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
can be changed to corrected link
However in some cases files had to be removed to reduce the size of my Website
Contact me at 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 


Wolfram Alpha (WA) Computational Knowledge Search Engine ---
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 ---

Wolfram Alpha (styled Wolfram|Alpha) 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 ---

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 one-step 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, 137-142 --- 

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 ---

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)


A few other things to try in Wolfram Alpha:


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, 137-142 --- 

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 vis-à-vis the marginal cost u(2) of the second unit.

Two online spreadsheets are available at 


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


September 22, 2010 message from Deeley, Chris []


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 closed-form 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 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 []



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 non-closed-form 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 left-hand 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.




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    
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

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


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
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 equally-spaced payments per year and m is the number equally-spaced 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 semi-annual 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:

by Chris Deeley 
Working Paper, Charles Sturt University, Australia, September 22, 2010

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 semi-annually, 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

July 31, 2012 message from David Albrecht

I'm sharing the link to my cost accounting syllabus for a couple of reasons.

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 --- 

For example, one module you might want to add in such a syllabus is "Lean Accounting" --- 


Personally, I would probably play down 80%  learning curves since the early work on learning curves in airplane manufacturing really don't extrapolate well to most other industries --- 

You can, however, bring in more recent evidence on learning curves.

"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


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


The extent and timing of cost-reducing improvements in low-carbon energy systems are important sources of uncertainty in future levels of greenhouse-gas 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.