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# Russell's Blog

## Makers do not make weapons

Posted by Russell on December 17, 2012 at 6:58 p.m.
Last Tuesday, I started writing an article about Thing 11770 on Thingiverse, a MakerBot Industries for sharing 3D printable objects. Thing 11770 is a reinforced 3D printable lower receiver for an AR-15 assault rifle. This is the part of the gun that feeds bullets from the magazine into upper receiver, which handles the cycling of the spent round and the insertion of the new round. With the right combination of upper and lower receiver, fresh rounds are cycled into the weapon using a portion of the kinetic energy from firing the previous round. When the trigger is pulled, this process happens continuously, firing one bullet after another. That is what it means to be an "automatic" weapon. Thing 11770 is particularly interesting because, legally speaking, the lower receiver is the gun itself. It is the engine that makes the gun a gun, rather than a movie prop. And you can 3D print it. And it works.

At very the moment I was hemming and hawing over how to articulate my feelings about this development, someone used an AR-15 to murder twenty seven people, including twenty children, ages six and seven at Sandy Hook Elementary School. Now I know exactly how I feel.

I love 3D printing. I love the maker movement. I love the idea of people building home-brew versions of all sorts of devices, and inventing entirely new classes of devices. 3D printing has played, and will continue to play, an important role in that.

When I was fourteen, like many boys at that age, I thought missiles and fighter planes and tanks were pretty awesome. I read a lot of Tom Clancy books, and I indulged in my interest by dragging my family to the Wright Patterson Air Force Base Museum, the Smithsonian Air & Space Museum, the California Science Center’s Air & Space Museum, and the Intrepid Air, Sea & Space Museum. At Wright Patterson, I visited the F-117 Nighthawk as many times as I could. The author of Thing 11770 calls himself "Have Blue," the codename for the Nighthawk demonstrator aircraft.

When I was sixteen, I went to boarding school, where I learned vector calculus and farming. I learned how to grub potatoes out of the freezing ground in the driving rain, how to make maple syrup, how to lay in beets and squash and onions for the winter. I stood on a windy mountain top and learned how to find the orbital ephemera of a comet. I learned how to milk cows, how to care for cows when they are sick, and how to make the most delicious yogurt and mozzarella cheese you could possibly imagine. I learned how to repair a tractor engine with a mallet and a wrench. One freezing night, I found myself covered in blood and shit and urine and fear as I helped bring a new life gasping and staggering into the world.

Farming also means slaughtering and butchering. One morning, I walked into the barn. I was handed a weapon. I was asked to take a life.

I found that I could not.

Not ever.

The instant my shoulders took up the weight of the strange, snub nosed machine, it felt like the weight of the metal hung from my heart, stretching and distorting it. I wanted the weight of it to tear me apart, but I knew it was a weight I could carry, if I wanted to. I quietly handed the gun back to the farm manager, and walked out into the thawing snow, and spent the rest of the black pre-dawn puking into the mud behind the water tower.

Many people have wondered why I do not eat meat. This is why. For the rest of my life, I will feel the weight of that terrible little machine.

There are reasons to make, to have and to use guns. To defend your country, yes. To humanely put down an animal before butchering it, perhaps. For vainglory? For entertainment? No.

Tools are sacred things. We are a tool-using species; our tools are projections of our hopes and aspirations. When we are filled with joy, we pick up our tools and hammer the air into music. We need to understand and to be understood, and so we shape our voices into language. We send our tools delicately probing into the bodies of our loved ones, seeking out cancers and blood clots and infections. We invest huge amounts of effort building and maintaining tools that allow us to speak to one another across great distances. We hurl our tools across the void to other planets to satisfy our craving for knowledge. When we grieve, we take up our tools and carve the names of those we have lost into the living rock of our planet. Our tools are our souls. They are our defining characteristic. Love may be what makes us alive, but our tools are what make us human.

A gun is a tool. It is a simple tool. Any man or woman or child can use one. A gun is not much more complicated than a can opener, and not nearly as sophisticated as cordless screwdriver. Like all tools, a gun reveals something fundamental about its maker, its wielder and its abuser. This is true for all weapons.

As a strong supporter of the maker movement, of free and open source software, of open science, I want people to have as much freedom as possible to make and remake and experiment. I also believe very, very strongly in the responsibly we have to one another. I believe that we each have a responsibility not make things that hurt and kill and destroy.

I am not yet prepared to call for a law to prohibit Have Blue from posting functional 3D printable assault rifle parts on the internet. The law is a blunt instrument, and would cause a great deal of collateral damage. However, I am prepared to say that Have Blue is a fucking asshole. I am prepared to call Justin Halford, who created the original CNC model, a fucking asshole. I am prepared to say that anyone who considers themselves a "gun enthusiast" and is older than about sixteen needs to grow the fuck up. The maker community should not tolerate this behavior. Meditate on the meaning of the word antisocial for a moment. What could be more antisocial than gleefully proliferating machines whose principal function is murder?

The maker community should not tolerate these designs, or the ideas and opinions of their designers until they show evidence of behaving like adults. It's clear that the CNC Gunsmithing community has a lot of talented, clever people. It's clear from reading his blog that Have Blue is neither ignorant nor stupid.

