I've been reading benchmarks that test the benefits of systems with Multiple Memory Channel Architectures. The general conclusion of most of these benchmarks is that the performance benefits of systems with greater numbers of memory channels over those systems with fewer channels are negligible.
However nowhere have I found an explanation of why this is the case, just benchmark results indicating that this is the real world performance attained.
The theory is that every doubling of the system's memory channels doubles the bandwidth of memory access, so in theory there should be a performance gain, however in real world applications the gains are negligible. Why?
My postulation is that when the NT Kernel allocates physical memory it is not disturbing the allocations evenly across the the memory channels. If all of a process's virtual memory is mapped to a single memory channel within a MMC system then the process will effectively only be able to attain the performance of having a single memory channel at its disposal. Is this the reason for negligible real world performance gains?
Naturally a process is allocated virtual memory and the kernel allocates physical memory pages, so is this negligible performance gain the fault of the NT Kernel not distributing allocations across the available channels?
related: Why is Skylake so much better than Broadwell-E for single-threaded memory throughput? two memory controllers is sufficient for single-threaded memory bandwidth. Only if you have multiple threads / processes that all miss in cache a lot do you start to benefit from the extra memory controllers in a big Xeon.
(e.g. your example from comments of running many independent image-processing tasks on different images in parallel might do it, depending on the task.)
Going from two down to one DDR4 channel could hurt even a single-threaded program on a quad-core if it was bottlenecked on DRAM bandwidth a lot of the time, but one important part of tuning for performance is to optimize for data reuse so you get at least L3 cache hits.
Matrix multiplication is a classic example: instead of looping over rows / columns of the whole matrix N^2 times (which is too big to fit in cache) (one row x column dot product for each output element), you break the work up into "tiles" and compute partial results, so you're looping repeatedly over a tile of the matrix that stays hot in L1d or L2 cache. (And you hopefully bottleneck on FP ALU throughput, running FMA instructions, not memory at all, because matmul takes O(N^3) multiply+add operations over N^2 elements for a square matrix.) These optimizations are called "loop tiling" or "cache blocking".
So well-optimized code that touches a lot of memory can often get enough work done as its looping that it doesn't actually bottleneck on DRAM bandwidth (L3 cache miss) most of the time.
If a single channel of DRAM is enough to keep up with hardware prefetch requests for how quickly/slowly the code is actually touching new memory, there won't be any measureable slowdown from memory bandwidth. (Of course that's not always possible, and sometimes you do loop over a big array doing not very much work or even just copying it, but if that only makes up a small fraction of the total run time then it's still not significant.)
The theory is that every doubling of the system's memory channels doubles the bandwidth of memory access, so in theory there should be a performance gain, however in real world applications the gains are negligible. Why?
Think of it as a hierarchy, like "CPU <-> L1 cache <-> L2 cache <-> L3 cache <-> RAM <-> swap space". RAM bandwidth only matters when L3 cache wasn't big enough (and swap space bandwidth only matters if RAM wasn't big enough, and ...).
For most (not all) real world applications, the cache is big enough, so RAM bandwidth isn't important and the gains (of multi-channel) are negligible.
My postulation is that when the NT Kernel allocates physical memory it is not disturbing the allocations evenly across the the memory channels.
It doesn't work like that. The CPU mostly only works with whole cache lines (e.g. 64 byte pieces); and with one channel the entire cache line comes from one channel; and with 2 channels half of a cache line comes from one channel and the other half comes from a different channel. There is almost nothing that any software can do that will make any difference. The NT kernel only works with whole pages (e.g. 4 KiB pieces), so whatever the kernel does is even less likely to matter (until you start thinking about NUMA optimizations, which is a completely different thing).
Let's assume an algorithm is repeatedly processing buffers of data, it may be accessing say 2 to 16 of these buffers, all having the same size. What would you expect to be the optimum size of these buffers, assuming the algorithm can process the full data in smaller blocks.
I expect the potential bottleneck of cache misses if the blocks are too big, but of course the bigger the blocks the better for vectorization.
Let's expect current i7/i9 CPUs (2018)
Any ideas?
