Often an NP-hard problem has a NP-hard solution and a trivial solution. And sometimes the trival solution is not very much worse than the NP-hard solution. Lets see an example.
Consider a job allocator that must allocate jobs to workers. Let each worker have a description $(r_1,...,r_n)$ listing the amount of each resource that it has. Jobs also have a resource description $(r_1,...,r_n)$ that indicates how much of each resource they consume while active. How the allocator must allocate jobs to workers so that no worker is overloaded. How should this be done, and for a given set of resources, how many workers are needed?
Well there is a hard optimal solution and a trivial non optimal solution to this problem. In the trivial solution:
Allocate the jobs one by one to a worker that has room for them
Have enough workers such that this allocation will always succeed.
To workout how many workers are needed let us consider the fail condition, that is the point at which each worker does not have room for the next job.
To detect the fail condition, take out $j_i$ from the jobs, now the space allocated is $sum_{k,k!=i}$ r(j_k)$ and the space remaining in each worker must be less than $j_i$, so an upper bound for the space remaining is $n r(j_k)$ where $n$ is the number of workers. The total space provided by the workers is $sum_{k} r(w_k)$ which for the fail condition to be satisfied must be less than allocated space plus the upper bound on the remaining space. Needless to say we need to satisfy this equation for all $j_i$.
So we have a mechanism to either (i) for a set of workers figure out whether or not they fit in a given number of workers using trival scheduling, or (ii) for a set of jobs figure out how many workers are required such that we never run out workers.
On a final note, sometimes a trivial solution is not satisfactory. For example in diagram layout it is seldom the case that a trivial placement of diagram nodes is satisfactory.