So, I'm calling you folks out. There are twenty children dead in Connecticut. Their bodies were ripped apart by the very machines you are "democratizing." As far as I know, nobody has used your designs to kill anyone. If you continue down this path, some future version of Thing 11770 will be used to murder little children. It's just a matter of time, and probably a lot less time than you think. However, there is still time to take a stand. Do the right thing. Take down the designs. Apologize for what you've done. Find a new project. Use your talents for something good. This will not stop people from murdering children with 3D printed guns, but perhaps you can buy us some time before that day comes. You know that this is true.

If making home-brew assault rifles is really what you want to do, there is perhaps one venue where this might actually make sense. Freight your CNC machine to Istanbul, and smuggle it into Homs or Aleppo. Help the Free Syrian Army get rid of Bashar Assad. Oh wait, what’s that? You don't want to get shot? Fancy that.

It takes courage to admit you are wrong. Show us some courage.

Update : It appears that MakerBot has decided to remove Thing 11770 from Thingiverse. If you follow the link to the item, the files have been removed and a message says, "This Thing is currently under moderation for violating the Thingiverse Terms of Service. Files and images for this Thing are currently unavailable." I'm glad it's no longer up, but I am disappointed in how this was handled. I'm disappointed that MakerBot left it up for so long, but I'm also disappointed that Have Blue didn't just take it down himself.

## A new Prometheus

Posted by Russell on December 08, 2012 at 2:33 a.m.
One year ago, UC Davis law student Megan Glanville was killed a stone's throw from my front door. She was crossing the street for a morning run. It was foggy. The driver didn't see her.

Since then, the intersection where she died has been redesigned. It is now a three-way stop with modern LED lighting. Watching over the scene, there is a new flashing red beacon.

This sort of infrastructure is easy to take for granted. As a Commissioner for the City of Davis, I suppose I pay closer attention to these things that most people do. I've payed particular attention to this little piece of city infrastructure because I pass through it several times a day.

Something has changed there since the red beacon went up. Up and down the boulevard, for almost a mile, there are crossings to access the bicycle path. Drivers now stop and let me cross. They never did that before. I am not exaggerating when I say that wherever the beacon's light falls, the feel of the street has changed. It's no longer the tail end of a lonely country road. It's a neighborhood street, and people act accordingly.

I would like to think that drivers feel the significance of the flashing beacon. I would like to think that they have noticed that the intersection has been redesigned. I would like to think that they know that Megan Glanville died there. In all likelihood, they are oblivious to these things. They stop and smile and waive me through anyway.

Why?

Good design matters. That's why.

The rocky ledge runs far into the sea,
And on its outer point, some miles away,
The Lighthouse lifts its massive masonry,
A pillar of fire by night, of cloud by day.

Even at this distance I can see the tides,
Upheaving, break unheard along its base,
A speechless wrath, that rises and subsides
In the white lip and tremor of the face.

And as the evening darkens, lo! how bright,
Through the deep purple of the twilight air,
Beams forth the sudden radiance of its light
With strange, unearthly splendor in the glare!

Not one alone; from each projecting cape
And perilous reef along the ocean's verge,
Starts into life a dim, gigantic shape,
Holding its lantern o'er the restless surge.

Like the great giant Christopher it stands
Upon the brink of the tempestuous wave,
Wading far out among the rocks and sands,
The night-o'ertaken mariner to save.

-- The Lighthouse, Henry Wadsworth Longfellow

On a superficial level, a flashing red beacon is a utilitarian thing. If you look more closely, you will see that it is also a thing of beauty. It is an avatar of the compulsion we all feel to protect, to warn, to guide. The humble beacon is one of the better angels of our nature, sculpted with massive limbs of galvanized steel and eyes of electrically exuberant gallium phosphide. It sends our message out into the world, again, and again, and again.

be careful

be careful

be careful

be careful

be careful

...

## Moving forward by stopping

Posted by Russell on April 02, 2012 at 4:06 p.m.
Just a three weeks after I was sworn in for my term on the City of Davis Safety and Parking Advisory Commission, UC Davis law student Megan Glanville was killed just a few dozen feet from my doorstep. She was out jogging on a foggy morning, and truck coming into town from the county road ran her down in the crosswalk. I never knew Megan, but her death deeply upsets me.

I've been worrying about pedestrian and bike safety ever since my little sister was nearly killed by a careless driver.

I find it extremely frustrating that most people do not look beyond the (usually imagined) behavior of the people involved in an accident like the one that almost killed my sister, or that did kill Megan Glanville. Either they identify with the frustrating experience of driving, and blame the victim, or they side with the law, and place the responsibility at the feet of the operator of the more dangerous vehicle. I will always side with the person who suffered more, but both views are myopic. When someone has been killed in an accident, the question of who was more "right" in that sliver of time is irrelevant. It is worse than irrelevant; it is an insult to the lives of all the people affected.

There are other, far more urgent questions that need to be raised. If you see a problem, the first question you should always ask is, "In what way am I responsible for this?" We are all bound together by bonds of mutual responsibility, and nothing happens among people, good or bad, for which each of us are not in some sense responsible. That is what words like "society," "community," and "civilization" mean. They describe the fact that the bonds that link us together are fundamentally inescapable. There is such a thing as integrity, but there is no such thing as self-reliance. Interdependence is the very essence of what makes us human. And so, if you see something that upsets you, the first thing you should look at is your own role in causing it. Through our choices, we were all present on morning that George Souza killed Megan Glanville. You. Me. Everyone. We all had a hand in it.

