PRINCIPLES OF COMPUTER ARCHITECTURE phần 10 pdf
... PROBLEMS 3084: 100 00001 1100 0011 1 1100 000 0000 0100 3088: 100 0101 0 100 00000 0100 0000 00000011 3092: 000 0101 0 100 00000 00000000 00000011 3096: 100 0101 0 100 00001 0 1100 000 00000001 3100 : 100 00001 1100 0011 1 1100 000 ... 2.17 (107 .15) 10 = 1101 011.0 0100 1100 1100 1100 0 100 0111 1101 0 110 0100 1100 1100 1100 2.18 (a) +1.011 × 2 4 (b) -1....
Ngày tải lên: 14/08/2014, 20:21
... to Binary (base 2) 0 1 10 11 100 101 110 111 100 0 100 1 101 0 101 1 1100 1101 1 110 1111 Octal (base 8) 0 1 2 3 4 5 6 7 10 11 12 13 14 15 16 17 Decimal (base 10) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Hexadecimal (base ... Network Architecture 396 9.6.4 Outlook on ATM 398 10 TRENDS IN COMPUTER ARCHITECTURE 403 10. 1 Q UANTITATIVE ANALYSES OF PROGRAM EXECUTION...
Ngày tải lên: 14/08/2014, 20:21
... base -10 numbers to base-2: 37 470 1220 17 99 100 101 1 1101 0 110 100 1100 0100 100 01 1100 011 3.6 Convert the following base-2 numbers to base -10: 00000111 7 101 0101 0 170 001 1100 1 57 0101 0101 85 00 1100 11 51 3.7 ... initials in capital letters. CHR C 67 100 0011 H 72 100 1000 R 82 101 0 010 3.9 Referring to the list of Intel x86 instructions in this chapter, arrange a...
Ngày tải lên: 12/08/2014, 21:22
Schaum’s Outline Series OF Principles of Computer Science phần 10 potx
... development of timesharing computers, and has remained an important part of scheduling for interactive computer systems? The round robin scheduler was the basis of the first timesharing computers. ... undemand- ing of high performance. ANSWERS TO REVIEW QUESTIONS 213 This page intentionally left blank 219 Aiken, Howard, 9 algorithm, 2, 14 order of growth of (see order of grow...
Ngày tải lên: 12/08/2014, 21:22
PRINCIPLES OF COMPUTER ARCHITECTURE phần 2 doc
... precision: x 3 = x 2 (2 − x 2 b) = (1.0 1100 10 1100 10 0101 1101 )(2 − (1.0 1100 10 1100 10 0101 1101 )( .101 1011)) = 1.0 1101 00000 0100 1 = (1.40652466) 10 . The precise value is (1.40659341) 10 , but our 16-bit value is as ... encoding of BCD digits. 90 CHAPTER 3 ARITHMETIC x 2 = x 1 (2 − x 1 b) = 1. 0100 101 (10 − (1. 0100 101)( .101 1011)) = 1.0 1100 10 1100 10 0101 1101 ....
Ngày tải lên: 14/08/2014, 20:21
PRINCIPLES OF COMPUTER ARCHITECTURE phần 3 pps
... 6176 6101 0 010 6a61 7661 2f6c 00b0 616e 672f 4f62 6a65 6374 0100 046d 6169 00c0 6e00 2100 0100 0200 0000 0000 0200 0900 00d0 1100 0600 0100 0800 0000 2d00 0200 0400 00e0 0000 0d10 0f3c 100 9 3d03 ... contents of register %r1 into memory location x. Object code: 1100 0 0100 0100 0000 0101 00000 0100 00 (x = 2064) Instruction: sethi Description: Set the high 22 bits and zero the low 10 b...
Ngày tải lên: 14/08/2014, 20:21
PRINCIPLES OF COMPUTER ARCHITECTURE phần 4 pptx
... 101 11111 0 0101 010 0 1101 010 1 0101 111 101 1 1101 11111 110 11111111 00001111 101 0101 0 11111111 00 0101 01 1101 0101 0 0101 010 1111 1101 11111111 011 1100 0 0 1101 001 0 0101 001 01111111 100 0 0100 1 1100 111 Figure ... generate mm2 mm2 mm0 ↓↓↓↓↓↓↓↓ 101 10011 100 0 1101 0 1100 110 1 0101 010 0 0101 011 0 1101 010 1101 0101 0 0101 010 10 1100 11 00000000 0 1100 110 1 01...
Ngày tải lên: 14/08/2014, 20:21
PRINCIPLES OF COMPUTER ARCHITECTURE phần 5 ppsx
... # 0 1 2 3 4 5 6 7 1 0 1 0 1 0 0 1 00 xx 01 xx 11 xx xx 10 Disk address Page frame 0100 101 1100 1 1101 1100 10 101 10 0101 11 000 0100 1111 0101 1100 101 101 001 1100 1 00 1101 0 1100 0101 00 0101 1 Figure 7-22 A page table for a virtual ... ADDRALU COND A M U X B M U X C M U X 0101 00 000001 00 0100 0 1100 00000000000000 000011 00 0101 00 0100 0100 0 1101 1100 000001 1 0 0 000 010...
Ngày tải lên: 14/08/2014, 20:21
PRINCIPLES OF COMPUTER ARCHITECTURE phần 6 pot
... are composed of sectors of typically 512 bytes in size, stored serially, as shown in (a) Tim e Voltage 100 0 110= ‘F’ (b) Tim e Voltage 100 0 110= ‘F’ Figure 8-17 (a) Straight amplitude (NRZ) encoding of ASCII ... sequence: 100 1 1101 . 8.4 A disk that has 16 sectors per track uses an interleave factor of 1:4. What is the smallest number of revolutions of the disk required to r...
Ngày tải lên: 14/08/2014, 20:21
PRINCIPLES OF COMPUTER ARCHITECTURE phần 7 potx
... frame T(x) = M(x) + R(x) gets corrupted dur- 1101 0 1101 10000 100 11 1100 0 0101 0 100 11 ⊕ 100 11 100 11 ⊕ 1 100 11 ⊕ 0101 0 100 11 ⊕ 1 110 0 1 1 0 0 G(x), of degree n = 4 M(x) n = 4 zeros Bitwise exclusive ... Figure 10- 10. An explanation of the more significant aspects of the assembled code is given in Figure 10- 10, which includes a number of features found only in...
Ngày tải lên: 14/08/2014, 20:21