Take the 2-minute tour ×
Super User is a question and answer site for computer enthusiasts and power users. It's 100% free, no registration required.

If the memory is byte addressable, shouldn't it be

2^64 / (2^10 * 2^10 * 2^10) = 2^34 GB ?

A general Google search says 8GB or 16GB max.

How is that true? Also, how does it depend on the Operating System?

share|improve this question

migrated from stackoverflow.com Jul 27 '10 at 2:47

This question came from our site for professional and enthusiast programmers.

    
8 GB is obviously off the wall as my work machine has 12 GB and can address it all, plus a GB of video RAM, plus all my hardware plus all my virtual memory just fine. –  Mark Peters Jul 25 '10 at 6:35
    
Newegg sells a 32GB package of ram, intended for one computer. 32-bit supported 3.2GB, and 128-bit will support millions of terabytes, so I think 64-bit is a lot more than double what 32-bit had. –  tkbx Feb 10 '13 at 15:38

5 Answers 5

up vote 24 down vote accepted

There are multiple interleaving factors.

First of all, you will never, ever be able to assemble a system that has 2^64 bytes (16 exabytes) of physical RAM.

Second, just because an architecture uses 64-bit pointers, doesn't mean that all the bits of those pointers are actually used. Notably, current x86-64 CPUs (aka AMD64 and Intel's current 64-bit chips) actually use 48-bit address lines (AMD64) and 42-bit address lines (Intel) (see http://en.wikipedia.org/wiki/X86_64#Virtual_address_space_details ), theoretically allowing 256 terabytes of physical RAM.

Second, motherboards have their own limits on how much RAM they can support, both physically and logically. Physically, there are only going to be so many slots available for RAM. As for the "logical" limits, don't fully understand why this is still the case for x86-64 (the memory controller having long since been migrated into the CPU itself), but there it is. Presumably corners are being cut on address lines to save a few bucks in design and manufacturing.

Third, an operating system may have internal limitations as to how much RAM it can efficiently support. In part, this is actually to prevent needing overly large data structures to keep track of usage for memory that isn't really there. Last I checked, Linux allows 128TB of virtual address space per process on x86-64, and can theoretically support 64TB of physical RAM.

Fourth, some operating systems (e.g. Windows) will artificially limit how much RAM can be used as a tactic to make users upgrade to more expensive versions if they want more RAM (Windows 7 Starter is limited to 2GB, Home Basic to 8, Home Premium to 16, and Professional and above are 192GB, and Windows Server releases have far higher limits).

share|improve this answer
    
+1 for mentioning physical address lines being typically << number of bits in an address register on 64-bit architectures –  Paul R Jul 25 '10 at 8:39
1  
A source for the 42-bit address lines (added by anonymous edit) would be appreciated. Until then, I personally don't consider that information trustworthy. –  Bob Jun 4 '12 at 17:06
8  
Never say never. Moore's Law is still in effect, the last I heard, so systems with 16 exabytes of RAM may well appear within our lifetimes. Then we'll all have to move to 128-bit operating systems... –  Michael Hampton Jul 24 '12 at 0:42
    
Physical addresses are also limited by the number of bits provided for physical addresses in the TLBs (and by the size of cache tags given physical address tagging) and by the bits available for physical addresses in the page table entries (if such are defined by the architecture as in x86-64 which supports 52-bit physical address space with 4KiB pages). The latter is a firmer (architectural not implementation-based) limit, though even architectures can be changed. –  Paul A. Clayton Oct 24 '13 at 19:02

If you use MS Windows, the maximum memory is constrained by the operating system.

For Windows 7, the maximums are:

  • Starter: 2GB
  • Home Basic: 8GB
  • Home Premium: 16GB
  • Professional: 192GB
  • Enterprise: 192GB
  • Ultimate: 192GB

Windows Server 2008 ranges from 8GB for entry level to 2TB for the top of the line.

See http://msdn.microsoft.com/en-us/library/aa366778%28VS.85%29.aspx for a full-ish list.

These limits are primarily for marketing reasons. By restricting the cheaper versions they make the high-end versions (comparatively) more attractive.

I think that recent Linux builds work with around 32TB of RAM, although not all hardware can support that.

share|improve this answer

A 64-bit machine should be able to address up to 264 addressable units (in architectures designed over the last few decades, addressable units are invariably bytes, a.k.a. octets). If you define a "gigabyte" as 230 addressable units, then yes, 234 gigabytes would be another way to express the same count.

What Google search produced those ridiculous tiny results? Maybe the page you found was talking about the amounts of RAM you can buy in reasonably cheap boxes of a certain model -- you can most definitely have much more RAM (all addressable, of course) even in boxes that are far from being very costly.

Operating systems most definitely can and do restrict the amount of addressable memory (for example, they can do so by limiting the size of page tables, perhaps to some configurable maximum that can be set by recompiling or otherwise reconfiguring the OS -- it's hard to be more specific without focusing on some specific group of OSs).

share|improve this answer

I think it can take more than exabyte "EB"

look at this 32-bit processor can take 2^32 B = 4 GB, but at most it take 3 GB "because they are 1 GB for reversed addresses"

that mean 64-bit processor can take 2^64 B = 16 EB, but I cant how addresses that locked for reversed used. so we can say 64-bit processor take more than 1 EB. and this specification does not depend of what the OS the PC used or what its model.

share|improve this answer
    
In practice, 64-bit processors usually have 48-bit address space which is enough for 256 terabytes of memory. –  465544 Jun 10 '13 at 8:41

You can have as much RAM as you'd like but it all depends on the operating system and the application you're running - as someone else stated.

If you're running an hungry CPU intensive application that's also intensive on other hardware like your hard disk then you best seriously consider a second CPU, in a completely separate machine. For example, running an database application and email server both on separate machines.

It doesn't matter how much memory you have - in today's market the whole system will grind to a halt. Single CPUs simply can't handle the multitasking the way you'd expect.

For a single machine with loads of memory it's better but not that brilliant if you have multiple applications running on the same system - regardless with the amount of memory installed. This is the very reason why big companies use not only 64bit architecture to address more memory but they also load balance the whole lot with two or more servers.

If you're really serious about performance then definitely go 64 bit but, also consider load balancing with a second server.

I've been down that road - configuring a single machine with maximum RAM. Yet when running multiple applications it still runs like a dog. That's because the CPU itself just can't handle it.

My advice is to save your bucks on memory and purchase a second server and then load balance the two. It's a lot simpler plus - once you get into the grips of it - it's easy to expand when you need more processing power and speed.

Also, a 64bit architecture machine can cope with 32bit software but with a 32 bit machine you're limited to only 32 bit or less applications - including the operating system.

Since the whole market is going 64bit in terms of both hardware and software I would advise to go 64bit for the long term but, forget the idea of squashing all onto one single machine with bucket loads of memory because you'll be disappointed.

If you are uncertain what "load balancing" is then google it.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.