Clearly, we failed. You failed. I failed. Someone is dead as a consequence of that failure.

So, let us set aside the choices of George Souza and Megan Glanville, and look at the choices we made that contributed to this terrible thing. They are easy enough to see :

This is the crosswalk where Megan was killed, which is part of a system of roads that belong to the City of Davis. The arrow on the yellow sign is pointing almost directly at the spot. The laws that govern the design of the road are a kaleidoscopic fugue of local, county, state, federal and international regulations. Within that often contradictory matrix of statutes, the city government has a small keyhole of authority within which it may choose what the road looks like and how it works.

From an engineering point of view, it's pretty clear what the problem is. The road on the left is just a stone's throw from the border of the city. Beyond the border, it is a wide county road that cuts a nearly straight line for miles among orchards and farms. When it crosses into the city, this road suddenly plunges into a dense residential neighborhood with no transition whatsoever. The intersection where Megan was killed is the very first intersection an eastbound driver encounters in the City of Davis. So, drivers come in from the county road going at county road speeds, and roar through this intersection where people are trying to cross to the bike path that parallels the road. Add a little darkness and bit of fog, and the accident was basically inevitable.

Why was this intersection designed this way? I don't know. According to the laws and statutes that regulate its engineering, there is nothing particularly wrong with it. But then again, houses that catch fire and burn people alive inside are often built to code. Compliance with the law is not enough. Only thoughtful design can keep people safe, and the absence of that thoughtfulness killed someone.

So, who is to blame? The legislators who wrote the statues describing how intersections should be designed? The engineers whose designs were constrained by those statutes? The City of Davis Public Works Department that built and maintained it? Surely, some of the responsibility falls to them. But not very much. If you've ever driven, walked or bicycled through the intersection of Lake and Russell, then a great deal of the responsibility falls on you. If you've ever felt uncomfortable or unsafe while passing through it, then you knew someone would get hurt there sooner or later.

The Council Chambers are open to the public. The meetings and agendas are available weeks in advance for all to see, at CityOfDavis.org. You can even submit your concerns in writing if you don't have time to come to the meetings. In other words, you had the reason and the means to get this fixed, or at least play a part in getting it fixed, before Megan Glanville was killed. I share in this responsibility; I serve on the commission charged with advising the City Council on these things, and I did not raise this issue either. And I use this intersection several times a day. And I always feel unsafe. It is my fault too.

So, here is what is going to happen. The City Council was asked, and agreed, to take steps to prevent anyone else from getting killed. The proposed changes will add stop signs on Russell Boulevard in both directions, a blinking red light in case drivers don't see the stop signs in the fog, and four new street lights for better illumination overall. It will cost about \$20,000.

This is a much better design. It's impossible to know if it would have saved Megan's life had it been in place in December, but it seems likely that it would have. I strongly support it.

Roads are not natural phenomena. They are public infrastructure, and they are designed and built and maintained in exactly the way the public asks them to be. Let's try to do a better job of holding up our end of that conversation.

## The Dover Train

Posted by Russell on February 14, 2012 at 4:47 a.m.
Well, it's Valentine's Day again. For all you lucky folks who have someone from whom deserts may be extorted, cheers!

As for me, I'm writing this post for the girl on the Dover train who gave me a funny look. Um... Hi.

Don't know if this will work, but stranger things have happened.

## Happy 60th, Mom and Dad!

Posted by Russell on February 10, 2012 at 2:24 p.m.
As is often the case, odd people had odd children. Readers of this blog have no doubt long noted a certain... peculiarity of the author, and hypothesized about the individuals who raised me. I can proudly confirm that they are every bit as odd as I am. Anyway, they both turned sixty in the last couple of weeks (my mom in December, and my dad a few days ago), and I wanted to post a little celebration of our great family tradition of crossgrainedness.

For my mom's birthday, she and my sister came over and slept on the floor of my apartment in Davis, despite the availability of perfectly comfortable and reasonably priced lodgings downtown. For her birthday, we found a fetid puddle of water near Lake Berryessa with some tadpoles in it. She was delighted.

For my dad's birthday, he's celebrating his election to the National Academy of Engineering (the sister organization to the Institute of Medicine. He also got this birthday card from his longtime college friend Bruce Reznick :

Bruce is a professor of mathematics at the University of Illinois. He studies the identities of high-order polynomials. So, this really is the a birthday card only he would think to send.

## Cat replenishment

Posted by Russell on January 22, 2012 at 3:45 p.m.
I hear the internet is running low on pictures of adorable cats. It's raining outside in Davis today. Buzz seems to be hoping that I'll put down the camera and stop playing with ISO settings, and rub his belly. Either that, or he's mugging for the camera. It's hard to tell.

That is all.

## New Equipment Thursday

Posted by Russell on August 25, 2011 at 7:40 p.m.
My vacuum desiccator arrived today, and so naturally I put it to productive use. You know. For science.

Haw! This thing is cool.

