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### HW1Sol

Course: COT 4501, Spring 2012
School: University of Florida
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Word Count: 291

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Spring COT4501 2012 Solution I Problem 1 Provide short answers to the following questions: 1.8 False 1,10 False 1.23 1. No. Propagated data error is measured by the conditioning of an algorithm and is not directly tied to stability. 2. Yes. Accuracy is affected by both the conditioning and stability of a problem 3. No. The conditioning and stability of an algorithm are separate qualities that both deal with...

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Spring COT4501 2012 Solution I Problem 1 Provide short answers to the following questions: 1.8 False 1,10 False 1.23 1. No. Propagated data error is measured by the conditioning of an algorithm and is not directly tied to stability. 2. Yes. Accuracy is affected by both the conditioning and stability of a problem 3. No. The conditioning and stability of an algorithm are separate qualities that both deal with error(computational and data error, respectively) 1.29. 1. Every number has a unique representation 2. No digits are wasted by leading 0s maximizing precision 1.32 (1/2) 1p 1.38 1/10 .1 in decimal form .00011001100110011 . . . in binary form Problem 2 (1.9.) 5.101 108 km2 . New surface area is 5.102 108 , so change is 1.0 105 With approximate formula, change in surface area is 1.601 105 With six-digit arithmetic, original surface area is 5.00906 108 ,new surface area is 5.10065 108 , difference between these is 1.59 105 ,and change in surface area from approximate formula is 1.60095 105 , so approximate formula is more nearly correct than exact formula. Problem 3 1.14. Assuming a and b have the same sign, formula is (b) better because the result is guaranteed to lie in the interval [a, b], and it also avoids the possibility of overow. With 2-digit arithmetic, for example, formula (a) gives a midpoint of 0.7 for the interval [0.67, 0.69], whereas formula (b) gives a midpoint of 0.68. Problem 4 1 1.18.(a) 2( 1) p1 (U L + 1) + 1. (b) 2( p1 1). 1.19.(a) OFL = 2128 (1 224 ), UFL = 2126 . (b) OFL unchanged, UFL = 2149 Problem 5 1.20. (a) 0.00011001100110011001100. (b) 0.00011001100110011001101. 1.21. (a) Not necessarily (see part b). For practical oating-point systems, however, mach is a machine number. 1.21 (b) The toy system of Example 1.9 is an example, as mach = 0.25 and UFL = 0.5. Problem 6 1.22. NOTE: In the posted homework c = 2.88 was typed instead of c = 2.28. Following solution is with respect question in the book. For c = 2.88 just replace this value in the following solution (a)b2 4ac = (3.34)2 4 1.22 2.28 = 11.2 11.1 = 0.1. (b) 0.0292. (c) (0.10.0292)/0.0292 2.425. Problem 7 1.23. (a) UFL = L = 1098 (b) x y= 6.87 1097 6.81 1097 = 0.06 1097 = 6 : 00 1099 0. (c) With gradual under ow, result is 0.60 1098 2
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University of Florida - COT - 4501
COT4501 Spring 2012Homework IIThis assignment has eight problems and they are equally weighted. The assignmentis due in class on Tuesday, February 14, 2012. There are six regular problems and twocomputer problems (using MATLAB). For the computer probl
University of Florida - COT - 4501
COT4501 Spring 2012Homework II SolutionsThis assignment has eight problems and they are equally weighted. The assignmentis due in class on Tuesday, Tuesday 14, 2012. There are six regular problems and twocomputer problems (using MATLAB). For the compu
University of Florida - COT - 4501
COT4501 Spring 2012Homework IIIThis assignment has six problems and they are equally weighted. The assignment is due inclass on Tuesday, February 21, 2012. There are four regular problems and two computer problems (using MATLAB). For the computer probl
University of Florida - COT - 4501
COT4501 Spring 2012Homework IIIThis assignment has six problems and they are equally weighted. The assignment is due inclass on Tuesday, February 21, 2012. There are four regular problems and two computer problems (using MATLAB). For the computer probl
University of Florida - COT - 4501
COT4501 Spring 2012Homework IVThis assignment has seven problems. The assignment is due in class on Thursday, March1, 2012. There are ve regular problems and two computer problems (using MATLAB). Forwritten problems, you need to show your work and it
University of Florida - COT - 4501
COT4501 Spring 2012Homework IV SolutionsProblem 11. FALSE. for example following matrix has only one real eigenvalue.2 1 1A = 1 1 1 ,11 22. False3. For a given matrix True False4. Which of the following conditions necessarily imply that a n n r
University of Florida - COT - 4501
COT4501 Spring 2012Homework VThis assignment has seven problems. The assignment is due in class on Thursday, March29, 2012. There are seven regular problems and two computer problems (using MATLAB).For written problems, you need to show your work and
University of Florida - COT - 4501
Review Questions for Midterm Exam1. Know the oating-point number systems. All problems in the four homework sets we havehad so far are fair questions to ask on the exam. The exam will cover the three chapterswe have covered so far: oating-point number
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UPenn - ECON - 010
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UPenn - ECON - 010
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UPenn - ECON - 010
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UPenn - ECON - 010
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UPenn - ECON - 010
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UPenn - ECON - 010
UPenn - ECON - 010
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UPenn - ECON - 010
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UPenn - ECON - 010
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UPenn - ECON - 010
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UPenn - ECON - 010
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In111afpfipc)'u . ) J L 6i.c5- )k2- ( ~ m p + i + iv,d0542n=o,',bb)qI,JPrpi &quot; ~ R I ( Q \$ ) = Ac250-\-Q+/IOO=J rbj-A[aq,/~o= 04Ia l G - - 0,Aojli/SOQ j = 6 0 0 0 ) = [ 3 , O CXJ~ S F lo0=I~C.'-4';t~A t K .2 ,: -L ,I100
UPenn - ECON - 010
Econ 001: Midterm 2March 21, 2007 Dr. SpiegelInstructions:!This is a 60-minute examination.!Write all answers on this question packet. Show all work. Use diagramswhereappropriate and label all diagrams carefully.!This exam is given under the rules
UPenn - STAT - 102
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UPenn - STAT - 102
STAT 102Homework 4Due Date: Wednesday,2012March 28 before 5:00pm in 400 JMHH.1. Problem 4.8 (p. 169) from the textbook. But ignore part f of the question. [Also, donot use the output from the text book. You should produce JMP output and use that in
UPenn - STAT - 102
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UPenn - STAT - 102
Stat 102 Spring 2012Homework #2 Solutions3.4a)Based on the original information, we have: xi = 5 + 6 + 7 + 8 + 9 + 10 + 11 = 562i x = 25 + 36 + 49 + 64 + 81 + 100 + 121 = 476 y = 12 + 11.5 + 14 + 15 + 15.4 + 15.3 + 17.5 = 100.7 x y = 60 + 69 + 9
UPenn - STAT - 102
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UPenn - STAT - 102
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UPenn - STAT - 102
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UPenn - STAT - 102
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