Do you have multiple threads? Can you arrange things so the same thread uses the same buffer repeatedly? (i.e. keep buffers associated with threads when possible).
Modern Intel CPUs have 32k L1d, 256k L2 private per-core. (Or Skylake-AVX512 has 1MiB private L2 caches, with less shared L3). (Which cache mapping technique is used in intel core i7 processor?)
Aiming for L2 hits most of the time is good. L2 miss / L3 hit some of the time isn't always terrible, but off-core is significantly slower. Remember that L2 is a unified cache, so it covers code as well, and of course there's stack memory and random other demands for L2. So aiming for a total buffer size of around half L2 size usually gives a good hit-rate for cache-blocking.
Depending on how much bandwidth your algorithm can use, you might even aim for mostly L1d hits, but small buffers can mean more startup / cleanup overhead and spending more time outside of the main loop.
Also remember that with Hyperthreading, each logical core competes for cache on the physical core it's running on. So if two threads end up on the same physical core, but are touching totally different memory, your effective cache sizes are about half.
Probably you should make the buffer size a tunable parameter, and profile with a few different sizes.
Use perf counters to check if you're actually avoiding L1d or L2 misses or not, with different sizes, to help you understand whether your code is sensitive to different amounts of memory latency or not.
Given a cache size with constant capacity and associativity, for a given code to determine average of array elements, would a cache with higher block size be preferred?
[from comments]
Examine the code given below to compute the average of an array:
total = 0;
for(j=0; j < k; j++) {
sub_total = 0; /* Nested loops to avoid overflow */
for(i=0; i < N; i++) {
sub_total += A[jN + i];
}
total += sub_total/N;
}
average = total/k;
Related: in the more general case of typical access patterns with some but limited spatial locality, larger lines help up to a point. These "Memory Hierarchy: Set-Associative Cache" (powerpoint) slides by Hong Jiang and/or Yifeng Zhu (U. Maine) have a graph of AMAT (Average Memory Access Time) vs. block size showing a curve, and also breaking it down into miss penalty vs. miss rate (for a simple model I think, for a simple in-order CPU that sucks at hiding memory latency. e.g. maybe not even pipelining multiple independent misses. (miss under miss))
There is a lot of good stuff in those slides, including a compiler-optimization section that mentions loop interchange (to fix nested loops with column-major vs. row-major order), and even cache-blocking for more reuse. A lot of stuff on the Internet is crap, but I looked through these slides and they have some solid info on how caches are designed and what the tradeoffs are. The performance-analysis stuff is only really accurate for simple CPUs, not like modern out-of-order CPUs that can overlap some computation with cache-miss latency so more shorter misses is different from fewer longer misses.
Specific answer to this question:
So the only workload you care about is a linear traversal of your elements? That makes cache line size nearly irrelevant for performance, assuming good hardware prefetching. (So larger lines mean less HW complexity and power usage for the same performance.)
With software prefetch, larger lines mean less prefetch overhead (although depending on the CPU design, that may not hurt performance if you still max out memory bandwidth.)
Without any prefetching, a larger line/block size would mean more hits following every demand-miss. A single traversal of an array has perfect spatial locality and no temporal locality. (Actually not quite perfect spatial locality at the start/end, if the array isn't aligned to the start of a cache line, and/or ends in the middle of a line.)
If a miss has to wait until the entire line is present in cache before the load that caused the miss can be satisfied, this slightly reduces the advantage of larger blocks. (But most of the latency of a cache miss is in the signalling and request overhead, not in waiting for the burst transfer to complete after it's already started.)
A larger block size means fewer requests in flight with the same bandwidth and latency, and limited concurrency is a real limiting factor in memory bandwidth in real CPUs. (See the latency-bound platforms part of this answer about x86 memory bandwidth: many-core Xeons with higher latency to L3 cache have lower single-threaded bandwidth than a dual or quad-core of the same clock speed. Each core only has 10 line-fill buffers to track outstanding L1 misses, and bandwidth = concurrency / latency.)
