this post was submitted on 23 Aug 2024
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[–] wholookshere@lemmy.blahaj.zone 26 points 2 months ago (2 children)

If still expect a discount for worse parts.

[–] DebatableRaccoon@lemmy.ca 9 points 2 months ago

This is Nvidia we're talking about. Lowered prices isn't a tactic in those scumbags' wheelhouse

[–] SpaceNoodle@lemmy.world 4 points 2 months ago (2 children)

It's not worse if it doesn't perform any differently. Besides, you don't actually know the BOM cost.

[–] RightHandOfIkaros@lemmy.world 12 points 2 months ago* (last edited 2 months ago) (2 children)

If it was overspecced before, then that means it was using parts more expensive than it needed to. Nobody makes RAM that is slower and also more expensive for the same capacity. Logically, this should translate to lowered prices for the GPUs using the cheaper parts.

[–] Someonelol@lemmy.dbzer0.com 7 points 2 months ago (1 children)

But think of Nvidia's shareholders! /s

[–] filister@lemmy.world 6 points 2 months ago

Honestly NVIDIA shareholders don't give a shit about the discrete GPU market as long as NVIDIA is able to overcharge the datacenters and reek of insane profits.

Unfortunately, the crypto boom normalised those prices and now there is no turning back.

[–] SpaceNoodle@lemmy.world 5 points 2 months ago

For all we know, they used overspecced RAM because it was what was available in the quantities needed, or they got a good price from the supplier - which is something that has specifically happened with hardware I've worked on before. Again, we don't actually know the specific pricing details. Higher speed does not inherently mean higher cost.

[–] wholookshere@lemmy.blahaj.zone 3 points 2 months ago (1 children)

It’s objectively worse. “Real world performance” might be the same, but I’m paying for performance AND parts.

[–] SpaceNoodle@lemmy.world -4 points 2 months ago (1 children)

You're not paying for the discrete parts. You're not gonna desolder that RAM and use it for something else.

[–] wholookshere@lemmy.blahaj.zone 4 points 2 months ago* (last edited 2 months ago) (1 children)

No but I am paying for the accumulation of those parts no? Otherwise I’m not buying hardware.

And we know shoe on the other foot, if there was no performance increase, but a fancy marketing label, they’d be all over increasing the price for it.

[–] SpaceNoodle@lemmy.world 3 points 2 months ago (1 children)

You're paying for the overall performance of the product, not for specs of each discrete component by itself.

Yes, you also pay for whatever they decide is relevant to marketing.

[–] wholookshere@lemmy.blahaj.zone 0 points 2 months ago* (last edited 2 months ago) (1 children)

How is buying hardware based on specs not doing both?

To that end, that’s like saying apple doesn’t need to offer higher base specs on things like ssds and internal storage because the performance is the same.

[–] SpaceNoodle@lemmy.world 1 points 2 months ago (1 children)

Correct, there would be no reason for them to purposefully overspec their parts.

[–] wholookshere@lemmy.blahaj.zone 0 points 2 months ago (1 children)

So then why wouldn’t I expect a discount on this v2 card?

[–] SpaceNoodle@lemmy.world 1 points 2 months ago (2 children)

Because you're paying for the card as a whole, not discrete components.

[–] wholookshere@lemmy.blahaj.zone 0 points 2 months ago (1 children)
[–] SpaceNoodle@lemmy.world 1 points 2 months ago (1 children)

No, you're not purchasing discrete components for use as discrete components.

[–] wholookshere@lemmy.blahaj.zone 0 points 2 months ago (1 children)

If im not buying the cumulation of parts I don’t know what im buying.

[–] SpaceNoodle@lemmy.world 1 points 2 months ago (2 children)

You don't understand how the overall system as a whole differs from the discrete components in isolation?

[–] wholookshere@lemmy.blahaj.zone 0 points 2 months ago

No…

You can’t buy a card without also buying all the individual composts though.

[–] wholookshere@lemmy.blahaj.zone 0 points 2 months ago

Except I’m buying a card with objectively worse materials.