lean-math - Basic Math
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enCombinations and Permutations. Count the Ways.
http://www.leanmath.com/blog-entry/combinations-and-permutations-count-ways
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even" property="content:encoded"><p><strong>Combinations and Permutations</strong> How many different poker hands are there? How many different pizza orders can be made? How many different ways can this work schedule be filled out? How many different ways are there to arrange your books in a bookshelf? These are all examples of combinations and permutations. And knowing how to calculate them is a helpful tool for decision making. The basic equations are: . <a href="/sites/lean-math/files/blog/wp-content/uploads/2015/01/Combination1a.jpg"><img alt="Combination1a" height="224" width="472" class="media-image alignnone wp-image-1275 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2015/01/Combination1a-300x139.jpg" /></a> <strong>Permutations (with repetitions)</strong> <strong>Example:</strong> How many different ways can this work schedule be filled out? This is very straightforward. Suppose there are three operators (i.e. <em>n</em> equals 3) who have to cover of two shifts (i.e. <em>r</em> equals 2), then 9 possible work schedules that can be created. </p>
<table><tbody><tr><td width="25"> </td>
<td width="164"><strong>Shift 1</strong></td>
<td width="213"><strong>Shift 2</strong></td>
</tr><tr><td width="25">1</td>
<td width="164">A</td>
<td width="213">A</td>
</tr><tr><td width="25">2</td>
<td width="164">A</td>
<td width="213">B</td>
</tr><tr><td width="25">3</td>
<td width="164">A</td>
<td width="213">C</td>
</tr><tr><td width="25">4</td>
<td width="164">B</td>
<td width="213">A</td>
</tr><tr><td width="25">5</td>
<td width="164">B</td>
<td width="213">B</td>
</tr><tr><td width="25">6</td>
<td width="164">B</td>
<td width="213">C</td>
</tr><tr><td width="25">7</td>
<td width="164">C</td>
<td width="213">A</td>
</tr><tr><td width="25">8</td>
<td width="164">C</td>
<td width="213">B</td>
</tr><tr><td width="25">9</td>
<td width="164">C</td>
<td width="213">C</td>
</tr></tbody></table><p> The number of permutations is simply: <a href="/sites/lean-math/files/blog/wp-content/uploads/2015/01/Combination2.jpg"><img alt="Combination2" height="127" width="431" class="media-image alignnone wp-image-1276 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2015/01/Combination2-300x80.jpg" /></a> <strong>Permutations (without repetitions) </strong><strong> </strong> <strong>Example:</strong> How many different ways are there to arrange your books in a bookshelf, if you have 3 books (called A, B, and C)? Again this is a straightforward question. it is easy to quickly determine that the correct answer is 6. For the first book, there are three choices, for the second there are only two choices and only one choice for the last choice. So the total number of permutations in this case is This simple example gives insight into constructing the equation for the general case. For the first choice there are options, for the second choice there are options, and this process repeats times. This can be written as: <a href="/sites/lean-math/files/blog/wp-content/uploads/2015/01/Combination3.jpg"><img alt="Combination3" height="98" width="453" class="media-image alignnone wp-image-1274 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2015/01/Combination3-300x58.jpg" /></a> <a href="/sites/lean-math/files/blog/wp-content/uploads/2015/01/Combinations4.jpg"><img alt="Combinations4" height="302" width="447" class="media-image alignnone wp-image-1277 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2015/01/Combinations4-300x199.jpg" /></a> <strong>Combinations (with repetitions)</strong> <strong>Example:</strong> How many different pizzas can be made? Assuming that are n different kinds of pizza and you are ordering r pizzas. <a href="/sites/lean-math/files/blog/wp-content/uploads/2015/01/Combinations5.jpg"><img alt="Combinations5" height="199" width="485" class="media-image alignnone wp-image-1278 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2015/01/Combinations5-300x116.jpg" /></a> </p>
<table><tbody><tr><td colspan="2" width="253"><b>The six pizza orders (pizza pies come in flavors A, B, or C) that can be made with two pies are:</b></td>
</tr><tr><td width="128">A</td>
<td width="125">A</td>
</tr><tr><td width="128">A</td>
<td width="125">B</td>
</tr><tr><td width="128">A</td>
<td width="125">C</td>
</tr><tr><td width="128">B</td>
<td width="125">B</td>
</tr><tr><td width="128">B</td>
<td width="125">C</td>
</tr><tr><td width="128">C</td>
<td width="125">C</td>
