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## The performance outcome cliff of early professionalism Russell Degnan

The last decade has seen professional contracts expand across both associate and women's cricket. The opportunities have improved player skills, and undoubtedly history will be kind to the first generation of professional players wherever that has occurred. But it also causes a problem. One that has manifested in many of the teams across the world over many years, both at international and first class level.

Namely: if you give a small number of elite players generous contracts to improve performance, the next generation won't be able to displace them until they are in steep decline.

It was a problem, misdiagnosed largely, when Australian first class cricket aged considerably in the early 2000s, with the Kenyan team of the same era, and more latterly, with the English women, and the Irish team, both of whom bemoan the lack of emerging youth.

The phenomenon is best demonstrated with a simulation. In the graph below, five careers overlap, with different starts, peaks and lengths. The blue lines represent the best outcome for a player (continually professionally contracted), the red lines represent a player contracted only if they were in the top two players in the previous year. They receive a 10% performance jump in the first year, and 25% in subsequent years, while contracted. The thick lines are the average of the top two players

In an optimum selection outcome, players A and B would be selected through years 7-12, A making way for C until year 14, B for D until year 16, C for E until year 18, when C returns for D at the end of their short but prolific peak. The selected pair maintaining an average of close to 50 from year 10 until 18.

In the selection outcome with limited contracts, A's declining performance is not matched by an under-developed C until year 14, when they are actually replaced by D. C comes into the side for B a year later, but the lack of experience means the average drops into the low 40s from year 13 to 17. E, similarly, does not emerge from D's shadow until year 19, causing another gap from optimum to selection to form.

The numbers in the simulation are made up - if hopefully in the appropriate ballpark. We don't know what the performance advantage of a professional contract is, and there is still a glaring hole in the research over peak performance years. But the general shape will be correct, and that points to a gap between potential performance and what teams get on the field.

This may not be the problem right now. There may be no players in Ireland with the potential to perform better than the golden generation in situ. But with only eighteen contracts on offer for the entire squad, there are certainly going to be young players that aren't getting the opportunity to reach their peak. Some, many, will drop out altogether, as non-professionals tend to do. Others will have a late career bloom as the simulated C and E did, but not achieve as much as they ought.

This is an under-appreciated aspect of professional development in sport. Across most sports a player will not reach that peak performance until their late-20s. Yet, their future must be decided by their early 20s, because they need to be in the system until they reach the top. Club programmes work through this via a process of drafting and recruiting, second teams and development leagues. They tend to be cut-throat, and not necessarily in a good way.

A centralised representative sport like cricket tends towards promoting only the top performers, letting late developing or second-tier talent drop off. Often this also manifests as the flip-side to letting talent drop: of prematurely promoting youth on potential, in the hope that they will develop faster in an elite environment.

Perversely, premature selection can be a good thing, as limited opportunities are best given to those with the greatest potential. But it will always be inferior to selecting the best option from a group that have all had the chance to reach their peak. The challenge for all teams, is ensuring they find, and more importantly, improve the talent they have available. Repeatedly, we've seen a strategy of giving contracts to the players you need now, and not the players you might need later, result in ageing stars wondering why noone was there to replace them.

Cricket - Analysis 24th August, 2016 12:57:20   [#] [0 comments]

## Automated no-ball detection, proof of concept Russell Degnan

The first test between Australia and New Zealand may not have hung on the non-dismissal of Voges at the end of the first day, but the 232 runs it cost New Zealand certainly made it a lot harder. Predictably, a noticeably wrong umpiring decision led to a renewed call for third umpire reviews on every delivery, the return of the back-foot no-ball rule, and some less predictable, non-sequiturs about punishment.

But there is an easy solution.

Tennis has, for over 30 years used electronic means to judge service calls, and more recently, detected let calls with motion sensors. These are relayed to the central umpire, and they use that in calling the point. No-balls in cricket are slightly more complex, as they depend on the position of the foot over (not necessarily on) a line, with confounding shadows and the curved surface of the ground preventing the use of light beams that worked (mostly) for tennis.

But there is an easy solution.

With a fixed side-on camera with a clear view of the line (two is preferable), it is incredibly easy to build a system that will detect a landing foot within an area, and decide if it fell in front of, or before the line.

Computer vision techniques, the sort used by path finding robots, have been around for several decades. I learned the basics (in 1998) and those and more advanced techniques have been developed into the free OpenCV library I used for the code outlined below. To give a sense of how ways it would be to implement automated no-ball checking: my code, using not-particularly high-res footage, sans any setup programs, a live stream, or communication device to the umpire (a phone will suffice for that though) took me around 14 hours. But that involved me learning, from scratch, the OpenCV library, installing Java and SBT, and relearning some coding techniques.

There is no excuse for no-balls not to be automated. It is a trivially easy application of computer technology to a glaring issue.

Step 1. The code [downloadable here] uses the VideoCapture to load the video, and the BackgroundSubtractorMOG to detect edges from the non-filled part of the crease.

This image shows white areas where there is movement from the previous sequence of frames; grey areas show where there is a change, but the same colour as before (indicating shadow). You can see the outline of the bowler as he moves through the crease, and the non-striker backing up.

Step 2. Each frame is examined within the space shown below, to look for objects that will land within it

In a real-world application it needs to specify the side of the pitch to view, and be turned on and off for each ball (as hawkeye also would, so the same operator could be used).

Step 3. A relatively simple formula was used to calculate if a foot was within the frame:

• There must be at least ten rows of pixels (out of 20) with a continuous line longer than 60 pixels (about 8 inches), and less than 120 pixels.
• A threshold is used to determine the continuity as there are often bits of noise at the edge
• Two edges are determined for the back foot - a hard edge (more than 10 rows, and a soft edge (more than 3 rows) to account for movement on the foot. That left a 3 pixel margin of error in this instance (around 10mm), but further testing could improve that. This is the blue box in the first image)
• That line is compared to the crease line, that is configured before hand (the red line in the first image).

Needless to say, on the ball in question, the bowler was unquestionably behind the line (by 9-12 pixels, or 3-4cm).

Assuming a stream from the fixed cameras could be obtained at the ground, a working and fully tested system could be in place in less than a month. Sometimes, there really is an easy solution.

Cricket - Analysis 22nd February, 2016 20:24:00   [#] [0 comments]

## Batting form and the "hot hand" Russell Degnan

There has been an interesting paper in circulation recently that deals with the idea of a "hot hand", which in cricket terms wed refer to as "good form". It is primarily interesting because for several decades, the idea of streaks being anything other than random luck has been derided. Attempts to measure it in cricket have been few and far between, but there was little to suggest a batsman was more likely to score runs having just done so - or indeed, that they werent largely replicable with a random number generator.

The paper in question upends a key piece of prior research because of a rather simple, but slightly counter-intuitive piece of statistics. The various explanationstend towards the counter-intuitive end of the scale but Ill try to explain.

The technique being used is very elegant: take an action that occurs roughly 50% of the time and measure the number of successes that follow a previous success. If they are completely independent, the subsequent attempt will continue to have a probability of 50%. If the successes are clustered around other successes, that number ought to be higher.

The quirk, is that because strings of attempts are being measured, the average probability found in those strings will not actually be 50%. This is the counter-intuitive part, but is relatively easily seen on a simple graph:

Here is shown strings of eight attempts broken down into the number of times a measure took place (that is, the previous attempt saw a success). The number of instances of measurements breaks down as a binomial distribution centred on 3.5: 2 of zero (ooooooox and oooooooo) and seven (xxxxxxxx and xxxxxxxo), 14 of one and six, and so on as follows:

AttemptsOpportunities usedInstancesPercentage
002
114147.14
2844221.43
32107035.71
42807050.00
52104264.29
6841478.57
714292.86

The distribution of opportunities to take a measurement is similar, but because it takes more successful attempts to generate higher opportunities, it is shifted slightly across, and centred around 4 (or n/2). The breakdown also demonstrates the key to the problem: if four opportunities are to be had, the attempts will be distributed in such a way that the average success rate is 50%. But the only way to generate 7 opportunities is to have succeeded in each of the first 7 measurements. The percentage will be either 6/7 (83.33%) or 7/7 (100%). And as a consequence, the average of multiple strings of measurements ought to sit not at 50% (the middle of the opportunity distribution), but at the centre of the instances of measurement distribution (plus a term for the two extras) - around 45% for strings of length 8.

All very fascinating, particularly as it implies that previous studies showed a "hot hand" after all. But what does it say about cricket? The short answer is that this is a very elegant way of measuring form: find the median score for a batsman, if they surpass it, then test their subsequent score.

For Tendulkar, who played so many innings that the expected percentage is close to 50, his test "form" saw a 53.1% success rate in innings where hed surpassed the median (excluding not outs below the median). In ODIs however (counting only matches where he opened) the figure drops to 50.5%.

That is only a single data point, and some batsmen are likely to be more prone to runs of form than others, but it also points to an issue. In ODI cricket, where multi-lateral series exist, a batsman tends to shuffle opposition quite quickly, and therefore face a reasonable variety of bowling strength from match to match. In test cricket, the subsequent innings is less likely to be independent from the first, without being held in identical conditions - the second innings being on a wearing pitch. Apparent runs of form may just be a string of matches against poor opposition.