Posted by Russell on August 15, 2010 at 4:59 p.m.
My apologies for getting behind on posting my updates from Uzon. After we returned from Uzon, we rested for a day, and then crammed ourselves and our equipment into a van and went to Peratunka for the Biodiversity, Molecular Biology and Biogeochemistry of Thermophiles international workshop, where I was scheduled to give a 20 minute talk.

The speaking docket got shuffled around a lot, and I ended up having to give my talk much earlier than planned. I suppose this is the inevitable downside of procrastination. While I was scrambling to finish it, I didn't have much time for blog updates!

I survived the talk. There were lots and lots of excellent questions, and I have a lot to think about now. Anyway, back to the updates from Uzon.

## Back from Uzon

Posted by Russell on August 10, 2010 at 8:49 p.m.

Panorama overlooking Orange Fields in Uzon Caldera

We just arrived back in Petropavlovsk after a week in the field. I was very sad to leave Uzon, and it was a privilege and an honor of the highest order to have spent those days there.

The expedition was, I think, a great success. We'll know for sure once we're back at our labs and can use more sophisticated methods to examine our samples. I am very confident, though.

It was a bit touch-and-go right at the end. Our high speed centrifuge crapped out last night, just as Sarah was in the middle of the last big run of DNA extractions. The Russian team brought their own centrifuge, but we couldn't run it on our generator. Much to our relief, Albert was able to magically get the thing working again by holding it at just the right angle. They worked through the night to finish processing the samples; I think Albert must have had his thumb wedged under the centrifuge for the entire run.

I'm sorry I wasn't able to send many Twitter updates toward the end of the expedition. Once I had identified my sampling targets, I suddenly had a lot less free time on my hands (and I didn't have much to begin with). Also, I'm sorry for updating in ALL CAPS. Iridium handsets are essentially 1993 technology. Composing text messages is extremely painful, and the battery only lasts long enough to compose two or three of them. This is a pain when you have to recharge on generator power, and the generator only cranks up for a few hours a night, and even then only to power lab equipment for DNA extractions. Hats off to my dad for relaying the messages!

Right now, I'm sitting in a friendly internet cafe in Petropavlovsk where they've let me use their wireless connection. When we arrived at our crowded little apartment, the hot water was broken, and thus no showers yet. A wide selection of interesting geologic samples are wedged under my fingernails, and I think I have wads of some sort of hardened liquid sulfur caked in my hair. The helicopter arrived ridiculously early, and we just barely get everything aboard. As a result, I'm still wearing my field clothes from yesterday, which are splattered with volcanic mud. I may actually be the worst-smelling person in Petropavlovsk. Perhaps it is fortunate that this internet cafe caters mainly to kids playing StarCraft.

I composed blog entries for each day we were in Uzon, and I'll be posting them as soon as I run them past the rest of the team. I also have almost two thousand photos to sort, tag and upload.

That said, I have a correction for one of my Twitter updates. I wrote :

YERTERDAY ALBERT & TEAM WERE CHASED AWAY FROM A SITE BY A BEAR THAT WAS ACTUALY A BUSH IN THE FOG.
Albert pointed out that they were interrupted for a few minutes, but not actually chased away. He stepped forward and shouted see if he bear (or bears) would go away, with his signal torch uncapped and ready. The bears were revealed to be bushes as the wind shifted and created a channel in the mist. It's funny, but given how foggy it was that day, it wasn't actually that surprising. We were at the same site the next day, and were surprised by an actual bear. It wandered pretty close to us before we could actually see it (the full story will come with the article for that day).

A bear interrupting important EisenLab work at Boiling Spring.

Update : Albert also says that I'm wrong about having to wedge his thumb under the centrifuge the whole time. It started working again after shaking it around in the air a bit, and placing it just so on the table. He only had his thumb wedged underneath it for a minute or two to check to see if it was overheating.

## I'm going to Kamchatka!

Posted by Russell on July 19, 2010 at 5:47 p.m.
I just got the reservations for my flight to Petropavlovsk-Kamchatsky for the International Workshop on Biodiversity, Molecular Biology and Biogeochemistry of Thermophiles, hosted by Moscow State University and Winogradsky Institute of Microbiology.

I've been working on the analysis of environmental samples from two sites at Uzon Caldera (about 10,000 Sanger reads from each sequenced at the JGI), and I'm hoping that I'll be able to reprocess the DNA here at the UC Davis Genome Center using some of our high-throughput machines. Licensing and customs restrictions will probably make it impossible to bring my own samples back, but I may be able to entrust them to a colleague with fancier credentials than my own.

I'll be arriving in Petropavlovsk on the 30th of July, with the help of a generous grant from the Carnegie Institution for Science Deep Carbon Observatory.

## Good luck, whoever you are

Posted by Russell on June 08, 2010 at 3:54 p.m.
Last week, I got an urgent call from the National Marrow Donor Program. Somewhere, there is a girl about to go into chemotherapy, and my tissue type matches hers. They needed to run some more detailed tissue typing, and screen for infections diseases. The NMDP sent a kit to the UC Davis student health center (the new building is right next to my house), and I had my blood drawn this morning.

I hate giving blood. They didn't need very much, but I don't get along very well with steel needles. I count it as a major victory that I didn't barf until I got home.

