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Some concept of Machine Organization are Anatomy, Cache Access Time, Instruction Formats, Instruction Formats, Instruction Formats, Multidimensional Meshes, Network Processors, Snooping Protocol. Main points of this lecture are: Increase, Accuracy, Analog Machines, Inputs to First, Output of First, Step, Effective Computability, Procedure Terminates, Computer, Added
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Chapter 1: 1.3 It is hard to increase the accuracy of analog machines. 1.5 (a) inputs to first (x) box are a and x output of first (x) box is ax inputs to second (+) box are ax and b output of second (+) box is ax + b (b) inputs to first (+) box are w and x output of first (+) box is w + x inputs to second (+) box are y and z output of second (+) box is y + z inputs to third (+) box are (w + x) and (y + z) output of third (+) box is w + x + y + z inputs to fourth (x) box are (w + x + y + z) and. output of fourth (x) box is 0:25(w + x + y + z), which is the average (c) The key is to factor a 2 + 2ab + b 2 = (a + b) 2 inputs to first (+) box are a and b output of first (+) box is a + b inputs to second (x) box are (a + b) and (a + b) output of second (x) box is (a + b) 2 = a 2 + 2ab + b 2 1.10 Definiteness: each step is precisely stated. Effective Computability: Each step can be carried out by a computer. Finiteness: the procedure terminates. 1.12 (a) Lacks definiteness, since it does not specify how two rows are to be added. Also, the 3rd or the 4th row could be added to the first row. So there are two possible answers. (b) This is not effectively computable, because there is no end to the number line. Anything involving infinity must not be effectively computable. This is also not finite, for the same reason. (c) This is an algorithm. (d) This is not finite, so it is not an algorithm. If, as Calvin suspects, the coin is weighted, they will be flipping that coin forever. (e) This is not finite, so it is not an algorithm. Steps 1 to 6 calculate, albeit in a long way, the number - 1. If the given number is negative or zero, then there will never be a time when you get 0 at the end of step 6. 1.15 Advantages of a higher level language: Fewer instructions are required to do the same amount of work. This usually means it takes less time for a programmer to write a program to solve a problem. High level language programs are generally easier to read and therefore know what is going on. Disadvantages of a higher level language: Each instruction has less control over the underlying hardware that actually performs the computation that the program frequently executes less efficiently. 1.18 A single microarchitecture typically implements only one ISA. However, many microarchitectures usually exist for the same ISA.
2.2 For 26 characters, we need at least 5 bits. For 52 characters, we need at least 6 bits. 2.4 2n integers can be represented. The range would be 0 to (2n) - 1. 2.8 The answers are: (a) 127 in decimal, 01111111 in binary. (b) -128 in decimal, 10000000 in binary. (c) (2n-1) - (d) - (2n-1) 2.10 The answers are: (a) - (b) 90 (c) - (d) 14803 2.11 (a) 01100110 (b) 01000000 (c) 00100001 (d) 10000000 (e) 01111111 2.13 (a) 11111010 (b) 00011001 (c) 11111000 (d) 00000001 2.48 (a) x (b) x6F (c) x75BCD Docsity.com