Problem 44655. Computational power of Cody servers

Created by David Verrelli

Solve Later

{"relationships":[{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/document","targetMode":"","relationshipId":"rId1","target":"/matlab/document.xml"},{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/output","targetMode":"","relationshipId":"rId2","target":"/matlab/output.xml"}],"parts":[{"partUri":"/matlab/document.xml","relationship":[],"contentType":"application/vnd.mathworks.matlab.code.document+xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n<w:document xmlns:w=\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\"><w:body><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>It has been</w:t></w:r><w:r><w:t> </w:t></w:r><w:hyperlink w:docLocation=\"https://en.wikipedia.org/wiki/Moore%27s_law#History\"><w:r><w:t>predicted</w:t></w:r></w:hyperlink><w:r><w:t> that the performance of integrated circuits would</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:i/></w:rPr><w:t>double</w:t></w:r><w:r><w:t> every 18 months. That suggests the time to perform a given computation should</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:i/></w:rPr><w:t>halve</w:t></w:r><w:r><w:t> roughly every 18 months.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>What about on Cody? Observational data is available from the final test case of</w:t></w:r><w:r><w:t> </w:t></w:r><w:hyperlink w:docLocation=\"https://www.mathworks.com/matlabcentral/cody/problems/963-we-love-vectorized-solutions-problem-1-remove-the-row-average\"><w:r><w:t>Problem 963</w:t></w:r></w:hyperlink><w:r><w:t> to help us quantify the improvement in performance! And thereby even make predictions for future computations.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[ Solution Date Runtime [s]\n ...\n 144393 04-Oct-12 4.230\n ...\n 654812 17-Apr-15 3.099\n ...\n 1272817 20-Sep-17 2.0402\n ...]]></w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>The complete data set will be provided to you as input. You should assume the general trend can be described by the following law:</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[ runtime = r0 - δ [1 - exp(-t/k)]\n runtime = r∞ + δ exp(-t/k)]]></w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>where</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>r0</w:t></w:r><w:r><w:t> is the runtime at the start of the period in seconds,</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>r∞</w:t></w:r><w:r><w:t> is the predicted runtime (in seconds) that will be approached far in the future,</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>δ = r0 - r∞</w:t></w:r><w:r><w:t>, and</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>t</w:t></w:r><w:r><w:t> is the time in nominal years since the start of the period, and</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>k</w:t></w:r><w:r><w:t> is a kinetic parameter (in nominal years).</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>The start of the period is defined by the earliest date in the series. Compute the number of days exactly, and assume that a nominal year comprises 365.24 days.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Your task is to</w:t></w:r><w:r><w:t> </w:t></w:r><w:hyperlink w:docLocation=\"https://au.mathworks.com/help/matlab/math/example-curve-fitting-via-optimization.html\"><w:r><w:t>fit the curve</w:t></w:r></w:hyperlink><w:r><w:t> and thereby predict the runtime for various future dates. Your output should be rounded to four decimal places.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>METHOD: You</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:i/></w:rPr><w:t>must</w:t></w:r><w:r><w:t> use</w:t></w:r><w:r><w:t> </w:t></w:r><w:hyperlink w:docLocation=\"https://au.mathworks.com/help/matlab/ref/fminsearch.html\"><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>fminsearch</w:t></w:r></w:hyperlink><w:r><w:t> to perform the non-linear regression, and you</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:i/></w:rPr><w:t>must</w:t></w:r><w:r><w:t> set the options using</w:t></w:r><w:r><w:t> </w:t></w:r><w:hyperlink w:docLocation=\"https://au.mathworks.com/help/matlab/ref/optimset.html\"><w:r><w:rPr><w:rFonts w:cs=\"monospace\"/></w:rPr><w:t>optimset</w:t></w:r></w:hyperlink><w:r><w:t> to ensure sufficient accuracy. The 'best' fit is defined —</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:i/></w:rPr><w:t>for this problem, as in the common convention</w:t></w:r><w:r><w:t> — as that which minimises the sum of the squares of the residuals.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>EXAMPLE:</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[ % Input\n futureDate = '21-Nov-2023';\n data.solutionNumber = [1036949, ..., 1272817];\n data.date = ['29-Oct-2016'; ...; '20-Sep-2017'];\n data.runtime = [1.2630, ..., 2.0402];\n % Output\n predictedRuntime = 0.3619; % seconds]]></w:t></w:r></w:p></w:body></w:document>"},{"partUri":"/matlab/output.xml","contentType":"text/xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"no\" ?><embeddedOutputs><metaData><evaluationState>manual</evaluationState><layoutState>code</layoutState><outputStatus>ready</outputStatus></metaData><outputArray type=\"array\"/><regionArray type=\"array\"/></embeddedOutputs>"}]}

