What Is The Average Rate For Computer Repair

Cycles Per Instruction

Profiling and timing

Jim Jeffers , ... Avinash Sodani , in Intel Xeon Phi Processor Loftier Performance Programming (Second Edition), 2022

Cycles per education — description and usage

Cycles per teaching, or CPI, as divers in Fig. xiv.2 is a metric that has been a part of the VTune interface for many years. It tells the average number of CPU cycles required to retire an instruction, and therefore is an indicator of how much latency in the system afflicted the running application. Since CPI is a ratio, information technology will be affected by either changes in the number of CPU cycles that an awarding takes (the numerator) or changes in the number of instructions executed (the denominator). For that reason, CPI is best used for comparison when only ane part of the ratio is changing. For instance, changes might exist made to a data construction in one part of the code that lower CPI in a (different) hotspot. "New" and "quondam" CPI could be compared for that hotspot as long as the code within it hasn't changed. The goal is to lower CPI, both in hotspots and for the application as a whole.

Fig. xiv.ii. CPI events.

In order to brand full utilise of the metric, information technology is of import to understand how to interpret CPI when using multiple hardware threads. For analysis of Knights Landing functioning, CPI can exist analyzed in two means: "per-core" or "per-thread." Each way of analyzing CPI tin exist useful. The per-thread analysis is the nigh straightforward. It is calculated from 2 events:  CPU_CLK_UNHALTED.THREAD (also known equally clock ticks or cycles) and  INST_RETIRED.Any.  CPU_CLK_UNHALTED.THREAD counts ticks of the CPU core'due south clock when a thread is active. The other issue used is  INST_RETIRED.ANY, and this event is also counted at the thread level. On a sample, each thread executing on a core could take a different value for this result, depending on how many instructions from each thread have really been retired. Calculating CPI per thread is easy: it is just the event of dividing  CPU_CLK_UNHALTED.THREAD by  INST_RETIRED.Whatsoever. For whatever given sample, this calculation will apply the thread's value for clock ticks and an individual hardware thread'southward value for instructions executed. This calculation is typically done at the function level, using the sum of all samples for each function, and and then volition calculate an boilerplate CPI per hardware thread, averaged across all hardware threads running for the function.

CPI per core is fairly straightforward every bit shown in Fig. 14.iii. To calculate an "Aggregate" CPI, or Average CPI per cadre, split the sum of all cadre's threads CPU_CLK_UNHALTED.THREAD value past the sum of all the threads'  INST_RETIRED.ANY values. For case, presume an awarding that is using one hardware thread per cadre on Knights Landing. One hot function in the application takes 1200 clock ticks to complete. During those 1200 cycles, the thread executed 600 instructions and hence the CPI of this thread is 2. But since the automobile is capable of a CPI 0.5, the thread is utilizing but ¼ of the capacity of the cadre. In such a state of affairs consider calculation some other thread to the application and so equally to utilize the actress capacity of the core. Now presume that this awarding is parallelized and the 2 threads each gets 450 clocks and are able to retire 300 instructions each. Now the CPI of private threads improved to i.5 and the overall CPI of the core also improved to 1.five. Now nosotros further parallelize this and add one more thread so that each thread is active for 200 clocks and executed 200 instructions each thereby improving the CPI to 1. We add one more than thread to this and now each of the four threads is active for 75 clocks and retired 150 instructions, each achieving a CPI of 0.v. These performance numbers are summarized in Fig. fourteen.iv.

Fig. 14.iii. CPI formulas and thresholds.

Fig. 14.4. CPI scaling across threads on a core.

This is a hypothetical example, and typically such scaling is not achievable in existent world simply illustrates how instructions from different hardware threads can exist interleaved on a single cadre to apply full capacity of the core. So the principle to exist followed is to put multiple threads on a core but if i thread cannot fully utilize the core. If one thread is able to fully utilize the cadre and is able to reach maximum CPI and then it is better to map the threads to other cores. The availability of 4 hardware threads on Knights Landing can exist useful for absorbing some of the latency of a workload's information access — while one hardware thread is waiting on data, another can be executing.

