Intel Pentium III 533B & 600B

Let’s take a look at the current bandwidth requirements of your standard PC operating on a BX motherboard with an AGP 2X graphics accelerator and using the 100MHz FSB.

In order for your CPU to communicate with your memory and your graphics card, as well as the other devices in your system, it needs to have some sort of means of "transportation" for its data. The same is true for your peripherals, your graphics card, and your memory. If we ignore the PCI/ISA peripherals for the time being since, comparatively speaking, they aren’t huge bandwidth hogs , we are left with three parts to your system that need to stay in communication with each other: your CPU, AGP card, and memory.

Imagine a bridge that has three roads connecting to it. Each road can transport a certain amount of cargo. If the road is very wide and has no speed limit, then more cargo can be sent over it. If the road is narrow and has a very low speed limit, then very little cargo can reach the bridge.

These "roads" are symbolic of the busses in your system going to and from the chipset, the "bridge". Dealing with the three components we just mentioned (CPU, AGP, RAM) the three busses that connect these three components to one another via the chipset are: the system bus, the AGP bus, and the memory bus, respectively. So how much traffic are each of these "roads" capable of handling?

Your CPU, regardless of what the clock speed, is capable of up to 800MB/s of transfers on a BX platform running at the 100MHz FSB. How is this figure determined? Take the 64-bit system bus, multiply it by the operating speed (100MHz FSB) and divide by 8 (8 bits in a byte) and you have the answer in Megabytes per second (MB/s).

So, in theory, your CPU could require up to 800MB/s worth of bandwidth in the most intense of conditions, thus it would make sense for your memory subsystem to be able to deliver at least that amount of data at any given instance.

Do the same calculation for your memory bus:

(64-bit memory bus x 100MHz memory bus frequency) / 8-bits in a byte = 800MB/s

This means that in any given instance your processor is capable of moving 800MB/s worth of data and, at the same time, your memory is capable of moving the same amount of data. Since the two are equal, your CPU can move 800MB/s worth of data to and from your memory without having to wait for the memory bus. So if we use the road analogy from above, the width and speed limit of the road coming from/going to the CPU is equal to the width and speed limit of that coming from/going to the memory. Ideally, there should be no traffic jams on either of those two roads.