In 2017 my Website was migrated to
the clouds and reduced in size.
Hence some links below are broken.
One thing to try if a “www” link is broken is to substitute “faculty” for “www”
For example a broken link
http://faculty.trinity.edu/rjensen/Pictures.htm
can be changed to corrected link
http://faculty.trinity.edu/rjensen/Pictures.htm
However in some cases files had to be removed to reduce the size of my Website
Contact me at rjensen@trinity.edu if
you really need to file that is missing
Some Things You Might Want to Know About the Wolfram Alpha
(WA) Search Engine: The Good and The Evil
as Applied to Learning Curves (Cumulative Average vs. Incremental Unit)
Bob Jensen
at
Trinity University
Introduction
Wolfram Alpha (WA) Computational Knowledge Search Engine 
http://www.wolframalpha.com/
You may enter a search term (such as "cumulative average") in the search box or
a formula (such as "(1000000)(8)^((ln(80/100)/ln(2))"that you want solved.
The search term may be disappointing relative to what you obtain from Google,
Bing, or other search engine.
But the formula solving hardly ever disappoints and may even yield very neat
publishable formulas, an answer, and a bonus graph of the function.
Wolfram Alpha  http://en.wikipedia.org/wiki/Wolfram_Alpha
Wolfram Alpha (styled WolframAlpha) is an answer engine developed by Wolfram Research. It is an online service that answers factual queries directly by computing the answer from structured data, rather than providing a list of documents or web pages that might contain the answer as a search engine would. It was announced in March 2009 by Stephen Wolfram, and was released to the public on May 15, 2009.[1] It was voted the greatest computer innovation of 2009 by Popular Science.
Wolfram Alpha for Educators  http://www.wolframalpha.com/educators/
One evil of WA is that students may enter complex formulas into the search box and receive instant answers (and sometimes graphs) without having to learn much of the analytics that they would otherwise learn by solving formulas in traditional/iterative process that is often valuable in mathematics lessons. Although solutions can be obtained from other software such as Excel, WA is more of a onestep process that takes much of the drudgery and learning process that comes even with the other software. WA presents problems similar to and well beyond the issue of whether to allow graphing calculators.
But WA need not be so evil if instructors simply control when and if students are allowed to use WA, such as by not allowing WA to be used on examinations. That could motivate students to learn how to solve formulas without the aid of WA as well as to use WA to early on better understand functions before taking examinations.
I am awestruck by the good features of WA for ease of computation, preparing papers for publication, and for better understanding complicated formulas.
Although I knew the general features of WA, this week I seriously used it when confronted with some formulas I needed to solve.
My SJM Paper
"Reconciliation of Learning Rates Between the Cumulative Average Versus
Incremental Learning Rate Models," by Robert E. Jensen,
Scandinavian Journal of Management, Vol. 7,
No. 2, 1991, 137142 
http://www.cs.trinity.edu/~rjensen/temp/WorkingPaper275.pdf
After nearly 20 years I completely forgot about this paper that I drafted in 1989. In 1989 I did not own a PC. Being an old teacher of FORTRAN and COBAL, I could've written a main frame computer program to solve the formulas in the SJM paper, but at the time Trinity University was having troubles with its main frame computer, and I did not own or use a PC until I purchased a PC in 1990. In 1989 I spent days on an HP memory calculator solving the formulas in Table 1 of the paper. The problem was not so much solving the formulas as it was iteratively finding the parameters that equated the cumulative average versus incremental unit learning models.
Out of the blue on September 14, 2010 I received a message from a very polite Australian professor named Chris Deeley who challenged the computations in my SJM paper published in 1991. My initial reaction was that he was probably correct since I'd never subsequently verified my 1989 HP calculator solutions with any modern software like Excel. Subsequently, I think I have verified the numbers in the paper by using Excel. However, just for the heck of it I also verified some of the key numbers using Wolfram Alpha (WA). I'm truly impressed with some of the things WA will generate that cannot be generated in Excel.