</tr></tbody></table><p><strong>Combinations (without repetitions)</strong> <strong>Example:</strong> How many different poker hands are there? Here the relevant equation is: <a href="/sites/lean-math/files/blog/wp-content/uploads/2015/01/Combinations6.jpg"><img alt="Combinations6" height="238" width="485" class="media-image alignnone wp-image-1279 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2015/01/Combinations6-300x141.jpg" /></a> </p>
</div></div></div><div class="field field-name-field-tags field-type-taxonomy-term-reference field-label-hidden"><div class="field-items"><div class="field-item even"><a href="/tags/basic-math" typeof="skos:Concept" property="rdfs:label skos:prefLabel" datatype="">Basic Math</a></div></div></div>Wed, 04 Feb 2015 20:42:47 +0000drmike125 at http://www.leanmath.comhttp://www.leanmath.com/blog-entry/combinations-and-permutations-count-ways#commentsANOVA, Two Chevys, and a Ford
http://www.leanmath.com/blog-entry/anova-two-chevys-and-ford
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even" property="content:encoded"><p>Analysis of Variance (most commonly referred to as ANOVA) is a common statistical test that compares the averages of several sets of data. It is unlikely for datasets to have the same average, and an ANOVA test quantifies how likely those observed differences in occurred by chance. It is a really useful test because we frequently find ourselves comparing averages of different datasets. We may, for example, compare the average customer demand for different months, or the average turnaround time for several job shops, or the average length of stay for hip replacement patients for several different doctors. These comparisons are useful because they can give us insight into the underlying value streams. If the length of stay varies from physician to physician, it might suggest an underlying variation in the treatment protocol. If customer demand varies from month to month, is the variation just noise or are the observed differences in average customer demand unlikely to have occurred by chance? And if they are unlikely to have occurred by chance, what is causing the change in demand? Are your competitors having sale events that you aren’t? An ANOVA analysis won’t tell you why the averages are different, but it will signal if something unusual is going on that warrants further investigation. The steps for performing an ANOVA analysis are as follows: <strong> Step 1: Look at the data.</strong> Do the data points look reasonable? Are the values high or low (good or bad)? How do the values compare to expectations? How do they compare to benchmarks? Are there any outliers or anomalous values? Do the averages appear different? What factors distinguish the datasets? <strong>Step 2: Make several different graphs comparing the different datasets</strong> (e.g. Dot Plot, Boxplot, Histogram, and Interval Plot). Do the datasets appear to have the same averages? Do the datasets appear to be normally distributed and have the same variance? <strong>Step 3: Verify the prerequisites of ANOVA analysis.</strong> Test the normality of the datasets. Test for equal variances. (If the prerequisites are not satisfied, there are several options: transform the data; proceed with the ANOVA analysis and use the results with appropriate caution; or use a different hypothesis test e.g. Mood’s Median or Kruskal-Wallis and compare medians instead of averages). <strong>Step 4: Run the ANOVA analysis and analyze the results.</strong> The null hypothesis (H<sub>0</sub>) of an ANOVA analysis is that the means are equal, if the p Value is low, the statistical conclusion is that the difference in the means is statistically significant (i.e. unlikely to have occurred by chance). <strong>Step 5: Examine the residuals, and summarize the results.</strong> <strong>Example:</strong> A few car enthusiasts are testing the impact that a fuel additive has on their fuel mileage. They ran 10 trials, with 10 gallons of gas each, and the change of their miles per gallon was as follows:</p>
<table><tbody><tr><td>Nova</td>
<td>Chevy 1</td>
<td>Chevy 2</td>
<td>Ford</td>
</tr><tr><td> -0.9</td>
<td> 0.2</td>
<td> 2.5</td>
<td> -0.8</td>
</tr><tr><td> -0.7</td>
<td> 0.2</td>
<td> 0.8</td>
<td> 1.0</td>
</tr><tr><td> -1.1</td>
<td> -0.2</td>
<td> 1.1</td>
<td> -0.6</td>
</tr><tr><td> 1.2</td>
<td> -1.1</td>
<td> 1.8</td>
<td> -0.2</td>
</tr><tr><td> -0.1</td>
<td> 0.3</td>
<td> 1.8</td>
<td> -2.1</td>
</tr><tr><td> -0.1</td>
<td> 0.4</td>
<td> 2.9</td>
<td> -0.3</td>
</tr><tr><td> 0.0</td>
<td> -0.8</td>
<td> 0.7</td>
<td> 0.6</td>
</tr><tr><td> 0.8</td>
<td> 0.8</td>
<td> 2.7</td>
<td> -2.7</td>
</tr><tr><td> 1.1</td>
<td> 0.0</td>
<td> 1.8</td>
<td> -0.2</td>
</tr><tr><td> -0.2</td>
<td> -0.9</td>
<td> 0.7</td>
<td> -0.6</td>
