What is the highest percentage of the memory bus capacity that could be consumed by the disk’s Direct Memory Access if there are 4 disk drives in use?

Question

Consider a computer system that has the following characteristics:

1. The memory bus runs at 400 MHz.
2. The memory bus is 8 bytes wide.
3. Accessing a memory address in memory requires 8 memory bus cycles.
4. The disk drives can each read a block every 4 milliseconds.
5. Disk blocks read are 8 kilobytes each.
6. Assume Direct Memory Access from the disk drive.

What is the highest percentage of the memory bus capacity that could be consumed by the disk’s Direct Memory Access if there are 4 disk drives in use?

Answer 1

I'll take a stab at this. To answer the question, we need to find out the bandwidth usage of the disk drive and divide it by the total available bandwidth of the memory bus. Let's start with the former. Disk block reads are 8KB each and can be read every 4ms. Assuming this is done continuously, the bandwidth usage is:

8 KB / 4ms = 2 KB/ms = 2048 bytes/ms = 2048000 bytes/second

Now the second part. The memory bus is 400 MHz, which means that it can process 400000000 cycles per second. Accessing a single memory address takes 8 cycles, so we have 400000000 / 8 = 50000000 memory address accesses per second. The bus is 8 bytes wide, so we can read 8 * 50000000 = 400000000 bytes per second.

So the percent of the bandwidth used by one disk drive is 2048000 / 400000000, or 0.512%. If we have four disk drives and all four are maxed out, then they would consume 4 * 0.512%, or 2.048%.