Fig. 14.5 shows the CPI per core for a real-workload run in our lab as the number of hardware threads/cadre is scaled from i to 4. For this application, the operation of the application was increasing with the addition of each thread, although the addition of the fourth thread did not add together as much performance equally did the second or third. The data shows that the CPI per core is decreasing overall, as expected since each thread adds functioning by effectively utilizing the spare capacity of the cadre. For this workload, the number of instructions executed was roughly abiding beyond all the hardware thread configurations, then the CPI directly afflicted execution time. When CPI per core decreased, that translated to a reduction in total execution time for the application.

Fig. 14.v. CPI example.

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Microarchitecture

Sarah 50. Harris , David Coin Harris , in Digital Design and Estimator Architecture, 2022

7.5.4 Operation Assay

The pipelined processor ideally would accept a CPI of one, considering a new instruction is issued every cycle. Withal, a stall or a affluent wastes a bike, so the CPI is slightly higher and depends on the specific program beingness executed.

Example 7.7

Pipelined Processor CPI

The SPECINT2000 benchmark considered in Example vii.5 consists of approximately 25% loads, 10% stores, xiii% branches, and 52% data-processing instructions. Presume that 40% of the loads are immediately followed by an teaching that uses the result, requiring a stall, and that fifty% of the branches are taken (mispredicted), requiring a flush. Ignore other hazards. Compute the average CPI of the pipelined processor.

Solution

The average CPI is the sum over each instruction of the CPI for that instruction multiplied by the fraction of time that pedagogy is used. Loads take ane clock cycle when in that location is no dependency and two cycles when the processor must stall for a dependency, and then they have a CPI of (0.6)(1)+(0.4)(ii)=ane.4. Branches take one clock cycle when they are predicted properly and iii when they are not, and so they have a CPI of (0.five)(ane)+(0.five)(three)=2.0. All other instructions have a CPI of 1. Hence, for this criterion, average CPI=(0.25)(ane.iv)+(0.1)(1)+(0.13)(2.0)+(0.52)(1)=1.23.

Nosotros can make up one's mind the cycle fourth dimension past because the critical path in each of the five pipeline stages shown in Figure 7.58. Recall that the register file is written in the offset half of the Writeback cycle and read in the second half of the Decode bike. Therefore, the wheel fourth dimension of the Decode and Writeback stages is twice the time necessary to practise the half-cycle of piece of work.

(7.5) $T_{c3}=ma\mathrm{10}\left[\begin{array}{ll}{t}_{pcq}+t_{\mathrm{grand}em}+t_{\mathrm{south}\mathrm{due\; east}tup}& F\mathrm{due\; east}tch\hfill \\ 2(t_{RFr\mathrm{eastward}ad}+t_{\mathrm{south}etup})\hfill & D\mathrm{eastward}code\hfill \\ t_{pcq}+2t_{mux}+t_{ALU}+t_{setup}\hfill & \mathrm{Eastward}x\mathrm{east}cute\hfill \\ t_{pcq}+t_{m\mathrm{east}m}+t_{\mathrm{southward}etup}\hfill & M\mathrm{east}\mathrm{thousand}ory\hfill \\ 2(t_{pcq}+t_{\mathrm{grand}ux}+t_{RFs\mathrm{east}tup})\hfill & Writ\mathrm{due\; east}back\hfill \end{array}\right]$

Instance 7.8

Processor Performance Comparison

Ben Bitdiddle needs to compare the pipelined processor performance with that of the unmarried-cycle and multicycle processors considered in Instance 7.vi. The logic delays were given in Table 7.5. Help Ben compare the execution time of 100 billion instructions from the SPECINT2000 benchmark for each processor.

Solution

According to Equation seven.v, the wheel time of the pipelined processor is T _c3=max[40+200+50, 2(100+50), xl+two(25)+120+50, 40+200+50, 2(40+25+lx)]=300   ps. According to Equation vii.ane, the total execution time is T₃=(100×10⁹ instructions)(ane.23 cycles   /   didactics)(300×ten⁻¹²  s   /bike)=36.9 seconds. This compares with 84 seconds for the single-wheel processor and 140 seconds for the multicycle processor.

