Morgan Kaufmann Publishers April 25, 2017 Technology Trends Electronics technology continues to evolve Increased capacity and performance Reduced cost DRAM capacity Year Technology Relative performance/cost 1951 Vacuum tube 1 1965 Transistor 35 1975 Integrated circuit (IC) 900 1995 Very large scale IC (VLSI) 2,400,000 2013 Ultra large scale IC 250,000,000,000 Chapter 1 — Computer Abstractions and Technology — 1 Chapter 1 — Computer Abstractions and Technology

Microprocessor Complexity Gordon Moore Intel Cofounder # of transistors on an IC 2X Transistors / Chip Every 1.5 years Called “Moore’s Law” Year

Memory Capacity (Single-Chip DRAM) year size (Mbit) 1980 0.0625 1983 0.25 1986 1 1989 4 1992 16 1996 64 1998 128 2000 256 2002 512 2004 1024 (1Gbit) 2006 2048 (2Gbit) Bits Now 1.4X/yr, or 2X every 2 years 8000X since 1980! Year

Computer Technology – Dramatic Change! Memory DRAM capacity: 2x / 2 years (since ‘96); 64x size improvement in last decade Processor Speed 2x / 1.5 years (since ‘85); [slowing!] 100X performance in last decade Disk Capacity: 2x / 1 year (since ‘97) 250X size in last decade

Performance Metrics Purchasing perspective given a collection of machines, which has the best performance ? least cost ? best cost/performance? Design perspective faced with design options, which has the best performance improvement ? Both require basis for comparison metric for evaluation Our goal is to understand what factors in the architecture contribute to overall system performance and the relative importance (and cost) of these factors Or smallest/lightest Longest battery life Most reliable/durable (in space)

Morgan Kaufmann Publishers April 25, 2017 Defining Performance Which airplane has the best performance? Chapter 1 — Computer Abstractions and Technology — 6 Chapter 1 — Computer Abstractions and Technology

Response Time and Throughput Morgan Kaufmann Publishers April 25, 2017 Response Time and Throughput Response time How long it takes to do a task Throughput Total work done per unit time e.g., tasks/transactions/… per hour How are response time and throughput affected by Replacing the processor with a faster version? Adding more processors? We’ll focus on response time for now… Chapter 1 — Computer Abstractions and Technology — 7 Chapter 1 — Computer Abstractions and Technology

Morgan Kaufmann Publishers April 25, 2017 Relative Performance Define Performance = 1/Execution Time “X is n time faster than Y” Example: time taken to run a program 10s on A, 15s on B Execution TimeB / Execution TimeA = 15s / 10s = 1.5 So A is 1.5 times faster than B Chapter 1 — Computer Abstractions and Technology — 8 Chapter 1 — Computer Abstractions and Technology

Measuring Execution Time Morgan Kaufmann Publishers April 25, 2017 Measuring Execution Time Elapsed time Total response time, including all aspects Processing, I/O, OS overhead, idle time Determines system performance CPU time Time spent processing a given job Discounts I/O time, other jobs’ shares Comprises user CPU time and system CPU time Different programs are affected differently by CPU and system performance Chapter 1 — Computer Abstractions and Technology — 9 Chapter 1 — Computer Abstractions and Technology

Morgan Kaufmann Publishers April 25, 2017 CPU Clocking Operation of digital hardware governed by a constant-rate clock Clock period Clock (cycles) Data transfer and computation Update state Clock period: duration of a clock cycle e.g., 250ps = 0.25ns = 250×10–12s Clock frequency (rate): cycles per second e.g., 4.0GHz = 4000MHz = 4.0×109Hz Chapter 1 — Computer Abstractions and Technology — 10 Chapter 1 — Computer Abstractions and Technology

Review: Machine Clock Rate Clock rate (clock cycles per second in MHz or GHz) is inverse of clock cycle time (clock period) CC = 1 / CR one clock period 10 nsec clock cycle => 100 MHz clock rate 5 nsec clock cycle => 200 MHz clock rate 2 nsec clock cycle => 500 MHz clock rate 1 nsec (10-9) clock cycle => 1 GHz (109) clock rate 500 psec clock cycle => 2 GHz clock rate 250 psec clock cycle => 4 GHz clock rate 200 psec clock cycle => 5 GHz clock rate A clock cycle is the basic unit of time to execute one operation/pipeline stage/etc.

