Question

A simple computer has a main memory and a disk but not a cache. If a referenced word is in main memory, 90ns are required to access it. If it is on disk but not in main memory, 10ms are needed to load it into main memory (this includes the time to originally check the main memory), and then reference is started again. The main memory hit ratio is 0.8. What is the average time in ns required to access a referenced word on this system

Answer 1

The average time in ns to reference a work will be of 2,000,090.

Step-by-step explanation:

80% of the time, you will need just to load from the main memory(when there is a hit), that is, a time of access of 90 ns.

20% of the time, you will have a miss, so you have to access the main memory(90 ns) plus the disk(10 ms = 10000000 ns).

So, the average time needed to reference a word will be:

`t = 0.8*90 + 0.2*(90 + 10000000) = 2000090`

The average time in ns to reference a work will be of 2,000,090.