The pipelined processor is substantially faster than the others. Nonetheless, its advantage over the unmarried-bike processor is nowhere most the five-fold speed-upwards 1 might hope to get from a five-phase pipeline. The pipeline hazards introduce a small CPI penalty. More than significantly, the sequencing overhead (clk-to-Q and setup times) of the registers applies to every pipeline stage, not just once to the overall datapath. Sequencing overhead limits the benefits ane can hope to achieve from pipelining. The pipelined processor is similar in hardware requirements to the single-cycle processor, but it adds eight 32-bit pipeline registers, forth with multiplexers, smaller pipeline registers, and control logic to resolve hazards.

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Microarchitecture

David Money Harris , Sarah L. Harris , in Digital Blueprint and Estimator Architecture (Second Edition), 2022

7.4.four Functioning Assay

The execution time of an instruction depends on both the number of cycles it uses and the cycle time. Whereas the single-bicycle processor performed all instructions in 1 cycle, the multicycle processor uses varying numbers of cycles for the various instructions. However, the multicycle processor does less work in a single wheel and, thus, has a shorter cycle time.

The multicycle processor requires iii cycles for beq and j instructions, four cycles for sw, addi, and R-blazon instructions, and five cycles for lw instructions. The CPI depends on the relative likelihood that each pedagogy is used.

Case seven.7

Multicycle Processor CPI

The SPECINT2000 benchmark consists of approximately 25% loads, x% stores, 11% branches, 2% jumps, and 52% R-type instructions. ³ Determine the average CPI for this benchmark.

Solution

The average CPI is the sum over each teaching of the CPI for that instruction multiplied by the fraction of the time that didactics is used. For this benchmark, Average CPI = (0.eleven + 0.02)(3) + (0.52 + 0.10)(4) + (0.25)(five) = four.12. This is better than the worst-case CPI of five, which would be required if all instructions took the aforementioned time.

Retrieve that we designed the multicycle processor then that each cycle involved one ALU performance, retention admission, or register file access. Let united states of america presume that the register file is faster than the memory and that writing memory is faster than reading retention. Examining the datapath reveals two possible disquisitional paths that would limit the bicycle time:

(7.4) $T_c=t_{pcq}+t_{mu\mathrm{10}}+\max (t_{ALU}+t_{mux},t_{\mathrm{thousand}em})+t_{\text{setup}}$

The numerical values of these times will depend on the specific implementation technology.

Example 7.8

Processor Functioning Comparison

Ben Bitdiddle is wondering whether he would exist better off building the multicycle processor instead of the single-cycle processor. For both designs, he plans on using a 65 nm CMOS manufacturing process with the delays given in Table 7.6. Help him compare each processor's execution time for 100 billion instructions from the SPECINT2000 benchmark (see Instance 7.7).

Solution

According to Equation vii.4, the bike time of the multicycle processor is T_c2 = xxx + 25 + 250 + 20 = 325 ps. Using the CPI of 4.12 from Example 7.7, the total execution fourth dimension is Τ₂ = (100 × x⁹ instructions)(four.12 cycles/instruction) (325 × 10⁻¹² southward/cycle) = 133.ix seconds. Co-ordinate to Instance vii.4, the unmarried-cycle processor had a cycle time of T_c1 = 925 ps, a CPI of 1, and a full execution fourth dimension of 92.5 seconds.

One of the original motivations for building a multicycle processor was to avoid making all instructions take as long as the slowest i. Unfortunately, this example shows that the multicycle processor is slower than the unmarried-cycle processor given the assumptions of CPI and circuit element delays. The fundamental problem is that even though the slowest instruction, lw, was broken into v steps, the multicycle processor cycle time was not near improved five-fold. This is partly because not all of the steps are exactly the same length, and partly because the 50-ps sequencing overhead of the annals clk-to-Q and setup time must now be paid on every step, non just once for the entire education. In general, engineers have learned that it is difficult to exploit the fact that some computations are faster than others unless the differences are large.

Compared with the single-cycle processor, the multicycle processor is likely to be less expensive because it eliminates two adders and combines the teaching and data memories into a unmarried unit of measurement. It does, however, require five nonarchitectural registers and additional multiplexers.

