Intel intentionally delayed release of Broadwell?

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Roland00Address

Platinum Member
Dec 17, 2008
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Going to a new node means higher fixed cost (the cost to develop the new node) yet it returns lower variable cost (the cost of each transistor is cheaper, plus better performance for the chip due to it requiring less energy / less voltage to perform similar work.) It is a trade off.

If we know how many chips a foundry is going to produce (and there is no foundry competition) then there in theory is going to be a point where a foundry can maximize profits by develop to a certain node and just resting on its laurels for it isn't worth going to the new node for the cost of developing a new node is not worth the return on investment.

If there is enough competition though, a foundry will not rest on its laurels for it must compete for contracts, thus silicon costs for the customer will go down, yet profit for the foundries will also go down to increased competition.
 
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OCGuy

Lifer
Jul 12, 2000
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Unfortunately due to lack of competition in the x86 marketplace, Intel has to make smart decisions in order to recoup their investment for each process. If it comes down to either cutting margins or slowing their release schedule, I would go with release schedule every time.

The worst thing would be for them to cut their large R&D budget.
 

jdubs03

Golden Member
Oct 1, 2013
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While there is no doubt that there are monumental R&D challenges in improving process nodes, the point is that the reduction in transistor costs will compensate for any additional increase in wafer cost. Going to 450mm will accelerate the cost reduction further, as wafer cost will be lower compared to where we are with 300mm.

Edit: well said Roland.
 

OCGuy

Lifer
Jul 12, 2000
27,224
37
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Something tells me you can't assume it is automatically cheaper without taking into account volume and yield.
 

mrmt

Diamond Member
Aug 18, 2012
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Unfortunately due to lack of competition in the x86 marketplace, Intel has to make smart decisions in order to recoup their investment for each process. If it comes down to either cutting margins or slowing their release schedule, I would go with release schedule every time.

And what would Intel do with its shiny new 14nm fabs? Let it sit idle adding to the depreciation bill while Intel management has to hit the magical IRR number for 22nm? That would wreak havoc in the economics of its 14nm node unless Intel delayed 10nm for a period bigger than 14nm.

Once you spend CAPEX, you are done. You are commited to a given schedule. If you need to slow things down you will do so to nodes where you didn't spend CAPEX, not in the node where you payed for pretty much everything.
 

mrmt

Diamond Member
Aug 18, 2012
3,974
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Something tells me you can't assume it is automatically cheaper without taking into account volume and yield.

This. For the same yields curve every new node needs higher volumes than the previous one to pay the bill in the adoption curve, so every node you either raise volumes or adopt your node further behind your competition.

This is what put AMD out of the fab race, and this is what is happening to UMC and SMIC for example, which will have a 28nm node, by the time nobody in the bleeding edge talks about it.
 

MrK6

Diamond Member
Aug 9, 2004
4,458
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The cost of the transistor includes the R&D costs normalized across the expected volume. If this graph didn't work like this Moores law wouldn't exist. The reason the industry moves forward is that it's actually cheaper to build devices on newer nodes with the shrink. Counter-intuitive but true.
And that is all built into the cost. Development and capital equipment costs are all calculated into the cost per transistor. I'm sorry if you can't understand this point or simply don't believe it but it is true. That graph is representative of the overall cost reduction by going to 14nm.
And how many transistors are in the new devices compared to the old? Again, read the graph and interpret its meaning, not just recycling definitions.
Going to a new node means higher fixed cost (the cost to develop the new node) yet it returns lower variable cost (the cost of each transistor is cheaper, plus better performance for the chip due to it requiring less energy / less voltage to perform similar work.) It is a trade off.

If we know how many chips a foundry is going to produce (and there is no foundry competition) then there in theory is going to be a point where a foundry can maximize profits by develop to a certain node and just resting on its laurels for it isn't worth going to the new node for the cost of developing a new node is not worth the return on investment.

If there is enough competition though, a foundry will not rest on its laurels for it must compete for contracts, thus silicon costs for the customer will go down, yet profit for the foundries will also go down to increased competition.
This guy gets it.
 

witeken

Diamond Member
Dec 25, 2013
3,899
193
106
Going to a new node means higher fixed cost (the cost to develop the new node) yet it returns lower variable cost (the cost of each transistor is cheaper, plus better performance for the chip due to it requiring less energy / less voltage to perform similar work.) It is a trade off.