In my particular circumstances the Goal Seek utility in Excel saved hours of
tedious computations on a calculator 
http://office.microsoft.com/engb/excelhelp/aboutgoalseekHP005203894.aspx
It turned out that both Jensen and Deeley computed their numbers correctly under their different models. Professor Deeley defined a somewhat different incremental learning rate model. Deeley assumes the I(j,q) learning rate varies when computing each marginal unit cost whereas Jensen assumes the I(q) marginal unit cost varies only with the number of units, q, produced in total and the total variable cost V(q)..
Hence comparing Deeley versus Jensen outcomes is like comparing apples and oranges, although they do get the same total variable cost. It's just that the patterns of marginal unit costs differ in arriving at total variable cost. It came to my great relief that the numbers that I computed on a calculator in 1989 held up in 2010 under both Excel and Wolfram Alpha.
Consider the following numbers for q=8 units of product and an incremental learning rate of 80% which means that the u(q) marginal per unit of product will be 80% each time q is doubled. This 80% learning curve in the aircraft production industry was shown by Theodore Paul Wright to be somewhat like the inexplicable Moore's Law is to the cost of computer memory. Both inexplicably hold true under different circumstances.
Suppose the marginal cost of the first unit is u(1)=$1,000,000.
Here are examples of three test numbers that I first checked in Excel for Jensen's 1989 model on Page 140 of the SJM paper:
0.321928 = i exponent of the 80% learning rate
$512,000 = u(8) marginal cost of Unit 8
$5,345,914 = V(8,80) cumulative variable cost of the first eight production units (e.g., airplanes)
The WA code formulas that I first used to test these calculations in Wolfram Alpha are as follows:
0.321928 = i = (ln(80/100))/(ln(2)) where ln( ) is a natural base e logarithm.
$512,000 = u(8) = 1000000*(8^(( ln(80/100))/(ln(2)))
$5,345,914 = V(8,80)= (1000000)Sum(j=1,8,j^(ln(80/100)/ln(2))) This WA code did not work in Excel
Wolfram Alpha Extensions
I derived the three numbers above easily in Excel, although in Excel I simply computed V(8,80) as the sum of u(1) thru u(8). Now comes the interesting part. Suppose that I was writing a paper and wanted to publish the three functions shown above. In code, these functions are very difficult for readers of a paper. However, when these codes are read into WA all sorts of neat things happen. If I wanted to publish these results or better explain to students what is being calculated, I could cut and paste the WA math interpretations generated by the incomprehensible code.
Firstly, I get Wolfram Alpha formatted formulas that I can cut and past into a paper.
0.321928 = i = (ln(80/100))/(ln(2))
= where log is a natural logarithm$512,000 = u(8) = 1000000*(8^(( ln(80/100))/(ln(2)))
= where log is a natural logarithm$5,345,914 = V(8,80 = (1000000)Sum(j=1,8,j^(ln(80/100)/ln(2)))
= where log is a natural logarithm
Secondly, I get some added WA information for extending the formulas to the limit in a continuum:
=
Thirdly for some functions, graphs and other information about those
functions will be generated (including regressions and probability distribution
graphs)
http://www.wolframalpha.com/examples/
Excel Solutions and Extensions of SJM Paper
"Reconciliation of Learning Rates Between the Cumulative Average Versus
Incremental Learning Rate Models," by Robert E. Jensen,
Scandinavian Journal of Management, Vol. 7,
No. 2, 1991, 137142 
http://www.cs.trinity.edu/~rjensen/temp/WorkingPaper275.pdf
Excel is still more useful than WA for repetitive calculations.
As mentioned above, the huge time saver in this instance is the
Goal Seek utility in Excel.