Conversely, ODI cricket may be less prone to form, being a format that requires a higher amount of risk-taking, and therefore more luck. Hence a discrepancy between test and ODI matches is feasible. Comparing all innings adds in time gaps when a player might fall out of form (and vice versa), and a proper study ought to remove them. The relative sparsity of innings means that when a player is really in form, it would be hard to distinguish between that and luck with any method. Most likely the effect is small - perhaps three or four runs on a batting average, but probably half that.

Hence measuring the effect, if any, of form remains difficult. On selection matters, - the only avenue where form might matter - there is a lot to be said for judging a player on technique, temperament and overall career trajectory and ignoring runs of form. Everything else is largely academic, albeit an interesting question.

Cricket - Analysis 16th October, 2015 00:45:10   [#] [0 comments]

## Short Stat: Stability and performance Russell Degnan

We know such an approach is a good thing. There is an obvious correlation between that and success, though which is the chicken and which the egg is debatable.

- Rob Smyth - The joy of selection roulette

The need for stability is always the catch-cry of teams struggling, and players fearful of their places. It has been an article of faith that Australia built their dynasty around youth in the 1980s, though even that might need some revision.

The longest period of batting stability for Australia immediately followed the 1989 Ashes, with only the substitution of one Waugh for another in 21 tests. But it was also a period marked by weak opposition, with the only losses being in NZ and to the West Indies (2-1). The 12 tests that followed the enforced retirement of Geoff Marsh, leading up to the 1993 Ashes, was anything but stable, with 6 different openers, 4 different players at first drop and 6 more players in the middle order - 7 of you include Greg Matthews. The results? Only three losses, one in NZ, and a 2-1 loss to the West Indies. Another two top-order changes were made for the first test in 1993; as in 1989, Australia were 4-0 up by the final test.

Perhaps results might have been better with more stability (a series lost by one run has a lot of what-ifs); or perhaps the opposition over-rides whatever difference might exist. It is reasonably unlikely that swapping the 6th best player for the 7th makes a big difference, though ongoing panic such that you select the 13th best, might.

To factor out the opposition, we can compare the expected margin against the actual result, and graph that against the number of changes made. There is a lot of noise:

There is also some indication that making zero changes is better. In the short term, the best side is probably the one you thought was the best side. But making one or two changes is still likely to produce a (very slightly) above-average result - note that 20 ratings points equates to 10 around runs, a fifth that advantage conferred by playing at home. Though this doesn't necessarily solve Smyth's chicken and egg quandary, as a result above expectations may merely represent below average expectations.

It gets more interesting when we look at changes per match over the previous two years. A side in constant flux ought to under-perform relative to expectations, if stability matters.

Actually, we don't see that. There is a lot of noise, and the difference is minimal, but sides making fewer than 1 1/2 changes per match do worse than expected than those making more.

I'd proffer two possible explanations. Firstly, that there is an information problem, in finding the best set of cricketers, and most likely some benefit in trying several out until one shines sufficiently to become more permanent. And secondly, that stable sides are more likely to be older sides - established, successful - and therefore more likely to be declining in performance. That doesn't mean an alternative player will perform better though, particularly in the short term. As the game to game suggests, more often than not, the best players a team has are those who've proven to be the best they have, even when they are losing.

Cricket - Analysis 12th June, 2014 19:35:33   [#] [0 comments]

## Short Stat: Adelaide and batting first Russell Degnan

Don't. There are exceptions, but the oft-told story of Richie Benaud's that as a captain he was told to "bat; if in doubt, think about it, then bat anyway" hasn't been true for 20 years.

S Rajesh noted as much a couple of weeks ago, but his analysis was based on the results obtained which has issues (amongst them, that Australia automatically bats) while other sides are a little more discerning. We can run a slightly more sophisticated analysis by comparing the expected margin (based on my ratings) against the actual margin and seeing whether the batting or fielding team beats expectations in each match. In short: for the last 20-odd years they have not.

In the 1930s - with uncovered pitches - the advantage in winning the toss and using the (most likely) best conditions was clear: it added as much as 40 runs per game. But that benefit has steadily eroded, and batting first is now a negative proposition, while fielding sides are regularly beating their expected margin. Interestingly, this is happening in both drawn (margin of 0) and result games:

A drawn games means the better side missed its expected victory. And for the better side, fielding first offers the advantage of time. By bowling there is no wasted runs from the need to set a target - such as last season when Australia still needed two wickets at close of play and had 172 runs, but also in 2003 and 2006 when the batting side had a first innings in excess of 500 and still went on to lose courtesy of a poor third innings. Even with the margin as large as it was, given the rain on the last day, England probably wouldn't have managed to beat Australia in Adelaide in 2010, because their bowlers would have been tired (had they enforced the follow-on), or they'd have run out of time.

Similarly, despite having to bat last on a potentially wearing pitch, if the match is heading for late on the fifth day, the need to buffer a margin by 100 or so runs when declaring helps the weaker side avoid a loss. Of the three recent bat-first-and-win games at Adelaide, all three went into the fifth day, despite the losers scoring less than 520 runs in total in the match. Australia and England both batted and lost in that period with more than 680 runs.

In general, a side that wins beats its expected margin, because the expected margin takes into account draws. In games with a result then, you'd expect any advantage from the pitch to accrue to the batting side, because they get the best conditions, and managed to exploit them. But in recent years we've not seen that; the new pitch has offered movement to the bowlers, and the old pitches haven't broken up significantly enough to negate that. There isn't a huge difference (and quite a bit of randomness), but taking into account the time benefits the bat-first approach is no longer valid, and actually unhelpful.

So unhelpful, in fact, that the expected margin for the toss winner was negative in the 1990s and first part of the 2010s, as well as negative for those batting first in the 2000s. By less than a dozen runs, but negative is negative. Any side with ambitions to win in Adelaide should bowl first; new pitch caveats aside, there is little to fear on the fifth day.

Australia chose to bat; but that is not a surprise. For reference, this graph depicts the number of total runs in the match for teams batting first and second since 1990; wins at the top, losses at the bottom, and draws in the middle.

For teams batting second, more than ~560 almost guarantees at least a draw, although it is possible to win with less (because obviously the opposition can be bowled out for less). Batting first, there has only been one victory with less than 590 (by a single run no less), and three losses with more than 600. The runs required to force a result in Adelaide are substantial.

Moreover, there is always pressure on the side batting first to keep batting well, because all results remain possible, even with very high totals. Whereas, the side batting second can, if they bat well enough, guarantee at least a draw and press for a victory.

Finally, the innings by innings runs per wicket for the top order: 1st: 48.5 2nd: 49.5 3rd: 31.5 4th: 28.9. That calculates to a total value in the top-order of batting first of 11.2 runs (miniscule in context). The Adelaide pitch clearly becomes harder to bat on, almost twice as hard: but it does so too late to gain advantage in the second innings, and too early to prevent a catastrophic third innings resulting in defeat. In Adelaide, it is the third innings that counts, and you are better off bowling when it does.

Cricket - Analysis 4th December, 2013 14:01:14   [#] [0 comments]

## Observations on Cricket Finance Russell Degnan

What follows is necessarily inexact. Perhaps very inexact. Cricket has many issues that confront it, but by far the biggest is a lack of transparency. The Woolf review, despite the resources available to it was forced to admit much the same:

We believe there is an overall lack of transparency around financial distribution in global cricket, which means certain aspects of the finances of global cricket are not well understood. We have been unable to obtain a full picture of the current financial position of global cricket. For instance, although there are various media estimates in circulation of the impact of tour cancellations (actual or threatened), it is not known with any degree of certainty the financial effect a tour by one Member has on another Member. It is clear that tours by certain Members (such as India) to other Members give a significant revenue boost to the host nation.

There are four points we know with relative certainty however, from which we can begin a deeper investigation. Again, to quote the Woolf review:

• The level of funds flowing through the ICC and global cricket has been completely transformed in the last ten years, with a significant increase in revenues, principally from growth in income from television rights.
• For ICC-run events, the additional funds from television rights flow through the ICC to the Members.
• For other competitions and matches, such as those under the FTP, the income flows directly to the Members.
• The most notable source and beneficiary of the greater revenue flowing into cricket is India. That nation's love of the game has combined with significant population and GDP growth in recent years, to make India the commercial hub of world cricket. There are unsubstantiated estimates that India generates between 60% and 80% of revenues flowing into global cricket. India also has a significant impact on the ICC's Commercial Revenue, as almost all of the ICC's major commercial partners have significant links with India.

The basic structure of cricket finance runs from TV markets - nominally constituted at a national level - to either the home board of a particular fixture, or the ICC. We therefore need to make three assessments: the size of a local cricket market; the flow of money generated from that market to various bodies; and the distributive flow from the ICC and others.

For most purposes here I'll be talking averages over four years, because this takes into account the cycle of both the FTP (give or take), and the ICC major events. Every cricket board exhibits substantial variation from year to year, depending on who is touring, and the dividends distributed by the ICC.

Cricket Market Size

This is the most inexact of all the estimates, not least because it isn't clear what percentage of the cricket market is actually being drawn on by various members. Empty stands and no push to fill them by sensible scheduling, and fixturing that makes an inefficient use of resources, and no attempt to contextualise the season means most boards make less money than they might.

A previous assessment of TV rights deals across sports in Australia indicated that cricket gets roughly what you'd expect, given its ratings, and total hours of programming. Taking into account sponsorship, merchandise and match-day attendance; then cross-checking against Cricket Australia's annual reports, and adjusting for income earned from overseas, puts the size of the Australian cricket market (or at least, the part of that that pays to watch professional players), at something like an average of AUD$150m (which for current purposes is nearly identical to USD) over the past four years. The most recent TV deal has probably inflated that to nearer AUD$200m . The continuing rapid inflation of sports rights makes this process harder than it might otherwise be.