</tr></tbody></table><p>Step 1: Looking at the data. Here, we don’t see much other than all of the values for Chevy 2 are positive, indicating that for each trial the miles per gallon increased. Step 2: Graph the data. Here again we see that the average for Chevy 2 appears to be the highest. <a href="/sites/lean-math/files/blog/wp-content/uploads/2014/07/ANOVA-individual-value-plot.jpg"><img alt="ANOVA individual value plot" height="359" width="533" class="media-image alignnone wp-image-1132 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2014/07/ANOVA-individual-value-plot-300x200.jpg" /></a> Step 3: Validate the assumptions for an ANOVA analysis. Two of the assumptions for an ANOVA analysis are that the datasets are normally distributed and have equal variances. The histograms are consistent with these assumptions, and statistical tests such as the Anderson-Darling normality test and Bartlett’s test for equal variances also yield results that are also consistent with these assumptions. <a href="/sites/lean-math/files/blog/wp-content/uploads/2014/07/ANOVA-histograms.jpg"><img alt="ANOVA histograms" height="358" width="531" class="media-image alignnone wp-image-1133 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2014/07/ANOVA-histograms-300x200.jpg" /></a> Step 4: Run the ANOVA analysis. Any reasonable statistical software can do this for you. Our software gave us a p-value (i.e. the likelihood that the observation occurred by chance), of 0.000 indicating that the results are statistically unusual. Step 5: Examine the residuals and summarizing the analysis. A residual is the part of each data value that is not explained by the model. Where the model is simply the observed data is equal to the overall data average plus a column average offset. <a href="/sites/lean-math/files/blog/wp-content/uploads/2014/07/ANOVA-residuals-4.jpg"><img alt="ANOVA residuals 4" height="389" width="572" class="media-image alignnone wp-image-1140 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2014/07/ANOVA-residuals-4-300x200.jpg" /></a> There are no obvious patterns to the residuals, so our conclusion is that the mpg increase for Chevy 2 was unexpected. Subsequent conversations with the Chevy 2 driver revealed that they emptied their trunk, properly inflated their tires, and accelerated more gently for the test. All of which may explain the observed result. The point being that with analysis we gain insight into hidden, underlying process dynamics. </p>
</div></div></div><div class="field field-name-field-tags field-type-taxonomy-term-reference field-label-hidden"><div class="field-items"><div class="field-item even"><a href="/tags/basic-math" typeof="skos:Concept" property="rdfs:label skos:prefLabel" datatype="">Basic Math</a></div></div></div>Tue, 08 Jul 2014 17:35:12 +0000drmike116 at http://www.leanmath.comhttp://www.leanmath.com/blog-entry/anova-two-chevys-and-ford#commentsCumulative Deviation
http://www.leanmath.com/blog-entry/cumulative-deviation
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even" property="content:encoded"><p>Maybe you have a day by hour chart, or sales plan vs. actual sales data and you want an easy way to decide when to take action - consider the simple Cumulative Deviation. First you need some data, two variables that you want to compare. Here in this spreadsheet example we have Plan and Actual for eight time periods. <a href="/sites/lean-math/files/blog/wp-content/uploads/2014/06/CumulativeDeviation1.png"><img alt="CumulativeDeviation1" height="131" width="400" class="media-image aligncenter size-full wp-image-1121 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2014/06/CumulativeDeviation1.png" /></a>Deviation is Plan - Actual. In the first period we fell short of the Plan by 3 units. I period 2 we did a little better and slightly exceed the plan (-1). Cumulative Deviation is adding the period deviation to the one preceding it. Here in period 2 we have (-1) + (3) = 2. Drag the formula across all the columns and you now can more easily see that the deviations are growing. <a href="/sites/lean-math/files/blog/wp-content/uploads/2014/06/CumulativeDeviation2.png"><img alt="CumulativeDeviation2" height="121" width="437" class="media-image aligncenter size-full wp-image-1122 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2014/06/CumulativeDeviation2.png" /></a>At some point, maybe we need to change the Plan, or do 5 Whys or Root Cause Analysis to better understand the cause and effect.</p>
</div></div></div><div class="field field-name-field-tags field-type-taxonomy-term-reference field-label-hidden"><div class="field-items"><div class="field-item even"><a href="/tags/basic-math" typeof="skos:Concept" property="rdfs:label skos:prefLabel" datatype="">Basic Math</a></div></div></div>Sun, 22 Jun 2014 20:12:10 +0000lloucka115 at http://www.leanmath.comhttp://www.leanmath.com/blog-entry/cumulative-deviation#commentsBasic Probability (or What Are the Chances?)
