quwy wrote on 2026-08-25, 01:22:

1. One 8-MB SIMM BIOS show as 7808 KB.
2. Two 8-MB SIMMs BIOS show as 16000 KB (not 16384 as usual).

The BIOS is choosing to ignore the 384KB of memory space used for hardware communication, nothing unusual. 16000KB + 384KB = 16,384KB

quwy wrote on 2026-08-25, 01:22:

4. BIOS summary table show 15360 KB of extended memory (most DOS tools repeat this value).

I think this is normal , the first 1024KB is used for the real mode memory map and can't be re-used for EMS/XMS. Nothing usnusual here either. 1024KB + 15360KB = 16,384KB

quwy wrote on 2026-08-25, 01:22:

4. Win98 says about 16.0 MB of system RAM.

Fine, 16.0MB is installed because it can use a different interface to query installed memory, but it probably can't use all of it due to the mentioned reservations

quwy wrote on 2026-08-25, 01:22:

5. WinNT4 on the boot screen and system props says about 14 MB of RAM (and it feels like that's how it is, it's a bit slow).
6. Speedsys also show 14 MB of RAM.

Windows NT and speedsys are probably using the same memory calculation method, and are reporting how much memory they have available minus any reservations

quwy wrote on 2026-08-25, 01:22:

7. Memtest86 says about exactly 13368 KB of RAM

Since memtest86 needs to reside inside memory, it cannot ever test the full amount lest it delete itself while testing. So the difference is likely available memory plus a location for itself to reside.

I don't think there are any issues here. You can check if "Memory Hole at 15M" is enabled in the BIOS, that will remove 1MB of RAM from being available.

Hello,

In period-correct times I wondered about this also.

AMI BIOS subtracts "shadow RAM" from the amount of displayed RAM. Award BIOS does not. So you will see these oddball amounts on AMI, but an even number of megabytes with Award.

On PCI boards, like yours, the entire block from 640K-1024K (so 384K) is taken up as shadow RAM and is therefore subtracted from the allocation.
Older chipsets often have memory relocation, in which A0000-BFFFF and D0000-EFFFF (256K in total) gets relocated to the very top of memory, while C0000-CFFFF and F0000-FFFFF (128K) is consumed as shadow RAM. My UMC 481 board is like this, so AMI reports 8064 K.

NeoG_ wrote on 2026-08-25, 02:04:

The BIOS is choosing to ignore the 384KB of memory space used for hardware communication, nothing unusual. 16000KB + 384KB = 16,384KB

Ok, I think by same way. But it first time when I see suck calculation. All another AMI-based boards (from 386 to P6) shows traditional value.

I think this is normal , the first 1024KB is used for the real mode memory map and can't be re-used for EMS/XMS. Nothing usnusual here either. 1024KB + 15360KB = 16,384KB

Yes, in is normal, I wrote this value here only to compare with another sources.

Fine, 16.0MB is installed because it can use a different interface to query installed memory, but it probably can't use all of it due to the mentioned reservations

IMHO Win9x systems are not a model of correct interaction with RAM and there is a high probability what this value just calculated by same way and from the same source as DOS diagnostic tools.

Windows NT and speedsys are probably using the same memory calculation method, and are reporting how much memory they have available minus any reservations

2 MB of reservation on 486 system?! Only 1MB S3 PCI video card inserted, nothing else.
In 1995, for stealing two megabytes of RAM, any manufacturer would have been instantly thrown out of the market.
I just now removed one SIMM and as I suspected, there are only 6 MB left for NT. It was nonsense at that time.

Since memtest86 needs to reside inside memory, it cannot ever test the full amount lest it delete itself while testing. So the difference is likely available memory plus a location for itself to reside.

Ok, this still mean what only ~14 MB is available, regardless what Win98 think about this.

You can check if "Memory Hole at 15M" is enabled in the BIOS, that will remove 1MB of RAM from being available.

I know about this option, but it doesn't exist.

Windows 9x was known to round to nearest 4MB; see https://devblogs.microsoft.com/oldnewthing/20 … 06-00/?p=42283/ for the reasons why.

On DOS can you capture the output of MEM, MEM /C, and MEM /D

jakethompson1 wrote on 2026-08-25, 02:09:

On PCI boards, like yours, the entire block from 640K-1024K (so 384K) is taken up as shadow RAM and is therefore subtracted from the allocation.
Older chipsets often have memory relocation, in which A0000-BFFFF and D0000-EFFFF (256K in total) gets relocated to the very top of memory, while C0000-CFFFF and F0000-FFFFF (128K) is consumed as shadow RAM. My UMC 481 board is like this, so AMI reports 8064 K.

Around 1995 it was unthinkable to just give up two megabytes of memory.
I remember how much it cost and how we fought for every KB.

Windows 9x was known to round to nearest 4MB

This explains "16.0 MB" on the system properties window.

MEM

"Total memory: 16384 KB"