The simplest model for calculating market size is to multiply GDP (incomes, which includes population) but the level of cricket interest. There are various reasons why it won't correct: the distribution of cricket fans amongst income quartiles (particularly in England and to a lesser extent India where cricket is an upper-class sport); the size of disposable income which will make the sports market in wealthy nations much larger; and the difference between TV income and ground income, with the latter more easily captured in richer, but smaller nations. Nevertheless, I'll look at three methods of assessing cricket market size. Two are quite simple but (relatively) complete, the third complex, but stifled by the lack of annual reports from the most poorly governed members.[1]

 Announced Revenueavg. Last 4 years Population(2007 millions) GDP (millions USD) % Cricket Players % Cricket ArticlesKaufman + Patterson (2005) Market Estimate% News Articles Market Estimate% Cricket Players Market EstimateRegression Method India 168120 1169.0 1141346 7%* 46% 419711 669767 600000 England+Wales 185980 55.7 2440505 0.90% 17%# 207443 210265 204000 Australia 157041 20.7 1541797 2.41% 22%# 169598 191894 160000 Pakistan 163.9 231879 8%* 61% 70723 109469 South Africa 62065 48.6 384315 2.06% 20% 38432 49086 40000 Bangladesh 304.4 153000 2.52%* 49% 37485 27914 New Zealand 31399 4.2 169680 2.39% 17% 14423 26282 18000 Sri Lanka 19.3 59408 8%* 57% 16931 21685 West Indies 37219 4.5 50145 3%* 28% 7020 11701 20000 USA 701 305.83 15684750 0.01%+ 0.10%+ 7842 7948 Ireland 6.1 254500 0.31% 3%* 3156 6744 Canada 32.8 1819081 0.03% 1% 9095 4223 Scotland 2789 5.14 216000 0.20% 4%* 4320 3781 Afghanistan 27.1 19906 1.03%* 10%* 995 817 Zimbabwe 13.3 9802 0.37%* 5% 245 68 IPL 137915 ICC 205409 CLT20 100000*

* Estimate - by which I mean guessed and/or completely made up
+ Estimates of USA market size are very sensitive to assumptions. The K+P figure was close to zero, but also 8 years ago, and there has been a recent shift towards more cricket articles. Similarly, estimates of the number of players vary from the official figure of ~30,000 to ten times as many. Consider this a low figure; under reasonable assumptions the USA is cricket's fourth biggest market. Though almost none of that goes into US cricket.
# K+P give two figures for England - 8% normally and 17% in summer - I have used the higher one for obvious reasons (estimates outside the season are irrelevant). They don't offer similar figures for Australia, which makes all the figures in this column sensitive to this assumption.

Method 1: Estimate from GDP/news media

One of the more interesting pieces on cricket take-up in various nations is the 2005 piece by Kaufman and Patterson. It is worth reading the paper, but for my purposes its most useful feature is an estimate of cricket popularity by counting articles in the sporting press. The numbers have been relayed into the table above, and used to calculate, by multiplying that out by GDP and a factor that makes the markets I have good data (Australia, England, India and New Zealand) approximately the right size (around 120,000).

Method 2: Estimate from GDP/playing numbers

Playing numbers as a proportion of population ought to be a good measure, but are complicated by the fact that, although have a complete set of figures for associate/affiliate nations, I have none for most test nations. Estimates of the major nations should therefore be taken with very large grains of salt. Nevertheless, it gives some reasonable numbers, and those give a good indication of the size of markets where published annual reports are sparse, or the market is undeveloped for lack of matches.

Method 3: Estimate from Annual Reports

Estimating market size from actual revenues is complicated by the amount of revenue generated by most boards in external markets (either overseas tv rights or sales to spectators), and the lack of reporting on the source of that revenue.

India is perhaps the easiest market to estimate, because the BCCI generates relatively little profit from external sources. Their annual report puts average four year revenue (adjusted for currency) at USD$168m, of which$25m is dividends from the IPL or CLT20. Those two competitions have approximately $240m in revenue. The ICC brings in around$200m in revenue a year, of which India is the source of approximately 60% (by most accounts). Finally we must estimate the amount of revenue earnt by playing India at home, and which therefore goes to the local board. If that is assumed to be around $100m then the Indian market is roughly three times that of Australia and England: around$600-700m. This is less than most estimates, but consistent with their annual reporting.

What we really need to do though is understand where the money comes from, and where it goes.

Distributions

ICC Distributions are relatively easy to calculate, and the annual report is quite informative. 75% of profits - after events costs, TAPP payments and administration are removed - are paid as dividends to the ten full members, with the remaining 25% going into the development fund, the majority of which is distributed to the 96 associate and affiliate members. Dividends to full members over the last four years were USD$319.9m, with USD$41.4m being distributed to the development fund. That doesn't include prize money, or TAPP funds, from which relevant members took an extra million or so (relatively little, when considered over four years).

Non-full-members are paid according to a scorecard system - judged on 35% the men's ranking, 41% various participation figures, and 24% administrative development. The top-tier receive $300,000 USD, the bottom$5,000 USD, in addition to $100,000 for associates and$10,000 for affiliate members. High Performance Program members receive funds commensurate with the expected costs of transitioning from amateur to semi-professional cricket structures, depending on what events they play and qualify for. The Asian Cricket Council derives significant additional funding (approx $5m) from the Asia Cup, which in turn feeds into their member-base. Estimating value: a multi-variate regression approach A significant proportion of income for full members comes from the rights to host certain nations via the FTP - unless you hire Haroon Lorgat, then all bets are off. To estimate the size of these flows I went through every recently published annual report and noted the revenue in USD - converting by the exchange rate of the time - the year, the ICC grant (where noted, or estimated based on the ICC report where not), the number of home matches, and the number of those matches that were played against Australia, England, India or Other (meaning everyone else). A linear regression was then run, which produced some moderately accurate looking numbers: The top line of figures - in thousands USD - is the important one - the second is the standard deviation which is worth noting only because they are quite large (around$1.2m on each variable). By multiplying out the revenue earned for a day of cricket by the number of days played I've estimated the flow from markets to boards in the following figure (click for pdf version).

This is necessarily representative, which is why I left the figures off. The size of each board logo and market is proportioned to represent the relativity between comparable entities. Where arrows are not clear, remember the money flows from market to board, sometimes via intermediaries (such as the ICC). I have not noted any payments directly from board to board, though the Woolf report implied they might exist. [2] Nor have I accounted for county cricket (which earns perhaps $40-60m) nor other T20 leagues which aren't accounted independently. In the interests of readability, flows less than$1m have been left off, as have most associate members. I can't guarantee I caught everything. [3]

Observations

There are a number of points that can be made from this. In no particular order:

The BCCI earns around $4m more than average from home matches, as does England, with Australia on$2.7m. (All give or take $1.2m; the Ashes earns at the top-end of those numbers - i.e.. above$4-6m per day). This makes intuitive sense. It also shows the importance of attendance versus TV in the revenue streams of otherwise smaller markets, and the weakness of the BCCI in creating a home schedule that meets what they might earn. As the annual reports bear out, India earns less from their home internationals than England, which doesn't accord with their perceived financial muscle.

Notwithstanding that the IPL effectively doubles what the BCCI make from their home market, they ultimately end up with only around half the revenue generated locally, despite having a monopoly control over the team that market pays to watch. This is both quite surprising, and an indication of why they are increasingly bullish about increasing their share of global revenue.

Board revenue has increased by $6m a year; that is, the regression estimates base revenue of$29m in 2000, and $89m in 2010. Obviously only three boards actually earned this; the variables are best interpreted as relative amounts. Moreover, the large standard deviation hints at the growing disparity in how much teams have managed to gain from overall revenue increases. Playing India earns$1.6m for the home team more than Australia or England, and $2m more than any other team. This, in one sentence, explains most of what you need to know about the nexus of finance and the obnoxious chaotic scheduling of the FTP. South Africa make$450k above the base rate (around $1m per match) from home matches, but New Zealand and the West Indies are making less than that, which means matches against teams outside the big-3 are likely to be losing money. We know this, in relation to why test cricket costs these teams money to play, but add in Pakistan - whose market lies dormant with no tours possible; Sri Lanka, Bangladesh and Zimbabwe, and it is clear the bulk of fixtures are neither profitable nor generating particular interest. The total size of the non-full member market is around$90m; but it is almost certainly not being tapped. Zimbabwe has no market to speak of at all - less than \$0.5m. Their GDP is tiny, their population is small; cricket is a minority sport. You frequently hear commentators remark on the importance of building up existing markets, rather than chasing markets in nations cricket has a small profile in. This is basically nonsense. There are three really big cricket nations, each of which has a GDP in the top-15 in the world. There is limited scope for growth in the big-3. But amongst the rest, because they are already pushing against the point of market saturation; and because their GDP is relatively small - despite their population size in the case of Pakistan and Bangladesh - their potential for growth is weak.

If one attitude weakens cricket's case for globalisation it is the perception - largely because of the bias in the origin of the cricket media - that there is a certain standard to aspire to equal to India, England, Australia and perhaps South Africa. With the possible exception of the USA, and in the longer term, China, no nation will reach that standard in the next 20, or perhaps 40 years, without remarkable (unprecedented) growth. The nations currently in the HPP are too small, or too poor, or both; the G-20 nations that might open up new frontiers have tiny playing bases.