It has been <a href = "https://en.wikipedia.org/wiki/Moore%27s_law#History">predicted</a> that the performance of integrated circuits would double every 18 months. That suggests the time to perform a given computation should halve roughly every 18 months.What about on Cody? Observational data is available from the final test case of <a href = "https://www.mathworks.com/matlabcentral/cody/problems/963-we-love-vectorized-solutions-problem-1-remove-the-row-average">Problem 963</a> to help us quantify the improvement in performance! And thereby even make predictions for future computations.<pre> Solution Date	 Runtime [s]
 ...
 144393	 04-Oct-12	 4.230
 ...
 654812	 17-Apr-15	 3.099
 ...
 1272817 20-Sep-17	 2.0402
 ...</pre>The complete data set will be provided to you as input. You should assume the general trend can be described by the following law:<pre> runtime = r0 - δ [1 - exp(-t/k)]
 runtime = r∞ + δ exp(-t/k)</pre>where <tt>r0</tt> is the runtime at the start of the period in seconds, <tt>r∞</tt> is the predicted runtime (in seconds) that will be approached far in the future, <tt>δ = r0 - r∞</tt>, and <tt>t</tt> is the time in nominal years since the start of the period, and <tt>k</tt> is a kinetic parameter (in nominal years).The start of the period is defined by the earliest date in the series. Compute the number of days exactly, and assume that a nominal year comprises 365.24 days.Your task is to <a href = "https://au.mathworks.com/help/matlab/math/example-curve-fitting-via-optimization.html">fit the curve</a> and thereby predict the runtime for various future dates. Your output should be rounded to four decimal places.METHOD: You must use <a href = "https://au.mathworks.com/help/matlab/ref/fminsearch.html"><tt>fminsearch</tt></a> to perform the non-linear regression, and you must set the options using <a href = "https://au.mathworks.com/help/matlab/ref/optimset.html"><tt>optimset</tt></a> to ensure sufficient accuracy. The 'best' fit is defined — for this problem, as in the common convention — as that which minimises the sum of the squares of the residuals.EXAMPLE:<pre> % Input
 futureDate = '21-Nov-2023';
 data.solutionNumber = [1036949, ..., 1272817];
 data.date = ['29-Oct-2016'; ...; '20-Sep-2017'];
 data.runtime = [1.2630, ..., 2.0402];
 % Output
 predictedRuntime = 0.3619; % seconds</pre>

It has been <https://en.wikipedia.org/wiki/Moore%27s_law#History predicted> that the performance of integrated circuits would _double_ every 18 months. That suggests the time to perform a given computation should _halve_ roughly every 18 months.

What about on Cody? Observational data is available from the final test case of <https://www.mathworks.com/matlabcentral/cody/problems/963-we-love-vectorized-solutions-problem-1-remove-the-row-average Problem 963> to help us quantify the improvement in performance! And thereby even make predictions for future computations.

Solution    Date	   Runtime [s]
 ...
 144393	     04-Oct-12	   4.230
 ...
 654812	     17-Apr-15	   3.099
 ...
 1272817     20-Sep-17	   2.0402
 ...

The complete data set will be provided to you as input.  You should assume the general trend can be described by the following law:

runtime = r0 - δ [1 - exp(-t/k)]
 runtime = r∞ + δ exp(-t/k)

where |r0| is the runtime at the start of the period in seconds, |r∞| is the predicted runtime (in seconds) that will be approached far in the future, |δ = r0 - r∞|, and |t| is the time in nominal years since the start of the period, and |k| is a kinetic parameter (in nominal years).

The start of the period is defined by the earliest date in the series.  Compute the number of days exactly, and assume that a nominal year comprises 365.24 days.

Your task is to <https://au.mathworks.com/help/matlab/math/example-curve-fitting-via-optimization.html fit the curve> and thereby predict the runtime for various future dates. Your output should be rounded to four decimal places.

METHOD: You _must_ use <https://au.mathworks.com/help/matlab/ref/fminsearch.html |fminsearch|> to perform the non-linear regression, and you _must_ set the options using <https://au.mathworks.com/help/matlab/ref/optimset.html |optimset|> to ensure sufficient accuracy. The 'best' fit is defined — _for this problem, as in the common convention_ — as that which minimises the sum of the squares of the residuals.

EXAMPLE:

% Input
 futureDate = '21-Nov-2023';
 data.solutionNumber = [1036949, ..., 1272817];
 data.date = ['29-Oct-2016'; ...; '20-Sep-2017'];
 data.runtime = [1.2630, ..., 2.0402];
 % Output
 predictedRuntime = 0.3619;  % seconds

Solve

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Last Solution submitted on Jun 01, 2020

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