If your cache-miss handling has an early restart design, even that bit of extra latency can be avoided. (That's very common, but Paul says theoretically possible to not have it in a CPU design). The load that caused the miss gets its data as soon as it arrives. The rest of the cache line fill happens "in the background", and hopefully later loads can also be satisfied from the partially-received cache line.
Critical word first is a related feature, where the needed word is sent first (for use with early restart), and the burst transfer then wraps around to transfer the earlier words of the block. In this case, the critical word will always be the first word, so no special hardware support is needed beyond early restart. (The U. Maine slides I linked above mention early restart / critical word first and point out that it decreases the miss penalty for large cache lines.)
An out-of-order execution CPU (or software pipelining on an in-order CPU) could give you the equivalent of HW prefetch by having multiple demand-misses outstanding at once. If the CPU "sees" the loads to another cache line while a miss to the current cache line is still outstanding, the demand-misses can be pipelined, again hiding some of the difference between larger or smaller lines.
If lines are too small, you'll run into a limit on how many outstanding misses for different lines your L1D can track. With larger lines or smaller out-of-order windows, you might have some "slack" when there's no outstanding request for the next cache line, so you're not maxing out the bandwidth. And you pay for it with bubbles in the pipeline when you get to the end of a cache line and the start of the next line hasn't arrived yet, because it started too late (while ALU execution units were using data from too close to the end of the current cache line.)
Related: these slides don't say much about the tradeoff of larger vs. smaller lines, but look pretty good.
The simplistic answer is that larger cache blocks would be preferred since the workload has no (data) temporal locality (no data reuse), perfect spacial locality (excluding the potentially inadequate alignment of the array for the first block and insufficient size of the array for the last block, every part of every block of data will be used), and a single access stream (no potential for conflict misses).
A more nuanced answer would consider the size and alignment of the array (the fraction of the first and last cache blocks that will be unused and what fraction of the memory transfer time that represents; for a 1 GiB array, even 4 KiB blocks would waste less than 0.0008% of the memory bandwidth), the ability of the system to use critical word first (if the array is of modest size and there is no support for early use of data as it becomes available rather than waiting for the entire block to be filled, then the start-up overhead will remove much of the prefetching advantage of larger cache blocks), the use of prefetching (software or hardware prefetching reduces the benefit of large cache blocks and this workload is extremely friendly to prefetching), the configuration of the memory system (e.g., using DRAM with an immediate page close controller policy would increase the benefit of larger cache blocks because each access would involve a row activate and row close, often to the same DRAM bank preventing latency overlap), whether the same block size is used for instructions and page table accesses and whether these accesses share the cache (instruction accesses provide a second "stream" which can introduce conflict misses; with shared caching of a two-level hierarchical page table TLB misses would access two cache blocks), whether simple way prediction is used (a larger block would increase prediction accuracy reducing misprediction overhead), and perhaps other factors.
From your example code we can't say either way as long as the hardware pre-fetcher can keep up a memory stream at maximum memory throughput.
In a random access scenario a shorter cache-line might be preferable as you then don't need to fill all the line. But the total amount of cached memory would go down as you need more circuits for tags and potentially more time for comparing.
So a compromise must be made Intel has chosen 64-bytes per line (and fetches 2 lines) others has chosen 32-bytes per line.
From a previous question on this forum, I learned that in most of the memory systems, L1 cache is a subset of the L2 cache means any entry removed from L2 is also removed from L1.
So now my question is how do I determine a corresponding entry in L1 cache for an entry in the L2 cache. The only information stored in the L2 entry is the tag information. Based on this tag information, if I re-create the addr it may span multiple lines in the L1 cache if the line-sizes of L1 and L2 cache are not same.
Does the architecture really bother about flushing both the lines or it just maintains L1 and L2 cache with the same line-size.
I understand that this is a policy decision but I want to know the commonly used technique.
Cache-Lines size is (typically) 64 bytes.
Moreover, take a look at this very interesting article about processors caches:
Gallery of Processor Cache Effects
You will find the following chapters:
Memory accesses and performance
Impact of cache lines
L1 and L2 cache sizes
Instruction-level parallelism
Cache associativity
False cache line sharing
Hardware complexities
In core i7 the line sizes in L1 , L2 and L3 are the same: that is 64 Bytes.