There is, nevertheless, strong encouragement to the idea that cricket could have 20 nations of a standard somewhere between that historically maintained by New Zealand and currently by Bangladesh. With the assumption that, given that, at least a half dozen of those teams will have a transcendent talent (ala Hadlee or Muralitharan) that will allow them to compete with the big-4. A future post may look into this; as some nations will surprise.

If it wasn't obvious, cricket's finances are fundamentally unstable. The wealth available to three boards, and their local competitions means that noone else can afford the market rates for their players. While we haven't seen mass defections, it is increasingly clear that international cricket, as currently structured, cannot support the existing nations, let alone provide the investments needed to promote and grow the game elsewhere. Either a substantially larger proportion of the money moving from market to boards needs to be routed through the ICC (which means them taking ownership and control of tournaments), or a substantially larger proportion of the money must direct itself into competitions that will pay players from all nations, with a reduced emphasis on international cricket.

This would not be historically unusual; it has been the case for West Indies cricketers from the turn of the 20th century in English league cricket, through Constantine and Sobers; and onto the Packer years. There are also various ways both these scenarios could come to pass. Some are outlined in my manifesto on test cricket; others ideas will have to wait for another day. But don't be surprised if the CSA-BCCI spat is a harbinger of things to come. There are too many opportunities for the BCCI to redirect money currently exiting the Indian market back into their own pockets, and too much inequality, for things to stay as they are.

[1] It is interesting the cricket's largest markets come under both the best and worst governed nations but relatively few in the middle. The crisis of governance at ICC level is exacerbated by very different philosophies of action by its board members.
[2] The late Ronald Coase would have found this interesting. There is no good reason why boards couldn't bid for tours, thus maximising both BCCI income and cricket's overall revenue by playing the most desirable fixtures (albeit not those that make the best competition/product). Transaction costs at the ICC are high though, and we are far from man efficient touring structure.
[3] Also, apparently cutting a google map means I get abused by nationalist idiots over J+K. I don't care. Don't bother me over your craziness. It isn't remotely relevant to cricket.

Cricket - Analysis 11th October, 2013 01:52:36   [#] [0 comments]

## Short Stat: On the Back of a Collapse Russell Degnan

One of the standout aspects of Australia's collapse in Durham was the tentative batting; admittedly it was what begun the collapse - Khawaja and Clarke's half hearted footwork - and not what continued it - Haddin and Watson's playing across the line. But it raises an interesting question over whether players play worse in the midst of a collapse, or much the same. Is there a drop in performance from the psychological pressure, in other words.

I tested this proposition using a technique Chris at Declaration Game used, by comparing the runs scored by the 5th and 6th wickets against the other innings, and matched that against the difference in runs between the fall of the 2nd and 4th wickets (the collapse amount - though most aren't a collapse).

As it turns out, there is no effect. The average scoring for the 5th-6th wickets is 59, which is consistent with one exception across the range of collapse amounts. Not only that, but there is so much randomness in the difference between the two innings, and the previous run-scoring, that even sample sizes over a hundred for low collapse amounts end up with reverse effects from one to the next.

You can see from all the data points that the difference remains resolutely centred at zero for all low amounts. This is actually doubly odd, because it indicates that even where several wickets have fallen for other reasons - a crumbling pitch or new ball - the difference between that and the previous innings was negligible.

Data clumping over shows some of the randomness, and don't be confused by the jump around 30; a different division produces a completely different result.

What both graphs do show though, is that where the previous two wickets have put on 200+ the average of the 5th-6th wickets combined drops to 45. It isn't clear why this is - the bowlers, presumably are tired - but perhaps one or more large preceding partnerships make it harder for an incoming batsman. Something to look at another day.

It does bode badly for Australia though. There is a tendency after a collapse to attribute it to the moment, and assume that next time, more focus and hard-work will arrest the problem. The data suggests that even losing three wickets for not many makes almost no difference to the mind-set. If a team is in the habit of losing 6 or 7 for not many it is because they are poor, and just as likely to lose quick wickets when the previous stands have been productive or dismal.

Cricket - Analysis 14th August, 2013 19:16:20   [#] [1 comment]

## Short stat: 2013 Ashes Probabilities Russell Degnan

Inspired by Fake Ritzy's ICC rankings based analysis of Australia's Ashes chances, I ran a monte-carlo simulation of the series using my own (25% draw probability, matching the historic English average).

Australia is a roughly 1 in 8 chance of winning, and a 1 in 8 chance of drawing. 3-1 England is the only scoreline returning a positive net return on the betting markets. Australia's most probable winning scoreline, 1-2, is very (very, very) slightly more likely than losing 5-0. Australia wins 5-0 in about 0.04% of series. I think Australia's Indian tour has caused their rating to under-estimate their chances. With a decent team selection there is grounds for no more than mild pessimism, but given Watson is locked in to open, things are very bleak. Very.

Cricket - Analysis 29th June, 2013 01:13:21   [#] [1 comment]

## Notes on Victorian Cricket Attendances Russell Degnan

The MCG is perhaps the best barometer of crowds because it has both the best data, and is almost impossible to sell out, even when being renovated. There is actually less of interest here than you may think, but it is worth collating occasionally to see.

Start with two graphs. Figures are derived from the MCG, news reports and Austadiums. Not all sources are consistent, but the figures are close enough for this purpose. The first three seasons of the Big Bash are estimates on the Austadium site, but probably within a thousand or so.

• Test crowds are basically constant within the bounds of opposition. This season was down, as Sri Lanka is a weak draw-card, but next season will be up, as the Ashes adds 25-30,000 to the base average.
• One-day crowds are either trending down, or made up of two relatively flat lines, with the introduction of international T20 causing a drop of 5-10,000 per game. The combined aggregate attendance of international limited overs cricket is basically the same as it was before IT20: about 100,000
• One anomaly is the drop in support for English limited-overs cricket. 2006-07 had a unusually large ODI crowd (79,000) but the 2010-11 T20I was consistent with the average for other teams (58,000)
• The International T20 match has shown a significant drop in attendance, from 80,000+ against India in 2007-08, four years of ~60,000 and down to 40,000 this season. If any figure stands out, it is that one. Next year will be interesting.
• Domestic T20 has increased the number of games year on year without a significant drop in average (~24,000 at the MCG, 13,000 at Etihad). That has propelled the aggregate attendance to the top of any format, indicating significant pent-up demand for more local games.
• Conversely, the average BBL attendance dropped this season despite this being an under-whelming January of cricket. This doesn't match what happened in 2009-10 which saw a big increase in average attendance, followed by a big drop when England was in Australia.
• 2009-11 is better discussed in relation to show-piece games (Vic-NSW was followed by 43k) and performance related, consistent with the economic literature on domestic crowds. The season just gone is less clear: outside the two derby games, the Melbourne crowds were weaker than expected, particularly before Christmas, even with both teams doing well.
• It is very difficult to make any conclusions, good or bad, about BBL crowds or trends.
• The next two years will be interesting for attendance, but not necessarily representative either. There is a clear shift away from international limited-overs cricket to domestic cricket; but international attendance is strongly correlated with opposition, so both formats ought to recover with England and India in town.

A final thought: the aggregate attendance of all formats has jumped from around 250,000 to 400,000 in line with an increase in high-profile matches (with apologies to the Shield and ODD Cup) from four to more than a dozen. The average number of people at each event has decreased, but the fan in Victoria has a lot more cricket to go to, and does so. Definitely a good thing.

Update:

I chanced upon another source of daily crowd figures on the MCG site, and by using the wisden almanac, have produced a long-range graph of crowds in Melbourne over 30 years. The key take-away is that Boxing Day has become ever more important, and ODIs have steadily declined, starting in the mid-90s. Note that I didn't include the average for the non-Boxing Day test of 1989-90, for which the 5-day aggregate attendance was 68,865; though it was less than 19,000; figures missing: ODIs: 1990-91 (2), 1984-85 (1) and 1980-81.

The nadir of test match crowds was an almost washed-out match against South Africa (1993-94) although the worst years were behind. Note that ODI aggregates vary wildly from variations in the number of matches played by Australia. Nevertheless, the long-term trend for ODI crowds looks bleak.

Cricket - Analysis 14th February, 2013 21:47:29   [#] [2 comments]

## Short stat: BBL versus Test attendance 2012/13 Russell Degnan

A special BBL final related graphic on BBL and Test match attendances. The only city in Australia this year to record fewer patrons to the BBL than test cricket: Adelaide. And that was a Strikers semi-final away from being toppled too.

Attendance at test cricket was approximated for several days, but will be correct to a couple of thousand, given the sparse stands. Total attendance 490k in 34 matches for the BBL, ~400k for test cricket over 26 days.

