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Intel Core Ultra 9 275HX Defeats Core i9-14900HX In Cinebench R23 Multi-Core, While Losing Out in Single-Core

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The Intel Core Ultra 9 275HX was unveiled by Big Blue at CES 2025, packing 8 Lion Cove performance cores and 16 Skymont efficiency cores for a total of 24 cores and 24 threads, thanks to Intel doing away with Hyper-Threading support. As such, we already had quite low expectations for multicore performance from Arrow Lake-HX, although a recently listed Cinebench R23 run appears to confirm the contrary. However, considering the nature of leaks, be sure to accept this information with a fair amount of skepticism until more information comes our way.

To be honest, this result is quite a doozy, for the aforementioned reason. Like its predecessor, the Core Ultra 9 275HX will also boast the same number of performance and efficiency cores, without Hyper-Threading support. But somehow, the result indicates that the 275HX handily outperforms the Core i9-14900HX in multicore performance by ~11%, despite losing out in single-core performance by the skin of its teeth. Of course, there are plenty of desktop-replacement laptops that can pull off higher multicore scores, although it would be best to stick to median figures considering that the details of the tested system are not known. Clearly, something is going on here, that I am unable to wrap my head around. It is possible that Intel has massively improved multicore scaling, but that is not what we witnessed with a recent Passmark result, where the Core Ultra 7 255H portrayed healthy improvements over its predecessor in single-core, but not quite so in multicore.



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It may be able to boost higher due to the lower power usage and heat-load. That tends to hold back the mobile 13th/14th Gen cpus a lot.
 
they should throw 128 old atom cores int here, to really take the MT crown.
 
12% is not exactly "crushing" lets be real here.
 
More treading of water.

Regarding the removal of HT, is Intel using the "rentable units' tech with these ?
 
It may be able to boost higher due to the lower power usage and heat-load. That tends to hold back the mobile 13th/14th Gen cpus a lot.
Yeah, I suspect ARL is able to maintain higher clocks for the duration. The E-cores are already clocked higher in ARL, so that might be entirely responsible for the performance gain. Cinebench is a fun number and all, but a program designed to max out all the cores is not all that representative of daily usage. Where I see P+E struggle is on mixed use, where a program only occasionally needs as much multithreaded power as it can get, and that’s usually when P+E falls on its face.
 
12% is not exactly "crushing" lets be real here.

I've calculated the difference of the multi-threaded performance as follows: ( 31,854 * 100.0 / 35,481 ) - 100.0 = ~10.22%.

Note: In HPC world a difference greater than 5% is considered as a good improvement. On a Big Data set it will reduce processing time.
 
I like how there is different math from different people. From the same data set:

GGforever says 18%
Vya Domus says 12%
ScaLibBDP says 10%

Does anyone want to show math for 5%? 25%?
 
I like how there is different math from different people. From the same data set:

GGforever says 18%
Vya Domus says 12%
ScaLibBDP says 10%

Does anyone want to show math for 5%? 25%?

I've described how I've calculated the difference. Next, I've just used the Microsoft Copilot to verify if my way is wrong. Here is a log of my "conversation" with Microsoft Copilot.

The result depends on what number is used as a base to compare with another number. So two possible cases:

...31,854 is lower than 35,481...
or
...35,481 is greater than 31,854...

Both results are Valid!


...
[ ScaLibBDP ] Calculate for how many percent 31,854 is lower than 35,481

[ Copilot ]

To find how much percent 31,854 is lower than 35,481, you can use the following formula:
Percent Lower=DifferenceOriginal Number×100\text{Percent Lower} = \frac{ \text{Difference} }{ \text{Original Number} } \times 100
Let's plug in the values:
\text{Difference} = 35,481 - 31,854 = 3,627 $$ $$ \text{Original Number} = 35,481
Now, use the formula to calculate the percent lower:
Percent Lower=3,62735,481×100≈10.22%\text{Percent Lower} = \frac{3,627}{35,481} \times 100 \approx 10.22\%
So, 31,854 is approximately 10.22% lower than 35,481

CpuPerfDifference.jpg


...
[ ScaLibBDP ] Calculate for how many percent 35,481 is greater than 31,854

[ Copilot ]

To find how much percent 35,481 is greater than 31,854, you can use the following formula:
Percent Greater=DifferenceOriginal Number×100\text{Percent Greater} = \frac{ \text{Difference} }{ \text{Original Number} } \times 100
Let's plug in the values:
\text{Difference} = 35,481 - 31,854 = 3,627 $$ $$ \text{Original Number} = 31,854
Now, use the formula to calculate the percent greater:
Percent Greater=3,62731,854×100≈11.39%\text{Percent Greater} = \frac{3,627}{31,854} \times 100 \approx 11.39\%
So, 35,481 is approximately 11.39% greater than 31,854
 
I've calculated the difference of the multi-threaded performance as follows: ( 31,854 * 100.0 / 35,481 ) - 100.0 = ~10.22%.