http://www.leanmath.com/blog-entry/basic-probability-or-what-are-chances
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even" property="content:encoded"><p>Let’s suppose there are three BIG potential orders in your sales pipeline. What are your chances that you are going to win an order? This is a question that faces manufacturing and service industries all the time. If they chase too much business, they run the risk of winning the business and not being able to fulfill the request but if they don’t chase any business, they run the risk of being idle. To illustrate the math behind this important question let’s suppose the following: <a href="/sites/lean-math/files/blog/wp-content/uploads/2014/04/chances1.png"><img alt="chances1" height="194" width="787" class="media-image aligncenter size-full wp-image-1081 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2014/04/chances1.png" /></a>So what are your chances of winning an order? Clearly, it isn’t the sum of the probabilities of success. You can’t have a 110% chance of winning an order. And clearly the chance of success has to be greater than 50% because it has to be at least as large as you highest chance of winning. This is one of those interesting math questions that comes up frequently and fortunately has a simple solution. The trick is to not calculate the probability of winning an order, but instead to calculate the probability of not winning any orders and then take it from there. This is because if you know the probability of not winning any orders, then the probability of winning an order must be one minus that probability. If we assume that the orders are independent, the probability of not winning any orders is: <em><a href="/sites/lean-math/files/blog/wp-content/uploads/2014/04/chances11.png"><img alt="chances1" height="178" width="1284" class="media-image aligncenter size-full wp-image-1086 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2014/04/chances11.png" /></a></em>So therefore, there is a 76% chance of winning at least one order. Another way to solve this problem would have been to enumerate all the possible outcomes and then count how many times each outcome occurs and divide that number by the total number of outcomes to determine the probability for that outcome to occur (see Figure 1).</p>
<div style="width: 457px;display:block;margin:0 auto;"><a href="/sites/lean-math/files/blog/wp-content/uploads/2014/04/chances3.png"><img alt="chances3" height="342" width="447" class="media-image wp-image-1083 size-full media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2014/04/chances3.png" /></a> Figure 1: Illustration of all of the possible outcomes for orders A, B, and C. There are 200 cells shown, each cell represents one possible outcome. Half of the cells are shaded, representing the 50% chance of winning order B. 40% of the cells have a upper left to bottom right diagonal (representing the probability of winning of order A), and 20% of the cells have a bottom left to upper right diagonal representing those outcomes where order C is won. Cells with no shading or diagonals are those cases where no orders are won. There are 48 cases where that happens. Since there are 200 possible outcomes, the chance of winning no orders is 48/200 or 24%.