Cricket - Analysis 19th January, 2013 18:47:05   [#] [0 comments]


text:
# Online ad spending poised for 20-percent jump this year

Strong demand for U.S. display advertising is helping pump up projections for the overall online advertising market, which is now set to grow by 20 percent this year to $31.3 billion, according to new estimates from eMarketer. The firm updated a previous estimate of 10 percent growth for 2011, largely on the strength of display advertising which is now poised to overtake search advertising within five years. EMarketer said search advertising will grow by 19.8 percent this year to$14.4 billion, outpaced by online display ads, which will grow 24.5 percent to $12.3 billion. Increased video and banner ads are fueling the growth of display advertising with video ads expected to pull in$2.2 billion this year and banner ads estimated to get $7.6 billion. Overall, online advertising is expected to account for almost 20 of all major media ad buying this year, up from 17 percent last year and by 2015, online ads will represent 28 percent of all U.S. media ad spending. Online advertising is creeping up on TV, which is expected to hover at 38 percent for the coming years. Online advertising grew to$26 billion last year, a 14.9 percent increase over 2009. It eclipsed newspaper advertising for the first time to move into second place in market share. And with this latest projection, it looks like the good times are only getting better for online publishers, who can expect to get more dollars as advertising continues to migrate online.

format: plain


text:
Let $f(%1)=%0[1].$ Fill in the following table of exact values of $f$.
$\bold{%1\ }$ \t $\bold{f(%1) = %0[1]}$ \\ $-3$ \t BOX \\ $-2$ \t BOX \\ $-1$ \t BOX \\ $0$ \t BOX \\ $1$ \t BOX \\ $2$ \t BOX \\ $3$ \t BOX
BUTTONS
MESSAGE
In the graph of $f(%1)= %0[1]$, the $y$-intercept is BOX and the value of $y$ is multiplied by BOX for every 1-unit increase of $x$.

BUTTONS
MESSAGE
#[Now choose the correct graph of $f$.][Ahora escoge la gráfica correcta de $f$.]#
MULTIPLECHOICE

BUTTONS
MESSAGE
RANDOMIZE


text:
# Find the minimal polynomial for the roots $\sqrt{2}$ and $\sqrt{3}$ over $\mathbb{Q}$.

Find the minimal polynomial for the roots $\sqrt{2}$ and $\sqrt{3}$ over $\mathbb{Q}$.

I know the definition of a minimal polynomial is defined for a single root, but can we talk about minimal polynomials for multiple roots simultaneously?

I have only seen methods for finding the minimal polynomial of a single root.

Would it just be the product of the minimal polynomials for each root separately?

I am trying to better my understanding of finding the degree of field extensions such as $\left[ \mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q} \right]$

• What you are saying has the flavor of finding a primitive element $a$ of the extension $\mathbb{Q}(\sqrt{2},\sqrt{3})$ over $\mathbb{Q}$. This is $\mathbb{Q}(a)=\mathbb{Q}(\sqrt{2},\sqrt{3})$ and then finding the minimal polynomial of $a$ over $\mathbb{Q}$. If the final goal is just to find the degree, it looks to me simpler just to look at the intermediate extension $\mathbb{Q}\subset\mathbb{Q}(\sqrt{2})\subset\mathbb{Q}(\sqrt{2},\sqrt{3})$. And showing that $\sqrt{3}\notin \mathbb{Q}(\sqrt{2})$. – YAlexandrov Mar 13 '18 at 13:37
• I would (first) interpret the question as looking for 2 separate minimal polynomials, one for $\sqrt2$ and one for $\sqrt3$. I doubt a reducible polynomial would be called a 'minimal polynomial'. – Berci Mar 13 '18 at 13:37
• @YAlexandrov, Could you give a suggestion on how to find the primitive element for a given field extension? – Zed1 Mar 13 '18 at 13:46
• There's some theorem that in such a case, for $K(\alpha,\beta)$ with infinite $K$, one can always find a primitive element in the form of $\alpha+k\beta$. – Berci Mar 13 '18 at 13:51
• $a=\sqrt{2}+\sqrt{3}\in\mathbb{Q}(\sqrt{2},\sqrt{3})$. Then $\sqrt{3}=\frac{1}{2}a+\frac{1}{2}\frac{1}{a}$ and $\sqrt{2}=\frac{1}{2}a-\frac{1}{2}\frac{1}{a}$. Therefore, $\mathbb{Q}(\sqrt{2}+\sqrt{3})=\mathbb{Q}(\sqrt{2},\sqrt{3})$. – YAlexandrov Mar 13 '18 at 13:58

As previous comments point out, you should find a primitive element of the field extension first, in this case $\sqrt2+\sqrt3$ is the element you want to find. Next, it is a standard approach to find such element's minimal polynomial: \begin{align} \text{Let } \theta= \sqrt{2}+\sqrt{3}, (\sqrt{2}+\sqrt{3})^2 = 5+2\sqrt{6} &\Rightarrow \theta^2 - 5=2\sqrt{6} \Rightarrow (\theta^2-5)^2=24 \\ &\Rightarrow \theta^4-10\theta^2 + 1= 0 \end{align} It is easy to verify that $f(x)=x^4-10x^2+1$ is irreducible over $\mathbb{Q}$ by the rational root test and solving for a quadratic factor (which doesn't exist). So $f$ is the minimal polynomial for $\theta$.


text:
## anonymous 5 years ago what is half of 2^{40} ?

1. anonymous

$2^{40}$

2. anonymous

$2^{39}$

3. anonymous

could you show me how you got that ?

4. anonymous

sure.

5. anonymous

lets do a simpler one. $2^5=32$ half of 32 is 16 $2^4=16$

6. anonymous

ok

7. anonymous

now for the "math" half of $2^{40}$ means $\frac{2^{40}}{2}$

8. anonymous

when you divide you subtract the exponents, so you get $2^{40-1}=2^{39}$

9. anonymous

sorry im a little lost, why did u subtract 1 ?

10. anonymous

oh because the denominator was 2, and $2=2^1$

11. anonymous

oh i see

12. anonymous

and when you divide you subtract the exponents. $\frac{b^n}{b^m}=b^{n-m}$

13. anonymous

but forget about that for a second. power of two look like 2,4,8,16,32,64,128,...

14. anonymous

multiply by 2 to get the next one. so if you divide by 2 you get the previous one.

15. anonymous

alright then thank you :)!

16. anonymous

o wow


text:
The Ordinary Differential Equations Project

Section6.2Solving Initial Value Problems

There is no need for Laplace transforms when solving many initial value problems. If we consider the initial value problem
\begin{align*} y'' + 7y' + 10y & = 0\\ y(0) & = 1\\ y'(0) & = 1, \end{align*}
the methods described in Chapter 4 work quite well. The characteristic polynomial of $$y'' + 7y' + 10y = 0$$ is
\begin{equation*} r^2 + 7r + 10 = (r + 2)(r + 5), \end{equation*}
and the general solution of the differential equation must be
\begin{equation*} y(t) = c_1 e^{-2t} + c_2 e^{-5t}. \end{equation*}
Applying the initial conditions, we find that the solution to our initial value problem is
\begin{equation*} y(t) = 2 e^{-2t} - e^{-5t}. \end{equation*}
However, suppose that we have a harmonic oscillator with a discontinuous forcing term,
\begin{gather*} y'' + 2 y' + 5y = h(t)\\ y(0) = y'(0) = 0, \end{gather*}
where $$h(t)$$ is given by
\begin{equation*} h(t) = \begin{cases} 7 & t \lt 5 \\ 0 & t \geq 5. \end{cases} \end{equation*}
None of the previous techniques that we described are sufficient to solve such an initial value problem. Yet, we can easily imagine such equations arising in physics or engineering.
Fortunately, Laplace transforms forms behave very nicely with respect to derivatives. We can use Laplace transforms to transform an initial value problem into an algebraic equation. Once the algebraic equation is solved, we can use the inverse transform to obtain the solution to our original initial value problem.