The Excel verifications of the numbers in my SJM paper and some extensions are
shown below:
Equivalent C =  87.43  From Goal Seek  Fixed I =  80  The 87.43% learning rate agrees with C(80,8) on Page 140 of the paper  
c=  0.1938545  i=  0.321928095  The i= 0.321928 exponent appears on Page 140 of the paper  
u(1)=  $1,000,000  u(1)=  $1,000,000  
u(8)=  $545,574  u(8)=  $512,000  The u(8) = $512,000 cost of the 8th unit agrees with Page 140 of the paper  
Marginal  Variable  Cumulative  Double j  Marginal  Variable  Cumulative  Double j  
Cost  Cost  Average  Learning  Cost  Cost  Average  Learning  
u(j)  V(j)  v(j)  Rate  u(j)  V(j)  v(j)  Rate  
1000000  1000000  1000000  1000000  
1  1000000  1000000  1000000  1000000  1000000  1000000  1000000 


2  748534  1748534  874267  874267  0.8743  800000  1800000  900000  0.8000  
3  676005  2424539  808180  808180  702104  2502104  834035  
4  632831  3057370  764342  764342  0.8743  640000  3142104  785526  0.8000  
5  602550  3659920  731984  731984  595637  3737741  747548  
6  579468  4239388  706565  706565  0.8743  561683  4299424  716571  0.8000  
7  560952  4800340  685763  685763  534490  4833914  690559  
8  545574  5345914  668239  668239  0.8743  512000  5345914  668239  0.8000  
V(8)=  $5,345,914  =Sum  From Goal Seek  $5,345,914  =Sum = V(8)  This V(8) total variable cost agrees with Page 140 of the paper  
$668,239  =Average  $668,239  =Average  This average is incorrectly given as $664,489 on Page 140 of the paper  
Fixed C =  80  Equivalent I =  67.79  From Goal Seek  
c=  0.3219281  i=  0.560926783  
u(1)=  $1,000,000  u(1)=  $1,000,000  
u(8)=  $354,573  u(8)=  $311,482  
Marginal  Variable  Cumulative  Double j  Marginal  Variable  Cumulative  Double j  
Cost  Cost  Average  Learning  Cost  Cost  Average  Learning  
u(j)  V(j)  v(j)  Rate  u(j)  V(j)  v(j)  Rate  
1000000  1000000  1000000  1000000  
1  1000000  1000000  1000000  1000000  1000000  1000000  1000000 


2  600000  1600000  800000  800000  0.8000  677867  1677867  838933  0.6779  
3  506311  2106311  702104  702104  539970  2217837  739279  
4  453689  2560000  640000  640000  0.8000  459503  2677340  669335  0.6779  
5  418187  2978187  595637  595637  405442  3082782  616556  
6  391911  3370098  561683  561683  0.8000  366028  3448810  574802  0.6779  
7  371329  3741427  534490  534490  335708  3784518  540645  
8  354573  4096000  512000  512000  0.8000  311482  4096000  512000  0.6779 
Both models are
fairly close in marginal cost outcomes (Jensen in red, Deeley in blue) for q=8.
They will converge as q increases.
In my opinion, both the Jensen and Deeley models have a learning rate that is too high between the u(1) marginal cost of the first unit visŕvis the marginal cost u(2) of the second unit.
Two online spreadsheets are
available at
http://www.cs.trinity.edu/~rjensen/temp/WP275Table02b.xlsx
http://www.cs.trinity.edu/~rjensen/temp/WP275Table03b.xlsx
Having verified the main numbers that appear in the Table 1 of my 1991 SJM paper, I did find a few corrections.
On Page 138, the c in Equation 3 should've been upper case.
On Page 140, ln (0.80/100) should read ln (80)/100))
On Pages 140 and 141, $664,489 should read $668,239
Updates
September 22, 2010 message from Deeley, Chris [CDeeley@csu.edu.au]
Bob
I have now solved the riddle (as you already had) of why our two sets of numbers differ. Mine bases parity on incremental cost, while your's bases parity on total cost (and also, of course, cumulative average cost). This point is made perfectly clear below Table 1 where you refer to “what cumulative average model learning rate would achieve the same total cost”. I must have overlooked this earlier and mistakenly assumed that you were trying to establish parity with respect to incremental cost.