<p> </p>
</div>
<p>For cases where enumerating all the possible outcomes are difficult, one might rely on the basic rules of probability such as:</p>
<ol><li>The probability of an event happening is, one minus the probability of it not happening (i.e. Subtraction rule)</li>
<li>The probability of several independent events happening is the product of the probability of the events happening (i.e. Special multiplication rule)</li>
<li>The total probability for mutually exclusive events to occur is the sum of their probabilities (i.e. Addition rule)</li>
</ol><p>For more rules of basic probability, see: <strong><em>The Cartoon Guide to Statistics</em></strong>, by Larry Gonick and Woollcott Smith.</p>
</div></div></div><div class="field field-name-field-tags field-type-taxonomy-term-reference field-label-hidden"><div class="field-items"><div class="field-item even"><a href="/tags/basic-math" typeof="skos:Concept" property="rdfs:label skos:prefLabel" datatype="">Basic Math</a></div></div></div>Thu, 24 Apr 2014 06:09:28 +0000drmike109 at http://www.leanmath.comhttp://www.leanmath.com/blog-entry/basic-probability-or-what-are-chances#commentsNumeric Simulation, Apple Pie Sales, and Thanksgiving
http://www.leanmath.com/blog-entry/numeric-simulation-apple-pie-sales-and-thanksgiving
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even" property="content:encoded"><p>Accurately determining the appropriate stocking levels of perishable items is very important. Stock too few items and you will have disappointed customers. Stock too many and you’ll have unsold product - which has a high probability of turning into a loss. </p>
<p>A number of industries face this dilemma each and every day.</p>
<p>For example, how many seats should an airline sell for a flight? If a carrier sell exactly the same number of seats that are on the plane, then they are almost certain to have empty seats. If they oversell too much, they are likely to strand passengers and create some serious angst...especially around this time of year when people are trying to get home for Thanksgiving and their favorite apple pie.</p>
<p>There are a number of ways to model and address these questions, perhaps the easiest way is to simply do a numeric simulation.</p>
<p>Let’s say that a bakery has 1,000 customers and they expect 90% of their customers to purchase an apple pie. In this case, the number of customers that actually pick up an apple pie might look like this:</p>
<p><a href="/sites/lean-math/files/blog/wp-content/uploads/2013/11/apple-pie.png"><img alt="apple pie" height="491" width="736" class="media-image aligncenter size-full wp-image-927 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2013/11/apple-pie.png" /></a>Ten thousand cases were simulated for the above graph and not surprisingly, the average number of apple pies picked up is 900 (i.e. the number of customers multiplied by the likelihood that each customer will pick up an apple pie). But, the real interesting questions tie to how much variation there is in this distribution, because that determines how many pies the bakery needs to make. That is, if we just rearrange the data and plot it as a cumulative probability (see figure below), we can see that 99.99% of the time, customers picked up fewer than 937 apple pies.</p>
<p><a href="/sites/lean-math/files/blog/wp-content/uploads/2013/11/applie-pie2.png"><img alt="applie pie2" height="481" width="722" class="media-image aligncenter size-full wp-image-929 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2013/11/applie-pie2.png" /></a>And, if you do these numeric simulations enough, for different number of customers (<em>n</em>) and differently probabilities (<em>p</em>), you will notice that the average number of apple pies picked up (x-bar) is:</p>
<p><a href="/sites/lean-math/files/blog/wp-content/uploads/2013/11/apple-pie3.png"><img alt="apple pie3" height="31" width="69" class="media-image aligncenter size-full wp-image-931 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2013/11/apple-pie3.png" /></a> And the standard deviation of the distribution (σ) is:</p>
<p><a href="/sites/lean-math/files/blog/wp-content/uploads/2013/11/apple-pie4.png"><img alt="apple pie4" height="41" width="155" class="media-image aligncenter size-full wp-image-932 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2013/11/apple-pie4.png" /></a> which is simply a binomial process in action. </p>
<p>Happy Thanksgiving, everyone. Enjoy your apple pie!</p>
<p> </p>
</div></div></div><div class="field field-name-field-tags field-type-taxonomy-term-reference field-label-hidden"><div class="field-items"><div class="field-item even"><a href="/tags/basic-math" typeof="skos:Concept" property="rdfs:label skos:prefLabel" datatype="">Basic Math</a></div></div></div>Wed, 20 Nov 2013 19:49:51 +0000drmike95 at http://www.leanmath.comhttp://www.leanmath.com/blog-entry/numeric-simulation-apple-pie-sales-and-thanksgiving#commentsBayes' Theorem and the VMA's Were Outrageous (Again)
http://www.leanmath.com/blog-entry/bayes-theorem-and-vmas-were-outrageous-again
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even" property="content:encoded"><p>The VMA's were outrageous?!?</p>
