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## George’s dream

Posted in Kids, Travel with tags , , , , , on April 11, 2015 by xi'an

While I have shared this idea with many of my friends [in both senses that I mentioned it and that they shared the same feeling that it would be a great improvement], the first time I heard of the notion was in George Casella‘s kitchen in Ithaca, New York, in the early 1990’s… We were emptying the dishwasher together and George was reflecting that it would be so convenient to have a double dishwasher and remove the need to empty it altogether! Although, at the moral level, I think that we should do without dishwashers, I found this was a terrific idea and must have told the joke to most of my friends. I was nonetheless quite surprised and very pleased to receive the news from Nicole today that Fisher & Paykel (from Auckland, New Zealand) had gone all the way to produce a double dishwasher, or more exactly a double dishdrawer, perfectly suited to George’s wishes! (Pleased that she remembered the notion after all those years, not pleased with the prospect of buying a double dish washer for more than double the cost of [and a smaller volume than] a regular dishwasher!)

## 1500th, 3000th, &tc

Posted in Books, R, Statistics, University life with tags , , , , , , , , , , , , on January 8, 2012 by xi'an

As the ‘Og reached its 1500th post and 3000th comment at exactly the same time, a wee and only mildly interesting Sunday morning foray in what was posted so far and attracted the most attention (using the statistics provided by wordpress). The most visited posts:

Title Views
In{s}a(ne)!! 7,422
“simply start over and build something better” 6,264
Julien on R shortcomings 2,676
Sudoku via simulated annealing 2,402
Of black swans and bleak prospects 1,768
Solution manual to Bayesian Core on-line 1,628
Parallel processing of independent Metropolis-Hastings algorithms 1,625
Bayesian p-values 1,595
Bayes’ Theorem 1,537
#2 blog for the statistics geek?! 1,526
Do we need an integrated Bayesian/likelihood inference? 1,501
Coincidence in lotteries 1,396
Solution manual for Introducing Monte Carlo Methods with R 1,340
Julian Besag 1945-2010 1,293
The Search for Certainty 1,016

Hence, three R posts (incl. one by Julien and one by Ross Ihaka), three (critical) book reviews, two solution manuals, two general Bayesian posts, two computational entries, one paper (with Pierre Jacob and Murray Smith), one obituary, and one photograph news report… Altogether in line with the main purpose of the ‘Og. The most commented posts:

Not exactly the same as above! In particular, the posts about ABC model choice and our PNAS paper got into the list. At last, the top search terms:

Search Views
benidorm 914
introducing monte carlo methods with r 514
andrew wyeth 398
mistborn 352
abele blanc 350
nested sampling 269
particle mcmc 269
bayesian p-value 263
julian besag 257
rites of love and math 249
millenium 237
bayesian p value 222
marie curie 221
bonsai 200

(out of which I removed the dozens of variations on xian’s blog). I find it rather sad that both top entries are beach towns that are completely unrelated to my lifestyle and to my vacation places. Overall, more than a  half of those entries do not strongly relate to the contents of the ‘Og (even though I did post at length about Saunderson’s Mistborn and Larsson’s Millenium trilogies). At last, the most popular clicks are

which include links to my books on Amazon, Andrew Gelman’s, Terry Tao’s, Radford Neal’s and Romain François’s blogs, the CREST stat students collective blog, and a few arXiv papers of mine’s…

## WSC [2]011

Posted in Mountains, pictures, Running, Statistics, Travel, University life with tags , , , , , , , , , , , , on December 15, 2011 by xi'an

Last day at WSC 2011: as it was again raining, I could not run a second time into the South Mountain Preserve park. (I had a swim at 5am instead and ended up having a nice chat with an old man in the pool under the rain!) My first morning session was rather disappointing with two talks that remained at such a high level of generality as to be useless and a mathematical talk about new forms of stochastic approximation that included proofs and no indication on the calibration of its many parameters. During the coffee break, I tried to have a chat with a vendor of a simulation software but we were using so different vocabularies that I soon gave up. (A lot of the software on display was a-statistical in that users would build a network, specify all parameters, incl. the distributions at the different nodes and start calibrating those parameters towards a behaviour that suited them.) The second session was much more in my area of interest/expertise, with Paul Dupuis giving a talk in the same spirit as the one he gave in New York last September. using large deviations and importance sampling on diffusions. Both following talks were about specially designed importance sampling techniques for rare events and about approximating the zero variance optimal importance function: Yixi Shin gave a talk on cross-entropy based selection of mixtures for the simulation of tail events, connecting somehow with the talk on mixtures of importance sampling distributions I attended yesterday. Although I am afraid I dozed a while during the talk, it was an interesting mix with the determination of the weights by cross-entropy arguments reminded me of what we did for the population Monte Carlo approach (since it also involved some adaptive entropy optimisation). Zdravko Botev gave a talk on approximating the ideal zero variance importance function by MCMC and a sort of Rao-Blackwell estimator that gives an unbiased estimator of this density under stationarity. Then it was time to leave for the airport (and wait in a Starbucks for the plane to Minneapolis and then London to depart, as there is no such thing as a lounge in Phoenix airport…). I had an interesting exchange with a professional magician in the first plane, The Amazing Hondo!, who knew about Persi and was a former math teacher. He explained a few tricks to me, plus showed me his indeed amazing sleight of hands in manipulating cards. In exchange, I took Persi’s book on Magic and Mathematics out of my bag so that he could have look at it. (The trip to London was completely uneventful as I slept most of the way.)

Overall, WSC 2011 was an interesting experience in that (a) the talks I attended on zero variance importance simulation set me thinking again on potential applications of the apparently useless optimality result; (b) it showed me that most people using simulation do not, N.O.T., relate to Monte Carlo techniques (to the extent of being completely foreign to my domains of expertise); and (c) among the parallel sessions that cover military applications, health care simulation, &tc., there always is a theme connecting to mines, which means that I will find sessions to attend when taking part in WSC 2012 in Berlin next year (since I have been invited for a talk). This will be the first time WSC is held outside North America. Hopefully, this will attract simulation addicts from Europe as well as elsewhere.

## art brut

Posted in pictures, Travel with tags , on October 15, 2011 by xi'an

## water towers

Posted in pictures, Travel with tags , on October 1, 2011 by xi'an

## workshop in Columbia [day 3]

Posted in pictures, R, Running, Statistics, Travel, University life with tags , , , , , , , , , , on September 27, 2011 by xi'an

Although this was only a half-day of talks, the third day of the workshop was equally thought-challenging and diverse.  (I managed to miss the ten first minutes by taking a Line 3 train to 125th street, having overlooked the earlier split from Line 1… Crossing south Harlem on a Sunday morning is a fairly mild experience though.) Jean-Marc Azaïs gave a personal recollection on the work of Mario Wschebor, who passed away a week ago and should have attended the workshop. Nan Chen talked about the Monte Carlo approximation of quantities of the form

$\mathbb{E}[f(\mathbb{E}[Y|X])]$

which is a problem when f is non linear. This reminded me (and others) of the Bernoulli factory and of the similar trick we use in the vanilla Rao-Blackwellisation paper with Randal Douc. However, the approach was different in that the authors relied on a nested simulation algorithm that did not adapt against f. And did not account for the shape of f. Peter Glynn, while also involved in the above, delivered a talk on the initial transient that showed possibilities for MCMC convergence assessment (even though this is a much less active area than earlier). And, as a fitting conclusion, the conference organiser, Jingchen Liu gave a talk on non-compatible conditionals he and Andrew are using to handle massively-missing datasets. It reminded me of Hobert and Casella (1996, JASA) of course and also of discussions we had in Paris with Andrew and Nicolas. Looking forward to the paper (as I have missed some points about the difference between true and operational models)! Overall, this was thus a terrific workshop (I just wish I could have been able to sleep one hour more each night to be more alert during all talks!) and a fantastic if intense schedule fitting the start of the semester and of teaching (even though Robin had to teach my first R class in English on Friday). I also discovered that several of the participants were attending the Winter Simulation Conference later this year, hence another opportunity to discuss simulation strategies together.

