Minimum Cost

Problem:  The cost of operation of a unit consists of two components C1 and C2 which can be expressed as functions of a 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.

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 Fig.6.14 evaluates the cost at successive intervals of p (in steps of 0.1) and stops when the cost begins to increase.  The program employs break and continue statements to exit the loop.

PROBLEM OF MINIMUM COST

Program:                

   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