I suggest a sensible synthesis is to acknowledge that we have closedform equations for (1) converting incremental rates into cumulative average rates if parity is with reference to cumulative average costs, and (2) converting cumulative average rates into incremental rates if parity is with reference to incremental costs. That being the case, I don’t see why Table 1 cannot be constructed on that basis. I assume the main objective of the exercise is to determine which curve is more realistic (to date) and therefore presumed to be the more reliable for forecasting purposes. It doesn’t really matter what the points of parity and comparison are, so long as they are clearly identified.
Anyway, I’ll carry on with my table and send you the finished product. I suspect that the “hybrid” table will not take long to complete.
Chris
Chris Deeley
Senior Lecturer in Accounting & Finance
School of Accounting, Faculty of Business
Charles Sturt University, Locked Mail Bag 588
Wagga Wagga, NSW 2678
Ph: +612 69332694 Fax: +612 69332790
September 23, 2010 message from Deeley, Chris [CDeeley@csu.edu.au]
Bob
Thanks for this. I must say I find the material you’ve made available on the Internet most fascinating, especially the comments on retirement, which is a key interest of mine. In Australia one cannot be forced to retire because of age. I’m 66 and not planning to retire anytime soon. I switched out of a defined benefit pension scheme and into a defined contribution scheme some time ago, so I have a financial incentive to stay on a payroll. Fortunately my wife is on an indexed pension, so we have a good safety net.
I’ve now completed my “hybrid” table, which establishes parity between the two learning curves with respect to (1) cumulative average cost when the incremental learning rate is held constant, and (2) incremental cost when the cumulative average learning rate is held constant. This avoids the issue of nonclosedform algebraic solutions, as mentioned in your SJM article (although a suitable solution algorithm, if there is one, would overcome that problem). Unfortunately, I still struggle to compute accumulated costs from a given incremental learning rate for values of q above 200 (assuming that we start with q = 1). This explains the question marks in my table, which you may be able to quantify for me using Wolfram Alpha (if you have the time and inclination). You’ll see that the equivalency figures in the lefthand part of my table are identical to yours, which comes as no surprise after eliminating the apples v oranges element.
Re worthwhile retirement activities: how about reviewing submitted papers for journals/conferences? Maybe you already do this? I also find trading shares is a fascinating pastime.
My own research interests are mainly theoretical. It’s amazing how much stuff is out there which is wrong, but so difficult to change. For example, I contend that the conventional approach to solving general annuity problems is invalid when the frequency of payments exceeds the frequency of interest compounding. I’ll attach a draft paper on this topic, in case it may be of interest.
I hope I haven’t taken up too much of your time.
Chris
Table II. Reconciliation of incremental & cumulative average learning curves  
Incremental learning rate:  80%  Cumulative average learning rate:  80%  
Cumulative average learning  Incremental learning  
Cumulative  rate  Incremental  rate  
average cost parity  varies  cost parity  varies  
q  U  CAC  C %  U  CAC  CAC  U  I %  CAC  U  
1  1,000,000  1,000,000  1,000,000  1,000,000  1,000,000  1,000,000  1,000,000  1,000,000  
2  800,000  900,000  90.000  800,000  900,000  800,000  600,000  60.000  800,000  600,000  
3  702,104  834,035  89.181  718,482  834,035  702,104  506,311  65.089  719,067  506,311  
4  640,000  785,526  88.630  664,465  785,526  640,000  453,689  67.356  665,450  453,689  
5  595,637  747,548  88.223  624,457  747,548  595,637  418,187  68.696  625,713  418,187  
6  561,683  716,571  87.904  592,982  716,571  561,683  391,911  69.602  594,426  391,911  
7  534,490  690,559  87.644  567,232  690,559  534,490  371,329  70.266  568,811  371,329  