Subsection6.2.1Laplace Transforms of the Derivative

Suppose that we have linear differential equation with constant coefficients
\begin{equation*} a y''(t) + b y'(t) + c y(t) = f(t) \end{equation*}
and initial conditions $$y(0) = y_0$$ and $$y'(0) = y_0'\text{.}$$ We can take the Laplace transform of both sides to obtain
\begin{equation*} a {\mathcal L}(y''(t)) + b {\mathcal L}(y'(t)) + c {\mathcal L}(y(t)) = {\mathcal L}(f(t)). \end{equation*}
Notice that we have used the fact that the Laplace transform is a linear operator (Theorem 6.1.3). To proceed further, we need to know the Laplace transform of the derivative of a function.
We can evaluate the Laplace transform of $$y'$$ by using integration by parts,
\begin{align*} {\mathcal L}(y')(s) & = \int_0^\infty y'(t) e^{-st} \, dt\\ & = \lim_{b \to \infty}\left[ e^{-st} y(t) \Big|_0^b + s \int_0^b y(t) e^{-st} \, dt \right]\\ & = \lim_{b \to \infty}\left[ e^{-sb} y(b) -y(0) + s {\mathcal L}(y)(s) \right]. \end{align*}
We claim that $$\lim_{b \to \infty} e^{-sb} y(b) = 0\text{.}$$ Since, $$y$$ is exponentially bounded, there exist constants $$M \geq 0$$ and $$a$$ such that $$|y(t)| \leq Me^{at}\text{,}$$ for all $$t$$ in $$[0, \infty)\text{.}$$ Thus,
\begin{equation*} |e^{-sb} y(b)| \leq Me^{-(s -a)b}. \end{equation*}
The right-hand side of this inequality as $$b \to \infty$$ for $$s \gt a\text{.}$$ Thus,
\begin{equation*} {\mathcal L}(y')(s) = s {\mathcal L}(y)(s) - y(0). \end{equation*}
The Laplace transform also behave nicely with respect to higher order derivatives.
We can use Theorem 6.2.1 and Theorem 6.2.2 to solve initial value problems.
Consider the initial value problem
\begin{align*} y'' + 7y' + 10y & = 0\\ y(0) & = 1\\ y'(0) & = 1. \end{align*}
\begin{gather*} {\mathcal L}(y')(s) = s {\mathcal L}(y)(s) - y(0) = s Y(s) - y(0)\\ {\mathcal L}(y'')(s) = s^2 {\mathcal L}(y)(s) - sy(0) - y'(0)= s^2 Y(s) - sy(0) - y'(0). \end{gather*}
where $$Y(s)$$ is the Laplace transform of $$y\text{.}$$ If we take the Laplace transform of $$y'' + 7y' + 10y\text{,}$$ we have
\begin{align*} {\mathcal L}(y'' + 7y' + 10y) & = {\mathcal L}(y'') + 7{\mathcal L}(y') + 10{\mathcal L}(y)\\ & = [s^2 Y(s) - sy(0) - y'(0)] + 7[s Y(s) - y(0)] + 10Y(s)\\ & = [s^2 Y(s) - s - 1] + 7[s Y(s) - 1] + 10Y(s). \end{align*}
Since $${\mathcal L}(0)\text{,}$$ we have the algebraic equation
\begin{equation*} [s^2 Y(s) - s - 1] + 7[s Y(s) - 1] + 10Y(s) = 0. \end{equation*}
Solving for $$Y(s)\text{,}$$ we get
\begin{equation*} Y(s) = \frac{s + 8}{s^2 + 7s + 10} = \frac{s + 8}{(s + 5)(s + 2)} = \frac{2}{s + 2} - \frac{1}{s + 5}, \end{equation*}
where we have used partial fractions to get the last expression.
Since the Laplace transform of $$e^{at}$$ is $$1/(s - a)\text{,}$$ we know that
\begin{equation*} {\mathcal L}^{-1} \left( \frac{1}{s-a}\right) = e^{at}. \end{equation*}
We can now solve our initial value problem,
\begin{equation*} y(t) = 2 e^{-2t} - e^{-5t}, \end{equation*}
and our solution agrees with the one that we found at the beginning of this section.

Activity6.2.1.Solving Linear Differential Equations with Laplace Transforms.

Use Laplace Transforms to solve each of the following initial value problems.
(a)
$$x' - 5x = e^{5t}\text{,}$$ $$x(0) = 0$$
(b)
$$x'' - x' - 6x = 0\text{,}$$ $$x(0) = 2\text{,}$$ $$x'(0) = -1$$
(c)
$$x'' + 4x = \sin 3t\text{,}$$ $$x(0) = 0\text{,}$$ $$x'(0) = 0$$

Subsection6.2.2Discontinuous Functions

If $$c \geq 0$$ and we define the function
\begin{equation*} u_c(t) = \begin{cases} 0 & t \lt c \\ 1 & t \geq c, \end{cases} \end{equation*}
recall that the Laplace transform of $$u_c$$ is given by
\begin{equation*} {\mathcal L}(u_c(t))(s) = \int_0^\infty u_c(t) e^{-st} \, dt = \frac{e^{-cs}}{s}. \end{equation*}
We can use this inforamtion to solve initial value problems with discontinuous functions.
Consider the initial value problem
\begin{align*} y' + y& = u_3(t)\\ y(0) & = 1. \end{align*}
If we take the Laplace transform of both sides of $$y' + y = u_3(t)\text{,}$$ we obtain
\begin{equation*} sY(s) - y(0) + Y(s) = \frac{e^{-3s}}{s}. \end{equation*}
Using the fact that $$y(0) = 1$$ and solving for $$Y(s)\text{,}$$ we get
\begin{equation*} Y(s) = \frac{1}{s + 1} + \frac{e^{-3s}}{s(s+1)}. \end{equation*}
Therefore,
\begin{equation*} y(t) = {\mathcal L}^{-1} \left( \frac{1}{s + 1} \right) + {\mathcal L}^{-1} \left( \frac{e^{-3s}}{s(s+1)} \right). \end{equation*}
The inverse Laplace transform of the first term is
\begin{equation*} {\mathcal L}^{-1} \left( \frac{1}{s + 1} \right) = e^{-t}. \end{equation*}
To compute the inverse Laplace transform of the second term, recall from Example 6.1.7 that if $${\mathcal L}(f) = F(s)\text{,}$$ then
\begin{equation*} {\mathcal L}(u_a(t) f( t- a)) = e^{-as} F(s). \end{equation*}
Using partial fractions to obtain
\begin{equation*} \frac{1}{s(s+1)} = \frac{1}{s} - \frac{1}{s - 1}. \end{equation*}
Hence,
\begin{equation*} {\mathcal L}^{-1} \left( \frac{e^{-3s}}{s(s+1)} \right) = {\mathcal L}^{-1} \left( \frac{e^{-3s}}{s} \right) - {\mathcal L}^{-1} \left(\frac{e^{-3s}}{s+1} \right) = u_3(t) - {\mathcal L}^{-1} \left( \frac{e^{-3s}}{s+1} \right). \end{equation*}
If $$g(t) = u_3(t) e^{-(t - 3)}\text{,}$$ then the Laplace transform of $$g(t)$$ is
\begin{equation*} {\mathcal L}(g) = e^{-3s} {\mathcal L}(e^{-t}) = \frac{e^{-3s}}{s + 1}. \end{equation*}
Thus,
\begin{equation*} y(t) = e^{-t} + u_3(t)\left( 1 - e^{-(t - 3)} \right). \end{equation*}

Subsection6.2.3Forced Harmonic Oscillators

Consider the forced harmonic oscillator
\begin{gather*} y'' + 4y = 3 \cos t\\ y(0) = y'(0) = 0. \end{gather*}
Taking the Laplace transform of both sides of the equation $$y'' + 4y = 3 \cos t\text{,}$$ we obtain
\begin{equation*} {\mathcal L}(y'') + 4 {\mathcal L}(y) = 3{\mathcal L}(\cos t) \end{equation*}
or
\begin{equation*} s^2 Y(s) - sy(0) - y'(0) + 4Y(s) = \frac{3s}{s^2 + 1}, \end{equation*}
where $${\mathcal L}(y)(s) = Y(s)\text{.}$$ Substituting the initial conditions and solving for $$Y\text{,}$$ we have
\begin{equation*} Y(s) = \frac{3s}{(s^2 + 4)(s^2 + 1)} = \frac{-s}{s^2 + 4} + \frac{s}{s^2 + 1}, \end{equation*}
where the last expression was obtained using partial fractions. Taking the inverse Laplace transform, we have our solution
\begin{equation*} y(t) = {\mathcal L}^{-1} \left( \frac{-s}{s^2 + 4} \right) + {\mathcal L}^{-1} \left( \frac{s}{s^2 + 1} \right) = -\cos 2t + \cos t. \end{equation*}
Now let us consider a harmonic oscillator with discontinuous forcing,
\begin{gather*} y'' + 2 y' + 5y = h(t)\\ y(0) = y'(0) = 0, \end{gather*}
where $$h(t)$$ is given by
\begin{equation*} h(t) = \begin{cases} 5 & t \lt 7 \\ 0 & t \geq 7. \end{cases} \end{equation*}
That is, $$h(t) = 5 (1 - u_7(t))\text{.}$$ We may consider this to be a mass-spring system sliding on a table, where the mass is one unit, the spring constant is 5, and the damping coefficient is 2. When $$t \lt 7$$ the table is tilted so that gravity provides a force of 5 units when stretching the spring. At time $$t = 7\text{,}$$ the table is suddenly returned to the level position.
Taking the Laplace transform of both sides of $$y'' + 2 y' + 5y = h(t)\text{,}$$ we obtain
\begin{equation*} [s^2 Y(s) -sy(0) - y'(0)] + 2 [sY(s) - y(0)] + 5Y(s) = {\mathcal L}(h), \end{equation*}
where $${\mathcal L}(y)(s) = Y(s)\text{.}$$ Substituting the initial conditions and evaluating the Laplace transform on the right, we have
\begin{equation*} (s^2 + 2s + 5)Y(s) = 5 \left( \frac{1}{s} - \frac{e^{-7s}}{s} \right). \end{equation*}
Solving for $$Y(s)\text{,}$$ we have
\begin{equation*} Y(s) = \frac{5}{s(s^2 + 2s + 5)} - \frac{5e^{-7s}}{s(s^2 + 2s + 5)} \end{equation*}
and
\begin{equation*} y = {\mathcal L}^{-1} \left( \frac{5}{s(s^2 + 2s + 5)} - \frac{5e^{-7s}}{s(s^2 + 2s + 5)} \right). \end{equation*}
Using partial fractions, we can rewrite the first term as
\begin{equation*} \frac{5}{s(s^2 + 2s + 5)} = \frac{1}{s} - \frac{s + 2}{s^2 + 2s + 5}. \end{equation*}
The inverse Laplace transform of $$1/s$$ is 1. To find the inverse Laplace transform of the second term, we complete the square of the denominator,
\begin{align*} \frac{s + 2}{s^2 + 2s + 5} & = \frac{s + 2}{(s+ 1)^2 + 4}\\ & = \frac{s + 1}{(s+ 1)^2 + 4} + \frac{1}{(s+ 1)^2 + 4}\\ & = \frac{s + 1}{(s+ 1)^2 + 4} + \frac{1}{2} \frac{2}{(s+ 1)^2 + 4}. \end{align*}
Consequently,
\begin{equation*} {\mathcal L}^{-1} \left( \frac{s + 2}{s^2 + 2s + 5} \right) = e^{-t} \cos 2t + \frac{1}{2} e^{-t} \sin 2t = e^{-t} \left( \cos 2t + \frac{1}{2} \sin 2t \right). \end{equation*}
and
\begin{equation*} {\mathcal L}^{-1} \left( \frac{5}{s(s^2 + 2s + 5)} \right) = {\mathcal L}^{-1} \left( \frac{1}{s} - \frac{s + 2}{s^2 + 2s + 5} \right) = 1 - e^{-t} \left( \cos 2t + \frac{1}{2} \sin 2t \right). \end{equation*}
We can compute the inverse Laplace transform of
\begin{equation*} \frac{5e^{-7s}}{s(s^2 + 2s + 5)} \end{equation*}
using the Heaviside function $$u_7(t)$$ and the inverse Laplace transform that we just calculated to obtain
\begin{equation*} {\mathcal L}^{-1} \left( \frac{5e^{-7s}}{s(s^2 + 2s + 5)} \right) = u_7(t) \left( 1 - e^{-(t-7)} \left( \cos 2(t - 7) + \frac{1}{2} \sin 2(t - 7) \right) \right). \end{equation*}
Therefore, the solution to our original initial value problem is
\begin{equation*} y(t) = 1 - e^{-t} \left( \cos 2t + \frac{1}{2} \sin 2t \right) - u_7(t) \left( 1 - e^{-(t-7)} \left( \cos 2(t - 7) + \frac{1}{2} \sin 2(t - 7) \right) \right). \end{equation*}