<p>Well, if the media is to be believed, the news of the day was that it was an outrageous Video Music Award show.</p>
<p>Not true and true. Yes - the show was outrageous, no – it wasn’t news. </p>
<p>Let me explain. News is noteworthy, significant, and new. It changes your perspective, it is unusual. </p>
<p>If a fish swims in a fish tank all day and doesn’t drown it isn’t news because it happens all the time. But, if that same fish were to send text messages on a waterproof smartphone that would be news because that is something you don’t see every day. </p>
<p>The VMA's were outrageous and the VMA's were always outrageous. The only way for them to be newsworthy anymore is if they <b>were<b> NOT</b></b> outrageous.</p>
<p>The math behind this is pretty simple, it is called Bayes’ Theorem and it goes like this:</p>
<p><a href="/sites/lean-math/files/blog/wp-content/uploads/2013/08/VMA.1.png"><img alt="VMA.1" height="69" width="285" class="media-image aligncenter size-full wp-image-806 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2013/08/VMA.1.png" /></a><a href="/sites/lean-math/files/blog/wp-content/uploads/2013/08/VMA.21.png"><img alt="VMA.2" height="287" width="976" class="media-image aligncenter size-full wp-image-808 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2013/08/VMA.21.png" /></a>Now to estimate the probabilities let’s devise a model of MTV's programming. For our model we will say that MTV has three different types of shows (represented as circles in the figure below), videos (represented as triangles), reality TV shows, and award shows (represented as squares), and that some of the programming is outrageous (shaded in red) and some of the programming is not outrageous (not shaded).</p>
<p> <a href="/sites/lean-math/files/blog/wp-content/uploads/2013/08/VMA.3.png"><img alt="VMA.3" height="404" width="811" class="media-image aligncenter size-full wp-image-809 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2013/08/VMA.3.png" /></a>We are now in a position to calculate the probability that show was going to be outrageous (Event A) given that it was an awards show (Event B). By inspection we can see:</p>
<p><a href="/sites/lean-math/files/blog/wp-content/uploads/2013/08/VMA.4.png"><img alt="VMA.4" height="233" width="976" class="media-image aligncenter size-full wp-image-810 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2013/08/VMA.4.png" /></a>That is, based on our assumptions there was a 100% chance that the VMAs were going to be outrageous. Now admittedly the assumptions in our model are wildly off the mark, and Bayes’ Theorem is generally used for much more challenging calculations like what is the probability that someone actually has a disease given that they have tested positive for it, or how should an assessment of risk change based on the number accidents observed, but these applications can wait for another day and another blog post. </p>
<p>Today, it is enough to simply observe that the VMAs were outrageous, and that is not news. </p>
</div></div></div><div class="field field-name-field-tags field-type-taxonomy-term-reference field-label-hidden"><div class="field-items"><div class="field-item even"><a href="/tags/basic-math" typeof="skos:Concept" property="rdfs:label skos:prefLabel" datatype="">Basic Math</a></div></div></div>Thu, 29 Aug 2013 14:09:58 +0000drmike86 at http://www.leanmath.comhttp://www.leanmath.com/blog-entry/bayes-theorem-and-vmas-were-outrageous-again#commentsDemand Segmentation Graph
http://www.leanmath.com/blog-entry/demand-segmentation-graph
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even" property="content:encoded"><p>An interesting pattern often emerges when you graph the <a href="http://leanmath.com/blog/2013/02/18/coefficient-of-variation/" title="Coefficient of Variation">coefficient of variation</a> (Cv) vs. volume of a process time series.</p>
<p><a href="/sites/lean-math/files/blog/wp-content/uploads/2013/06/Demand-Segmentation.png"><img alt="Demand Segmentation" height="177" width="300" class="media-image alignnone size-medium wp-image-685 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2013/06/Demand-Segmentation-300x177.png" /></a></p>
<p>Often, but not always, you'll see a distribution where there are:</p>
<ol><li>A few high volume products that will have low demand or process volume variation as measured by having a low coefficient of variation (standard deviation/mean),</li>
<li>and many other products that have erratic demand or activity.</li>
<li>Occasionally you may see a few high volume and high Cv products (large one-time bulk orders for example).</li>
</ol><p>The math is straight forward - gather up orders, shipments, consumption, receipts or other transactions of interest and calculate the time series average, standard deviation, and Cv. Then plot the average vs. the Cv.</p>
<p>This 'demand segmentation' is a fundamental analysis component of many process capacity and material handling design efforts. Traditional ABC-Pareto 80/20 can be misleading when only volume is considered. Typically, high Cv items don't make good candidates for kanban or point of use replenishment. Low volume 'steady eddy' C items can be very predictable and make perfect candidates to take off of MRP reordering.</p>
