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## Today’s class

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**Today’s class**• Floating point numbers • Computer systems organization Computer Architecture I - Class 6**Principles of Floating Point**• Must separate range from precision • Use scientific notation n = f × 10e • f is the fraction or mantissa • e is the exponent (a positive or negative integer) • Examples3.14 = 0.314 × 101 = 3.14 × 1000.000001 = 0.1 × 10−5 = 1.0 × 10−61941 = 0.1941 × 104 = 1.941 × 103 Computer Architecture I - Class 6**Seven Regions of Real Number Line**• Large negative numbers less than −0.999 × 1099 • Negative numbers between −0.999 × 1099 and −0.100 × 10−99 • Small negative numbers, magnitudes less than 0.100 × 10−99 • Zero • Small positive numbers, magnitudes less than 0.100 × 10−99 • Positive numbers between 0.100 × 10−99 and 0.999 × 1099 • Large positive numbers greater than 0.999 × 1099 Computer Architecture I - Class 6**Seven Regions of Real Number Line**Computer Architecture I - Class 6**Approximate Bounds of Expressible Floating-Point Numbers**(Unnormalized) Computer Architecture I - Class 6**Examples of normalized floating-point numbers.**Normalized numbers Computer Architecture I - Class 6**IEEE Floating-Point Standard 754**IEEE floating-point formats. (a) Single precision. (b) Double precision. Computer Architecture I - Class 6**IEEE Floating-point Standard 754**Characteristics of IEEE floating-point numbers. Computer Architecture I - Class 6**IEEE Numerical Types**Computer Architecture I - Class 6**In-Class Exercise**• Convert the following numbers to IEEE single-precision format. Give the results as 8 hexadecimal digits. • 9.0 • -6.125 • Convert the following IEEE single-precision floating-point numbers from hex to decimal: • 42E48000 • 3F880000 Computer Architecture I - Class 6**Organization of a simple computer**Computer Architecture I - Class 6**CPU Organization**Computer Architecture I - Class 6**Instructions**• Register-memory • Allow memory words to be fetched into registers • Allow registers to be stored to memory words • Register-register • Fetches two operands from the registers • Brings them to the ALU input registers • Performs some operation on them • Stores the result back in one of the registers Computer Architecture I - Class 6**Instruction Execution Steps**• Fetch next instruction from memory into instruction register • Change program counter to point to next instruction • Determine type of instruction just fetched • If instruction uses a word in memory, determine where it is • Fetch word, if needed, into CPU register • Execute the instruction • Go to step 1 to begin executing the following instruction Computer Architecture I - Class 6**Design Principles for Modern Computers**• All instructions are directly executed by hardware • Maximize the rate at which instructions are issued • Instructions should be easy to decode • Only loads and stores should reference memory • Provide plenty of registers Computer Architecture I - Class 6**Instruction-Level Parallelism**Computer Architecture I - Class 6**Pipelining effect**• Suppose the cycle time of the machine is 2 ns • For the 5-stage pipeline shown in the previous slide, it takes 10 ns for an instruction to completely execute • It appears this machine can run at 100 MIPS • However, it’s much better than this • At every clock cycle, starting with the 5th, one instruction completes execution • This gives an actual processing rate of 500 MIPS Computer Architecture I - Class 6**Dual Pipelines**• If one pipeline is good, two must be better! Computer Architecture I - Class 6**Superscalar Architecture**Computer Architecture I - Class 6**Processor-Level Parallelism**Computer Architecture I - Class 6**Multiprocessors**A single-bus multiprocessor A multicomputer with local memories Computer Architecture I - Class 6**Memory**• Most basic unit of memory is the bit, a 0 or 1 (binary digit) • Smallest addressable unit of memory in a computer is the byte (containing 8 bits) • Words are groups of bytes, typically 4 bytes (32 bits) to a word Computer Architecture I - Class 6**Big- and Little-Endian**• Bytes can be numbered left-to-right or right-to left Computer Architecture I - Class 6**Byte Ordering**Computer Architecture I - Class 6**Parity**• Memories can have occasional errors • Errors will be bit inversions • Can detect one-bit errors by adding an extra bit called a parity bit • This bit is set or cleared so that the total number of 1’s in the word is either even or odd, depending on which type of parity is being used Computer Architecture I - Class 6**Codewords**• Let one memory word consist of m data bits • Let’s add r redundant (or check or parity) bits • Total length is n bits, where n = m + r • The n-bit unit is called a codeword • Given two codewords you can determine how many bits differ by computing the exclusive or (XOR) of the two codewords and counting how many 1s are in the result • This sum is called the Hamming distance Computer Architecture I - Class 6**How Many Check Bits?**• There are 2m valid memory words • Each valid memory word has n illegal codewords 1 bit away from it • Thus, each of the 2m valid words needs n+1 bit patterns dedicated to it • Total number of bit patterns is 2n • So, (n+1)2m≤ 2n, or (m+r+1) ≤ 2r Computer Architecture I - Class 6**Number of check bits for a code that can correct a single**error Computer Architecture I - Class 6**Error Correcting Codes**(a) Encoding of 1100 (b) Even parity added (c) Error in AC Computer Architecture I - Class 6**Hamming Codes**• 16-bit word needs 5 parity bits • The parity bits are the bits whose positions are powers of 2 • Bit b is checked by the parity bits whose positions sum to b • For example, bit 6 is checked by bits 2 and 4 Computer Architecture I - Class 6**Cache Memory**Computer Architecture I - Class 6**Memory Hierarchy**Computer Architecture I - Class 6**Magnetic Disk Format**Computer Architecture I - Class 6**Disk Platters**Computer Architecture I - Class 6**Disk Performance**• First, arm must be moved to the correct track; this is called a seek • Average seek times between random tracks are in the range 5-10 ms • Then we have the rotational delay to find the correct sector • Average rotational delay is 3-6 ms • Transfer time depends on the linear density and rotation speed • Typical transfer times for one sector are 13-26 μs • Clearly seek time and rotational delay dominate the transfer time Computer Architecture I - Class 6**Zones**• Outer tracks of a disk have larger circumferences than inner tracks • Thus linear density is lower the further out you go • To increase capacity the disk is divided into zones (there are 5 zones to the right) • The number of sectors per track is increased as you move from zone to zone going outward Computer Architecture I - Class 6**CD-ROM**Computer Architecture I - Class 6**CD-ROM Logical Data Layout**Computer Architecture I - Class 6**DVD**• Smaller pits than CD (0.4 microns vs. 0.8 microns for CD) • Tighter spiral (0.74 microns vs. 1.6 microns for CD) • Capacity is 4.7 GB Computer Architecture I - Class 6**Input/Output**Computer Architecture I - Class 6**Buses**Computer Architecture I - Class 6