Note: In HPC world a difference greater than 5% is considered as a good improvement. On a Big Data set it will reduce processing time.
Good improvement is Zen 4 to Zen 5, for example 9800X3D vs. 7800X3D under Linux where the improvement is ~25% using your formula.
Those 10% we're not even sure of because both Intel CPUs can scale from 45W to 160W, so unless both CPUs run at the same power limits with similar cooling it's not a useful comparison for laptop chips.
 
I like how there is different math from different people. From the same data set:

GGforever says 18%
Vya Domus says 12%
ScaLibBDP says 10%

Does anyone want to show math for 5%? 25%?
The median for the 14900HX is ~30k, thats what I used....
 
ok but does anyone care?
 
Didn’t Intel update the E-core u-arch? Isn’t it supposed to be significantly faster now?
 
A 14900HX can reach 35,000 points in R23. There is lots of room to go beyond that if you use ThrottleStop to do some undervolting.
 

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A 14900HX can reach 35,000 points in R23. There is lots of room to go beyond that if you use ThrottleStop to do some undervolting.
Sure, but better to stick with the median value when the tested config details are under wraps.
ThrottleStop is great btw, I use it to disable bd_prochot on my MBP (dead battery, lol).. can't help the igpu though...
 
Note: In HPC world a difference greater than 5% is considered as a good improvement. On a Big Data set it will reduce processing time.
Okay, but this is just a consumer processor. Yes, it's not from the Ultra low power consumption series. And yes, I know that workstations are also assembled with consumer processors. But still, when you say HPC, I somehow understand it as supercomputers with many orders of magnitude more productive than a laptop.
 
“The Needs of the Many Outweigh the Needs of the Few” -2% ST vs 11%+MT gee I wonder which makes more sense overall. Time to trash those E cores and get +5% ST single core chips at 6.5GHz!!!?
 
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"Crushes" with score barely faster than 7945hx and 7% faster than Raptor Lake (non refresh). When 9945hx come and wipe the floor with it, will we have article 9945hx crushes everything? I doubt
1739081087511.png
 
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12% is not exactly "crushing" lets be real here.
Um...
Yup..
12% is not exactly "crushing" lets be real here.
I think you might have misread..

However...
I like how there is different math from different people. From the same data set:

GGforever says 18%
Vya Domus says 12%
ScaLibBDP says 10%

Does anyone want to show math for 5%? 25%?
Because they aren't doing the math properly. This is basic 5th grade math people.

To the find the percent increase, first subtract the initial value from the final value. Then take the difference and divide it by the initial value. Finally, multiply this number by 100 to convert the number to a percentage. This final result will represent the percent increase between the two values. So;

35481 - 31854 = 3627
3627 / 31854 = 0.1138632511
0.1138632511 * 100 = 11.38632511

The difference is 11.38% aprox.

EDIT:
All that said. 11.3% is a solid increase gen on gen.
 
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Any real world work tested?
 
They weren't directly comparing to AMD it was a comparison to Intel previous CPU's and it's defiantly between the two and crushes the former. You're reading into it way too much. It crushes the previous Intel chip they didn't say it crushes a particular AMD chip, but certainly will a number of them just not better ones. A 11% MT uplift for a 2% ST drop off is a very good compromise. It's like a 9:1 ratio between ST/MT exchange in that scenario's how anyone might possibly see that as a bad compromise is beyond me and unfathomable.
 
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If you need a new windows laptop to do some rendering on and can’t find a zen5 laptop?
Why'd you wanna render anything on a Windows laptop :wtf:
I like how there is different math from different people. From the same data set:

GGforever says 18%
Vya Domus says 12%
ScaLibBDP says 10%

Does anyone want to show math for 5%? 25%?
I think some of the guys here need to go back to high school ~

So, 100 is 25% faster than 80 <=> 80 is 20% slower than 100 i.e. simple math :slap:
 
They weren't directly comparing to AMD it was a comparison to Intel previous CPU's and it's defiantly between the two and crushes the former. You're reading into it way too much. It crushes the previous Intel chip they didn't say it crushes a particular AMD chip, but certainly will a number of them just not better ones. A 18% MT uplift for a 2% ST drop off is a very good compromise. It's like a 9:1 ratio between ST/MT exchange in that scenario's how anyone might possibly see that as a bad compromise is beyond me and unfathomable.
Where you saw 18% multi? It is compared to the worst of the worst examples for the previous gen with unknown CPU that can be power limited or overclocked. When compared to not gimped samples of the previous gen its barely faster in multi so it is on par with the previous gen. Maybe you should learn the meaning of "crushing"
 
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