8  512,000  668,239  87.427  545,574  668,239  512,000  354,573  70.779  547,251  354,573  
9  492,950  648,763  87.241  526,974  648,763  492,950  340,546  71.190  528,724  340,546  
10  476,510  631,537  86.062  481,203  607,357  476,510  328,552  71.529  508,707  328,552  
20  381,208  524,247  86.121  413,497  524,247  381,208  260,614  73.261  415,429  260,614  
30  334,559  467,330  85.639  364,177  467,330  334,559  228,091  73.992  366,044  228,091  
40  304,966  429,836  85.329  332,409  429,836  304,966  207,631  74.425  334,190  207,631  
50  283,827  402,434  85.106  309,523  402,434  283,827  193,080  74.721  311,222  193,080  
60  267,647  381,130  84.933  291,913  381,130  267,647  181,975  74.942  293,538  181,975  
70  254,690  363,869  84.794  277,757  363,869  254,690  173,097  75.115  279,317  173,097  
80  243,973  349,465  84.679  266,021  349,465  243,973  165,766  75.256  267,521  165,766  
90  234,895  337,176  84.581  256,059  337,176  234,895  159,562  75.374  257,508  159,562  
100  227,062  326,508  84.496  247,451  326,508  227,062  154,213  75.474  248,852  154,213  
200  181,649  263,600  83.994  197,389  263,600  181,649  123,271  76.044  198,487  123,271  
300  159,421  232,211  83.741  172,841  232,211  159,421  108,157  76.316  ?  108,157  
400  145,319  212,122  83.578  157,275  212,122  145,319  98,577  76.488  ?  98,577  
500  135,246  197,694  83.460  146,166  197,694  135,246  91,736  76.610  ?  91,736  
1bn*  1,267  ?  ?  ?  ?  1,267  859  78.967  859  
* Total costs in $million  
q = units of accumulated prd'n  U = Incremental unit cost  CAC = cumulative average cost  
******************
Chris also seen me his latest working paper on an entirely different topic.
September 23, 2010 message from Chris Deeley cdeeley@csu.edu.au
Bob
Yes, by all means post my working paper on general annuities on the AECM
website. I suspect that Wolfram Alpha may not be able to handle this sort of
thing. In fact, I wouldn’t be surprised if the application of standard maths has
created and entrenched the error. Chris
Chris Deeley
Senior Lecturer in Accounting & Finance
School of Accounting,
Faculty of Business
Charles Sturt University,
Locked Mail Bag 588
Wagga Wagga, NSW 2678
Ph: +612 69332694 Fax: +612 69332790
Email:
cdeeley@csu.edu.au
Web:
www.csu.edu.au
I put his paper on one of my Web servers. I'm sure that Chris will appreciate any comments that you have regarding this technical topic. It may be a good exercise for accounting and finance students to study this paper.
"IDENTIFICATION AND CORRECTION OF A COMMON ERROR IN GENERAL ANNUITY CALCULATIONS," by Chris Deeley, Charles Sturt University, Australia., September 23, 2010 Working Draft 
Chris Deeley in Australia and I have been corresponding regarding an antique
learning curve paper that I published nearly 20 years ago. You can read some of
our correspondence at
http://faculty.trinity.edu/rjensen/theorylearningcurves.htm
In that correspondence I discuss the good and evil of the Wolfram Alpha
computational search engine.
Instructors might want to consider adding this to their teaching modules on time value of money and annuity mathematics of finance.
Working Paper 440
Annuities With Unequal Compounding and Payment Periods: The
CFA Deconstruction Analysis
Bob Jensen at
Trinity University
Financial calculators and Excel financial formulas for computing present value, interim payments, and rates of return assume that p=m where p is the number of equallyspaced payments per year and m is the number equallyspaced interest compoundings per year. Complications introduced by p not being equal to m are not trivial problems. These complications are overlooked in many (probably almost all) mathematics of finance modules in both high school and college courses.
This note will demonstrate how to deal with complications when the number of payments per year is unequal to the number in times interest is compounded per year. This is not a purely academic problem. Companies buying and selling annuities often do not want to change the number of times interest is compounded every time they change the number of payments per year in a contract such as semiannual payments versus quarterly payments versus monthly payments.