Now we all wait for the results.

Good luck, whoever you are.

## 20s

Posted by Russell on May 31, 2010 at 9:17 p.m.
To steal the idea from John Scalzi, here is the last photograph of me in my 20s.

My twenties; better than my teens. Some good times, and some pretty awful times, and on average kind of meh. If the trend holds, my thirties should be in the tolerable to nice range. Hopefully the underlying process is geometric, and not linear or logarithmic.

Hence the awkward sort of half-smile.

## Holly Allan-Young

Posted by Russell on May 26, 2010 at 10:57 p.m.

Terry Young: She is gone
Sent: 2:54PM

I will miss her terribly.

Posted by Russell on April 28, 2010 at 11:26 a.m.
After six months of using Google Latitude, I've amassed about 7108 location updates, or about 38 a day. It would probably be a lot more if I hadn't managed on occasion to break the GPS or automatic updating by fiddling with the software.

Not surprisingly, I spent most of my time in California, mostly in Davis and the Bay Area, with a few trips to Los Angeles via I-5, the Coast Starlight, and the San Joaquin (the density of points along those routes is indicative of the data service along the way).

The national map shows my trip to visit my dad's family in New Jersey and Massachusetts, as well as a layover in Denver that I'd completely forgotten about.

I have somewhat mixed feelings about this dataset. On one hand, it's very useful to have, and sharing it with my friends and with Google is very useful. It's also cool to have this sort of quantitative insight into my recent past so easily accessible. On the other hand, I'm not particularly happy with the idea that Google controls this data. I chose the word controls deliberately. I don't mind that they have the data -- after all, I did give it to them. As far as I know, Google has been a good citizen when it comes to keeping personal location data confidential. The Latitude documentation makes their policy pretty clear :

### Privacy

Google Location History is an opt-in feature that you must explicitly enable for the Google Account you use with Google Latitude. Until you opt in to Location History, no Latitude location history beyond your most recently updated location if you aren't hiding is stored for your account. Your location history can only be viewed when you're signed in to your Google Account.

You may delete your location history by individual location, date range, or entire history. Keep in mind that disabling Location History will stop storing your locations from that point forward but will not remove existing history already stored for your Google Account.

...

If I delete my history, does Google keep a copy or can I recover it?

No. When you delete any part of your location history, it is deleted completely and permanently within 24 hours. Neither you nor Google can recover your deleted location history.

So, that's what they'll do with it, and I'm happy with that. What bothers me is this: Who owns this data?

This question leads directly to one of the most scorchingly controversial questions you could ask for, and there are profound legal, social, economic and moral outcomes riding on how we answer it. This isn't just about figuring out what coffee shops I like. If you want to see how high the stakes go, buy one of 23andMe's DNA tests. You're giving them access to perhaps the most personal dataset imaginable. In fairness, 23andMe has a very strong confidentiality policy.

But therein lays the problem -- it's a policy. Ambiguous or fungible confidentiality policies are at the heart of an increasing number of lawsuits and public snarls. For example, there is the case of the blood samples taken from the Havasupai Indians for use in diabetes research that turned up in research on schizophrenia. The tribe felt insulted and misled, and sued Arizona State University (the case was recently settled, the tribe prevailing on practically every item).

You can't mention informed consent and not revisit HeLa, the first immortal human cells known to science. HeLa was cultured from a tissue biopsy from Henrietta Lacks and shared among thousands of researchers -- even sold as a commercial product -- making her and her family one of the most studied humans in medical history. The biopsy, the culturing, the sharing and the research all happened without her knowledge or consent, or the knowledge or consent of her family.

And, of course, there is Facebook -- again. Their new "Instant Personalization" feature amounts to sharing information about personal relationships and cultural tastes with commercial partners on an op-out basis. Unsurprisingly, people are pissed off.

Some types of data are specifically protected by statute. If you hire a lawyer, the data you share with them is protected by attorney-client privilege, and cannot be disclosed even by court order. Conversations with a psychiatrist are legally confidential under all but a handful of specifically described circumstances. Information you disclose to the Census cannot be used for any purpose other than the Census. Nevertheless, there are many types of data that have essentially no statutory confidentiality requirements, and these types of data are becoming more abundant, more detailed, and more valuable.

While I appreciate Google's promises, I'm disturbed that the only thing protecting my data is the goodwill of a company. While a company might be full of a lots of good people, public companies are always punished for altruistic behavior sooner or later. There is always a constituency of assholes among shareholders who believe that the only profitable company is a mean company, an they'll sue to get their way. Managers must be very mindful of this fact as they navigate the ever changing markets, and so altruistic behavior in a public company can never be relied upon.

We cannot rely on thoughtful policies, ethical researchers or altruistic companies to keep our data under our control. The data we generate in the course of our daily lives is too valuable, and the incentives for abuse are overwhelming. I believe we should go back to the original question -- who owns this data? -- and answer it. The only justifiable answer is that the person described by the data owns the data, and may dictate the terms under which the data may be used.

People who want the data -- advertisers, researchers, statisticians, public servants -- fear that relinquishing their claim on this data will mean that they will lose it. I strongly disagree. I believe that people will share more freely if they know they can change their mind, and that the law will back them up.