## workshop in Columbia [day 2]

Posted in Kids, pictures, Statistics, Travel, University life with tags , , , , on September 26, 2011 by xi'an

The second day at the workshop was closer to my research topics and thus easier to follow, if equally enjoyable compared with yesterday: Jun Liu’s talk went over his modification of the Clifford-Fearnhead particle algorithm in great details, Sam Kou explained how a simulated annealing algorithm could make considerable improvement in the prediction of the 3D structure of molecules, Jeff Rosenthal showed us the recent results on and applications of adaptive MCMC, Gareth Roberts detailed his new results on the exact simulation of diffusions, and Xiao-Li Meng went back to his 2002 Read Paper to explain how we should use likelihood principles in Monte Carlo as well. And convince me I was “too young” to get the whole idea! (As I was a discussant of this paper.) All talks were thought-provoking and I enjoyed very much Gareth’s approach and description of the algorithm (as did the rest of the audience, to the point of asking too many questions during the talk!). However, the most revealing talk was Xiao-Li’s in that he did succeed in convincing me of the pertinence of his “unknown measure” approach thanks to a multiple mixture example where the actual mixture importance sampler

$\dfrac{1}{n}\sum_{i=1}^n \dfrac{q(x_i)}{\sum \pi_j p_j(x_i)}$

gets dominated by the estimated mixture version

$\dfrac{1}{n}\sum_{i=1}^n \dfrac{q(x_i)}{\sum \hat\pi_j p_j(x_i)}$

Even though I still remain skeptical by the group averaging perspective, for the same reason as earlier that the group is not acting in conjunction with the target function. Hence averaging over transforms of no relevance for the target. Nonetheless, the idea of estimating the best “importance function” based on the simulated values rather than using the genuine importance function is quite a revelation, linking with an earlier question of mine (and others) on the (lack of) exploitation of the known values of the target at the simulated points. (Maybe up to a constant.) Food for thought, certainly… In memory of this discussion, here is a picture [of an ostrich] my daughter drew at the time for my final slide in London:


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# If $f(x)\in F[x]$ is solvable by radicals then its Galois group $Gal(E/F)$ is solvable.

In all the books that I've checked the use of roots of unity (as hypothesis) is very crucial to prove that if $f(x)\in F[x]$ is solvable by radicals then its Galois group $Gal(E/F)$ is solvable, but the professor said that it's not needed that the field must have enough roots of unity. I think this is curious, and I want to prove it. She also gave three steps to prove this and I'm struggling with them.

(1) Let $F=B_0\subseteq B_1\subseteq\dots\subseteq B_t$ a radical extension with $E\subseteq B_t$ and $B_t$ the splitting field of $f(x)\in F[x]$ with $deg(f(x))=m$. We defined $\mathcal E$ as the splitting field of $h(x)=x^m-1$ over $E$ and $\mathcal F=F(\Psi)$, where $\Psi$ are all the $m$th roots of unity. Prove that $\mathcal E$ is the splitting field of $f(x)$ over $\mathcal F$, and there for $Gal(\mathcal E/\mathcal F)$ is solvable.

I don't see why do we need that function $h(x)$ if what we are trying to do is to take the roots of unity out. Also, for $\mathcal E,\mathcal F$ we are basically adding the same elements, $\Psi$, to $E,F$, but by hypothesis we already have that $E=F(\alpha_1,\dots,\alpha_m)$ with $\alpha_i$ the roots of $f(x)$, and $\mathcal E= E(\Psi)=F(\alpha_1,\dots,\alpha_m,\xi_1,\dots,\xi_m)$ so if anything we are adding things and $f(x)$ can still get decomposed into linear factors, so what are we suppose to do?

(2) Prove that $Gal(\mathcal E/\mathcal F)\unlhd Gal(\mathcal E/ F)$ and $Gal(\mathcal E/ F)/Gal(\mathcal E/\mathcal F)\cong Gal(\mathcal F/F)$ and $Gal(\mathcal E/F)$ is solvable.

To prove that $Gal(\mathcal E/\mathcal F)\unlhd Gal(\mathcal E/ F)$, we show that $\mathcal F/F$ is a Galois extension: lets take $f(x)\in F[x]$, then by definition all the roots of $f(x)$ are in $\mathcal F$, then it is a splitting field, hence $\mathcal F/F$ is a Galois extension.
Now we can talk about the quotient $Gal(\mathcal E/ F)/Gal(\mathcal E/\mathcal F)$ and $Gal(\mathcal F/F)$, but I can't find the right isomorphism to fit this, and I'm starting to think that maybe this isn't right.
And to prove $Gal(\mathcal E/F)$ is solvable, I wanted to prove that there exists a radical extension, but to use that I would need that $F$ is of characteristic $0$, and I don't have as a hypothesis.

(3) Prove that $Gal(\mathcal E/E) \unlhd Gal(\mathcal E/F)$ and $Gal(\mathcal E/ F)/Gal(\mathcal E/ E)\cong Gal(E/F)$, and conclude that $Gal(E/F)$ is solvable.

The prove is pretty much the same as before, isn't it?

(p.s. sorry for the long post)

EDIT: I think I came up with te proof for (1). We know that $\alpha_i\in\mathcal E$ and $\mathcal F\subseteq\mathcal E$ then by the definition of $\mathcal E$ it is the splitting field of $f(x)$ over $\mathcal F$. Now lets take the tower of fields: $\mathcal F=B_0(\Psi)\subseteq B_1(\Psi)\subseteq\dots\subseteq B_t(\Psi)$ with $\mathcal E\subseteq B_t(\Psi)$ then it is a radical extension and $\mathcal E/\mathcal F$ is Galois, where $\mathcal F$ contains all the roots of unity, then $Gal(\mathcal E/\mathcal F)$ is solvable

For (2),(3) I still can't figure oit how to prove the isomorphism.

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A minor comment: it does not make sense to say that a group is solvable by radicals. Groups are just solvable, or not. –  Derek Holt May 18 '14 at 12:10
@DerekHolt, yes you're right, I've corrected it... –  Ana Galois May 18 '14 at 16:47

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Cost behavior is associated with learning how costs change... 1 answer below »

What is cost behavior? Cost behavior is associated with learning how costs change when there is a change in an organization's level of activity. The costs which vary proportionately with the changes in the level of activity are referred to asvariable costs. The costs that are unaffected by changes in the level of activity are classified asfixed costs.
Cost behavior is not required for external reporting under U.S.GAAP. However, the understanding of cost behavior is very important for management's efforts to plan and control its organization's costs.Budgetsand variance reports are more effective when they reflect cost behavior patterns.
The understanding of cost behavior is also necessary for calculating a company'sbreak-even pointand for any other cost-volume-profit analysis.

Introduction to Activity Based Costing Activity based costing (ABC) assigns manufacturing overhead costs to products in a more logical manner than the traditional approach of simply allocating costs on the basis of machine hours. Activity based costing first assigns costs to the activities that are the real cause of the overhead. It then assigns the cost of those activities only to the products that are actually demanding the activities.
Let's discuss activity based costing by looking at two products manufactured by the same company. Product 124 is a low volume item which requires certain activities such as special engineering, additional testing, and many machine setups because it is ordered in small quantities. A similar product, Product 366, is a high volume product—running continuously—and requires little attention and no special activities. If this company used traditional costing, it might allocate or "spread" all of its overhead to products based on the number of machine hours. This will result in little overhead cost allocated to Product 124, because it did not have many machine hours. However, it did demand lots of engineering, testing, and setup activities. In contrast, Product 366 will be allocated an enormous amount of overhead (due to all those machine hours), but it demanded little overhead activity. The result will be a miscalculation of each product's true cost of manufacturing overhead. Activity based costing will overcome this shortcoming by assigning overhead on more than the one activity, running the machine.
Activity based costing recognizes that the special engineering, special testing, machine setups, and others are activities that cause costs—they cause the company to consume resources. Under ABC, the company will calculate the cost of the resources used in each of these activities. Next, the cost of each of these activities will be assigned only to the products that demanded the activities. In our example, Product 124 will be assigned some of the company's costs of special engineering, special testing, and machine setup. Other products that use any of these activities will also be assigned some of their costs. Product 366 will not be assigned any cost of special engineering or special testing, and it will be assigned only a small amount of machine setup.
Activity based costing has grown in importance in recent decades because (1) manufacturing overhead costs have increased significantly, (2) the manufacturing overhead costs no longer correlate with the productive machine hours or direct labor hours, (3) the diversity of products and the diversity in customers' demands have grown, and (4) some products are produced in large batches, while others are produced in small batches. Activity Based Costing with Two Activities Let's illustrate the concept of activity based costing by looking at two common manufacturing activities: (1) the setting up of a production machine for running batches of products, and (2) the actual production of the units of product.
We will assume that a company has annual manufacturing overhead costs of $2,000,000—of which$200,000 is directly involved in setting up the production machines. During the year the company expects to perform 400 machine setups. Let's also assume that thebatch sizesvary considerably, but the setup efforts for each machine are similar.
The cost per setup is calculated to be $500 ($200,000 of cost per year divided by 400 setups per year). Under activity based costing, $200,000 of the overhead will be viewed as abatch-level cost.This means that$200,000 will first be allocated to batches of products to be manufactured (referred to as aStage 1 allocation), and then be assigned to the units of product in each batch (referred to asStage 2 allocation). For example, if Batch X consists of 5,000 units of product, the setup cost per unit is $0.10 ($500 divided by 5,000 units). If Batch Y is 50,000 units, the cost per unit for setup will be $0.01 ($500 divided by 50,000 units). For simplicity, let's assume that the remaining $1,800,000 of manufacturing overhead is caused by the production activities that correlate with the company's 100,000 machine hours. For our simple two-activity example, let's see how the rates for allocating the manufacturing overhead would lookwithactivity based costing andwithoutactivity based costing: Next, let's see what impact these different allocation techniques and overhead rates would have on the per unit cost of a specific unit of output. Assume that a company manufactures a batch of5,000units and it produces 50 units per machine hour, here is how the cost assigned to the units with activity based costing and without activity based costing compares: If a company manufactures a batch of50,000units and produces 50 units per machine hour, here is how the cost assigned to the units with ABC and without ABC compares: As the tables above illustrate, with activity based costing the cost per unit decreases from$0.46 to $0.37 because the cost of the setup activity is spread over 50,000 units instead of 5,000 units. Without ABC, the cost per unit is$0.40 regardless of the number of units in each batch. If companies base their selling prices on costs, a companynotusing an ABC approach might lose the large batch work to a competitor who bids a lower price based on the lower,more accurateoverhead cost of $0.37. It's also possible that a company not using ABC may find itself being the low bidder for manufacturing small batches of product, since its$0.40 is lower than the ABC model of $0.46 for a batch size of 5,000 units. With its bid price based on manufacturing overhead of$0.40—but a true cost of $0.46—the company may end up doing lots of production for little or no profit. Activity Based Costing with Four Activities Let's add two more activities to our example:procurementand material handling. The costs of these two activities are not caused by—nor do they correlate with—machine hours. Rather, we will assume that both of these activities are related to thephysical weightof the direct material used in making the product. The company determines that$300,000 of its annual manufacturing overhead is associated with procurement and material handling. As a result, the company removes $300,000 from the manufacturing overhead that will be allocated via machine hours, and instead plans to allocate the$300,000 to the products based on the weight of the materials used. The company expects that during the year it will procure and handle 3,000,000 pounds of material. Under activity based costing, the company will assign $0.10 ($300,000 divided by 3,000,000 pounds) per pound of product weight to each unit manufactured. The end result is that the heavier parts will not only have more direct material cost, they will also be assigned more factory overhead than the lighter parts. By assigning some manufacturing overhead to a product based on the product's weight, the remaining manufacturing overhead assigned via machine hours will be reduced. These points are illustrated in the following table:

In the table below we can see how ABC would assign costs to the following:
1. A product that weighs 0.5 pound and is produced in a batch of 50,000 units at a rate of 50 per hour.
2. A product that weighs 1.5 pounds and is produced in a batch of 50,000 units at a rate of 50 per hour.
3. No activity based costing allocations—all manufacturing overhead costs are allocated entirely via machine hours.

If the manufacturing overhead costs are caused by a number of activities such as setup, procurement, handling, and production, then using the activity based costing method of determining costs will give you a result that is closer to the true costs. As you can see, the product that weighs 0.5 pound is assigned $0.36 of manufacturing overhead, while the product weighing 1.5 pounds is assigned$0.46 of manufacturing overhead. Under the traditional costing allocations the procurement and handling costs would be assigned on production hours. Keep in mind that whenever manufacturers have a diverse lineup of products, allocating costs on a single basis (such as machine hours) will result in inaccurate per-unit manufacturing overhead costs.
Activity Based Costing and Indirect Manufacturing Costs
Direct Costs
Some manufacturing costs are directly traceable to products. For example, consider a company that makes golf clubs and tennis rackets by assembling components purchased from suppliers.
Golf clubs and tennis rackets are made from the following components.
 Golf Clubs Tennis Rackets Club Heads Frames Shafts Strings Grips Grips

For golf clubs the costs of the club heads, shafts and grips can be traced and accounted for as raw material costs. For tennis rackets the costs of the frames, grips and strings can also be traced and accounted for as raw material costs. So direct raw material costs can be traced to the product lines of golf clubs and tennis rackets.
Similarly, the direct labor costs of assembling golf clubs and tennis rackets can be traced to the product lines of golf clubs and tennis rackets. The time cards of factory workers can identify how many hours each worker spent on each product line.
Indirect Costs
Imagine that the assembly of golf clubs and tennis rackets takes place in a rented facility. The costs of this facility are rent, insurance, security, cleaning, water, electric power, telephone, postage and delivery, and depreciation of machinery and equipment. These combined costs are known as manufacturing overhead[1]. Most of these costs are fixed, meaning that they do not increase or decrease according to the level of manufacturing activity.
It is very difficult to trace these manufacturing overhead costs to the product lines of golf clubs and tennis rackets. For example, it is hard to know how much of the telephone expense relates to golf clubs and how much to tennis rackets. It is even harder to distinguish the telephone expense for golf clubs with steel shafts versus graphite shafts, or tennis rackets with gut strings versus nylon strings.
So direct product costs like raw materials and factory labor can readily be traced to product lines. But indirect product costs cannot be traced to product lines. Therefore indirect product costs must be allocated to product lines. The term “allocated” means assigned to product lines either by some formula or by arbitrary means.
Traditional cost systems allocate manufacturing overhead costs to individual product lines in an arbitrary manner, usually by means of a single allocation base, such as direct labor hours. For example, imagine that manufacturing overhead costs total $800,000 for the month of May. Direct labor hours in May total 6,450, of which 2,560 hours are for tennis rackets and 3,890 hours are for golf clubs. Then manufacturing overhead costs are allocated as follows: Allocation of Manufacturing Overhead to Tennis Rackets and Golf Clubs  Item Tennis Golf Total Direct Labor Hours 2,560 3,890 6,450 Direct Labor Hours % 39.7% 60.3% 100.0% Manufacturing Overhead Allocation 39.7% of$800,000 = $317,600 60.3% of$800,000 = $482,400$800,000 # of units produced 12,000 rackets 20,000 clubs Manufacturing Overhead Allocation per unit $26.47$24.12

It is typically unknown whether or not this allocation is accurate. In fact, it can only be accurate if manufacturing overhead costs are actually driven by direct labor hours. This seems unlikely, because direct labor hours vary according to the level of manufacturing activity, but manufacturing overhead costs tend to include fixed costs as well as variable costs. So manufacturing overhead costs do not vary according to the level of manufacturing activity because they include fixed costs. Therefore traditional cost allocations are frequently inaccurate and unreliable. As a result, product decisions by management can likewise be inaccurate and unreliable.
Activity Based Costing
In order to provide more accurate and reliable allocations of manufacturing overhead costs, activity based costing was developed. Instead of a single allocation base, activity based costing uses separate cost pools and allocation bases for different activities. Each cost pool is allocated to individual product lines based on whatever driver causes the costs in each pool to be incurred by each individual product line.
For example, one cost pool may consist of all costs relating to machinery and equipment, such as maintenance and depreciation of machinery and equipment, plus a share of the rent, insurance, and electric power costs. Assume that all machinery and equipment is used to assemble golf clubs, and that tennis rackets are gripped and strung completely by hand. Then clearly, all machinery and equipment costs will be allocated to golf clubs, and none to tennis rackets. This is a more accurate allocation than traditional costing can provide. Under traditional costing, using direct labor hours to allocate manufacturing overhead costs to product lines, machinery and equipment costs would partly be allocated to golf clubs, and partly to tennis rackets – which would be incorrect.
Activity based costing is more accurate and reliable than traditional costing because:
• It uses multiple cost pools,
• Based on different activities,
• To allocate costs to product lines,
• Using appropriate cost drivers.
Summary
Costs of direct materials and direct labor are not difficult to trace to different product lines. But indirect manufacturing overhead costs are very difficult to trace to different product lines. Traditional cost systems use arbitrary single allocators (like direct labor hours) to assign indirect manufacturing overhead costs to different product lines. The resulting allocations are often inaccurate and unreliable.
Activity based costing is more accurate and reliable than traditional costing in allocating indirect manufacturing overhead costs to different product lines. Therefore, the greater the proportion of indirect manufacturing overhead costs to direct material and labor costs, the greater the need for activity based costing. And, since indirect manufacturing overhead costs tend to be fixed costs, the greater the proportion of fixed costs to variable costs, the greater the advantage of activity based costing over traditional costing.
Capital Budgeting
Payback, Discounted Payback, NPV, Profitability Index, IRR and MIRR are all capital budgeting decision methods.
Cash Flow- We are going to assume that the project we are considering approving has the following cash flow. Right now, in year zero we will spend 15,000 dollars on the project. Then for 5 years we will get money back as shown below.
 Year Cash flow 0 -15,000 1 +7,000 2 +6,000 3 +3,000 4 +2,000 5 +1,000
Payback -When exactly do we get our money back, when does our project break even. Figuring this is easy. Take your calculator.
 Year Cash flow Running Total 0 -15,000 -15,000 1 +7,000 -8,000 (so after the 1st year, the project has not yet broken even) 2 +6,000 -2,000 (so after the 2nd year, the project has not yet broken even) 3 +3,000 +1,000 (so the project breaks even sometime in the 3rd year)
But when, exactly? Well, at the beginning of the year we had still had a -2,000 balance, right? So do this.
 Negative Balance / Cash flow from the Break Even Year = When in the final year we break even -2,000 / 3,000 = .666
So we broke even 2/3 of the way through the 3rd year. So the total time required to payback the money we borrowed was 2.66 years.
Discounted Payback -is almost the same as payback, but before you figure it, you first discount your cash flows. You reduce the future payments by yourcost of capital. Why? Because it is money you will get in the future, and will be less valuable than money today. (SeeTime Value of Moneyif you don't understand). For this example, let's say the cost of capital is 10%.
 Year Cash flow Discounted Cash flow Running Total 0 -15,000 -15,000 -15,000 1 7,000 6,363 -8,637 2 6,000 4,959 -3,678 3 3,000 2,254 -1,424 4 2,000 1,366 -58 5 1,000 621 563