If we know how many chips a foundry is going to produce (and there is no foundry competition) then there in theory is going to be a point where a foundry can maximize profits by develop to a certain node and just resting on its laurels for it isn't worth going to the new node for the cost of developing a new node is not worth the return on investment.

If there is enough competition though, a foundry will not rest on its laurels for it must compete for contracts, thus silicon costs for the customer will go down, yet profit for the foundries will also go down to increased competition.
No, Moore's law will still work without competition. You can develop a new node and you'll still get the benefits of Moore's law, and get even higher profits. Lower nodes are better for both companies and consumers.

However, if there is no competition, it would probably go slower because the company would minimize its risks.
 

ShintaiDK

Lifer
Apr 22, 2012
20,378
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No, Moore's law will still work without competition. You can develop a new node and you'll still get the benefits of Moore's law, and get even higher profits. Lower nodes are better for both companies and consumers.

However, if there is no competition, it would probably go slower because the company would minimize its risks.

Competition actually ruins it. Since you need to distribute the same volume at multiple companies. Meaning the node you can go down to profitable is not the same due to increased cost and reduced income. And competition makes companies minimize the risk, not the other way around. Its the race to the bottom.
 

Homeles

Platinum Member
Dec 9, 2011
2,580
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And how many transistors are in the new devices compared to the old? Again, read the graph and interpret its meaning, not just recycling definitions.

This guy gets it.
You've just dodged the fact that you've been proven wrong, and are now asking a useless question.

The transistor count of new devices is completely irrelevant -- if a new node is more economical, it's more economical. Period. What if you had kept those new devices on 22nm? They'd be even more expensive than on 14nm.

Fixed costs also aren't a problem here. This is Intel we're talking about -- their volume is tremendous, and is still very much in line with their historical norm:
exh99144.jpg


Moving to a new node also increases demand for their products, since their new products will perform better and use less power. This is more true for Intel now than it has been for quite some time. They haven't had competition like this. They have the ability to leapfrog their competition with new nodes, and it is critical for them to utilize it as soon as that node is ready.

Delaying a new node would actually be very bad for financials. If the node is ready and just sitting there unused, you've just wasted lots of money. Doubly so when you consider that the lower variable costs could have been leveraged during this time.

Say 22nm is still in the red right now. So what? The financials will balance out eventually. They don't cease production on 22nm just because 14nm is available. They'll still use 22nm for years to come.

So even if 22nm has still not turned a profit (unlikely), it simply does not make sense to delay 14nm.

And here's another chart to ruin your morning:
exh99134.jpg

Even with exponentially rising wafer processing costs, the exponential density increases are still more than enough to offset them. Not only that, but 14nm and 10nm are even more economical than past nodes. Intel's also projecting this trend to continue with 7nm.
 

Roland00Address

Platinum Member
Dec 17, 2008
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No, Moore's law will still work without competition. You can develop a new node and you'll still get the benefits of Moore's law, and get even higher profits. Lower nodes are better for both companies and consumers.

However, if there is no competition, it would probably go slower because the company would minimize its risks.

You are correct the actual law parts of Moore's law will still work without competition (that the science of making transistors gets cheaper over time), and the actual cost of node development gets cheaper over time. For example it is cheaper to move to 32nm type process today then it would have been in earlier (intel demostrated 32nm chips in 2007, started selling them in Jan 2010, amd/gf started selling 32nm chips in Mid 2011, TMSC shipped 28nm chips in 2012 etc).

You can't just assume though we would get the speed of development the same without competition. Moore's law is not a true law, it just an observation that at current speeds we will get approximately double the transistors every two years. Moore's law is not just a question on the science of making transistors smaller it was also a question on the economics of making transistors smaller.

The cost of going to new nodes is getting more expensive (higher fixed costs), thus we are seeing less competition with the foundry space. We are going to see foundries going to lower nodes but it is going to be harder for the foundries to do so. Foundries with lots of cash (such as intel) will throw money at the problem getting to these lower nodes first. Foundries with low amounts of cash will take longer time to achieve these nodes. Eventually some of the foundries with low amounts of cash will be so behind in process node technology that they can't attract any orders with a high enough margin to sustain themselves and these foundries will go bankrupt.

The foundries with lots of cash will eventually go to the lower nodes for you are right over the long term beter process nodes pay for themselves, it is just expensive to get there.
 