### Update

The EFF put together a very sad timeline of Facebook's privacy policies as they've evolved from 2005 to now. They conclude, depressingly :
Viewed together, the successive policies tell a clear story. Facebook originally earned its core base of users by offering them simple and powerful controls over their personal information. As Facebook grew larger and became more important, it could have chosen to maintain or improve those controls. Instead, it's slowly but surely helped itself — and its advertising and business partners — to more and more of its users' information, while limiting the users' options to control their own information.

## A desirable extinction

Posted by Russell on March 25, 2010 at 3:19 p.m.
Some weeks ago, Buzz (my cat) escaped out my front door while I was carrying my bicycle into the apartment. For ten or twenty minutes, he romped through the ivy and bushes around my apartment while I followed him around rattling a bag of cat treats. Eventually, he let me pick him up and take him back inside. Naturally, he picked up a few fleas. Naturally, they have multiplied.

Oddly, the fleas don't seem to like Neil very much, nor do they like me. It's just poor Buzz that's beset by the nasty little critters.

### Figure 1: A flea.

As it happens, I've been thinking about endogenous metrics for estimating the sampling quality of an environmental shotgun sequencing dataset, and Buzz's little problem presented an opportunity to play with a simplified problem. So, I have decided to make Buzz, or rather his fleas, into a small experiment in ecology. I am going to try to see if I can drive them into extinction.

Now, this is normally what a pet owner does when they discover their pet has contracted some sort of annoying parasite, but I decided to take a more quantitative approach.

### Figure 2: A cat.

It's simple enough to count fleas on a cat, if the cat is willing to cooperate. Buzz loves the flea comb, and will gleefully hop onto the coffee table and wait to be combed if you show it to him. So, in the interest of science, I convinced my roommate to count the number of passes I made with the flea comb and how many fleas I captured (posterity will remember your efforts, Mehdi). Using his tally, I plotted the cumulative number of passes verses the cumulative number of fleas.

### Figure 3: Fleas captured

As expected, it became somewhat more difficult to capture the next flea as more fleas were captured, suggesting a depletion curve. The value of the asymptote should be the actual number of fleas on Buzz at the time, and reaching that number would imply local extinction for the fleas. Of course, there are probably other fleas lurking about that would recolonize Buzz. In principle, if I were to repeat the exercise frequently enough, Buzz would become a sink for fleas, and their migration to his fur would gradually deplete them from the environment.

There are a couple of different ways to model the impact of the combing on the flea population, with various advantages and disadvantages. All we really want to do here is to estimate the value of the asymptote, and so a simple model is probably sufficient. I showed this data to my fried Sharon Shewmake, an economics graduate student. Sharon, after editorializing on the endeavor ("Ew."), suggested this very simple model.

Assume that Buzz is not going to sit still long enough for the fleas to reproduce, for more fleas to migrate to his fur, and that the fleas already on his fur are going to stay put unless captured. Thus, there is a fixed initial population which only changes as a result of capturing fleas. Next, we assume that any given flea is equally likely to be captured on a single pass of the comb. So, the expectation value for number of fleas captured on a single pass is the product of the current population and the probability of capturing a flea.

$N_{i+1}=N_i-pN_i$

where N is the population of fleas and p is the probability of any particular flea being captured on a single pass. One could tart this up a bit by modeling it as a stochastic process and executing a bunch of Monte Carlo trials until the outcomes converge, but that seems like overkill for a simple single variable problem like this. We will put up with the intellectual inconvenience of capturing fractional fleas.

This is a little easier to see if we let N represent the number of fleas remaining on the cat, rather than the number of fleas captured.

$N_{i+1}=qN_i$

If we stretch our credulity far enough to imagine this as a continuous function, we can express it as a differential equation.

$\frac{d N}{d t}=\alpha N$

Sorry if this bothers you. Not only are we extracting fractional fleas, but we are now modeling the combing process as a sort of flea-killing-combine continuously mowing its way through the fur. This is a model, so you shouldn't be surprised to find massless rope and spherical cows. Anyway, it has a nice easy solution.

$N(t)=e^{\alpha t}$

Well, what the heck. This is a decaying function, so let's pluck a minus sign out of the exponential factor, and maybe tack on a scale factor for the initial population.

$N(t)=N_0 e^{-\alpha t}$

While we're at it, why don't we go back to letting the function stand for the number of fleas captured, rather than the fleas on the cat.

$N(t)=N_0(1-e^{-\alpha t})$

This gives us a nice function to use for a linear regression. A little help from scipy, and we find that the initial population is estimated at 39.7 fleas, and the decay factor is 0.011.

### Figure 4: Flea population

I captured 34 fleas, so that means I missed about five or six. In order to be reasonably confident that I'd captured all 39 fleas, I would have had to continued for about 400 passes with the comb, instead of 173. Buzz is a patient cat, but he started to loose interest around 120 passes, and had to be fetched back onto the coffee table a few time times during the last 50 passes. My guess is that 400 passes would require some kind of sedative. On the other hand, he does seem to like Guinness, so there may be something to that.

Science has been served. I'm going to the pet store to buy some flea collars.