Morgan Kaufmann Publishers April 25, 2017 CPU Time Performance improved by Reducing number of clock cycles Increasing clock rate Hardware designer must often trade off clock rate against cycle count Chapter 1 — Computer Abstractions and Technology — 12 Chapter 1 — Computer Abstractions and Technology

Morgan Kaufmann Publishers April 25, 2017 CPU Time Example Computer A: 2GHz clock, 10s CPU time Designing Computer B Aim for 6s CPU time Can do faster clock, but causes 1.2 × clock cycles How fast must Computer B clock be? Chapter 1 — Computer Abstractions and Technology — 13 Chapter 1 — Computer Abstractions and Technology

Instruction Count and CPI Morgan Kaufmann Publishers April 25, 2017 Instruction Count and CPI Instruction Count for a program Determined by program, ISA and compiler Average cycles per instruction Determined by CPU hardware If different instructions have different CPI Average CPI affected by instruction mix Chapter 1 — Computer Abstractions and Technology — 14 Chapter 1 — Computer Abstractions and Technology

Morgan Kaufmann Publishers April 25, 2017 CPI Example Computer A: Cycle Time = 250ps, CPI = 2.0 Computer B: Cycle Time = 500ps, CPI = 1.2 Same ISA Which is faster, and by how much? A is faster… …by this much Chapter 1 — Computer Abstractions and Technology — 15 Chapter 1 — Computer Abstractions and Technology

Morgan Kaufmann Publishers April 25, 2017 CPI in More Detail If different instruction classes take different numbers of cycles Weighted average CPI Relative frequency Chapter 1 — Computer Abstractions and Technology — 16 Chapter 1 — Computer Abstractions and Technology

Morgan Kaufmann Publishers April 25, 2017 CPI Example Alternative compiled code sequences using instructions in classes A, B, C Class A B C CPI for class 1 2 3 IC in sequence 1 IC in sequence 2 4 Sequence 1: IC = 5 Clock Cycles = 2×1 + 1×2 + 2×3 = 10 Avg. CPI = 10/5 = 2.0 Sequence 2: IC = 6 Clock Cycles = 4×1 + 1×2 + 1×3 = 9 Avg. CPI = 9/6 = 1.5 Chapter 1 — Computer Abstractions and Technology — 17 Chapter 1 — Computer Abstractions and Technology

A Simple Example  = Op Freq CPIi Freq x CPIi ALU 50% 1 . Load 20% 5 Store 10% 3 Branch 2  = How much faster would the machine be if a better data cache reduced the average load time to 2 cycles? How does this compare with using branch prediction to shave a cycle off the branch time? What if two ALU instructions could be executed at once?

A Simple Example  = Op Freq CPIi Freq x CPIi ALU 50% 1 Load 20% 5 Store 10% 3 Branch 2  = .5 1.0 .3 .4 .5 .4 .3 .5 1.0 .3 .2 .25 1.0 .3 .4 2.2 1.6 2.0 1.95 How much faster would the machine be if a better data cache reduced the average load time to 2 cycles? How does this compare with using branch prediction to shave a cycle off the branch time? What if two ALU instructions could be executed at once? For lecture CPU time new = 1.6 x IC x CC so 2.2/1.6 means 37.5% faster CPU time new = 2.0 x IC x CC so 2.2/2.0 means 10% faster CPU time new = 1.95 x IC x CC so 2.2/1.95 means 12.8% faster

Determinates of CPU Performance CPU time = Instruction_count x CPI x clock_cycle Instruction_ count CPI clock_cycle Algorithm Programming language Compiler ISA Core organization Technology For class handout

Determinates of CPU Performance CPU time = Instruction_count x CPI x clock_cycle Instruction_ count CPI clock_cycle Algorithm Programming language Compiler ISA Core organization Technology X X X X X X For lecture X X X X X X

Morgan Kaufmann Publishers April 25, 2017 Performance Summary The BIG Picture Performance depends on Algorithm: affects IC, possibly CPI Programming language: affects IC, CPI Compiler: affects IC, CPI Instruction set architecture: affects IC, CPI, Tc Chapter 1 — Computer Abstractions and Technology — 22 Chapter 1 — Computer Abstractions and Technology

Morgan Kaufmann Publishers April 25, 2017 Power Trends In CMOS IC technology ×40 5V → 1V ×1000 Chapter 1 — Computer Abstractions and Technology — 23 Chapter 1 — Computer Abstractions and Technology

Morgan Kaufmann Publishers April 25, 2017 Reducing Power Suppose a new CPU has 85% of capacitive load of old CPU 15% voltage and 15% frequency reduction The power wall We can’t reduce voltage further We can’t remove more heat How else can we improve performance? Chapter 1 — Computer Abstractions and Technology — 24 Chapter 1 — Computer Abstractions and Technology