Activity6.2.2.Solving Differential Equations with Laplace Transforms with Discontinuous Forcing Functions.

Use Laplace Transforms to solve each of the following initial value problems.
(a)
$$x' + x = u_3(t)\text{,}$$ $$x(0) = 2$$
(b)
$$x'' + x = h(t)\text{,}$$ $$x(0) = 0\text{,}$$ $$x'(0) = 0$$ where
\begin{equation*} h(t) = \begin{cases} 0 & t \lt 1 \\ 2 & t \geq 1. \end{cases} \end{equation*}

Subsection6.2.4Important Lessons

• Using the Laplace transform including how the transform behaves, we can solve initial value problems such as
\begin{align*} a y''(t) + b y'(t) + c y(t) \amp = f(t)\\ y(0) \amp = y_0\\ y'(0) \amp = y'_0 \end{align*}
even when $$f(t)$$ is discontinuous.
• Let $$y = y(t)$$ be a piecewise continuous, exponentially bounded function and assume that $$y'$$ is also exponentially bounded. Then for large values of $$s$$
\begin{equation*} {\mathcal L}(y')(s) = s {\mathcal L}(y)(s) - y(0) = s Y(s) - y(0), \end{equation*}
where $$Y(s)$$ is the Laplace transform of $$y\text{.}$$
• Let $$y = y(t)$$ and $$y'(t)$$ be piecewise continuous, exponentially bounded functions and assume that $$y''$$ is exponentially bounded. Then for large values of $$s$$
\begin{equation*} {\mathcal L}(y'')(s) = s^2 {\mathcal L}(y)(s) - sy(0) - y'(0)= s^2 Y(s) - sy(0) - y'(0), \end{equation*}
where $$Y(s)$$ is the Laplace transform of $$y\text{.}$$ In general, if $$y = y(t)$$ and all of its derivatives up to order $$k - 1$$ are piecewise continuous, exponentially bounded functions and $$y^{(k)}$$ is piecewise continuous, then
\begin{gather*} {\mathcal L}(y^{(k)})(s) = s^k {\mathcal L}(y)(s) - s^{k - 1}y(0) - \cdots - s y^{(k -2)}(0) - y^{(k -1)}(0)\\ = s^k Y(s) - s^{k - 1}y(0) - s^{k - 1}y(0)- \cdots - s y^{(k -2)}(0) - y^{(k -1)}(0). \end{gather*}

1.

Describe in your own words how the Laplace transform can be used to solve an initial value problem.

2.

Explain what an exponentially bounded function is. Give an example.

Exercises6.2.6Exercises

Solving Initial Value Problems.

Solve the initial problems in Exercise Group 6.2.6.1–8 using the Laplace transform.
1.
$$y'' - 2y' - 3y = 3 e^{2t}\text{,}$$ $$y(0) = 1\text{,}$$ $$y'(0) = 0$$
2.
$$y'' - y' - 2y = 4x^2\text{,}$$ $$y(0) = -1\text{,}$$ $$y'(0) = 1$$
3.
$$\dfrac{d^2x}{dx^2} - 6 \dfrac{dx}{dt} + 25 x = 64e^{-t}\text{,}$$ $$x(0) = 1\text{,}$$ $$x'(0) = -2$$
4.
$$y'' + 16y = 2 \sin 2t\text{,}$$ $$y(0) = 1\text{,}$$ $$y'(0) = 0$$
5.
$$y'' + 16y = 2 \sin 4t\text{,}$$ $$y(0) = 1\text{,}$$ $$y'(0) = 0$$
6.
$$y'' + 2y' + y = 2e^{-t}\text{,}$$ $$y(0) = -1\text{,}$$ $$y'(0) = 3$$
7.
$$y'' + 6y' + 8y = \cos 3t\text{,}$$ $$y(0) = -2\text{,}$$ $$y'(0) = 1$$
8.
$$u'' + \omega_0^2 y = \cos \omega t\text{,}$$ $$\omega^2 \neq \omega_0^2\text{,}$$ $$u(0) = 1\text{,}$$ $$u'(0) = -1$$

9.

Find the solution of the initial value problem
\begin{gather*} 2 y'' + y' + 2y = g(t)\\ y(0) = y'(0) = 0 \end{gather*}
where $$g(t)$$ is defined by
\begin{equation*} g(t) = u_5(t) - u_{20}(t) = \begin{cases} 1, & 5 \leq t \lt 20, \\ 0, & 0 \leq t \lt 5 \text{ and } t \geq 20. \end{cases} \end{equation*}

10.

Let $$y = y(t)$$ and $$y'(t)$$ be piecewise continuous, exponentially bounded functions and assume that $$y''$$ is exponentially bounded.
1. Prove that
\begin{equation*} {\mathcal L}(y'')(s) = s^2 {\mathcal L}(y)(s) - sy(0) - y'(0)= s^2 Y(s) - sy(0) - y'(0) \end{equation*}
for large values of $$s\text{,}$$ where $$Y(s)$$ is the Laplace transform of $$y\text{.}$$
2. If $$y = y(t)$$ and all of its derivatives up to order $$k - 1$$ are piecewise continuous, exponentially bounded functions and $$y^{(k)}$$ is piecewise continuous, prove that
\begin{gather*} {\mathcal L}(y^{(k)})(s) = s^k {\mathcal L}(y)(s) - s^{k - 1}y(0) - \cdots - s y^{(k -2)}(0) - y^{(k -1)}(0)\\ = s^k Y(s) - s^{k - 1}y(0) - s^{k - 1}y(0)- \cdots - s y^{(k -2)}(0) - y^{(k -1)}(0). \end{gather*}

12.

Let $$f(t)$$ be a peicewise continuous function for $$t \geq 0\text{.}$$ In addition, suppose that $$f(t)$$satisfies the condition $$|f(t)| \leq M e^{ct}$$ with $$t \geq 0\text{.}$$
1. Prove that
\begin{equation*} {\mathcal L} \left( \int_0^t f(\tau) \, d\tau \right) = \frac{1}{s} {\mathcal L}(f(t) = \frac{F(s)}{s}. \end{equation*}
2. Show that
\begin{equation*} {\mathcal L}^{-1} \left( \frac{F(s)}{s} \right) = \int_0^t f(\tau) \, d\tau \end{equation*}
follows from (a).


text:
## CryptoDB

### Paper: Ate pairing for $y^{2}=x^{5}-\alpha x$ in characteristic five

Authors: Ryuichi Harasawa Yutaka Sueyoshi Aichi Kudo URL: http://eprint.iacr.org/2006/202 Search ePrint Search Google Recently, the authors proposed a method for computing the Tate pairing using a distortion map for $y^{2}=x^{5} -\alpha x$ ($\alpha = \pm2$) over finite fields of characteristic five. In this paper, we show the Ate pairing, an invariant of the Tate pairing, can be applied to this curve. This leads to about $50\%$ computational cost-saving over the Tate pairing.
##### BibTeX
@misc{eprint-2006-21695,
title={Ate pairing for $y^{2}=x^{5}-\alpha x$ in characteristic five},
booktitle={IACR Eprint archive},
keywords={Tate/Ate pairing, Hyperelliptic curves},
url={http://eprint.iacr.org/2006/202},
note={The full version, entitled "Tate and Ate Pairings for $y^{2} = x^{5} - \alpha x$ in Characteristic Five", is published in Japan Journal of Industrial and Applied Mathematics (JJIAM), Vol. 24, No. 3, pp. 251 - 274, 2007. harasawa@cis.nagasaki-u.ac.jp 13887},
author={Ryuichi Harasawa and Yutaka Sueyoshi and Aichi Kudo},
year=2006
}


text:
# Linear Programming-Problem

1. Mar 31, 2006

### love_physics

Hi there!
I have gott a problem with this task. My hope is that someone could help me out of this :=)
we have two products: A and B. There are three moments of processing in order to get the our products:

Number of tools avaible for each moments : 1 1 2
Time to use per day and tool: 8 8 8
Capacity per tool for product A: 500 250 250
Capacity per tool for product B: 200 500 125

The profit made from each manufactured and sold unit of product A is 2,2 $and from product B is 4,1$. Because of limitation of market vi can not have more than 1800 of product A and more than 1500 of product B. Number of product B shuld not be less than 20 % of total number of both products.