## Espresso

Posted by Russell on March 13, 2010 at 3:56 p.m.
A few weeks ago, my dad sent me this really nice espresso machine to cheer me up. Actually, he sent it to me because it was it was his birthday. He's a pretty awesome dad that way -- I only sent him a book.

I'm still getting the hang of getting a decent pull of espresso out of it. I've found that my burr grinder doesn't quite go fine enough for espresso, so I'm going to have to take it apart and see if I can adjust the grinding wheels so they're closer together. Anyway, here is my latest effort :

## Facepaw

Posted by Russell on February 10, 2010 at 3:42 a.m.
Spent the day writing a piece of code I already wrote six months ago. Not sure how I managed to forget. The new code wasn't very good, so I threw it away. Day down the tube.

Even asleep, Neil seems to understand.

## Loss

Posted by Russell on February 03, 2010 at 11:48 p.m.

I normally don't talk a lot about my personal life on my blog, and except for the occasional announcement, I'd like to keep it that way. People's little triumphs and tragedies are mainly interesting to those directly involved, and are at best kind of boring to everyone else. A lot of my friends and family do read this blog, but by and large most of you are strangers or acquaintances. I try to respect that.

Those of you who are close to me know that I'm going through a sad time in my life right now. Those of you who work or study with me have probably noticed that I've not been my usual cheerful self. In deference to the many people who aren't here to read about that, and the fact that I can barely think about it (never mind write about it), I'm not going to discuss what's happened on my blog.

The one thing that has helped has been hearing about all the cool things that other people are doing. So, even though I'm not exactly Mr. Social right now, please don't take that as a sign that I want to be left alone.

On the contrary. Now would be a great time to tell me about whatever is on your mind, especially if it's cool.

To those of you who've been kind enough to treat me like a normal person over the last two weeks despite my melancholy behavior, I owe you guys. Really.


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# A deduction question for perfect logicians

A colleague of mine recently messaged me a riddle. It is about a deduction process between two perfect logicians.

The riddle is as follows:

A person writes two non-zero digits on a paper. He tells person A the product of the digits, and person B the sum of the digits. The both persons have the following conversation:

A: "I don't know the digits"

B: "I don't know the digits"

A: "I don't know the digits"

B: "I don't know the digits"

A: "I don't know the digits"

B: "I don't know the digits"

A: "I don't know the digits"

B: "I don't know the digits"

Then person A says "I know the digits"

From this conversation I'm supposed to obtain the digits which were written. I can express this as a mathematical statemente but I lack the idea of how to deduce this riddle.

I know that the first person chose two numbers $x,y\in [9]:=\{1,\cdots,9\}$ and that person A knows $x+y$ and person B knows $xy$. If $x,y$ are different then $x+y \le 17$ and $xy\le 72$. Otherwise, $x+y\le 18$ and $xy\le 81$.

Now I think I must search for numbers below 81 such that any of $[9]$ divides. Definitely any prime is not such a number and I'm sure there are a couple more. Also, for numbers below 18, it is possible to count the partitions of each number reducing the number of cases which $x,y$ can meet.

I think that this is not a correct approach to this problem and I would like to hear a good approach, albeit without giving the complete answer if possible. Also if there is anything wrong with my reasoning, please tell me so I can correct it.

• $A$ knows $xy$, while $B$ knows $x+y$ according to the question. – Hirak Nov 24 '17 at 6:31
• OP knows that ;) – Cornman Nov 24 '17 at 6:38
• I hate to be obvious here… Could the key to the problem be how many times the line is repeated? If the logicians are perfect, they’ll use the fastest method possible to communicate the data each has to the other. – gen-z ready to perish Nov 24 '17 at 6:46
• Yes! My thoughts go along the lines of: Person A knows "such and such" are all the possibilities of the numbers then since A knows, B reduces the amount of his possibilities. Since there are many (9) deductions in total, that might tell me something as well. But I can't yet grasp it. – Ignacio Rojas Nov 24 '17 at 6:51
• They'll never be able to deduce $ab$ from $ba$ so we know it must be $aa$ or $a0$. If it is $aa$ then B is given an even number A is given a square. If it is $aa\ge 22$ B will always have the possibility is was an ab so A never knowing will never tell him anything. So B not knowing will not tell A anything. So it was $a0$. And as B being told an even number greater than 4 will never let B know, we know that didn't happen. I think it can be deduced from there. – fleablood Nov 24 '17 at 7:41

We may suppose that $x\ge y$.

A: "I don't know the digits", so we see that $(x,y)$ is either of the followings where red numbers, green numbers represent $xy,x+y$ respectively :

$\qquad\qquad\qquad$

B: "I don't know the digits", so we see that $(x,y)=(2,2),(9,4),(6,6)$ were impossible since each of $\color{green}{4,12,13}$ appears at only one place.

A: "I don't know the digits", so we see that $(x,y)=(4,1)$ was impossible since $\color{red}{4}$ appears at only one place.

B: "I don't know the digits", so we see that $(x,y)=(3,2)$ was impossible since $\color{green}{5}$ appears at only one place.

A: "I don't know the digits", so we see that $(x,y)=(6,1)$ was impossible since $\color{red}{6}$ appears at only one place.

B: "I don't know the digits", so we see that $(x,y)=(4,3)$ was impossible since $\color{green}{7}$ appears at only one place.