I guess this simplifies maintaining the inclusive property, and coherence.
See page 10 of: https://www.aristeia.com/TalkNotes/ACCU2011_CPUCaches.pdf
The most common technique of handling cache block size in a strictly inclusive cache hierarchy is to use the same size cache blocks for all levels of cache for which the inclusion property is enforced. This results in greater tag overhead than if the higher level cache used larger blocks, which not only uses chip area but can also increase latency since higher level caches generally use phased access (where tags are checked before the data portion is accessed). However, it also simplifies the design somewhat and reduces the wasted capacity from unused portions of the data. It does not take a large fraction of unused 64-byte chunks in 128-byte cache blocks to compensate for the area penalty of an extra 32-bit tag. In addition, the larger cache block effect of exploiting broader spatial locality can be provided by relatively simple prefetching, which has the advantages that no capacity is left unused if the nearby chunk is not loaded (to conserve memory bandwidth or reduce latency on a conflicting memory read) and that the adjacency prefetching need not be limited to a larger aligned chunk.
A less common technique divides the cache block into sectors. Having the sector size the same as the block size for lower level caches avoids the problem of excess back-invalidation since each sector in the higher level cache has its own valid bit. (Providing all the coherence state metadata for each sector rather than just validity can avoid excessive writeback bandwidth use when at least one sector in a block is not dirty/modified and some coherence overhead [e.g., if one sector is in shared state and another is in the exclusive state, a write to the sector in the exclusive state could involve no coherence traffic—if snoopy rather than directory coherence is used].)
The area savings from sectored cache blocks were especially significant when tags were on the processor chip but the data was off-chip. Obviously, if the data storage takes area comparable to the size of the processor chip (which is not unreasonable), then 32-bit tags with 64-byte blocks would take roughly a 16th (~6%) of the processor area while 128-byte blocks would take half as much. (IBM's POWER6+, introduced in 2009, is perhaps the most recent processor to use on-processor-chip tags and off-processor data. Storing data in higher-density embedded DRAM and tags in lower-density SRAM, as IBM did, exaggerates this effect.)
It should be noted that Intel uses "cache line" to refer to the smaller unit and "cache sector" for the larger unit. (This is one reason why I used "cache block" in my explanation.) Using Intel's terminology it would be very unusual for cache lines to vary in size among levels of cache regardless of whether the levels were strictly inclusive, strictly exclusive, or used some other inclusion policy.
(Strict exclusion typically uses the higher level cache as a victim cache where evictions from the lower level cache are inserted into the higher level cache. Obviously, if the block sizes were different and sectoring was not used, then an eviction would require the rest of the larger block to be read from somewhere and invalidated if present in the lower level cache. [Theoretically, strict exclusion could be used with inflexible cache bypassing where an L1 eviction would bypass L2 and go to L3 and L1/L2 cache misses would only be allocated to either L1 or L2, bypassing L1 for certain accesses. The closest to this being implemented that I am aware of is Itanium's bypassing of L1 for floating-point accesses; however, if I recall correctly, the L2 was inclusive of L1.])
Typically, in one access to the main memory 64 bytes of data and 8 bytes of parity/ECC (I don't remember exactly which) is accessed. And it is rather complicated to maintain different cache line sizes at the various memory levels. You have to note that cache line size would be more correlated to the word alignment size on that architecture than anything else. Based on that, a cache line size is highly unlikely to be different from memory access size. Now, the parity bits are for the use of the memory controller - so cache line size typically is 64 bytes. The processor really controls very little beyond the registers. Everything else going on in the computer is more about getting hardware in to optimize CPU performance. In that sense also, it really would not make any sense to import extra complexity by making cache line sizes different at different levels of memory.
Why is the size of L1 cache smaller than that of the L2 cache in most of the processors ?
L1 is very tightly coupled to the CPU core, and is accessed on every memory access (very frequent). Thus, it needs to return the data really fast (usually within on clock cycle). Latency and throughput (bandwidth) are both performance-critical for L1 data cache. (e.g. four cycle latency, and supporting two reads and one write by the CPU core every clock cycle). It needs lots of read/write ports to support this high access bandwidth. Building a large cache with these properties is impossible. Thus, designers keep it small, e.g. 32KB in most processors today.
