C Program to find Minimum Cost With Example

We can explain C Program for Minimum Cost With Example with simple Examples. 

The cost of operation of a unit consists of two components C1 and C2 which can be expressed as functions of parameter p as follows:

C1 = 30 – 8p

C2 = 10 + p2

The parameter p ranges from 0 to 10. Determine the value of p with an accuracy of + 0.1 where the cost of operation would be minimum.

C Program for Minimum Cost With Example

Problem Analysis:

Total cost = C1 + C2 = 40 – 8p + p2

The cost is 40 when p = 0, and 33 when p = 1 and 60 when p = 10. The cost, therefore, decreases first and then increases. The program in below evaluates the cost at successive intervals of p (in steps of 0.1) and stops when the cost begins to increase. The C program employs break and continue statements to exit the loop.

main()
{
     float p, cost, p1, cost1;
     for (p = 0; p <= 10; p = p + 0.1)
  {
     cost = 40 - 8 * p + p * p;
     if(p == 0)
  {
     cost1 = cost;
     continue;
}
     if (cost >= cost1)
break;
     cost1 = cost;
     p1 = p;
}
    p = (p + p1)/2.0;
    cost = 40 - 8 * p + p * p;
    printf("\nMINIMUM COST = %.2f AT p = %.1f\n",
    cost, p);
}

Output

MINIMUM COST = 24.00 AT p = 4.0

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