Roland00Address

Platinum Member
Dec 17, 2008
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To use an analogy to describe the economics of node development.
You live on a nice street in a nice house that you currently rent. There is a better house (it is bigger and more beautiful) on the same street, and it for sale, you want to move into said house and only that house (or the one you currently living in) would do, only a house on this exclusive street.
To move into the nice new house for sale you would have to buy it outright with cash (there is no mortgages in this analogy.) Over the long haul buying the better house across the street will save you money for over the long haul the cheaper variable expenses (no needing to pay rent) is cheaper than the fixed cost of buying the house outright.
If the house that you want to buy is the same price throughout all of time then it makes sense to buy it as soon as possible so you can start receiving the benefits of lower variable cost (no rent) as soon as possible. But what happens if the seller of the nice house will drop the price of the house 30% each year, due to house selling being a tough market and no one has yet bought this nice pretty house. If the house price drops 30% each year, it may be advantageous to wait and buy the house later. Sure you don’t get the benefit of lower variable cost (no rent) but you get the benefit of lower fixed cost instead (cheaper house). If there is no competition for this house it may make sense to wait.
If there is competition for buying this house, other potential buyers, then you may want to buy the house as fast as possible, for once the house is bought there is no turning back and you get the best house on the pretty street.

---------------------------------------

Moving to a lower process node for foundries is much the same thing, there is always a better node (a better house on the same street), but you must be able to afford moving into the new node. As long as there is competition in the foundry space it makes sense to move into the better node as fast as possible, but if there is no or little competition in the foundry space then you let science develop a little further so it is cheaper to move into the new node. Business go bankrupt due to lack of cash reserves, either immediately (many business with the 2008 recession) or over the long term (AMD before selling their foundries to global foundries). There is no point in lighting that cash reserves on fire, if there is no return on investment for using that cash.

What will happen to the $/mm² when Intel moves to 450mm wafers?
The variable cost of transistors is cheaper with 450mm wafers. With small dies there is less of a benefit going to 450mm wafers, with big dies their is a huge benefit of going to 450mm wafers.

That said there is a large fixed cost on transitioning to 450mm wafers, intel is going to build 3 new foundries for 450mm, with each foundry costing a couple of billion dollars each for just the building and not the equipment inside the building. Intel is building new foundries instead of repurposing old foundries (which if possible would have been cheaper) for the 450mm requires new buildings for some of the equipment is not reusable.

This is a quick summary of the benefits of moving to 450mm instead of 300mm
Intel breaks ground on first 450mm fab
 

Fjodor2001

Diamond Member
Feb 6, 2010
4,270
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Competition actually ruins it. Since you need to distribute the same volume at multiple companies. Meaning the node you can go down to profitable is not the same due to increased cost and reduced income. And competition makes companies minimize the risk, not the other way around. Its the race to the bottom.

With that communism style of logic logic there should only be one car model, one mobile phone model, one TV model, and so on. Then you could spread the development costs over many more units so it would be cheaper.

But in reality it doesn't work that way, and market capitalism, diversity, consumer choice, and competition brings down cost more effectively in the long run.
 

Homeles

Platinum Member
Dec 9, 2011
2,580
0
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With that communism style of logic logic there should only be one car model, one mobile phone model, one TV model, and so on. Then you could spread the development costs over many more units so it would be cheaper.

But in reality it doesn't work that way, and market capitalism, diversity, consumer choice, and competition brings down cost more effectively in the long run.
This is perhaps one of the most arguable beliefs in today's society, yet you pass it off as fact. It is impossible to know whether a communist economy is inferior to a capitalist one, because the former has never been truly executed.
 
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MrK6

Diamond Member
Aug 9, 2004
4,458
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You've just dodged the fact that you've been proven wrong, and are now asking a useless question.
This is inane: by sputtering "buh buh buh MOORE'S LAW" you haven't proven a thing.

The transistor count of new devices is completely irrelevant -- if a new node is more economical, it's more economical. Period. What if you had kept those new devices on 22nm? They'd be even more expensive than on 14nm.
Who says how, when, or even if they have to develop new devices? Honestly, your thinking is so one dimensional it's astounding.