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Memory optimization and video processing

Jason D. Bakos , in Embedded Systems, 2022

4.13 Performance Results

Effigy 4.18 shows the cache miss rate and CPI for the filter function on the Raspberry Pi. The filter function includes the tile conversion lawmaking and frame buffer output lawmaking. Notation that some tile sizes are unreasonably large equally compared to the frame size of 640   ×   480.

Figure iv.xviii. Cache miss charge per unit and CPI for 2D Sobel edge detection filter for a range of square tile sizes, from viii   ×   8   pixels to 352   ×   352   pixels, and without tiling. This operation includes the role that converts and buffers each frame and the function that writes the frame buffer. Cache miss rate roughly correlates with average CPI. The highest-performing tile was viii   ×   viii, which provided a speedup of 1.7 in miss charge per unit as compared to the nontiled version.

The worst enshroud miss charge per unit occurs when there is no tiling, merely the worst CPI occurs with tile size 288   ×   288. CPI improves slightly when tiling is discontinued. This is likely due to lower education CPI that results from the reduction of executed branch instructions from needing fewer iterations of the tile loops.

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Microarchitecture

Sarah Fifty. Harris , David Harris , in Digital Design and Computer Architecture, 2022

7.5.4 Functioning Assay

The pipelined processor ideally would have a CPI of 1 because a new instruction is issued—that is, fetched—every cycle. Still, a stall or a flush wastes 1 to 2 cycles, and then the CPI is slightly higher and depends on the specific program being executed.

Example vii.9

Pipelined Processor CPI

The SPECINT2000 benchmark considered in Example vii.4 consists of approximately 25% loads, 10% stores, eleven% branches, 2% jumps, and 52% R- or I-type ALU instructions. Assume that forty% of the loads are immediately followed by an instruction that uses the event, requiring a stall, and that fifty% of the branches are taken (mispredicted), requiring 2 instructions to be flushed. Ignore other hazards. Compute the average CPI of the pipelined processor.

Solution

The average CPI is the weighted sum over each teaching of the CPI for that instruction multiplied past the fraction of time that didactics is used. Loads take i clock cycle when there is no dependency and two cycles when the processor must stall for a dependency, so they have a CPI of (0.6)(one) + (0.4)(ii) = 1.4. Branches have one clock cycle when they are predicted properly and three when they are non, so they take a CPI of (0.5)(1) + (0.v)(3) = ii. Jumps take iii clock cycles (CPI = 3). All other instructions have a CPI of 1. Hence, for this benchmark, the average CPI = (0.25)(ane.4) + (0.1)(1) + (0.eleven)(2) + (0.02)(3) + (0.52)(1) = 1.25.

The critical path analysis for the Execute stage assumes that the Chance Unit delay for calculating ForwardAE and ForwardBE is less than or equal to the delay of the Effect multiplexer. If the Gamble Unit delay is longer, it must exist included in the critical path instead of the Result multiplexer delay.

Nosotros tin determine the bicycle time past considering the disquisitional path in each of the v pipeline stages shown in Figure 7.61. Think that the register file is used twice in a single cycle: it is written in the outset half of the Writeback cycle and read in the second half of the Decode cycle; so these stages tin apply only one-half of the cycle time for their disquisitional path. Another way of saying it is this: twice the critical path for each of those stages must fit in a cycle. Effigy 7.62 shows the critical path for the Execute stage. It occurs when a co-operative is in the Execute stage that requires forwarding from the Writeback phase: the path goes from the Writeback pipeline annals, through the Result, ForwardBE, and SrcB multiplexers, through the ALU and AND-OR logic to the PC multiplexer and, finally, to the PC annals.

Figure vii.62. Pipelined processor critical path

(7.five) $T_{c\_pipeli\mathrm{due\; north}\mathrm{east}d}=\mathrm{yard}ax\left[\begin{array}{cc}{t}_{pcq}+t_{me\mathrm{1000}}+t_{\mathrm{southward}\mathrm{eastward}tup}\hfill & Fetch\\ \mathrm{ii}(t_{RFr\mathrm{due\; east}ad}+t_{\mathrm{due\; south}etup})\hfill & D\mathrm{east}cod\mathrm{due\; east}\\ t_{pcq}+4t_{\mathrm{thou}ux}+t_{ALU}+t_{A\mathrm{Northward}D-OR}+t_{\mathrm{southward}etup}\hfill & \mathrm{Eastward}xecute\\ t_{pcq}+t_{mem}+t_{\mathrm{south}\mathrm{east}tup}\hfill & \mathrm{1000}e\mathrm{grand}ory\\ 2(t_{pcq}+t_{\mathrm{thousand}u\mathrm{10}}+t_{RFsetup})\hfill & Writ\mathrm{due\; east}bacm\end{array}\right]$