So we break even sometime in the 5th year. When?
 Negative Balance / Cash flow from the Break Even Year = When in the final year we break even -58 / 621 = .093
So using the Discounted Payback Method we break even after 4.093 years.
Net Present Value (NPV) -Once you understand discounted payback, NPV is so easy! NPV is the final running total number. That's it. In the example above the NPV is 563. That's all. You're done, baby. Basically NPV and Discounted Payback are the same idea, with slightly different answers. Discounted Payback is a period of time, and NPV is the final dollar amount you get by adding all the discounted cash flows together. If the NPV is positive, then approve the project. It shows that you are making more money on the investment than you are spending on your cost of capital. If NPV is negative, then do not approve the project because you are paying more in interest on the borrowed money than you are making from the project.
Profitability Index
 Profitability Index equals NPV divided by Total Investment plus 1 PI = 563 / 15,000 + 1
So in our example, the PI = 1.0375. For every dollar borrowed and invested we get back $1.0375, or one dollar and 3 and one third cents. This profit is above and beyond our cost of capital. Internal Rate of Return- IRR is the amount of profit you get by investing in a certain project. It is a percentage. An IRR of 10% means you make 10% profit per year on the money invested in the project. To determine the IRR, you need your good buddy, the financial calculator.  Year Cash flow 0 -15,000 1 +7,000 2 +6,000 3 +3,000 4 +2,000 5 +1,000 Enter these numbers and press these buttons.  -15000 g CFo 7000 g CFj 6000 g CFj 3000 g CFj 2000 g CFj 1000 g CFj f IRR After you enter these numbers the calculator will entertain you by blinking for a few seconds as it determines the IRR, in this case 12.02%. It's fun, isn't it! Ah, yes, but there are problems. • Sometimes it gets confusing putting all the numbers in, especially if you have alternate between a lot of negative and positive numbers. • IRR assumes that the all cash flows from the project are invested back into the project. Sometimes, that simply isn't possible. Let's say you have a sailboat that you give rides on, and you charge people money for it. Well you have a large initial expense (the cost of the boat) but after that, you have almost no expenses, so there is no way to re-invest the money back into the project. Fortunately for you, there is the MIRR. Modified Internal Rate of Return - MIRR- Is basically the same as the IRR, except it assumes that the revenue (cash flows) from the project are reinvested back into the company, and are compounded by the company's cost of capital, but are not directly invested back into the project from which they came. WHAT? OK, MIRR assumes that the revenue is not invested back into the same project, but is put back into the general "money fund" for the company, where it earns interest. We don't know exactly how much interest it will earn, so we use the company's cost of capital as a good guess. Why use the Cost of Capital? Because we know the company wouldn't do a project which earned profits below the cost of capital. That would be stupid. The company would lose money. Hopefully the company would do projects which earn much more than the cost of capital, but, to play it safe, we just use the cost of capital instead. (We also use this number because sometimes the cash flows in some years might be negative, and we would need to 'borrow'. That would be done at our cost of capital.) How to get MIRR - OK, we've got these cash flows coming in, right? The money is going to be invested back into the company, and we assume it will then get at least the company's-cost-of-capital's interest on it. So we have to figure out thefuture value(not the present value) of the sum of all the cash flows. This, by the way is called theTerminal Value. Assume, again, that the company's cost of capital is 10%. Here goes...  Cash Flow Times = Future Valueof that years cash flow. Note 7000 X (1+.1)4 = 10249 compounded for 4 years 6000 X (1+.1)3 = 7986 compounded for 3 years 3000 X (1+.1)2 = 3630 compounded for 2 years 2000 X (1+.1)1 = 2200 compounded for 1 years 1000 X (1+.1)0 = 1000 not compounded at all becausethis is the final cash flow TOTAL = 25065 this is the Terminal Value OK, now get our your financial calculator again. Do this.  -15000 g CFo 0 g CFj 0 g CFj 0 g CFj 0 g CFj 25065 g CFj f IRR Why all those zeros? Because the calculator needs to know how many years go by. But you don't enter the money from the sum of the cash flows until the end, until the last year. Is MIRR kind of weird? Yep. You have to understand that the cash flows are received from the project, and then get used by the company, and increase because the company makes profit on them, and then, in the end, all that money gets 'credited' back to the project. Anyhow, the final MIRR is 10.81%. Decision Time- Do we approve the project? Well, let's review.  Decision Method Result Approve? Why? Payback 2.66 years Yes well, cause we get our money back Discounted Payback 4.195 years Yes because we get our money back, even after discounting our cost of capital. NPV$500 Yes because NPV is positive (reject the project if NPV is negative) Profitability Index 1.003 Yes cause we make money IRR 12.02% Yes because the IRR is more than the cost of capital MIRR 10.81% Yes because the MIRR is more than the cost of capital

[1] Note that manufacturing overhead does not include Selling, General or Administrative expenses, and does not include interest expense or income tax expense. Only manufacturing costs are allocated to products.

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# Converter with parameter

81 2 3

How to parse additional parameters to a converter attribute? Per default a new TypeConverter instance is created.

Is it possible to call the method coerceToUi with further parameters or is it possible to use a pre-instantiated class?

What is the optimal approach for this requirement?

delete retag edit

robertpic71
1275 1

First, there is no way to extend the TypeConverter directly.

Here are my work-a-rounds:

1. use the object instead of the object.attribute
my real world example:
It want to show "customer.sales" but the sales should have another colour when the is > 10 % higher than the last years sales.
The get the whole information i need the customer so i use

<rows>
<row self="@{each=customer}">
<label value="@{customer.number}"/>
..
<label value="@{customer, converter='com.at.od.converters.sales'}"/>
...

Now i get the customer inside my TypeConverter and have no problem to evaluate the color.

2. use custom-attributes to communicate with the TypeConverter
See this side for some examples with "static" parameters.

I should be possible to setup i.e. a converterprogram in the attribute.
<label value="@{customer, converter='com.at.od.converters.spring'}">
<custom-attributes convert="\${springconverter}"/></label>
I've not tested this, but it should work.

Note that the timing for all kinds of listbinding (listbox, rows, combobox...) is important. The single-row/item is used as template for the databinding.
If you need the attribute in all items you have to set this before the binder renders and copy the template.

3. use databinding for custom-attributes
Since ZK 5RC it is possible to use databinding for custom-attributus. So you could pass all your parameters inside custom-attributes and the last one render the item.

/Robert

[hide preview]


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# Is finding a path with more red vertices than blue vertices NP-hard?

Given a connected, directed graph $$G=(V,E)$$, vertices $$s,t \in V$$ and a coloring, s.t. $$s$$ and $$t$$ are black and all other vertices are either red or blue, is it possible to find a simple path from $$s$$ to $$t$$ with more red than blue vertices in polynomial time?

I think it should be possible but our TA said this was NP-hard.

Idea for a solution:

From $$G$$ create $$G'=(V',E')$$ as follows:

• Split all $$v \in V\setminus \{s,t\}$$ in two vertices $$v_{in}$$ and $$v_{out}$$. $$V'$$ is made up of the split vertex pairs and $$s$$ and $$t$$.

• For all $$e=(u,v) \in E$$ introduce an edge $$(u_{out},v_{in})$$. (For edge $$(x,v)$$ or $$(u,x)$$ where $$x \in \{s,t\}$$ create edge $$(x,v_{in})$$ or $$(u_{out},x)$$ resp.). Also, introduce an edge $$(v_{in},v_{out})$$ for any of the split vertices. So $$E'$$ contains two types of edges: the ones that correspond to edges from $$E$$ and the ones that correspond to vertices from $$V$$.