Fixed costs also aren't a problem here. This is Intel we're talking about -- their volume is tremendous, and is still very much in line with their historical norm:
exh99144.jpg


Moving to a new node also increases demand for their products, since their new products will perform better and use less power. This is more true for Intel now than it has been for quite some time. They haven't had competition like this. They have the ability to leapfrog their competition with new nodes, and it is critical for them to utilize it as soon as that node is ready.

Delaying a new node would actually be very bad for financials. If the node is ready and just sitting there unused, you've just wasted lots of money. Doubly so when you consider that the lower variable costs could have been leveraged during this time.

Say 22nm is still in the red right now. So what? The financials will balance out eventually. They don't cease production on 22nm just because 14nm is available. They'll still use 22nm for years to come.

So even if 22nm has still not turned a profit (unlikely), it simply does not make sense to delay 14nm.
Cool, you have an opinion. Where's the support of that? Oh right, there is none so you're just running your mouth.

And here's another chart to ruin your morning:
exh99134.jpg

Even with exponentially rising wafer processing costs, the exponential density increases are still more than enough to offset them. Not only that, but 14nm and 10nm are even more economical than past nodes. Intel's also projecting this trend to continue with 7nm.
Except that Intel sells chips, not transistors. If you don't understand what that means, shut up, and come back to the discussion after you've read and educated yourself. Clearly you have no idea what you're talking about since you have yet to make a coherent argument and are hoping that posting semi-relevant slides while being boorish is enough to leverage you despite your ignorance. Too bad this is the real world where you have to prove your points instead of getting a pat on the back for opening your mouth.

Moving to a lower process node for foundries is much the same thing, there is always a better node (a better house on the same street), but you must be able to afford moving into the new node. As long as there is competition in the foundry space it makes sense to move into the better node as fast as possible, but if there is no or little competition in the foundry space then you let science develop a little further so it is cheaper to move into the new node. Business go bankrupt due to lack of cash reserves, either immediately (many business with the 2008 recession) or over the long term (AMD before selling their foundries to global foundries). There is no point in lighting that cash reserves on fire, if there is no return on investment for using that cash.
Excellent analogy, great post! :thumbsup:
 

Concillian

Diamond Member
May 26, 2004
3,751
8
81
So even if 22nm has still not turned a profit (unlikely), it simply does not make sense to delay 14nm.

And here's another chart to ruin your morning:
exh99134.jpg

Even with exponentially rising wafer processing costs, the exponential density increases are still more than enough to offset them. Not only that, but 14nm and 10nm are even more economical than past nodes. Intel's also projecting this trend to continue with 7nm.

Wall street doesn't care if you simply turn "a profit" They care about how much profit, how much revenue, % margins and the forward looking expectations of all of those. Businesses aren't just out to make money. In this day and age, if a business isn't expecting growth, it's stock takes a significant dump. Unrealistic long term, but I guarantee that Intel execs don't care to simply turn "a profit". The OP noted that yields weren't there yet, and this implies a delay for business reasons. Acceptable yield is a sliding scale, depending on the opportunities to leverage product into $$$. With opportunities for increasing margin looking low to nil, why would they jump on a low yield node?

As for the chart, 32nm to 22nm saw a growth in # of transistors in quads from ~1billion transistors (SB) to ~1.4 billion (IB). Does the cost difference cover adding 40% more transistors? Probably. What about ~1.6B transistor Haswell? Does it cover 60% more transistors? Maybe... I mean it had to make business sense for them to move to Haswell. The point though is that scales on those charts don't have enough resolution to make any determination that has any semblance of confidence.
 
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witeken

Diamond Member
Dec 25, 2013
3,899
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As for the chart, 32nm to 22nm saw a growth in # of transistors in quads from ~1billion transistors (SB) to ~1.4 billion (IB). Does the cost difference cover adding 40% more transistors? Probably. What about ~1.6B transistor Haswell? Does it cover 60% more transistors? Maybe... I mean it had to make business sense for them to move to Haswell. The point though is that scales on those charts don't have enough resolution to make any determination that has any semblance of confidence.

Sandy Bridge has 995M schematic transistors and 1.16B real transistors. Ivy Bridge has 1.2B schematic and 1.4B real transistors. Haswell has 1.4B schematic and 1.6B real transistors.

That means 20% more transistors, not 40%.
 