L2 is accessed only on L1 misses, so accesses are less frequent (usually 1/20th of the L1). Thus, L2 can have higher latency (e.g. from 10 to 20 cycles) and have fewer ports. This allows designers to make it bigger.
L1 and L2 play very different roles. If L1 is made bigger, it will increase L1 access latency which will drastically reduce performance because it will make all dependent loads slower and harder for out-of-order execution to hide. L1 size is barely debatable.
If we removed L2, L1 misses will have to go to the next level, say memory. This means that a lot of access will be going to memory which would imply we need more memory bandwidth, which is already a bottleneck. Thus, keeping the L2 around is favorable.
Experts often refer to L1 as a latency filter (as it makes the common case of L1 hits faster) and L2 as a bandwidth filter as it reduces memory bandwidth usage.
Note: I have assumed a 2-level cache hierarchy in my argument to make it simpler. In many of today's multicore chips, there's an L3 cache shared between all the cores, while each core has its own private L1 and maybe L2. In these chips, the shared last-level cache (L3) plays the role of memory bandwidth filter. L2 plays the role of on-chip bandwidth filter, i.e. it reduces access to the on-chip interconnect and the L3. This allows designers to use a lower-bandwidth interconnect like a ring, and a slow single-port L3, which allows them to make L3 bigger.
Perhaps worth mentioning that the number of ports is a very important design point because it affects how much chip area the cache consumes. Ports add wires to the cache which consumes a lot of chip area and power.
There are different reasons for that.
L2 exists in the system to speedup the case where there is a L1 cache miss. If the size of L1 was the same or bigger than the size of L2, then L2 could not accomodate for more cache lines than L1, and would not be able to deal with L1 cache misses. From the design/cost perspective, L1 cache is bound to the processor and faster than L2. The whole idea of caches is that you speed up access to the slower hardware by adding intermediate hardware that is more performing (and expensive) than the slowest hardware and yet cheaper than the faster hardware you have. Even if you decided to double the L1 cache, you would also increment L2, to speedup L1-cache misses.
So why is there L2 cache at all? Well, L1 cache is usually more performant and expensive to build, and it is bound to a single core. This means that increasing the L1 size by a fixed quantity will have that cost multiplied by 4 in a dual core processor, or by 8 in a quad core. L2 is usually shared by different cores --depending on the architecture it can be shared across a couple or all cores in the processor, so the cost of increasing L2 would be smaller even if the price of L1 and L2 were the same --which it is not.
#Aater's answer explains some of the basics. I'll add some more details + an examples of the real cache organization on Intel Haswell and AMD Piledriver, with latencies and other properties, not just size.
For some details on IvyBridge, see my answer on "How can cache be that fast?", with some discussion of the overall load-use latency including address-calculation time, and widths of the data busses between different levels of cache.
L1 needs to be very fast (latency and throughput), even if that means a limited hit-rate. L1d also needs to support single-byte stores on almost all architectures, and (in some designs) unaligned accesses. This makes it hard to use ECC (error correction codes) to protect the data, and in fact some L1d designs (Intel) just use parity, with better ECC only in outer levels of cache (L2/L3) where the ECC can be done on larger chunks for lower overhead.
It's impossible to design a single level of cache that could provide the low average request latency (averaged over all hits and misses) of a modern multi-level cache. Since modern systems have multiple very hungry cores all sharing a connection to the same relatively-high latency DRAM, this is essential.
Every core needs its own private L1 for speed, but at least the last level of cache is typically shared, so a multi-threaded program that reads the same data from multiple threads doesn't have to go to DRAM for it on each core. (And to act as a backstop for data written by one core and read by another). This requires at least two levels of cache for a sane multi-core system, and is part of the motivation for more than 2 levels in current designs. Modern multi-core x86 CPUs have a fast 2-level cache in each core, and a larger slower cache shared by all cores.