The paper was inspired by the following working paper sent to me by an Australian professor named Chris Deeley. Chris subsequently allowed me to put his paper on one of my Web servers:
"IDENTIFICATION AND CORRECTION OF A COMMON ERROR
IN GENERAL ANNUITY CALCULATIONS"
by Chris Deeley
cdeeley@csu.edu.au
Working Paper, Charles Sturt University, Australia, September 22, 2010
http://www.cs.trinity.edu/~rjensen/temp/DeeleyAnnuityCorrections.pdf
For illustrative purposes I will focus on the following example on Page 11 of Professor Deeley's working paper:
Example 2
A loan of $1million is to be repaid in equal monthly installments over four
years. If the annual interest rate is 10% compounded semiannually, how much is
the monthly repayment?
The two solutions given by Professor Deeley for p=12 payments per year and m=2 interest compoundings per year are as follows:
Deeley Solution 1 PMT = $25,265.60 per month which Professor Deeley claims the "conventional solution"
Deeley Solution 2 PMT = $25,260.70 per month which Professor Deeley claims is his "proposed better solution"
I contend that there is a CFA Deconstruction and Rate Equivalence solution that I offer as an "alternate conventional solution" used of Certified Financial Analyst (CFA) examinations.
CFA Deconstruction PMT = $25,483 per month which conforms to David Frick's solution tutorial
I show how to calculate this $25,483 using both Wolfram
Alpha and Excel in my Working Paper 440 
Bob Jensen's analysis of Annuities With Unequal Compounding and Payment
Periods: The CFA Deconstruction Analysis
http://faculty.trinity.edu/rjensen/TheoryAnnuity01.htm
I'm sharing the link to my cost accounting syllabus for a couple of reasons.
http://profalbrecht.wordpress.com/2012/07/31/mycostaccount ingsyllabus/
First, it shows what leading educators say should be in a syllabus. Second, I've spent a lot of time making it pleasing to the eyes.
Dave Albrecht
Hi David,
This is a good start, and I commend you for the effort.
You should add a link to MAAW's great cost accounting modules 
http://maaw.info/JITMain.htm
For example, one module you might want to add in such a syllabus is "Lean
Accounting" 
http://maaw.info/JITMain.htm
"Ethanol learning Curve—the Brazilian experience," by José Goldemberga, Suani Teixeira Coelhob, Plinio Mário Nastaric, Oswaldo Lucond, Biomass and Bioenergy, Volume 26, Issue 3, March 2004, Pages 301–304
Abstract
Economic competitiveness is a very frequent argument against renewable energy (RE). This paper demonstrates, through the Brazilian experience with ethanol, that economies of scale and technological advances lead to increased competitiveness of this renewable alternative, reducing the gap with conventional fossil fuels.
Jensen Comment
Unlike corn ethanol, sugar cane ethanol is a viable renewable energy
alternative. Corn ethanol, however, just does not get enough energy out for the
energy going into its refining. Corn ethanol can only survive on the basis of
government subsidies and tariff. If we want ethanol in our fuel, we should shift
to cane sugar ethanol production and importing rather than raise tariff barriers
against cane sugar ethanol imports.
There is quite a lot of learning curve literature in the energy field. For
example, look at
http://www.sciencedirect.com/science/article/pii/S0301421505001795
Abstract
The extent and timing of costreducing improvements in lowcarbon energy systems are important sources of uncertainty in future levels of greenhousegas emissions. Models that assess the costs of climate change mitigation policy, and energy policy in general, rely heavily on learningcurves to include technology dynamics. Historically, no energy technology has changed more dramatically than photovoltaics (PV), the cost of which has declined by a factor of nearly 100 since the 1950s. Which changes were most important in accounting for the cost reductions that have occurred over the past three decades? Are these results consistent with the notion that learning from experience drove technical change? In this paper, empirical data are assembled to populate a simple model identifying the most important factors affecting the cost of PV. The results indicate that learning from experience, the theoretical mechanism used to explain learning curves, only weakly explains change in the most important factors—plant size, module efficiency, and the cost of silicon. Ways in which the consideration of a broader set of influences, such as technical barriers, industry structure, and characteristics of demand, might be used to inform energy technology policy are discussed.