In the past few days, I have been intensively reviewing "The Principles of Computer Composition". There are a lot of calculation problems. The more difficult to understand are the two mentioned in the title (of course there are others). It's actually pretty simple.
1s=1000ms= 1000 000μs, 1GHz = 1000 MHz = 1000 000 000 Hz, 1s = 1/1Hz
will be used later.
pipeline throughput
(Total number of instructions/Total pipeline execution time) x100%
Example: The main frequency is 1.03GHz, a four-stage instruction pipeline is used, and each pipeline requires 1 clock cycle to execute. Execute 100 instructions and find the throughput.
Analysis: The total number of instructions is 100, and the title has been given; the clock cycle is the reciprocal of the main frequency, pay attention to the unit of frequency, 100 executions require a total of 103 clock cycles in the pipeline.
result:
maximum data transfer rate
Note that the unit of maximum data transfer rate is Mbps, which is also Mbit/s
unit number of bytes/unit time
Example: The clock frequency is 100MHz, the transmission cycle is 4 clock cycles, and the bus width is 32 bits. Find the data transmission rate
Analysis: The number of bytes is 32/8 = 4 B; the time is the reciprocal of the frequency multiplied by the number of clock cycles; then divide to get 4 B/ (0.01μs) = 100Mbps, you can try the unit conversion, that is, with 4MB /0.01s to compare whether it is the same~