L1 hit-rate is still very important, so L1 caches are not as small / simple / fast as they could be, because that would reduce hit rates. Achieving the same overall performance would thus require higher levels of cache to be faster. If higher levels handle more traffic, their latency is a bigger component of the average latency, and they bottleneck on their throughput more often (or need higher throughput).
High throughput often means being able to handle multiple reads and writes every cycle, i.e. multiple ports. This takes more area and power for the same capacity as a lower-throughput cache, so that's another reason for L1 to stay small.
L1 also uses speed tricks that wouldn't work if it was larger. i.e. most designs use Virtually-Indexed, Physically Tagged (VIPT) L1, but with all the index bits coming from below the page offset so they behave like PIPT (because the low bits of a virtual address are the same as in the physical address). This avoids synonyms / homonyms (false hits or the same data being in the cache twice, and see Paul Clayton's detailed answer on the linked question), but still lets part of the hit/miss check happen in parallel with the TLB lookup. A VIVT cache doesn't have to wait for the TLB, but it has to be invalidated on every change to the page tables.
On x86 (which uses 4kiB virtual memory pages), 32kiB 8-way associative L1 caches are common in modern designs. The 8 tags can be fetched based on the low 12 bits of the virtual address, because those bits are the same in virtual and physical addresses (they're below the page offset for 4kiB pages). This speed-hack for L1 caches only works if they're small enough and associative enough that the index doesn't depend on the TLB result. 32kiB / 64B lines / 8-way associativity = 64 (2^6) sets. So the lowest 6 bits of an address select bytes within a line, and the next 6 bits index a set of 8 tags. This set of 8 tags is fetched in parallel with the TLB lookup, so the tags can be checked in parallel against the physical-page selection bits of the TLB result to determine which (if any) of the 8 ways of the cache hold the data. (Minimum associativity for a PIPT L1 cache to also be VIPT, accessing a set without translating the index to physical)
Making a larger L1 cache would mean it had to either wait for the TLB result before it could even start fetching tags and loading them into the parallel comparators, or it would have to increase in associativity to keep log2(sets) + log2(line_size) <= 12. (More associativity means more ways per set => fewer total sets = fewer index bits). So e.g. a 64kiB cache would need to be 16-way associative: still 64 sets, but each set has twice as many ways. This makes increasing L1 size beyond the current size prohibitively expensive in terms of power, and probably even latency.
Spending more of your power budget on L1D cache logic would leave less power available for out-of-order execution, decoding, and of course L2 cache and so on. Getting the whole core to run at 4GHz and sustain ~4 instructions per clock (on high-ILP code) without melting requires a balanced design. See this article: Modern Microprocessors: A 90-Minute Guide!.
The larger a cache is, the more you lose by flushing it, so a large VIVT L1 cache would be worse than the current VIPT-that-works-like-PIPT. And a larger but higher-latency L1D would probably also be worse.
According to #PaulClayton, L1 caches often fetch all the data in a set in parallel with the tags, so it's there ready to be selected once the right tag is detected. The power cost of doing this scales with associativity, so a large highly-associative L1 would be really bad for power-use as well as die-area (and latency). (Compared to L2 and L3, it wouldn't be a lot of area, but physical proximity is important for latency. Speed-of-light propagation delays matter when clock cycles are 1/4 of a nanosecond.)
Slower caches (like L3) can run at a lower voltage / clock speed to make less heat. They can even use different arrangements of transistors for each storage cell, to make memory that's more optimized for power than for high speed.
There are a lot of power-use related reasons for multi-level caches. Power / heat is one of the most important constraints in modern CPU design, because cooling a tiny chip is hard. Everything is a tradeoff between speed and power (and/or die area). Also, many CPUs are powered by batteries or are in data-centres that need extra cooling.
L1 is almost always split into separate instruction and data caches. Instead of an extra read port in a unified L1 to support code-fetch, we can have a separate L1I cache tied to a separate I-TLB. (Modern CPUs often have an L2-TLB, which is a second level of cache for translations that's shared by the L1 I-TLB and D-TLB, NOT a TLB used by the regular L2 cache). This gives us 64kiB total of L1 cache, statically partitioned into code and data caches, for much cheaper (and probably lower latency) than a monster 64k L1 unified cache with the same total throughput. Since there is usually very little overlap between code and data, this is a big win.