Now, we introduce weights as follows:

• $$w((v_{in},v_{out})) = -1$$ if the corresponding vertex $$v$$ was red.
• $$w((v_{in},v_{out})) = +1$$ if the corresponding vertex $$v$$ was blue.
• $$w(e) = 0$$ for all other edges, i.e. the ones that correspond to edges of $$G$$ rather than vertices.

Now, conduct an algorithm for shortest paths of your choice like Dijkstra, Bellman-Ford,... , check whether the length of the given path is $$<0$$ and you are done.

Why does this not work? Is it because we may have negative cycles? We could detect those with Bellman Ford but then we'd have to find the desired path with non-efficient means rendering this decision problem NP-hard? Is there an elegant reduction to show NP-hardness?

Your solution does not work because Dijkstra and Bellman-Ford cannot interpret "simple path" feature. And they will indeed fall in any negative cycle.

I think the best to show NP-completeness, is to use the Hamiltonian path problem. Let's take a graph $$G'$$ of $$N$$ red vertices.

Then you build a graph $$G$$, adding $$s$$, $$t$$ and $$N-1$$ blue vertices to $$G'$$. You first chain with edges all the blues vertices from the source to the last blue one ($$s$$->$$b_1$$->$$b_2$$->...->$$b_{N-1}$$). Then you put edges from $$b_{N-1}$$ to every red vertex and an edge from every red vertex to $$t$$.

So a single path from $$s$$ to $$t$$ passes necessarly through all blue nodes ($$N-1$$) and then have to pass to all red nodes ($$N$$) to answer to

Is there a simple path in $$G$$ from $$s$$ to $$t$$ with more red than blue vertices ?

which is thus like answer to:

Is there an Hamiltonian path in $$G'$$

So your problem is indeed NP-complete.

• Neat, thanks. So if I wanted to formally prove $DHP \preceq ProblemAbove$ where I have to start with an instance $G,s,t$ for the Hamiltonian Path and map it to an instance $G',s',t'$ of this problem, could I do the following: Color all existing vertices red, add $|V\setminus\{s,t\}|-1$ blue vertices and connect them in a chain $(b_1)\rightarrow \ldots \rightarrow(b_n)$. Add an edge $(s,b_1)$, and for each $(s,v)$ in $G$ an edge $(b_n,v)$ in $G'$, the rest stays as it is. – Valerie Poulain Apr 3 at 14:02
• Yes I am not very familiar with NP-completeness demonstration but this way to present it is clearly better. I would nevertheless start with an instance of $G$ without $s$ & $t$ as initial Hamiltonian path problem has no specific vertex. – Vince Apr 3 at 14:26
• Upvoted for (I think) correctness; but this argument would be WAY easier to understand if you STARTED with the arbitrary graph $G'$ and then embedded it within the specially constructed $G$. Right now the reduction is hiding entirely in the single sentence "Finally, you put any number of edges between red vertices," which I missed on my first reading. "Any number of edges" is code for "you pick any arbitrary graph $G'$ whose Hamiltonian path you'd like to discover, and embed it in there." That's what makes this a valid reduction. – Quuxplusone Apr 3 at 18:45
• @Quuxplusone you are right I edit to make it clearer. – Vince Apr 4 at 7:46
• @ValeriePoulain: Vince is right that starting and ending vertices are not given for the usual HP problem, but if they are given, your reduction needs a slight tweak: You must insert a chain of $|V \setminus \{s, t\}|-1$ blue vertices in a line along each edge between $s$ and any other red vertex. This is to prevent, e.g., a single edge from $s$ to some red vertex from being a valid solution to the constructed instance. – j_random_hacker Apr 4 at 8:23


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# [tex4ht] Problem using mjcli on Cygwin

Francis Wright f.j.wright at live.co.uk
Tue Mar 30 19:16:03 CEST 2021

I have set up the native Windows version of node.js and installed mjcli, which appears to work using both the cmd.exe and Cygwin bash shells. For example, using Cygwin bash, I get

$node -v v14.16.0$ mjcli --version
1.0.0

I have also installed TeX Live (2020 but with a lot of fairly recent updates) for Cygwin, which I have been using for some time. But when I run

\$ make4ht -f html5+mjcli manual reduce-sf,3,xhtml,mathjax " -cunihtf -utf8" -cvalidate

I get

[WARNING] mkutils: Cannot load extension: mjcli

Without the mjcli extension, this command runs fine. I don't understand why mjcli seems to run directly from the shell but not via make4ht. Could it be a path issue? Cygwin bash presumably passes a Cygwin-style path to make4ht, but since I'm using native Windows node.js it might expect a Windows-style path. But if so, why does it run directly from Cygwin bash? I would be grateful for any suggestions how to fix or debug this.

If necessary, I'll try building node.js for Cygwin, but I'm hoping to avoid that.

Thanks, Francis
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# How do you differentiate f(x)= (x-1)/(x+1)?

$f ' \left(x\right) = \frac{2}{x + 1} ^ 2$

#### Explanation:

Use $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \cdot \frac{d}{\mathrm{dx}} u - u \cdot \frac{d}{\mathrm{dx}} \left(v\right)}{{v}^{2}}$

Let $u = x - 1$ and $v = x + 1$

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \cdot \frac{d}{\mathrm{dx}} u - u \cdot \frac{d}{\mathrm{dx}} \left(v\right)}{{v}^{2}}$

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\frac{x - 1}{x + 1}\right)$

$f ' \left(x\right) = \frac{\left(x + 1\right) \cdot \frac{d}{\mathrm{dx}} \left(x - 1\right) - \left(x - 1\right) \cdot \frac{d}{\mathrm{dx}} \left(x + 1\right)}{{\left(x + 1\right)}^{2}}$

$f ' \left(x\right) = \frac{\left(x + 1\right) \left(1\right) - \left(x - 1\right) \left(1\right)}{{\left(x + 1\right)}^{2}}$

$f ' \left(x\right) = \frac{x + 1 - x + 1}{x + 1} ^ 2$

$f ' \left(x\right) = \frac{2}{x + 1} ^ 2$

God bless....I hope the explanation is useful.


text:
# Density of states

In solid-state and condensed matter physics, the density of states (DOS) of a system describes the number of states per interval of energy at each energy level that are available to be occupied. Unlike isolated systems, like atoms or molecules in gas phase, the density distributions are not discrete like a spectral density but continuous. A high DOS at a specific energy level means that there are many states available for occupation. A DOS of zero means that no states can be occupied at that energy level. In general a DOS is an average over the space and time domains occupied by the system. Local variations, most often due to distortions of the original system, are often called local density of states (LDOS). If the DOS of an undisturbed system is zero, the LDOS can locally be non-zero due to the presence of a local potential.

## Introduction

In quantum mechanical (QM) systems, waves, or wave-like particles can occupy modes or states with wavelengths and propagation directions dictated by the system. Often only specific states are permitted. In some systems, the interatomic spacing and the atomic charge of the material allows only electrons of certain wavelengths to exist. In other systems, the crystalline structure of the material allows waves to propagate in one direction, while suppressing wave propagation in another direction. Thus it can happen that many states are possible at a specific wavelength, and therefore at this associated energy, while no states are available at other energy levels: this distribution is characterized by the density of states. Depending on the QM system the density of states can be calculated for electrons, photons, or phonons, and can be given as a function of either energy or the wave vector k. The DOS is usually represented by one of the symbols g, ρ, D, n, or N. To convert between the DOS as a function of the energy and the DOS as a function of the wave vector, the system-specific energy dispersion relation between E and k must be known.

For example, the density of states for electrons in a semiconductor is shown in red in Fig. 4 (in section 5). For electrons at the conduction band edge, very few states are available for the electron to occupy. As the electron increases in energy, the electron density of states increases and more states become available for occupation. However, because there are no states available for electrons to occupy within the bandgap, electrons at the conduction band edge must lose at least ${\displaystyle E_{g}}$ of energy in order to transition to another available mode.

In general, the topological properties of the system have a major impact on the properties of the density of states. The most well-known systems, like neutronium in neutron stars and free electron gases in metals (examples of degenerate matter and a Fermi gas), have a 3-dimensional Euclidean topology. Less familiar systems, like 2-dimensional electron gases (2DEG) in graphite layers and the Quantum Hall effect system in MOSFET type devices, have a 2-dimensional Euclidean topology. Even less familiar are Carbon nanotubes, the quantum wire and Luttinger liquid with their 1-dimensional topologies. Systems with 1D and 2D topologies are likely to become more common, assuming developments in nanotechnology and materials science proceed.