<p> </p>
</div></div></div><div class="field field-name-field-tags field-type-taxonomy-term-reference field-label-hidden"><div class="field-items"><div class="field-item even"><a href="/tags/basic-math" typeof="skos:Concept" property="rdfs:label skos:prefLabel" datatype="">Basic Math</a></div></div></div>Mon, 24 Jun 2013 20:53:49 +0000lloucka78 at http://www.leanmath.comhttp://www.leanmath.com/blog-entry/demand-segmentation-graph#commentsTo Celebrate, or Not to Celebrate, the t-Test Will Tell
http://www.leanmath.com/blog-entry/celebrate-or-not-celebrate-t-test-will-tell
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even" property="content:encoded"><p>So you cut your cycle time by 10%.</p>
<p>...Or did you? </p>
<p><a href="/sites/lean-math/files/blog/wp-content/uploads/2013/06/t-test-table.png"><img alt="t-test table" height="527" width="452" class="media-image aligncenter size-full wp-image-668 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2013/06/t-test-table.png" /></a></p>
<p>We are constantly experimenting. We make improvements, measure our performance and update standard work. In comparing the cycle time before and after a kaizen, you calculate a 10% improvement. Is it time to do declare victory and move on (as much as that's possible) to the next challenge? </p>
<p>How do you know if the measured improvement is “real”? That is, how do you know if the observed 10% improvement is significant or if it could have happened by chance? A simple t-test can help answer that question. It compares, in this situation, the average cycle time before the kaizen to the average cycle time after the kaizen and determines how likely it is that the change in cycle time occurred by chance.</p>
<p>If the 10% improvement has an 85% likelihood of occurring by chance, then we probably want to hold off on the celebration party. But if the improvement only has a 2% chance of occurring by chance, then we can be fairly confident that the change is real.</p>
<p>The interval plot below shows the 95% confidence interval for the mean cycle time before and after the kaizen. Based on our sample, we are 95% confident that the actual average cycle time before the kaizen is somewhere between 28.2 and 31.7 (by the way, this is purely an example, the unit of measure could be seconds, minutes, days, weeks, etc.). Likewise, we are 95% confident that the average cycle time after the kaizen is somewhere between 25.2 and 28.8. So, it is possible that the average cycle time after the kaizen is actually not less than the average cycle time before the kaizen, but clearly that scenario is unlikely.</p>
<p>How unlikely? That is exactly what the t-test calculates.</p>
<p> <a href="/sites/lean-math/files/blog/wp-content/uploads/2013/06/t-Test-plot.png"><img alt="t-Test plot" height="600" width="900" class="media-image aligncenter size-full wp-image-664 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2013/06/t-Test-plot.png" /></a></p>
<p>There are a number of software packages that perform t-tests. Two decisions need to be made prior to making the calculation: 1) should the calculation assume that the two samples have equal variances, and 2) are you interested in determining if the averages are not equal or if one is greater than or less than the other (i.e. what is the alternative hypothesis for the t-test)? For this dataset, the variances are sufficiently close such that they can be treated as equal and we are interested in determining if the average cycle time after the kaizen is less than the cycle time before the kaizen. Given these two assumptions, the likelihood that the observed differences in the average cycle time before and after the kaizen occurred by chance is 0.009 or 0.9%. (Statisticians refer to this as the p-value.)</p>
<p>So, given that it is unlikely that the observed 10% reduction in cycle time occurred by chance, the team can reasonably celebrate their improvement and move on to the next challenge.</p>
</div></div></div><div class="field field-name-field-tags field-type-taxonomy-term-reference field-label-hidden"><div class="field-items"><div class="field-item even"><a href="/tags/basic-math" typeof="skos:Concept" property="rdfs:label skos:prefLabel" datatype="">Basic Math</a></div></div></div>Wed, 19 Jun 2013 03:53:05 +0000drmike77 at http://www.leanmath.comhttp://www.leanmath.com/blog-entry/celebrate-or-not-celebrate-t-test-will-tell#commentsSummation notation and sizing the kanban bus route
http://www.leanmath.com/blog-entry/summation-notation-and-sizing-kanban-bus-route
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even" property="content:encoded"><p>Σ or Sigma is the Greek upper case capital letter S and is used in mathematics to represent summation or addition of a series of elements or set of data values.<br />
x<sub>1</sub>, x<sub>2</sub>, x<sub>3</sub>, ... x<sub>n</sub> is a set of numbers. x<sub>1</sub> is the first number in the set, x<sub>i</sub> is the 'i'th number, x<sub>n</sub> is the last of n numbers.<br />
Elements can be simply adding up a set, or more involved such as summing squares or other algebraic equations.</p>