L1I can be placed physically close to the code-fetch logic, while L1D can be physically close to the load/store units. Speed-of-light transmission-line delays are a big deal when a clock cycle lasts only 1/3rd of a nanosecond. Routing the wiring is also a big deal: e.g. Intel Broadwell has 13 layers of copper above the silicon.
Split L1 helps a lot with speed, but unified L2 is the best choice.
Some workloads have very small code but touch lots of data. It makes sense for higher-level caches to be unified to adapt to different workloads, instead of statically partitioning into code vs. data. (e.g. almost all of L2 will be caching data, not code, while running a big matrix multiply, vs. having a lot of code hot while running a bloated C++ program, or even an efficient implementation of a complicated algorithm (e.g. running gcc)). Code can be copied around as data, not always just loaded from disk into memory with DMA.
Caches also need logic to track outstanding misses (since out-of-order execution means that new requests can keep being generated before the first miss is resolved). Having many misses outstanding means you overlap the latency of the misses, achieving higher throughput. Duplicating the logic and/or statically partitioning between code and data in L2 would not be good.
Larger lower-traffic caches are also a good place to put pre-fetching logic. Hardware pre-fetching enables good performance for things like looping over an array without every piece of code needing software-prefetch instructions. (SW prefetch was important for a while, but HW prefetchers are smarter than they used to be, so that advice in Ulrich Drepper's otherwise excellent What Every Programmer Should Know About Memory is out-of-date for many use cases.)
Low-traffic higher level caches can afford the latency to do clever things like use an adaptive replacement policy instead of the usual LRU. Intel IvyBridge and later CPUs do this, to resist access patterns that get no cache hits for a working set just slightly too large to fit in cache. (e.g. looping over some data in the same direction twice means it probably gets evicted just before it would be reused.)
A real example: Intel Haswell. Sources: David Kanter's microarchitecture analysis and Agner Fog's testing results (microarch pdf). See also Intel's optimization manuals (links in the x86 tag wiki).
Also, I wrote up a separate answer on: Which cache mapping technique is used in intel core i7 processor?
Modern Intel designs use a large inclusive L3 cache shared by all cores as a backstop for cache-coherence traffic. It's physically distributed between the cores, with 2048 sets * 16-way (2MiB) per core (with an adaptive replacement policy in IvyBridge and later).
The lower levels of cache are per-core.
L1: per-core 32kiB each instruction and data (split), 8-way associative. Latency = 4 cycles. At least 2 read ports + 1 write port. (Maybe even more ports to handle traffic between L1 and L2, or maybe receiving a cache line from L2 conflicts with retiring a store.) Can track 10 outstanding cache misses (10 fill buffers).
L2: unified per-core 256kiB, 8-way associative. Latency = 11 or 12 cycles. Read bandwidth: 64 bytes / cycle. The main prefetching logic prefetches into L2. Can track 16 outstanding misses. Can supply 64B per cycle to the L1I or L1D. Actual port counts unknown.
L3: unified, shared (by all cores) 8MiB (for a quad-core i7). Inclusive (of all the L2 and L1 per-core caches). 12 or 16 way associative. Latency = 34 cycles. Acts as a backstop for cache-coherency, so modified shared data doesn't have to go out to main memory and back.
Another real example: AMD Piledriver: (e.g. Opteron and desktop FX CPUs.) Cache-line size is still 64B, like Intel and AMD have used for several years now. Text mostly copied from Agner Fog's microarch pdf, with additional info from some slides I found, and more details on the write-through L1 + 4k write-combining cache on Agner's blog, with a comment that only L1 is WT, not L2.
L1I: 64 kB, 2-way, shared between a pair of cores (AMD's version of SMD has more static partitioning than Hyperthreading, and they call each one a core. Each pair shares a vector / FPU unit, and other pipeline resources.)