## Symmetry and density of states

First Brillouin zone of FCC lattice, a truncated octahedron, showing symmetry labels for high symmetry lines and points

There are a large variety of systems and types of states for which DOS calculations can be done. An important property of a condensed matter system is the symmetry of the structure on its microscopic scale. Fluids, glasses or amorphous solids have dispersion relations with a rotational symmetry. In spherically symmetric systems the integrals of functions, for instance, are one-dimensional because all variables in the calculation depend only on the radial parameter of the dispersion relation.

Angular dependent calculations or measurements on systems consisting of a single crystal of a compound, for example, are anisotropic, meaning the density of states will be different in one crystallographic direction than in another. Anisotropic problems are more difficult to calculate, and the anisotropic density of states is more difficult to visualize, so methods such as calculating the DOS for particular points or directions only, or calculating the projected density of states (PDOS), are often used.

Octahedron.

Measurements on powders or polycrystalline samples require evaluation and calculation functions and integrals over the whole domain, most often a Brillouin zone, of the dispersion relations the system of interest. Sometimes the symmetry of the system is high. The shape of the functions describing the dispersion relations of the system appears many times over the whole domain of the dispersion relation. In such cases the effort to calculate the DOS can be reduced by a great amount when the calculation is limited to a reduced zone or fundamental domain.[1] The Brillouin zone of the FCC lattice in the figure on the right has the 48 fold symmetry of the point group Oh with full octahedral symmetry. This means that the integration over the whole domain of the Brillouin zone can be reduced to a 48-th part of the whole Brillouin zone. As a crystal structure periodic table shows, there are many elements with a FCC crystal structure, like Diamond, Silicon and Platinum and their Brillouin zones and dispersion relations have this 48 fold symmetry.

First Brillouin zone of BCC lattice, a rhombic dodecahedron.

Two other familiar crystal structures are the BCC and HCP structures with cubic and hexagonal lattices. The BCC structure has the 24 fold pyritohedral symmetry of the point group Th. The HCP structure has the 12 fold prismatic dihedral symmetry of the point group D3h. A complete list of symmetry properties of a point group can be found in point group character tables.

In general it is easier to calculate a DOS when the symmetry of the system is higher and the number of topological dimensions of the dispersion relation is lower. The DOS of dispersion relations with rotational symmetry can often be calculated analytically. This is fortunate, since many materials of practical interest, such as steel and silicon, have high symmetry.

## k-space topologies

Figure 1: Spherical surface in k-space for electrons in three dimensions.

The density of states is dependent upon the dimensional limits of the object itself. The role dimensions play is evident from the units of DOS (Energy−1Volume−1). In the limit that the system is 2 dimensional, a volume becomes an area and in the limit of 1 dimension it becomes a length. It is important to note that the volume being referenced is the volume of k-space, the space enclosed by the constant energy surface of the system derived through a dispersion relation that relates E to k. The dispersion relation for electrons in a solid is given by the electronic band structure. An example of a 3-dimensional k-space is given in Fig. 1. It can be seen that the dimensionality of the system itself will confine the momentum of particles inside the system.

### Density of wave vector states (sphere)

The calculation for DOS starts by counting the N allowed states at a certain k that are contained within [k, k+dk] inside the volume of the system. This is done by differentiating the whole k-space volume ${\displaystyle \Omega _{n,k}}$ in n-dimensions at an arbitrary k, with respect to k. The volume, area or length in 3, 2 or 1-dimensional spherical k-spaces are expressed by

${\displaystyle \Omega _{n}(k)=c_{n}k^{n}}$

for a n-dimensional k-space with the topologically determined constants

${\displaystyle c_{1}=2,{\text{ }}c_{2}=\pi ,{\text{ }}c_{3}={\frac {4\pi }{3}}}$

for linear, disk and spherical symmetrical shaped functions in 1, 2 and 3-dimensional Euclidean k-spaces respectively.

According to this scheme the density of wave vector states N is, through differentiating ${\displaystyle \Omega _{n,k}}$ with respect to k, expressed by

${\displaystyle N_{n}(k)={\frac {d\Omega _{n}(k)}{dk}}=n{\text{ }}c_{n}{\text{ }}k^{(n-1)}}$

The 1, 2 and 3-dimensional density of wave vector states for a line, disk, or sphere are explicitly written as

${\displaystyle N_{1}(k)=2}$
${\displaystyle N_{2}(k)=2\pi k}$
${\displaystyle N_{3}(k)=4\pi k^{2}}$

One state is large enough to contain particles having wavelength λ. The wavelength is related to k through the relationship.

${\displaystyle k={\frac {2\pi }{\lambda }}}$

In a quantum system the length of λ will depend on a characteristic spacing of the system L that is confining the particles. Finally the density of states N is multiplied by a factor ${\displaystyle s/V_{k}}$, where s is a constant degeneracy factor that accounts for internal degrees of freedom due to such physical phenomena as spin or polarization. If no such phenomenon is present then ${\displaystyle s=1}$. Vk is the volume in k-space containing all possible states whose wavevectors are smaller than a certain k serving as a parameter.

### Density of energy states

To finish the calculation for DOS find the number of states per unit sample volume at an energy ${\displaystyle E}$ inside an interval ${\displaystyle [E,E+dE]}$. The general form of DOS of a system is given as

${\displaystyle D_{n}\left(E\right)={\frac {d\Omega _{n}(E)}{dE}}}$

The scheme sketched so far only applies to monotonically rising and spherically symmetric dispersion relations. In general the dispersion relation ${\displaystyle E(k)}$ is not spherically symmetric and in many cases it isn't continuously rising either. To express D as a function of E the inverse of the dispersion relation ${\displaystyle E(k)}$ has to be substituted into the expression of ${\displaystyle \Omega _{n}(k)}$ as a function of k to get the expression of ${\displaystyle \Omega _{n}(E)}$ as a function of the energy. If the dispersion relation is not spherically symmetric or continuously rising and can't be inverted easily then in most cases the DOS has to be calculated numerically. More detailed derivations are available.[2][3]

## Dispersion relations

The kinetic energy of a particle depends on the magnitude and direction of the wave vector k, the properties of the particle and the environment in which the particle is moving. For example, the kinetic energy of an electron in a Fermi gas is given by

${\displaystyle E=E_{0}+{\frac {(\hbar k)^{2}}{2m}}\ ,}$

where m is the electron mass. The dispersion relation is a spherically symmetric parabola and it is continuously rising so the DOS can be calculated easily.

Figure 2: Monatomic chain phonon dispersion relation

For longitudinal phonons in a string of atoms the dispersion relation of the kinetic energy in a 1-dimensional k-space, as shown in Figure 2, is given by

${\displaystyle E=2\hbar \omega _{0}|\sin(ka/2)|}$

where ${\displaystyle \omega _{0}={\sqrt {k_{F}/m}}}$ is the oscillator frequency, ${\displaystyle m}$ the mass of the atoms, ${\displaystyle k_{F}}$ the inter-atomic force constant and ${\displaystyle a}$ inter-atomic spacing. For small values of ${\displaystyle k\ll \pi /a}$ the dispersion relation is rather linear:

${\displaystyle E=\hbar \omega _{0}ka}$

When ${\displaystyle k\approx \pi /a}$ the energy is

${\displaystyle E=2\hbar \omega _{0}|\cos(\pi /2-ka/2)|}$

With the transformation ${\displaystyle q=k-\pi /a}$ and small ${\displaystyle q}$ this relation can be transformed to

${\displaystyle E=2\hbar \omega _{0}[1-(qa/2)^{2}]}$

### Isotropic dispersion relations

The two examples mentioned here can be expressed like

${\displaystyle E=E_{0}+c_{k}k^{p}}$

This is a kind of dispersion relation because it interrelates two wave properties and it is isotropic because only the length and not the direction of the wave vector appears in the expression. The magnitude of the wave vector is related to the energy as:

${\displaystyle k=\left({\frac {E-E_{0}}{c_{k}}}\right)^{1/p}\ ,}$

Accordingly, the volume of n-dimensional k-space containing wave vectors smaller than k is:

${\displaystyle \Omega _{n}(k)=c_{n}k^{n}}$

Substitution of the isotropic energy relation gives the volume of occupied states

${\displaystyle \Omega _{n}(E)={\frac {c_{n}}{c_{k}^{n/p}}}\left(E-E_{0}\right)^{n/p}\ ,}$

Differentiating this volume with respect to the energy gives an expression for the DOS of the isotropic dispersion relation

${\displaystyle D_{n}\left(E\right)={\frac {d}{dE}}\Omega _{n}(E)={\frac {nc_{n}}{pc_{k}^{n/p}}}\left(E-E_{0}\right)^{(n/p-1)}}$

### Parabolic dispersion

Figure 3: Free-electron DOS in 3-dimensional k-space

In the case of a parabolic dispersion relation (p = 2), such as applies to free electrons in a Fermi gas, the resulting density of states, ${\displaystyle D_{n}\left(E\right)}$, for electrons in a n-dimensional systems is

${\displaystyle D_{1}\left(E\right)={\frac {1}{\sqrt {c_{k}(E-E_{0})}}}\ .}$
${\displaystyle D_{2}\left(E\right)={\frac {\pi }{c_{k}}}}$
${\displaystyle D_{3}\left(E\right)=2\pi {\sqrt {\frac {E-E_{0}}{c_{k}^{3}}}}\ .}$

for ${\displaystyle E>E_{0}}$, with ${\displaystyle D(E)=0}$ for ${\displaystyle E.