Homeles

Platinum Member
Dec 9, 2011
2,580
0
0
This is inane: by sputtering "buh buh buh MOORE'S LAW" you haven't proven a thing.
When you can act like an adult and have a decent discussion, let me know.
Wall street doesn't care if you simply turn "a profit" They care about how much profit, how much revenue, % margins and the forward looking expectations of all of those. Businesses aren't just out to make money. In this day and age, if a business isn't expecting growth, it's stock takes a significant dump. Unrealistic long term, but I guarantee that Intel execs don't care to simply turn "a profit". The OP noted that yields weren't there yet, and this implies a delay for business reasons. Acceptable yield is a sliding scale, depending on the opportunities to leverage product into $$$. With opportunities for increasing margin looking low to nil, why would they jump on a low yield node?
They wouldn't, hence the delay we've already seen. That's completely different than withholding it because it's "too expensive." The majority of the costs have already been incurred.
As for the chart, 32nm to 22nm saw a growth in # of transistors in quads from ~1billion transistors (SB) to ~1.4 billion (IB). Does the cost difference cover adding 40% more transistors? Probably. What about ~1.6B transistor Haswell? Does it cover 60% more transistors? Maybe... I mean it had to make business sense for them to move to Haswell. The point though is that scales on those charts don't have enough resolution to make any determination that has any semblance of confidence.
By the time Haswell launched, much of the fixed costs of moving to 22nm had already been recouped. More transistors were inconsequential.

The idea behind Haswell was to have mobile products with such improved battery life that the increased cost would be outweighed by higher demand, and more so than the increased demand there would have been if Intel stuck with Ivy Bridge and had just lowered the price. The same idea drives pretty much any situation where you have a larger die being released on the same node.
 

hans007

Lifer
Feb 1, 2000
20,212
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intel has more capacity than it needs now. and unlike say global foundries they can't really use it for other things.

they are just trying to get back the capex that they spent to upgrade those fabs to 22nm. its not like AMD will ever catch up. 14nm WOULD cost less after you recoup the initial billions in investment. if you don't sell enough 14nm chips to make up for the initial capex you lose money. why do that when the 22nm stuff is already paid for
 

MrK6

Diamond Member
Aug 9, 2004
4,458
4
81
[redacted]

intel has more capacity than it needs now. and unlike say global foundries they can't really use it for other things.

they are just trying to get back the capex that they spent to upgrade those fabs to 22nm. its not like AMD will ever catch up. 14nm WOULD cost less after you recoup the initial billions in investment. if you don't sell enough 14nm chips to make up for the initial capex you lose money. why do that when the 22nm stuff is already paid for
Exactly. 22nm development is done with, it's a sunk cost. Everything now is profit and moving away from that profitable model before it's time is foolish.
 
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Homeles

Platinum Member
Dec 9, 2011
2,580
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[redacted]

Exactly. 22nm development is done with, it's a sunk cost. Everything now is profit and moving away from that profitable model before it's time is foolish.
No one is suggesting otherwise. Had you taken the time to read what's being said here, you'd understand this.

Both of you are acting like children. For that you can go sit in time out for a while, and both of you are hereby thrown out of this thread
-ViRGE
 
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Abwx

Lifer
Apr 2, 2011
11,892
4,876
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And here's another chart to ruin your morning:
exh99134.jpg

Even with exponentially rising wafer processing costs, the exponential density increases are still more than enough to offset them. Not only that, but 14nm and 10nm are even more economical than past nodes. Intel's also projecting this trend to continue with 7nm.

Ahem, if the y axys is a logarithmic scale then an exponential
curve and a straight curve cant be both exponentials , but whatever
keep arguing about thoses curves and their significance or lack of,
if it s not accurate scientificaly speaking at least it has some
entertainement value....
 

nenforcer

Golden Member
Aug 26, 2008
1,780
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And what would Intel do with its shiny new 14nm fabs? Let it sit idle adding to the depreciation bill while Intel management has to hit the magical IRR number for 22nm? That would wreak havoc in the economics of its 14nm node unless Intel delayed 10nm for a period bigger than 14nm.

That is exactly what they are doing.

http://www.extremetech.com/computin...-42-in-arizona-but-its-nothing-to-worry-about

They are putting the only 14nm production (Oregon?) into mobile Intel Atom / Quark / Aveton chips to better compete with the ARM Army while the PC market as a whole slows down and instead coming out with Haswell Refresh chips still at .22nm. The demand no longer justifies the PC being the leading edge process node.