<p><a href="/sites/lean-math/files/blog/wp-content/uploads/2013/05/summation-notation.png"><img alt="summation notation" height="416" width="553" class="media-image alignnone size-full wp-image-608 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/blog/wp-content/uploads/2013/05/summation-notation.png" /></a></p>
<p>Example - in kanban or point of use replenishment planning it's important to make sure that the material handler or water spider has enough time to travel from point to point and complete an entire circuit or bus route in a predictable period of time. It may not be possible, practical, or even necessary for parts to be delivered to or removed from the production cell or point of use every single takt cycle, but we do want to be regular and predictable, and so we will make deliveries and pickups in a multiple of takt time, otherwise known as a pitch.</p>
<p>Pitch = Takt Time * N</p>
<p>T = transport or walk time, A = number of stops<br />
R = replenishment picking time, B = number of pick points<br />
P = point of use restocking time, C = number of restock points<br />
N = number of takt that elapse during water spider’s circuit or route</p>
<p>Pitch might be every 45 minutes, or twice a shift, or however long it takes to empty or fill a container. The Lean Lexicon defines pitch as the amount of time needed in a production area to make one container of products. We'll expand that definition to be simply a multiple of takt.</p>
<p>Other factors: shopping cart (or train) size, number of different parts being picked, demand or replenishment time variation.</p>
<p><a href="/sites/lean-math/files/blog/wp-content/uploads/2013/05/bus-route2.png"><img alt="bus route" height="149" width="584" class="media-image alignnone size-large wp-image-621 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2013/05/bus-route2-1024x263.png" /></a></p>
<p>Translated in to English ... add up the time T that it takes to travel from stop to stop, then add up the times R in the supermarket that it takes to pick each part for each stop, and the same with the times P to restock every item needed at each stop, and finally make sure the total time is less than the Pitch Time.</p>
<p> Footnote: This formula is for a coupled route. For a decoupled route just eliminate the <i>R<sub>j</sub></i>summation.</p>
</div></div></div><div class="field field-name-field-tags field-type-taxonomy-term-reference field-label-hidden"><div class="field-items"><div class="field-item even"><a href="/tags/basic-math" typeof="skos:Concept" property="rdfs:label skos:prefLabel" datatype="">Basic Math</a></div><div class="field-item odd"><a href="/tags/inventory" typeof="skos:Concept" property="rdfs:label skos:prefLabel" datatype="">Inventory</a></div></div></div>Tue, 04 Jun 2013 16:37:44 +0000lloucka74 at http://www.leanmath.comhttp://www.leanmath.com/blog-entry/summation-notation-and-sizing-kanban-bus-route#commentsAverage - Not Always Plain and Simple
http://www.leanmath.com/blog-entry/average-not-always-plain-and-simple
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even" property="content:encoded"><p>When most people hear the word average they think normal, typical, boring. But, in many cases that is very far from the truth. </p>
<p>Averages can not only be interesting, but also impossible!</p>
<p>Consider the call center that measures average response time. The average response time is less than their target, so they must be doing pretty good, right??? </p>
<p>Wrong! </p>
<p>When an average is less than a target value, without additional information, you have almost no idea how you are doing. You know that at least one value is less than the target, but it could be that only one value is less than the target, or it could be that all of the values are less than the target. So, our imagined call center’s comparison of their actual response time to their target value actually yields very little information. A much better approach would be if they performed a capability analysis on their response times.</p>
<p>Consider the following two data sets, the data sets are clearly very different, but the averages are the same.</p>
<div style="width: 310px;display:block;margin:0 auto;"><a href="/sites/lean-math/files/blog/wp-content/uploads/2013/05/average-pic1.png"><img alt="average pic" height="200" width="300" class="media-image size-medium wp-image-572 media-element file-media-large" typeof="foaf:Image" src="http://www.leanmath.com/sites/lean-math/files/wp-content/uploads/2013/05/average-pic1-300x200.png" /></a> click to enlarge
<p> </p>
</div>
<p> Finally, when are averages impossible? Consider the case of the average family with 1.86 children. </p>
<p> Before you fall into the trap of thinking averages are boring, think again.</p>
</div></div></div><div class="field field-name-field-tags field-type-taxonomy-term-reference field-label-hidden"><div class="field-items"><div class="field-item even"><a href="/tags/basic-math" typeof="skos:Concept" property="rdfs:label skos:prefLabel" datatype="">Basic Math</a></div></div></div>Thu, 02 May 2013 01:10:45 +0000drmike71 at http://www.leanmath.comhttp://www.leanmath.com/blog-entry/average-not-always-plain-and-simple#comments