Question

A simulation of a factory production line is to be written in C++. There are 3 workplaces named station1, station2, station 3, where components of the widget are produced Station 1 needs the screwdriver to complete its part, and updates parts[0] of the parts list array. Station 2 needs both the screwdriver and saw to complete its part, and updates parts[1]. Station 3 needs the saw and hammer to construct its part, and updates parts[2]. For this problem, it is your job to efficiently utilize the mutexes (hammer, screwdriver, and saw) to correctly coordinate the production of the parts. Be aware that we must be careful to avoid deadlock and your solution should not reduce the problem to sequential execution. This program is due no later than 5:30 PM on 10-12-2020. No late submissions will be accepted. This program will carefully be examined for comments, efficiency, and accuracy. Name the program factory.cpp and place it in your pgm directory.

Input: From command line:

The input for this program will be the total number of parts each thread must produce. We are assuming that each thread will produce the same number of parts.

./a.out <number>

Output: To console.

Contents of parts array are:

parts[0] is: <number>

parts[1] is: <number>

parts[2] is: <number>

Answer 1

Answer :

include <bits/stdc++.h>

using namespace std;

// Function to find minimum cost

// to be incurred

int findMinCost(int arr[], int n, int choc_cost) {

  // To reach first station, initial

  // chocolates required

  int choc_buy = arr[0];

  int curr_choc = 0;

  // Start traversing

  for (int i = 0; i < n - 1; i++) {

    // Find no. of chocolates

    // lose or gain

    int choc = arr[i] - arr[i + 1];

    // Add into curr_coc

    curr_choc += choc;

    // if no. of chocolates becomes

    // negative that means we have

    // to buy that no. of chocolates

    if (curr_choc < 0) {

      choc_buy += abs(curr_choc);

      curr_choc = 0;

    }

  }

  // Return cost required

  return choc_buy * choc_cost;

}

// Drivers code

int main() {

  int arr[] = {10, 6, 11, 4, 7, 1};

  int n = sizeof(arr) / sizeof(arr[0]);

  // Price of each candy

  int p = 5;

  cout << findMinCost(arr, n, p);

  return 0;

}