Question

We find a way of improving the floating point instruction performance of a machine and make it run 15 times faster.

1. In a given program, how many of the instructions must be floating point to achieve an overall improvement of 4?
2. We have another way of achieving the required speedup of 4 in our computer. Instead of improving the floating point instruction performance, we improve the clock rate. We would like to obtain that magical speedup of 4 for our test program, considering that the CPI with the new clock cycle time will be 1.5 times higher. By how much do we need to improve the clock rate?

Answer 1

Time can be defined in different ways, depending on what we are measuring: Response time : The time between the start and completion of a task. It includes time spent executing on the CPU, accessing disk and memory, waiting for I/O and other processes, and operating system overhead. This is also referred to as execution time. Throughput :The total amount of work done in a given time. CPU execution time : Total time a CPU spends computing on a given task (excludes time for I/O or running other programs). This is also referred to as simply CPU time.

cycle time=1.5 time

speed up =4

improvement=4*1.5=9

I am explaining you all the details

with an example

machine A runs a program in 20 seconds machine B runs the same program in 25 seconds how many times faster is machine A? 25 20 = 1.25

Comparing Machines Metrics Execution time Throughput CPU time MIPS – millions of instructions per second MFLOPS – millions of floating point operations per second Comparing Machines Using Sets of Programs Arithmetic mean, weighted arithmetic mean Benchmarks

The clock cycle time is the amount of time for one clock period to elapse (e.g. 5 ns). The clock rate is the inverse of the clock cycle time. For example, if a computer has a clock cycle time of 5 ns, the clock rate is: 1 ---------------------- = 200 MHz 5 x 10-9 sec