What shuld I do to maximize the profit?!

I need help the figure out the how to write a LP-program with the problem above.

Can anybody HELP ME with this?!


text:
Lay Up Technique, Farmhouse Fireplace Screen, Anastasia Beverly Hills Brow, Trainee In Japan Rules, Thule Universal Key, Sportrack Fit Guide, Our Lady Of Sorrows Parish Vancouver, Cuny Nursing Programs, Jeep Cherokee Check Engine Light No Codes, Fuel Oil Tank Removal Near Me, " />

### fundamental theorem of calculus properties


text:
A BORSUK-ULAM TYPE THEOREM OVER ITERATED SUSPENSIONS OF REAL PROJECTIVE SPACES

Title & Authors
A BORSUK-ULAM TYPE THEOREM OVER ITERATED SUSPENSIONS OF REAL PROJECTIVE SPACES
Tanaka, Ryuichi;

Abstract
A CW complex B is said to be I-trivial if there does not exist a $\small{\mathbb{Z}_2}$-map from $\small{S^{i-1}}$ to S($\small{{\alpha}}$) for any vector bundle $\small{{\alpha}}$ over B a any integer i with i > dim $\small{{\alpha}}$. In this paper, we consider the question of determining whether $\small{\Sigma^k\mathbb{R}P^n}$ is I-trivial or not, and to this question we give complete answers when k $\small{\neq}$ 1, 3, 8 and partial answers when k
Keywords
sphere bundle;$\small{\mathbb{Z}_2}$-map;index;
Language
English
Cited by
References
1.
P. E. Conner and E. E. Floyd, Fixed point free involutions and equivariant maps, Bull. Amer. Math. Soc. 66 (1960), 416-441.

2.
P. E. Conner and E. E. Floyd, Fixed point free involutions and equivariant maps II, Trans. Amer. Math. Soc. 105 (1962), 222-228.

3.
M. Fujii, KO-groups of projective spaces, Osaka J. Math. 4 (1967), 141-149.

4.
F. Hirzebruch, Neue Topologische Methoden in der Algebraischen Geometrie, Ergebnisse der Math., Springer, 1956.

5.
J. Milnor, Some consequences of a theorem of Bott, Ann. of Math. (2) 68 (1958), 444-449.

6.
R. Tanaka, On the index and co-index of sphere bundles, Kyushu J. Math. 57 (2003), no. 2, 371-382.

7.
R. Tanaka, A stability theorem for the index of sphere bundles, Bull. Belg. Math. Soc. Simon Stevin 14 (2007), no. 1, 177-182.

8.
R. Tanaka, A berwise analogue of the Borsuk-Ulam theorem for sphere bundles over a 2-cell complex, Topology Appl. 154 (2007), no. 8, 1827-1833.

9.
R. Tanaka, A Borsuk-Ulam type theorem for sphere bundles over stunted projective spaces, Topology Appl. 156 (2009), no. 5, 932-938.

10.
R. Tanaka, On trivialities of Stiefel-Whitney classes of vector bundles over iterated suspension spaces, Homology, Homotopy Appl. 12 (2010), no. 1, 357-366.

11.
W.-T. Wu, Les i-carres dans une variete grassmannienne, C. R. Acad. Sci. Paris 230 (1950), 918-920.


text:
Question

# For the molecule of $I{{F}_{7}}$.(a)State the hybridisation of the central atom.(b)State the geometry of the molecule.

Hint: Hybridization of a molecule is determined using the postulates of the Valence band theory and its geometry can be formulated by using the VSEPR rule. The geometry depends on the no. of lone pairs and bond pairs.

(a)Hybridization is when the concept of atomic orbitals fuse to make new hybridized orbitals, which successively influences molecular geometry and bonding properties. Hybridization is also an expansion of the valence bond theory.
According to Valence band theory the metal atom or ion in the influence of ligands uses its (n-1)d, ns, np, or ns, np, nd orbitals for hybridization to yield a group of equivalent orbitals of definite geometry like octahedral, tetrahedral, square planar then on.(Here ligands refers to ion or molecule, which donates a pair of electron to the central metal These hybrid orbitals are allowed to overlap with ligand orbitals which will donate electron pairs for bonding.
Hybridisation = Number of valence electrons on central atom + Number of monovalent surrounding atoms (like hydrogen and halogen)- positive charge (if on complex) + anionic charge (if on complex) and divide this whole by 2.
Now, for $I{{F}_{7}}$: Hybridisation = 7 (number of valence electrons on I-atom) + 7 (number of surrounding halogen atoms) -0+0 and divide this whole by 2. The element Iodine has 7 lone electrons. When it is bonded to seven Fluorine atoms through sigma bonds, the total number of electron pairs around Iodine becomes 7 with no lone pair of el
(Ground State)
(Excited State)
$s{{p}^{3}}{{d}^{3}}$- Hybridization ($I{{F}_{7}}$)

(Ground State)
(Excited State)
$s{{p}^{3}}{{d}^{3}}$- Hybridization ($I{{F}_{7}}$)

Thus, the hybridisation of $I{{F}_{7}}$is $s{{p}^{3}}{{d}^{3}}$ with no lone pair of electrons and this corresponds to pentagonal bipyramidal geometry.

(b) Iodine heptafluoride, $I{{F}_{7}}$, may be an example of a pentagonal bipyramidal geometry. VSEPR rules suggest seven electron pairs. These are made up from six bonding pairs and one lone pair.

Note:
-VSEPR theory has certain postulates which one must follow to deduce the geometrical structure of a compound.
-It can be noted that sigma bonds involve the head-to-head overlapping of atomic orbitals whereas pi bonds involve parallel overlapping.


text:
# Monster Capturing

About a week ago, one of my friends (who I met at stage two of this year’s CCC – fun times, fun times) posed an interesting problem to me, which I will try to summarize below:

Alice and Bob are professional monster trappers, but occasionally, even they get stuck in difficult scenarios. In this world, Alice is currently at a location denoted by $(0,0)$ (think of a grid). $P$ monsters exist at certain locations, and there’s a maximum of one monster at a given location. Bob is on a plane right now, and he wants to ask: if I fall down and land at location $(x,y)$, how many monsters will lie between me and Alice (where “between” means if a monster is at $(m,n)$, then it’s “between” if $0 \leq m \leq x$ and $0 \leq n \leq y)$? Note that Bob will ask the question many, many times ($Q$ times), all with different values of $(x,y)$. (Also: $0 \leq x,y,m,n,P,Q \leq 30000$)

For the seasoned contest programming veteran, the question immediately reduces down to: given 30 000 of coordinates that fall within a range of 30 000 by 30 000, we must efficiently perform 30 000 queries, such that each query returns the number of coordinates below-or-at and left-or-at of the query location (we should always assume the biggest numbers). Thus, the super-naive $O(PQ)$ algorithm (~900 million calculations), where we iterate over all coordinates for each query, is too slow for the conventional 1-CPU-second time limit.

I loved this question, because there’s such an obviously sub-optimal classic data structure to solve it – the 2D binary index tree, or Fenwick tree (if you’re not sure what that is, here’s an excellent tutorial). The naive solution is simple, as follows:

First, make a 2D binary index tree of size 30 000-by-30 000. Proceed to add all the monsters into that tree (this takes $O(P\log ^2 P$), for $P$ insertions, each taking $\log ^2 P$ time). Then, we can run $Q$ queries on the tree, each taking $\log ^2 P$ time, so this takes $O(Q\log ^2 P)$. Overall, $O((Q+P)\log ^2 P)$, which is about 12 million primitive operations, a very respectable time (about 0.2 s, by my estimate).

However, the problem is that you can’t really make a 30 000-by-30 000 tree – not unless you’re given close to a gigabyte of memory! Since we’re usually given only 200 MB or so, a 2D tree fails spectacularly.

I admit, this problem puzzled me quite a bit, because I simply couldn’t see a solution without a 2D tree. Coordinate compression came to mind, but that didn’t seem to help significantly either. But then I thought back to my USACO days, and an “aha!” moment came in a flash. Here’s my solution:

First, we put everything in an array of size 60 000, monsters and queries, each in a 3-tuple of form $(x, y, r)$, where $x,y$ are the coordinates and $r$ is a number: 0 if a monster, 1 if a query. We sort it, with the leftmost coordinates first, and if tied, bottommost first, and so on (C++’s std::sort is a wonderful way to sort a pair<pair<int,int>, int> type like this, by the way). We also form a 1D binary index tree of size 30 000, which will keep track of how many monster below a certain y-coordinate we have seen already. Now, we do a plane sweep (shout-out to USACO for teaching me this!) by iterating over the array: if we see a monster, we add its y-coordinate to the tree. If we see a query, then we simply query the tree for the number of monsters, since we know that the monsters already seen are to the left of the query. Thus, the runtime of this algorithm is $O((P+Q)\log(P+Q))$, dominated by the sorting (array processing is “only” $O((P+Q)\log P))$. This solution clearly runs under the memory and time limits.