UPT) for short. The right figure in purple and green illustrates the memory layout more precisely. Anything below the address 0x3000 is considered non-virtual. The address space is not paged. The memory in the region 0x2400 through 0x3000 is reserved for the RPTs for up to thirty-two LC-3 tasks. These tables are again always present in memory and are not paged. Accessing any RPT does not require any type of address translation.

The addresses that reside above 0x3000 require an address translation. The memory area is in the virtual address space of the program. This virtual address space means that a UPT belonging to any given task is accessed using a virtual address. You must use the RPT in the system memory to keep track of the correct physical address for the UPT location. Once you have the physical address of the UPT you can complete the address translation by finding the data frame and combining it with the page offset to arrive at your final absolute physical address.

A Two-level page table for virtual memory management.

x7000 123F x7000 0042

x7001 6534 x7001 6534

x7002 300F x7002 300F

x7003 4005 after the program is run, memory x7003 4005

x7004 3F19

7 0
3 years ago
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Two cars A and B leave an intersection at the same time. Car A travels west at an average speed of x miles per hour and car B tr
drek231 [11]

Answer:

Here is the C++ program:

#include <iostream>  // to use input output functions

#include <cmath>  // to use math functions like sqrt()

#include <iomanip>  //to use setprecision method

using namespace std;   //to access objects like cin cout

int main ()  {  //start of main function

  double speedA;  //double type variable to store average speed of car A

  double speedB;  //double type variable to store average speed of car B

  int hour;  //int type variable to hold hour part of elapsed time

  int minutes;  //int type variable to hold minutes part of elapsed time

  double shortDistance;  // double type variable to store the result of shortest distance between car A and B

  double distanceA;  //stores the distance of carA

  double distanceB;  //stores the distance of carB

  double mins,hours;   //used to convert the elapsed time

cout << "Enter average speed of car A: " << endl;  //prompt user to enter the average speed of car A

cin >> speedA;   //reads the input value of average speed of car A from user

cout << "Enter average speed of car B: " << endl ;  //prompt user to enter the average speed of car B

cin >> speedB;   //reads the input value of average speed of car A from user

cout << "Enter elapsed time (in hours and minutes, separated by a space): " << endl;  //prompts user to enter elapsed time

cin>> hour >> minutes;    //reads elapsed time in hours and minutes

  mins = hour * 60;  //computes the minutes using value of hour

  hours = (minutes+mins)/60;     //computes hours using minutes and mins

distanceA = speedA * (hours);  // computes distance of car A

distanceB = speedB * (hours);   //computes distance of car B

   shortDistance =sqrt((distanceA * distanceA) + (distanceB * distanceB));   //computes shortest distance using formula √[(distanceA)² + (distanceB)²)]

cout << "The (shortest) distance between the cars is: "<<fixed<<setprecision(2)<<shortDistance;

//display the resultant value of shortDistance up to 2 decimal places

Explanation:

I will explain the program with an examples:

Let us suppose that the average speeds of cars are:

speedA = 70

speedB = 55

Elapsed time in hours and minutes:

hour = 2

minutes = 30

After taking these input values the program control moves to the statement:

mins = hour * 60;  

This becomes

mins = 2 * 60

mins = 120

Next

hours = (minutes+mins)/60;

hours = (30 + 120) / 60

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distanceA = speedA * (hours);  

this becomes

distanceA = 70 * (2.5)

distanceA = 175

distanceB = speedB * (hours);

distanceB = 55 * (2.5)

distanceB = 137.5

Next the shortest distance between car A and car B is computed:

shortDistance = sqrt((distanceA * distanceA) + (distanceB * distanceB));

shortDistance = sqrt((175 * 175) + (137.5 * 137.5))

                        = sqrt(30625 + 18906.25)

                        = sqrt(49531.25)

                        =  222.556173

shortDistance =  222.56

 

Hence the output is:

The (shortest) distance between the cars is: 222.56        

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Explanation:

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