We were discussing virtualization and software defined stacks. Software defined technology stack aims to virtualize compute, network, storage and security aspects. The primary benefit of SDN is that it makes provisioning dynamic as opposed to the static configuration from physical resources. It helps to manage complexity. It performs segmentation, workload monitoring, conditional forwarding, and automated switching for dynamic optimization and scaling.

We discussed an example of SDDC as an enabler to build a sustainable analytics platform for an insurer. Analytics require a lot of storage. And the insurer wanted rapid realization of benefits from analytics by folding timelines. In this case, a software defined storage was paired with a platform-as-a-service to help the insurer test and deploy new ideas.

If we ask the question where does it make sense not to use Software defined services, there is usually a reference to data that becomes big and unwieldy and requires dedicated or static resources. For example we might need a big product catalog or we might need to store social  engineering data. Similarly we refer to highend transactional database servers. In all these cases I could argue that we could have dedicated physical resources or fenced resources on a shared fabric but still host applications and their data on software defined services. A software host is easy to replace just like a container and provides very little overhead

One of the advantages of software defined services is that it can be repeated over and over again in different underlying layers with no change or impact to existing workloads.

1. Another benefit of software defined services is that reconfiguration is super easy

#codingexercise

int GetCountSquareSubMatrixSizekCountOnes (int[,] A, int rows, int cols, int k)  

{  

int total = 0;

for ( int I = 0; I  < rows; i++) {  

    for ( int j = 0; j < cols; j++) {  

            // use this as the start of submatrix  

            int count = 0;  

            for ( int x = i; x < k; x++)  

                for ( int y = j; y < k; y++)  

                       If  ( x < rows && y < cols && A[x,y] == 1) 

                            count += 1;   

If (count == k * k) total += 1; 

      }  

}  

return total;  

}