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Lecture 2: Computer Performance Examples

& Cost

Dr. Norafida Ithnin Advanced Computer System & Architecture

MCC 2313

ComparingCPUTimeCPU time = Seconds = Instructions x Cycles x Seconds

Program Program Instruction Cycle

–  A 500 MHz Pentium III processor takes 2 ms to run a program with 200,000 instructions.

–  A 300 MHz UltraSparc processor takes 1.8 ms to run the same program with 230,000 instructions.

–  What is the CPI for each processor for this program? CPI = Cycles / Instruction Count

= CPU time X Clock Rate / Instruction Count CPIPentium = 2*10-3 X 500*106 / 2*105 = 5.00 CPISPARC = 1.8*10-3 X 300*106 / 2.3*105 = 2.35

–  Which processor is faster and by how much? The UltraSparc is 2/1.8 = 1.11 times as fast, or 11% faster.

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CyclesPerInstruc6on

CPU time = CycleTime * Σ CPI * IC i = 1

n

i i

CPI = Σ CPI * F where F = IC i = 1

n

i i i i Instruction Count

“Instruction Frequency”

Investresourceswhere6meisspent!

CPU Time = Cycle Time * Number of cycles

“Average Cycles per Instruction”

Example:Calcula6ngCPI

Typical Mix

Base Machine (Reg / Reg) Op Freq Cycles Fi*CPIi (% Time) ALU 50% 1 .5 (33%) Load 20% 2 .4 (27%) Store 10% 2 .2 (13%) Branch 20% 2 .4 (27%) 1.5

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Base Machine (Reg / Reg) Op Freq Cycles ALU 50% 1 Load 20% 2 Store 10% 2 Branch 20% 2

Typical Mix

ExampleAdd register / memory ALU operations:

–  One source operand in memory –  One source operand in register –  Cycle count of 2

Branch cycle count to increase to 3. What fraction of the loads must be eliminated for this to pay off, assuming the clock rate is not affected?

ExampleSolu6onExecTime=InstrCntxCPIxClock

Op Freq Cycles ALU .50 1 .5 Load .20 2 .4 Store .10 2 .2 Branch .20 2 .4 Reg/Mem 1.00 1.5

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ExampleSolu6onExec Time = Instr Cnt x CPI x Clock Op Freq Cycles Freq Cycles ALU .50 1 .5 .5 – X 1 .5 – X Load .20 2 .4 .2 – X 2 .4 – 2X Store .10 2 .2 .1 2 .2 Branch .20 2 .4 .2 3 .6 Reg/Mem X 2 2X

1.00 1.5 1 – X (1.7 – X)/(1 – X)

CPINew must be normalized to new instruction frequency X is the fraction of register-memory instructions.

CyclesNew

InstructionsNew

ExampleSolu6onExec Time = Instr Cnt x CPI x Clock Op Freq Cycles Freq Cycles ALU .50 1 .5 .5 – X 1 .5 – X Load .20 2 .4 .2 – X 2 .4 – 2X Store .10 2 .2 .1 2 .2 Branch .20 2 .4 .2 3 .6 Reg/Mem X 2 2X

1.00 1.5 1 – X (1.7 – X)/(1 – X)

Instr CntOld x CPIOld x ClockOld = Instr CntNew x CPINew x ClockNew

1.00 x 1.5 = (1 – X) x (1.7 – X)/(1 – X)

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ExampleSolu6onExec Time = Instr Cnt x CPI x Clock Op Freq Cycles Freq Cycles ALU .50 1 .5 .5 – X 1 .5 – X Load .20 2 .4 .2 – X 2 .4 – 2X Store .10 2 .2 .1 2 .2 Branch .20 2 .4 .2 3 .6 Reg/Mem X 2 2X

1.00 1.5 1 – X (1.7 – X)/(1 – X)

InstrCntOldxCPIOldxClockOld=InstrCntNewxCPINewxClockNew1.00x1.5=(1–X)x(1.7–X)/(1–X)1.5=1.7–X0.2=XALLloadsmustbeeliminatedforthistobeawin!

IC cost = Die cost + Testing cost + Packaging cost Final test yield

Final test yield: The fraction of packaged dies which pass the final testing state.

IntegratedCircuitsCosts

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ICcost=Diecost+Tes6ngcost+PackagingcostFinaltestyieldDiecost=WafercostDiesperWafer*Dieyield

Dieyield:Thefrac6onofgooddiesonawafer,beforepackaging.

IntegratedCircuitsCosts

IC cost = Die cost + Testing cost + Packaging cost Final test yield Die cost = Wafer cost Dies per Wafer * Die yield Dies Π*( Wafer_diam/2)2 Π*Wafer_diam) Wafer Die Area Sqrt(2*Die Area)

IntegratedCircuitsCosts

– Test dies – =

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IC cost = Die cost + Testing cost + Packaging cost Final test yield Die cost = Wafer cost Dies per Wafer * Die yield Dies per wafer = Π * ( Wafer_diam / 2)2 – Π * Wafer_diam – Test dies Die Area Sqrt(2*Die Area) Die Yield = Wafer yield* {1+Defects_per_unit_area * Die_Area}

IntegratedCircuitsCosts

α

- α

Defects per unit area (0.6 to 1.2), Fabrication complexity: a (about 3) Die cost increases roughly as die area4

