Calculate the performance of a multicore architecture?

Question

Cal a multicore architecture with 10 computing cores: 2 processor cores and 8 coprocessors. Each processor core can deliver 2.0 GFlops, while each coprocessor can deliver 1.0 GFlops. All computing cores can perform calculation simultaneously. Any instruction can execute in either processor or coprocessor cores unless there are any explicit restrictions.

If 70% of dynamic instructions in an application are parallelizable, what is the maximum average performance (Flops) you can get in the optimal situation? Please note that the remaining 30% instructions can be executed only after the execution of the parallel 70% is over.

Consider another application where all the dynamic instructions can be partitioned into 6 groups (A, B, C, D, E, F) with the following dependency. For example, A --> C implies that all the instructions in A need to be completed before starting the execution of instructions in C. Each of the first four groups (A, B, C and D) contains 20% of the dynamic instructions whereas each of the remaining two groups (E and F) contains 10% of the dynamic instructions. All the instructions in each group must be executed sequentially on the same processor or coprocessor core. How to schedule them on the multicore architecture to achieve the best possible performance? What is the maximum average performance (Flops) now?

            A(20%) --> C(20%) -->
                                    E(10%)-->F(10%)
            B(20%) --> d(20%) -->

Answer 1

For the first part, you need to use Amdahl's Law, which is:

max speed-up = 1/(1-p+p/n)

where p is the parallelizable part. n is the improvement factor in executing the parallel portion.

(Note that the Amdahl's Law formula can be used for first order estimates on other types of changes. E.g., given a factor of N reduction in ALU energy use and P fraction of energy used by the ALU, one can find the improvement in total energy use.)

In your case, since the serial portion would be executed on the higher performance (2 GFLOPS) processor core, n is 6 ([8 coprocessor cores * 1 GFLOPS/core + 2 processor cores * 2 GFLOPS/core]/ 2 GFLOPS/processor core).

A quick calculation shows the max speed-up you can get is 2.4 related to 1 processor core. The maximum FLOPS would therefore be the speed-up times the speed if the whole program was executed serially on one processor core, i.e., 2.4 * 2 GFLOPS = 4.8 GFLOPS.

For the second part, note that initially there are two independent instruction streams: A -> C and B -> C. Since the system has two processor cores, both can be executed in parallel on the higher performance processor cores. Furthermore, both have the same amount of work (40% of total for each stream), so one the same performance core they will complete at the same time.

Since E depends on results from both C and D, it must be started after both finish. E and F would execute on a processor core (which core is arbitrary since E must wait for the tasks running on both processor cores to complete).

As you can see 80% of the program (40% for A+C; 40% for B+D) can be parallelized by a factor of 2 and 20% of the program (E+F) is serial. You can then just plug the numbers into the Amdahl's Law formula (p=0.8, n=2).