Example 7.ten

Pipelined Processor Performance Comparing

Ben Bitdiddle needs to compare the pipelined processor performance with that of the single-bicycle and multicycle processors considered in Examples 7.4 and vii.8. The logic delays were given in Table 7.7 (on page 415). Help Ben compare the execution fourth dimension of 100 billion instructions from the SPECINT2000 benchmark for each processor.

Solution

According to Equation 7.5, the cycle fourth dimension of the pipelined processor is T _{c_pipelined} = max[40 + 200 + l, 2(100 + l), 40 + 4(30) + 120 + twenty + fifty, xl + 200 + 50, 2(40 + 30 + 60)] = 350   ps. The Execute stage takes the longest. Co-ordinate to Equation seven.ane, the total execution time is T _pipelined = (100 × x⁹ instructions)(1.25 cycles/pedagogy)(350 × 10⁻¹²  s/cycle) = 44 seconds. This compares with 75 seconds for the single-bike processor and 155 seconds for the multicycle processor.

Our pipelined processor is unbalanced, with branch resolution in the Execute stage taking much longer than whatever other stage. The pipeline could be balanced meliorate by pushing the Result multiplexer back into the Memory stage, reducing the cycle time to 320 ps.

The pipelined processor is essentially faster than the others. However, its reward over the single-bike processor is nowhere nigh the fivefold speedup one might hope to get from a five-stage pipeline. The pipeline hazards introduce a pocket-sized CPI penalty. More than significantly, the sequencing overhead (clk-to-Q and setup times) of the registers applies to every pipeline stage, not just once to the overall datapath. Sequencing overhead limits the benefits one can hope to achieve from pipelining. Imbalanced filibuster in pipeline stages also decreases the benefits of pipelining. The pipelined processor is like in hardware requirements to the single-cycle processor, but information technology adds many 32-bit pipeline registers, along with multiplexers, smaller pipeline registers, and control logic to resolve hazards.

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Durable Phase-Alter Retentivity Architectures

Marjan Asadinia , Hamid Sarbazi-Azad , in Advances in Computers, 2022

7.v Metrics

The metrics used are memory access latency, arrangement performance (Cycles per Educational activity, CPI), energy dissipation, and lifetime, for a broad range of device densities including two-bit, 3-bit, and 4-bit MLCs (2-bit prototypes are now getting pop and 3-fleck and iv-fleck products are projected to be released in near future).

For energy dissipation, CACTI gives the static ability and energy dissipation per each access. So, nosotros can multiply all accesses by the energy of each access, and then split up it by the simulation cycles and get the dynamic power of the retention system. Again, Table 5 illustrates the label of the evaluated workloads based on intensity of their value locality for the baseline arrangement in Table 3.

(one) ${\text{Total Power}}_{\text{retention unit}}={\text{Leakage Power}}_{\text{memory unit}}+\frac{\#\mathit{Write}\mathit{Acc}.\times \mathit{Energy}\mathit{per}\mathit{Write}\mathit{Acc}.+\#\mathit{Read}\mathit{Acc}.\times \mathit{Energy}\mathit{per}\mathit{Read}\mathit{Acc}.}{\mathit{\text{Simulation Cycles}}}.$

The main endurance metrics used for the evaluated systems are time-to-failure in synthetic analysis and memory lifetime in real workloads. Time-to-failure is defined as the time elapsed between the arrangement startup time to the fourth dimension PCM capacity is reduced to less than 50% of its maximum capacity (i.e., 2   GB in our 4   GB arrangement). For memory lifetime, we have the same limit for defining organisation downtime. We presume the number of reliable writes onto a 2-bit MLC PCM cell is limited to 10⁶ [11] and past having perfect clothing-leveling to simplify the lifetime assay; and then, we have:

(2) $\text{Lifetime}{\left(\text{Twelvemonth}\right)}_{\text{Memoryunit}}=\frac{\#\text{Maximum reliable write counts}}{\left(\frac{\#\text{Measured writes}}{\text{Simulation cycles}}\right)}\times \frac{365\times 24\times 3600}{\mathrm{f}\left(\mathrm{Hz}\right)}$

where f is the processor frequency stock-still to 2.v   ×   10⁹  Hz in our experiments.