RealWorldExamples

Chip Metal Line Wafer Defect Area Dies/ Yield Die Cost layers width cost /cm2 mm2 wafer 386DX 2 0.90 $900 1.0 43 360 71% $4 486DX2 3 0.80 $1200 1.0 81 181 54% $12 PowerPC 601 4 0.80 $1700 1.3 121 115 28% $53 HP PA 7100 3 0.80 $1300 1.0 196 66 27% $73 DEC Alpha 3 0.70 $1500 1.2 234 53 19% $149 SuperSPARC 3 0.70 $1700 1.6 256 48 13% $272 Pen6um 3 0.80 $1500 1.5 296 40 9% $417

– From "Estimating IC Manufacturing Costs,” by Linley Gwennap, Microprocessor Report, August 2, 1993, p. 15 –  New products end up being much more expensive to manufacture

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DieTestCost=TestCost*Ave.TestTimeDieYieldPackagingCost:dependsonpins,heatdissipa6on,appearance,...

OtherCosts

Chip Die Package Test & Total

cost pins type cost Assembly 386DX $4 132 QFP $1 $4 $9 486DX2 $12 168 PGA $11 $12 $35 PowerPC 601 $53 304 QFP $3 $21 $77 HP PA 7100 $73 504 PGA $35 $16 $124 DEC Alpha $149 431 PGA $30 $23 $202 SuperSPARC $272 293 PGA $20 $34 $326 Pentium $417 273 PGA $19 $37 $473

•  Assume purchase 10,000 units

ChipPrices(August1993):AnExamples

Chip Area Mfg. Price Multi- Comment mm2 cost plier

386DX 43 $9 $31 3.4 Intense Competition 486DX2 81 $35 $245 7.0 No Competition PowerPC 601 121 $77 $280 3.6 Gain market share DEC Alpha 234 $202 $1231 6.1 Recoup R&D Pentium 296 $473 $965 2.0 Early in shipments

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386 386

486 486

Pentium(R) Pentium(R) MMX

Pentium Pro (R)

Pentium II (R)

1

10

100

1.5µ 1µ 0.8µ 0.6µ 0.35µ 0.25µ 0.18µ 0.13µ

Max

Pow

er (W

atts

) ?

PowerDissipa6on

➊  Lead processor power increases every generation ➋  Compactions provide higher performance at lower power

Sour

ce: I

ntel

Worksta6onCosts

•  DRAM: 50%to55%•  ColorMonitor: 15%to20%•  CPUboard: 10%to15%•  Harddisk: 8%to10%•  CPUcabinet: 3%to5%•  Video&otherI/O: 3%to7%•  Keyboard,mouse: 1%to2%

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Volumevs.Cost•  Rule of thumb on applying learning curve to

manufacturing: “When volume doubles, costs reduce 10%”

A DEC View of Computer Engineering by C. G. Bell, J. C. Mudge, and J. E. McNamara, Digital Press, Bedford, MA., 1978.

•  40 MPPs @ 200 nodes = 8,000 nodes/year vs. 100,000 Workstations/year

2X = (100,000/8,000) => x = 3.6 •  Since doubling value reduces cost by 10%, costs

reduces to (0.9)3.6 = 0.68 of the original price (about 1/3 less expensive).

Volumevs.Cost:PCsvs.Worksta6ons

1990 1992 1994 1997PC23,880,89833,547,589 44,006,000 65,480,000WS 407,624 584,544 679,320 978,585Ra6o 59 57 65 67•  2x=65=>X=6.0and(0.9)6.0=0.53

-PCcostsare47%lessthanworksta6oncostsforwholemarket.

Singlecompany:20%WSmarketvs.10%PCmarketRa6o 29 29 32 33•  2x=32=>X=5.0and(0.9)5.0=0.59

-PCscost41%lessthanworksta6onsforsinglecompany.

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LearningCurve

Years

production costs

volume

time to introduce new product

HighMarginsonHigh-EndMachines•  R&D considered return on investment

–  Most companies spend 4% to 12% of income on R&D. –  Every $1 R&D must generate $8 to $25 in sales

•  High end machines need more $ for R&D •  Sell fewer high end machines

–  Fewer to amortize R&D –  Much higher cost margins

•  Cost of 1 MB Memory (January 1994): PC $40 (Mac Quadra) WS $42 (SS-10) Mainframe $1920 (IBM 3090) Supercomputer $600 (M90 DRAM)

$1375 (C90 15 ns SRAM)

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Cost/PerformanceWhatisRela6onshipofCosttoPrice?

•  ComponentCosts•  DirectCosts(add25%to40%)recurringcosts:labor,purchasing,scrap,

warranty•  GrossMargin(add82%to186%)nonrecurringcosts:

R&D,marke6ng,sales,equipmentmaintenance,rental,financingcost,pretaxprofits,taxes

•  AverageDiscounttogetListPrice(add33%to66%):volumediscountsand/orretailermarkup

Component Cost

Direct Cost

Gross Margin

Average Discount

Avg. Selling Price

List Price

15% to 33% 6% to 8% 34% to 39%

25% to 40%

Summary:Pricevs.Cost

0%

20%

40%

60%

80%

100%

Mini W/S PC

Average Discount

Gross Margin

Direct Costs

Component Costs

0

1

2

3

4

5

Mini W/S PC

Average Discount

Gross Margin

Direct Costs

Component Costs

4.73.8

1.8

3.52.5

1.5

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