In 1-dimensional systems the DOS diverges at the bottom of the band as ${\displaystyle E}$ drops to ${\displaystyle E_{0}}$. In 2-dimensional systems the DOS turns out to be independent of ${\displaystyle E}$. Finally for 3-dimensional systems the DOS rises as the square root of the energy.[4]

Including all the prefactors, the expression for the 3D DOS is

${\displaystyle N(E)={\frac {V}{2\pi ^{2}}}\left({\frac {2m}{\hbar ^{2}}}\right)^{3/2}{\sqrt {E-E_{0}}}}$,

where ${\displaystyle V}$ is the total volume, and ${\displaystyle N(E-E_{0})}$ includes the 2-fold spin degeneracy.

### Linear dispersion

In the case of a linear relation (p = 1), such as applies to photons, acoustic phonons, or to some special kinds of electronic bands in a solid, the DOS in 1, 2 and 3 dimensional systems is related to the energy as:

${\displaystyle D_{1}\left(E\right)={\frac {1}{c_{k}}}}$
${\displaystyle D_{2}\left(E\right)={\frac {2\pi }{c_{k}^{2}}}\left(E-E_{0}\right)}$
${\displaystyle D_{3}\left(E\right)={\frac {4\pi }{c_{k}^{3}}}\left(E-E_{0}\right)^{2}}$

## Density of states and distribution functions

The density of states plays an important role in the kinetic theory of solids. The product of the density of states and the probability distribution function is the number of occupied states per unit volume at a given energy for a system in thermal equilibrium. This value is widely used to investigate various physical properties of matter. The following are examples, using two common distribution functions, of how applying a distribution function to the density of states can give rise to physical properties.

Figure 4: The Fermi-Dirac probability distribution, density of states, and their product for a semiconductor. The lower green lobe depicts hole energy, and thus uses ${\displaystyle f(-x)}$ as the distribution function.

Fermi–Dirac statistics: The Fermi–Dirac probability distribution function, Fig. 4, is used to find the probability that a fermion occupies a specific quantum state in a system at thermal equilibrium. Fermions are particles which obey the Pauli exclusion principle (e.g. electrons, protons, neutrons). The distribution function can be written as

${\displaystyle f_{\mathrm {FD} }(E)={\frac {1}{\exp \left({\frac {E-\mu }{k_{\mathrm {B} }T}}\right)+1}}}$

${\displaystyle \mu }$ is the chemical potential (also denoted as EF and called the Fermi level), ${\displaystyle k_{\mathrm {B} }}$ is the Boltzmann constant, and ${\displaystyle T}$ is temperature. Fig. 4 illustrates how the product of the Fermi-Dirac distribution function and the three-dimensional density of states for a semiconductor can give insight to physical properties such as carrier concentration and Energy band gaps.

Bose–Einstein statistics: The Bose–Einstein probability distribution function is used to find the probability that a boson occupies a specific quantum state in a system at thermal equilibrium. Bosons are particles which do not obey the Pauli exclusion principle (e.g. phonons and photons). The distribution function can be written as

${\displaystyle f_{\mathrm {BE} }(E)={\frac {1}{\exp \left({\frac {E-\mu }{k_{B}T}}\right)-1}}}$

From these two distributions it is possible to calculate properties such as the internal energy ${\displaystyle U}$, the number of particles ${\displaystyle n}$, specific heat capacity ${\displaystyle C}$, and thermal conductivity ${\displaystyle k}$. The relationships between these properties and the product of the density of states and the probability distribution, denoting the density of states by ${\displaystyle g(E)}$ instead of ${\displaystyle D(E)}$, are given by

${\displaystyle U=\int E\,f(E)\,g(E)\,dE}$
${\displaystyle n=\int f(E)\,g(E)\,dE}$
${\displaystyle C={\frac {\partial }{\partial T}}\int E\,f(E)\,g(E)\,dE}$
${\displaystyle k={\frac {1}{d}}{\frac {\partial }{\partial T}}\int Ef(E)\,g(E)\,\nu (E)\,\Lambda (E)\,dE}$

${\displaystyle d}$ is dimensionality, ${\displaystyle \nu }$ is sound velocity and ${\displaystyle \Lambda }$ is mean free path.

## Applications

The density of states appears in many areas of physics, and helps to explain a number of quantum mechanical phenomena.

### Quantization

Calculating the density of states for small structures shows that the distribution of electrons changes as dimensionality is reduced. For quantum wires, the DOS for certain energies actually becomes higher than the DOS for bulk semiconductors, and for quantum dots the electrons become quantized to certain energies.

### Photonic crystals

The photon density of states can be manipulated by using periodic structures with length scales on the order of the wavelength of light. Some structures can completely inhibit the propagation of light of certain colors (energies), creating a photonic bandgap: the DOS is zero for those photon energies. Other structures can inhibit the propagation of light only in certain directions to create mirrors, waveguides, and cavities. Such periodic structures are known as photonic crystals. In nanostructured media the concept of Local density of states (LDOS) is often more relevant than that of DOS, as the DOS varies considerably from point to point.

## Calculation of the density of states

Interesting systems are in general complex, for instance compounds, biomolecules, polymers, etc. Because of the complexity of these systems the analytical calculation of the density of states is in most of the cases impossible. Computer simulations offer a set of algorithms to evaluate the density of states with a high accuracy. One of these algorithms is called the Wang and Landau algorithm.[5]

Within the Wang and Landau scheme any previous knowledge of the density of states is required. One proceeds as follows: the cost function (for example the energy) of the system is discretized. Each time the bin i is reached one updates a histogram for the density of states, ${\displaystyle g(i)}$, by

${\displaystyle g(i)\rightarrow g(i)+f}$

where f is called the modification factor. As soon as each bin in the histogram is visited a certain number of times (10-15), the modification factor is reduced by some criterion, for instance,

${\displaystyle f_{n+1}\rightarrow (1/2)f_{n}}$

where n denotes the n-th update step. The simulation finishes when the modification factor is less than a certain threshold, for instance ${\displaystyle f_{n}<10^{-8}}$.

The Wang and Landau algorithm has some advantages over other common algorithms such as multicanonical simulations and Parallel tempering. For example, the density of states is obtained as the main product of the simulation. Additionally, Wang and Landau simulations are completely independent of the temperature. This feature allows to compute the density of states of systems with very rough energy landscape such as proteins.[6]

Mathematically the density of states is formulated in terms of a tower of covering maps.[7]

## References

1. ^ Walter Ashley Harrison (1989). Electronic Structure and the Properties of Solids. Dover Publications. ISBN 0-486-66021-4.
2. ^ Sample density of states calculation
3. ^ Another density of states calculation
4. ^ Charles Kittel (1996). Introduction to Solid State Physics (7th ed.). Wiley. Equation (37), p. 216. ISBN 0-471-11181-3.
5. ^ Wang, Fugao; Landau, D. P. (Mar 2001). "Efficient, Multiple-Range Random Walk Algorithm to Calculate the Density of States". Phys. Rev. Lett. (American Physical Society) 86 (10): 2050–2053. arXiv:cond-mat/0011174. Bibcode:2001PhRvL..86.2050W. doi:10.1103/PhysRevLett.86.2050. PMID 11289852.
6. ^ Ojeda, P.; Garcia, M. (2010). "Electric Field-Driven Disruption of a Native beta-Sheet Protein Conformation and Generation of a Helix-Structure". Biophysical Journal 99 (2): 595–599. Bibcode:2010BpJ....99..595O. doi:10.1016/j.bpj.2010.04.040. PMC 2905109. PMID 20643079.
7. ^ Adachi T. and Sunada. T (1993). "Density of states in spectral geometry of states in spectral geometry". Comment. Math. Helvetici 68: 480–493.


text:
I am trying to reproduce one of my PowerPoint template with beamer.

I would like to link each colored box in a equation with an other box of similar color. Also, I am trying to adjust the position of the boxes behind the equation such as they are aligned with the margin. (like in the PowerPoint).

I am using tabular because I would like to have a kind of title associated with each small box. I know that tikz is highly recommended for boxing equations but if it's possible to use it in this case I would be curious to know.