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Analysis of Toll and Performance

Bruce Jacob , ... David T. Wang , in Memory Systems, 2008

28.3.4 The Moral of the Story

So what is the real answer? The auto travels 160 miles, consuming 6.i gallons; it is not difficult to find the actual miles per gallon achieved.

(EQ 28.6) $\frac{160\text{miles}}{\mathrm{six.one}\text{gallons}}=26.2\text{mpg}$

The approach that is perhaps the least intuitive (sampling over the space of gallons?) does give the correct answer. We see that, if the metric we are measuring is miles per gallon,

•

Sampling over minutes (fourth dimension) is bad.

•

Sampling over miles (distance) is bad.

•

Sampling over gallons (consumption) is good.

Moreover (and perhaps most importantly), in this context, bad means "can be off by a factor of 2 or more."

The moral of the story is that if you are sampling the following metric:

then you must sample that metric in equal steps of dimension unit of measurement. To wit, if sampling the metric miles per gallon, you lot must sample evenly in units of gallon; if sampling the metric cycles per instruction, you must sample evenly in units of instruction (i.e., evenly in instructions committed, non instructions fetched or executed ¹ ); if sampling the metric instructions per bike, you lot must sample evenly in units of cycle; and if sampling the metric cache-miss rate (i.eastward., enshroud misses per cache access), you must sample evenly in units of cache access.

What does information technology mean to sample in units of instruction or bike or cache access? For a microprocessor, it means that one must have a inaugural timer that decrements every unit of measurement, i.east., one time for every instruction committed or in one case every cycle or one time every time the cache is accessed, and on every epoch (i.due east., whenever a predefined number of units accept transpired) the desired average must be taken. For an automobile providing existent-time fuel efficiency, a sensor must be placed in the gas line that interrupts a controller whenever a predefined unit of volume of gasoline is consumed.

What determines the predefined amounts that set the epoch size? Clearly, to take hold of all interesting beliefs 1 must sample oft enough to measure out all important events. Higher sampling rates lead to better accuracy at a higher toll of implementation. How does sampling at a lower rate touch on one's accuracy? For example, past sampling at a rate of once every 1/xxx gallon in the previous case, we were assured of communicable every segment of the trip. All the same, this was a contrived example where we knew the desired sampling rate ahead of time. What if, as in normal cases, one does non know the advisable sampling rate? For example, if the example algorithm sampled every gallon instead of every minor fraction of a gallon, we would have gotten the following results:

(EQ 28.8) $\frac{2}{\mathrm{half\; dozen}}30+\frac{2}{\mathrm{half-dozen}}10+\frac{2}{6}\mathrm{thirty}=\mathrm{23.iii}\text{mpg}$

The answer is off the true event, just it is not as bad every bit if we had generated the sampled boilerplate incorrectly in the first place (east.g., sampling in minutes or miles traveled).

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The Linux/ARM embedded platform

Jason D. Bakos , in Embedded Systems, 2022

1.xi Performance Results

Compile the code using the –O3 flag and run it on your platform.

Tabular array 1.three shows the memory bandwidth results for an ARM11, ARM Cortex A9, and ARM Cortex A15. For each processor, write bandwidth is approximately three times that of read bandwidth. The differences in CPI and miss charge per unit shed some light on the reasons for this deviation. The higher CPI and miss rate of the read examination indicates that the enshroud does not block the CPU or register a cache miss as often when writing, probably because the cache does non allocate infinite in the cache on a write miss, and a write miss is only triggered when all the write buffers are full.