L1D: 16 kB, 4-way, per core. Latency = 3-4 c. (Notice that all 12 bits below the page offset are still used for index, so the usual VIPT trick works.) (throughput: two operations per clock, up to one of them being a store). Policy = Write-Through, with a 4k write-combining cache.
L2: 2 MB, 16-way, shared between two cores. Latency = 20 clocks. Read throughput 1 per 4 clock. Write throughput 1 per 12 clock.
L3: 0 - 8 MB, 64-way, shared between all cores. Latency = 87 clock. Read throughput 1 per 15 clock. Write throughput 1 per 21 clock
Agner Fog reports that with both cores of a pair active, L1 throughput is lower than when the other half of a pair is idle. It's not known what's going on, since the L1 caches are supposed to be separate for each core.
The other answers here give specific and technical reasons why L1 and L2 are sized as they are, and while many of them are motivating considerations for particular architectures, they aren't really necessary: the underlying architectural pressure leading to increasing (private) cache sizes as you move away from the core is fairly universal and is the same as the reasoning for multiple caches in the first place.
The three basic facts are:
The memory accesses for most applications exhibit a high degree of temporal locality, with a non-uniform distribution.
Across a large variety of process and designs, cache size and cache speed (latency and throughput) can be traded off against each other1.
Each distinct level of cache involves incremental design and performance cost.
So at a basic level, you might be able to say double the size of the cache, but incur a latency penalty of 1.4 compared to the smaller cache.
So it becomes an optimization problem: how many caches should you have and how large should they be? If memory access was totally uniform within the working set size, you'd probably end up with a single fairly large cache, or no cache at all. However, access is strongly non-uniform, so a small-and-fast cache can capture a large number of accesses, disproportionate to it's size.
If fact 2 didn't exist, you'd just create a very big, very fast L1 cache within the other constraints of your chip and not need any other cache levels.
If fact 3 didn't exist, you'd end up with a huge number of fine-grained "caches", faster and small at the center, and slower and larger outside, or perhaps a single cache with variable access times: faster for the parts closest to the core. In practice, rule 3 means that each level of cache has an additional cost, so you usually end up with a few quantized levels of cache2.
Other Constraints
This gives a basic framework to understand cache count and cache sizing decisions, but there are secondary factors at work as well. For example, Intel x86 has 4K page sizes and their L1 caches use a VIPT architecture. VIPT means that the size of the cache divided by the number of ways cannot be larger3 than 4 KiB. So an 8-way L1 cache as used on the half dozen Intel designs can be at most 4 KiB * 8 = 32 KiB. It is probably no coincidence that that's exactly the size of the L1 cache on those designs! If it weren't for this constraint, it is entirely possible you'd have seen lower-associativity and/or larger L1 caches (e.g., 64 KiB, 4-way).
1 Of course, there are other factors involved in the tradeoff as well, such as area and power, but holding those factors constant the size-speed tradeoff applies, and even if not held constant the basic behavior is the same.
2 In addition to this pressure, there is a scheduling benefit to known-latency caches, like most L1 designs: and out-of-order scheduler can optimistically submit operations that depend on a memory load on the cycle that the L1 cache would return, reading the result off the bypass network. This reduces contention and perhaps shaves a cycle of latency off the critical path. This puts some pressure on the innermost cache level to have uniform/predictable latency and probably results in fewer cache levels.
3 In principle, you can use VIPT caches without this restriction, but only by requiring OS support (e.g., page coloring) or with other constraints. The x86 arch hasn't done that and probably can't start now.
For those interested in this type of questions, my university recommends Computer Architecture: A Quantitative Approach and Computer Organization and Design: The Hardware/Software Interface. Of course, if you don't have time for this, a quick overview is available on Wikipedia.
I think the main reason for this is, that L1-Cache is faster and so it's more expensive.
https://en.wikichip.org/wiki/amd/microarchitectures/zen#Die
Compare the size of the L1, L2, and L3 caches physical size for an AMD Zen core, for example. The density increases dramatically with the cache level.
logically, the question answers itself.
If L1 were bigger than L2 (combined), then there would be no need of L2 Cache.
Why would you store your stuff on tape-drive if you can store all of it on HDD ?