RISC (Reduced Instruction Set) Computers often fail to have a Divide routine imple- mented in hardware....

 

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RISC (Reduced Instruction Set) Computers often fail to have a Divide routine imple- mented in hardware. Instead, we can apply the Newton method with ONLY subtractions and multiplications to perform division for RISC computers. Here is the procedure. A non-zero real number in its, e.g., IEEE standard single precision floating point representation, is and its reciprocalis x=(-1) (1aa2 a23)2 2 =(-1)-2-x I =(-1)-2 The mantissa (1.a1a2a23)2 is actually in the interval [1,2). Thus in the following discussion, we assume that a is in [1,2). To compute 1/a with ONLY subtractions and multiplications, we can consider = 1/a as a root of the function = f(x)-a (1). To solve the equation f(x) = 0 by the Newton method, let's derive the associated fixed point function g(x) explicitly. Verify that 9(x)=x- =x (2-ax). f(x) f'(x) Then the fixed point iteration method gives us the following scheme Pat Pa(2-a pa), for n 0, with an initial value po for approximating the reciprocal of a in [1,2). (1) (2). Computer 1/a using the fixed point iteration method given in (1) for a 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, and 1.99 with the initial value po 1 and tolerance = 107. (Use the code "alg022Fixpoint.m", not the code "alg023Newton.m".) (3). How does the number of iterations behave for the cases done in part (2)? Explain why those cases need more iterations for larger values of a by discussing the term (a-1) used in the convergent analysis (see Appendix below). (4). Find another initial value of po such that the number of iterations can be reasonalbe (about 5 to 6 iterations) for all values of a given in part (2). RISC (Reduced Instruction Set) Computers often fail to have a Divide routine imple- mented in hardware. Instead, we can apply the Newton method with ONLY subtractions and multiplications to perform division for RISC computers. Here is the procedure. A non-zero real number in its, e.g., IEEE standard single precision floating point representation, is and its reciprocalis x=(-1) (1aa2 a23)2 2 =(-1)-2-x I =(-1)-2 The mantissa (1.a1a2a23)2 is actually in the interval [1,2). Thus in the following discussion, we assume that a is in [1,2). To compute 1/a with ONLY subtractions and multiplications, we can consider = 1/a as a root of the function = f(x)-a (1). To solve the equation f(x) = 0 by the Newton method, let's derive the associated fixed point function g(x) explicitly. Verify that 9(x)=x- =x (2-ax). f(x) f'(x) Then the fixed point iteration method gives us the following scheme Pat Pa(2-a pa), for n 0, with an initial value po for approximating the reciprocal of a in [1,2). (1) (2). Computer 1/a using the fixed point iteration method given in (1) for a 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, and 1.99 with the initial value po 1 and tolerance = 107. (Use the code "alg022Fixpoint.m", not the code "alg023Newton.m".) (3). How does the number of iterations behave for the cases done in part (2)? Explain why those cases need more iterations for larger values of a by discussing the term (a-1) used in the convergent analysis (see Appendix below). (4). Find another initial value of po such that the number of iterations can be reasonalbe (about 5 to 6 iterations) for all values of a given in part (2).

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