Tabular array 1.three. Results of Memory Bandwidth Test

	Raspberry Pi	Avnet Zedboard	NVIDIA Jetson Tegra TK1
CPU	ARM11	Dual Cortex A9	Quad Cortex A15
Read B/W	140   MB/due south	347   MB/s	two.94   GB/south
CPI	4.69	1.83	0.72
Miss rate	9.21%	xi.9%	six.26%
Write B/West	325   MB/southward	i.67   GB/south	11.2   GB/s
CPI	2.70	0.51	0.67
Miss rate	1.64%	28.seven%	0.00%

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Retentiveness Systems

Sarah 50. Harris , David Money Harris , in Digital Blueprint and Figurer Architecture, 2022

(a)

The educational activity cache is perfect (i.eastward., always hits) just the data cache has a xv% miss charge per unit. On a enshroud miss, the processor stalls for 200   ns to access master memory, then resumes normal operation. Taking cache misses into business relationship, what is the average memory access fourth dimension?

(b)

How many clock cycles per pedagogy (CPI) on average are required for load and store give-and-take instructions because the non-platonic memory system?

(c)

Consider the benchmark awarding of Instance 7.7 that has 25% loads, 10% stores, 11% branches, 2% jumps, and 52% R-type instructions. Taking the non-ideal retention organisation into business relationship, what is the average CPI for this benchmark?

(d)

Now suppose that the pedagogy enshroud is too non-ideal and has a 10% miss charge per unit. What is the boilerplate CPI for the criterion in part (c)? Take into account both pedagogy and data enshroud misses.

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Multicore and information-level optimization

Jason D. Bakos , in Embedded Systems, 2022

ii.half-dozen Functioning Analysis

Table ii.ane shows the performance results of our naïve kernel implementation on all three platforms. Even when using maximum compiler optimization the compiler only achieves x-18% of the performance bound! The performance counters can provide some insight into the program'southward implementation problems.

Table 2.1. Operation of Horner Code without Programmer-Supplied Optimizations

	Raspberry Pi	Xilinx Zedboard	NVIDIA Jetson TK1
CPU	ARM11	Dual Cortex A9	Quad Cortex A15
Average B/Westward	233   MB/southward	i.01   GB/s	7.07   GB/s
B/W jump	408   Mflops/southward	i.77   Gflops/s	12.37   Gflops/s
No optimization
Observed throughput/efficiency	12.xiii   Mflops	27.91   Mflops	63.07   Mflops
	2.97% efficiency	one.58% efficiency	0.51% efficiency
Effective memory B/W	6.61   MB/south	15.21   MB/southward	63.07   MB/s
CPI	2.78	1.84	2.91
Cache miss rate	23.61%	3.27%	1.91%
Instructions per bomb	20.8	25.86	26.47
Maximum optimization (-O3)
Observed throughput/efficiency	74.01   Mflops	212.59   Mflops	2209.38   Mflops
	18.i% efficiency	12.0% efficiency	17.9% efficiency
Effective memory B/W	40.33   MB/due south	115.85   MB/southward	1204.02   MB/due south
CPI	four.73	i.77	1.15
Cache miss charge per unit	38.nine%	0.77%	0.46%
Instructions per flop	2.00	iii.55	3.51

Memory bandwidth: Since the kernel is retentiveness bandwidth spring, the performance efficiency will friction match our memory bandwidth efficiency, so the effective retentivity bandwidth is not shown in subsequent tables.

CPI: The platonic CPI is 1 for the ARM11 and 0.v for the Cortex A9/A15. Our observed CPIs are 3 to six times this. This may exist caused past unsatisfactory cache operation or unsatisfactory instruction scheduling the compiler, processor, or both.

Cache miss rate: Miss rate measures cache operation and determines the average latency of a memory instruction. Every bit such, it gives an thought of how much the CPI is influenced by cache performance. Miss charge per unit is adamant past the locality of the kernel's admission pattern. Both the d and x arrays are accessed with both spatial and temporal locality (each element is accessed repeatedly and consecutively), so information technology is reasonable to expect the data cache to perform well for this kernel.

Instructions per bomb: This metric is some other manner to express the number of instructions executed, and is affected by how efficiently the compiler translates the high-level lawmaking.

In order to better the kernel the programmer requires more than control over its implementation.

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