P.S : Is there any way to make borders of the box less thicker ? Because it's interfering with parenthesis of the main equation...

Thank you for your kind help,

\documentclass[french]{beamer}

%%%%%% ENCODAGE %%%%%%%%%%%
\usepackage[utf8]{inputenc}

%%%%%% TIKZ %%%%%%%%%%%%%%%
\usepackage[beamer,customcolors,norndcorners]{hf-tikz}

%%%%%% OTHERS %%%%%%%%%%%%%
\usepackage{booktabs,calligra}
\usepackage{listings,stackengine}

\author{XXX}
\title{XXX}
\subtitle{XXX}
\institute [XXX] {XXX \\ XXX}
\date{\today}

%%%%%% DEFINITIONS %%%%%%%%%
\def\cmd#1{\texttt{\color{red}\footnotesize $\backslash$#1}}
\def\env#1{\texttt{\color{blue}\footnotesize #1}}
\definecolor{deepblue}{rgb}{0,0,0.5}
\definecolor{deepred}{rgb}{0.6,0,0}
\definecolor{deepgreen}{rgb}{0,0.5,0}
\definecolor{halfgray}{gray}{0.55}

\lstset{
basicstyle=\ttfamily\small,
keywordstyle=\bfseries\color{deepblue},
emphstyle=\ttfamily\color{deepred},    % Custom highlighting style
stringstyle=\color{deepgreen},
numbers=left,
numberstyle=\small\color{halfgray},
rulesepcolor=\color{red!20!green!20!blue!20},
}

%%%%%% VIDEO %%%%%%%%%%%%%%
\usepackage{multimedia}

%%%%%% BOX %%%%%%%%%%%%%%%%
\usepackage{fancybox}
\usepackage{varwidth}
\usepackage{subcaption}

\hfsetbordercolor{blue!50!black}

%%%%%% PGFPLOTS %%%%%%%%%%%%
\usepackage{pgfplots}

\definecolor{mygreen}{RGB}{28,172,0} % color values Red, Green, Blue
\definecolor{mylilas}{RGB}{170,55,241}
\definecolor{BgYellow}{HTML}{FFF59C}
\definecolor{FrameYellow}{HTML}{F7A600}
\usetikzlibrary{spy}
\usepgfplotslibrary{fillbetween}
\usetikzlibrary{patterns, matrix, positioning}
\usetikzlibrary{arrows.meta,
patterns.meta
}

\begin{document}

\begin{frame}{Equation des films minces générale}
\begin{exampleblock}{Cas cylindrique}
\begin{equation*}
\frac{\partial h}{\partial t} = \frac{1}{12 \mu r}  \frac{\partial}{\partial r}  \left(r h^{3} \left(\tikzmarkin<1->[set fill color=red!20, set border color=red]{a1}(0.075,-0.4)(-0.045,0.65) \rho g \frac{\partial h}{\partial r} \tikzmarkend{a1}+ \tikzmarkin<1->[set fill color=blue!20, set border color=blue]{e}(0.05,-0.4)(-0.045,0.65) D \frac{\partial}{\partial r}(\Delta^2_r h)  \tikzmarkend{e}  \right) \right)
\end{equation*}

\begin{tabular}{ccc}
Pesanteur &  & Elastique \\
\fcolorbox{red}{white}{\begin{aligned} R \propto t^{\frac{1}{2}} \end{aligned}} &
\fcolorbox{black}{white}{\begin{aligned} \lambda_{E-G} = \left(\frac{Ed^3}{12 \rho g (1 - \nu)} \right)^{\frac{1}{4}} \end{aligned}} &
\fcolorbox{blue}{white}{\begin{aligned} R \propto t^{\frac{1}{3}} \end{aligned}} \\
\end{tabular}
\end{exampleblock}
\end{frame}

\end{document}

I have this :

And I am trying to reproduce this :

Here one possible approach using tcolorbox:

\documentclass[french]{beamer}

%%%%%% ENCODAGE %%%%%%%%%%%
\usepackage[utf8]{inputenc}

%%%%%% TIKZ %%%%%%%%%%%%%%%
\usepackage[beamer,customcolors,norndcorners]{hf-tikz}

%%%%%% OTHERS %%%%%%%%%%%%%
\usepackage{booktabs,calligra}
\usepackage{listings,stackengine}

\author{XXX}
\title{XXX}
\subtitle{XXX}
\institute [XXX] {XXX \\ XXX}
\date{\today}

%%%%%% DEFINITIONS %%%%%%%%%
\def\cmd#1{\texttt{\color{red}\footnotesize $\backslash$#1}}
\def\env#1{\texttt{\color{blue}\footnotesize #1}}
\definecolor{deepblue}{rgb}{0,0,0.5}
\definecolor{deepred}{rgb}{0.6,0,0}
\definecolor{deepgreen}{rgb}{0,0.5,0}
\definecolor{halfgray}{gray}{0.55}

\lstset{
basicstyle=\ttfamily\small,
keywordstyle=\bfseries\color{deepblue},
emphstyle=\ttfamily\color{deepred},    % Custom highlighting style
stringstyle=\color{deepgreen},
numbers=left,
numberstyle=\small\color{halfgray},
rulesepcolor=\color{red!20!green!20!blue!20},
}

%%%%%% VIDEO %%%%%%%%%%%%%%
\usepackage{multimedia}

%%%%%% BOX %%%%%%%%%%%%%%%%
\usepackage{fancybox}
\usepackage{varwidth}
\usepackage{subcaption}

\hfsetbordercolor{blue!50!black}

%%%%%% PGFPLOTS %%%%%%%%%%%%
\usepackage{pgfplots}

\definecolor{mygreen}{RGB}{28,172,0} % color values Red, Green, Blue
\definecolor{mylilas}{RGB}{170,55,241}
\definecolor{BgYellow}{HTML}{FFF59C}
\definecolor{FrameYellow}{HTML}{F7A600}
\usetikzlibrary{spy}
\usepgfplotslibrary{fillbetween}
\usetikzlibrary{patterns, matrix, positioning}
\usetikzlibrary{arrows.meta,
patterns.meta
}

\usepackage[most]{tcolorbox}
\tcbset{highlight math style={enhanced,colframe=red,colback=red!10!white,boxsep=0pt,sharp corners,
equal height group=C,
minimum for equal height group=C:1.5cm,
valign=center,
}}

\begin{document}

\begin{frame}{Equation des films minces générale}
\begin{exampleblock}{Cas cylindrique}
\begin{equation*}
\frac{\partial h}{\partial t} = \frac{1}{12 \mu r}  \frac{\partial}{\partial r}  \left(r h^{3} \left(
%
\tcbhighmath[remember as=a,tcbox raise=-7mm]{
\rho g \frac{\partial h}{\partial r}
}
+
\tcbhighmath[remember as=b,colback=blue!10,colframe=blue,tcbox raise=-7mm]{
D \frac{\partial}{\partial r}(\Delta^2_r h)
}
\right) \right)
\end{equation*}

\vskip1.5cm

\begin{equation*}
\tcbhighmath[
tcbox raise=0mm,
remember as=c,
overlay={
\node[anchor=north] at (frame.south) {Pesanteur};
\draw[red,-latex,thick] (a.south) -- (frame.north);
}
]{
\displaystyle R \propto t^{\frac{1}{2}}
\tcbhighmath[
tcbox raise=0mm,
remember as=d,
colback=black!10,
colframe=black,
overlay={
\draw[black,-latex,thick] (c.east) -- (frame.west);
}
]{
\lambda_{E-G} = \left(\frac{Ed^3}{12 \rho g (1 - \nu)}  \right)^{\frac{1}{4}}
\tcbhighmath[
tcbox raise=0mm,
remember as=e,
colback=blue!10,
colframe=blue,
overlay={
\draw[blue,latex-,thick] (d.east) -- (frame.west);
\draw[blue,-latex,thick] (b.south) -- (frame.north);
\node[anchor=north] at (frame.south) {Elastique};
}
]{
R \propto t^{\frac{1}{3}}
}
\end{equation*}
\end{exampleblock}
\end{frame}

\end{document}

• Very nice solution using only tcolorbox. What a powerful package it is! Sep 15 at 11:51
• @SebGlav Very powerful, indeed! Sep 15 at 11:54
• Thank you for the answer : do you know in this case what is the difference between tcbhighmath and tcboxmath ? Because I have tried the second option and it doesn't work compared to the first one. Thanks !
– Wiss
Sep 16 at 9:13
• @Wiss \tcboxmath works as well, but you'll have to specify the options separately because it does not listen to highlight math style Sep 16 at 9:18
• @Wiss You also need the other options from highlight math style, in particular enhanced Sep 16 at 9:23