Today we continue discussing the best practice from storage engineering:

560) The number of applications using the same storage is usually not a concern. The ability to serve them with storage classes is noteworthy

561) When an application wants to change the workload on the storage, architects to prefer to swap the storage product with something more suitable. However, a performance engineer can circumvent the approach with optimizations that leverage the existing product. It is always a good practice to give this a try.

562) System architecture holds in favor of changing business needs from the smallest components to the overall product. However, it is rather centralized and sometimes using another instance of the product with customizations can mitigate the urgency while giving ample time for consolidation.

563) The use of storage product also depends on the developer community. Many products such as time series databases and graph databases have generated greater acceptance by endearing the product to developers.

564) Sales and support need to be armed with the latest information and remain current on all features from the customers. They need to have those features work exactly as they say it would.

565) The focus on business value does not remain confined to the people on the border with the customers. It comes from deep within product development and engineering.

#codingexercise

when we have to select groups as well as individuals, we use stars and bars methods
We put n objects in k bins with (n-1) Choose  (k-1)
int getGroups ( int n, int k) {
      return GetNChooseK (n-1, k-1);
}

double GetNChooseK(double n, double k)
{
if (k <0 || k > n || n = 0) return 0;
if ( k == 0 || k == n) return 1;
return Factorial(n) / (Factorial(n-k) * Factorial(k));
}
Alternatively,

 static int GetNChooseKDP(int n, int k)
        {
            if ( k <0 || k > n || n = 0)
                return 0;
            if (k == 0 || k == n)
                return 1;
            return GetNChooseKDP(n - 1, k - 1) + GetNChooseKDP(n - 1, k);
        }