A: "I don't know the digits", so we see that $(x,y)=(6,2)$ was impossible since $\color{red}{12}$ appears at only one place.

B: "I don't know the digits", so we see that $(x,y)=(4,4)$ was impossible since $\color{green}{8}$ appears at only one place.

A: "I know the digits", so we see that $(x,y)=(8,2)$ was possible since $\color{red}{16}$ appears at only one place, and that $(8,1),(9,1),(4,2),(9,2),(3,3),(6,3),(8,3),(6,4)$ were impossible since if $(x,y)$ were either of them, then A would have said "I don't know the digits").

It follows that the written digits are $2$ and $8$.

• Thanks for your help, with help of Adam's answer and Robb's comment I understood why you excluded 5 and 7. Very nice deduction! – Ignacio Rojas Nov 25 '17 at 16:44

To illustrate a possible approach, let's consider the simpler case in which the digits are limited to $1,2,3,4,6,8$. (It's convenient for illustrative purposes to omit the primes $5$ and $7$ but include the composites $6$ and $8$).

The possible products told to A are $1,2,3,4,6,8,9,12,16,18,24,32,36,48,64$. The only products which can be formed in more than one way, and in brackets the corresponding implied sums, are: $$4 = 1*4 = 2*2\quad (5,4)$$ $$8 = 1*8 = 2*4\quad (9,6)$$ $$12 = 2*6 = 3*4\quad (8,7)$$ $$16 = 2*8 = 4*4\quad (10,8)$$ $$24 = 3*8 = 4*6\quad(11,10)$$

So if A says "I don't know ..." B will know the product must be one of $4,8,12,16,24$. Note the assumptions here and below that B knows that A knows the product, A knows that B knows the sum, and both always say the truth.

If B then says "I don't know ...", the sum must be either $8$ or $10$, the only sums which occur above more than once. The possible products are therefore $12,16,24$.

If A again says "I don't know ...", the product cannot be $12$ or $24$, as the only product with possible sums $8$ and $10$ is $16$.

So B can then say "I know the digits" ($(4,4)$ if the sum is $8$, $(2,8)$ if it's $10$).

This sort of approach with digits $1,\dots,9$ should solve the riddle, although the details will probably be quite tedious.

• $5$ and $7$ cannot contribute to the list of products which can be formed in more than one way but $9$ can. You add in products $9$ and $18$ with sums $(10,6)$ and $(11,9)$ These sums duplicate some of the ones on the list, which explains why it takes longer to get to an answer. – Ross Millikan Nov 24 '17 at 15:15

format: plain


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# Thread: [SOLVED] quotient rule on holomorphic functions

1. ## [SOLVED] quotient rule on holomorphic functions

Clearly, this is not fun.

If we decompose f(z) and g(z) into real valued functions u + iv and s + it, it seems like a reasonable first step to multiply the numerator and denominator by the complex conjugate of g (s - it), which yields something that looks very similar to the desired result. However, when we take the derivative of these real valued functions, the denominator is no longer $g^2$ and the numerator looks more like f' multiplied by the complex conjugate of g, minus g'f.

Any hints on better steps to take?

Thank you!

2. I suggest you first solve the chain rule problem. Then you can show that $\frac{d}{dz}\frac{1}{g(z)} = \frac{-g'(z)}{g(z)^2}$ and then you can apply the product rule to $\frac{f(z)}{g(z)} = f(z) \times \frac{1}{g(z)}$.

3. Originally Posted by Bruno J.
I suggest you first solve the chain rule problem. Then you can show that $\frac{d}{dz}\frac{1}{g(z)} = \frac{-g'(z)}{g(z)^2}$ and then you can apply the product rule to $\frac{f(z)}{g(z)} = f(z) \times \frac{1}{g(z)}$.
Thank you very much.

This question was actually ordinally before the chain rule question. Is there any other way to approach it? Thanks for your help!

4. Once again you are very welcome!

There might be a way to prove it without the chain rule, but it would just be equivalent to proving the chain rule for the specific case $F \circ g$ where $F(z)=z^{-1}$. Therefore I think it really makes more sense to do the other problem first in all its generality.


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## Adobe Support Community

• Why is it so difficult to make an arrow in PSD?

# Why is it so difficult to make an arrow in PSD?

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Why is it so difficult to make an arrow in PSD? It is so simple in ILL and InD... Related note: making a line WAS easy before. Now the whole jump through hoops etc. Make it simple and easy again, please!

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# Probability of 0 bit in ASCII text files

by Cylab
Tags: ascii, files, probability, text
P: 54
 Quote by haruspex That's all true, but I'm not sure where it gets you. What are you trying to solve here? Are we done with the N≤8 case?
How about investigation of N > 8 case?
Just cant quiet figure out how N=8A+B works.
Homework
HW Helper
Thanks ∞
P: 9,215
 Quote by Cylab How about investigation of N > 8 case? Just cant quiet figure out how N=8A+B works.
N=8A+B, B<8, means there are A whole bytes and B odd bits, so either A or A+1 MSBs included.
See what gaps you can fill in here:
Prob that this includes A+1 MSB's = ....?; if A+1 MSBs, prob that all N bits are 0 is ...?
Prob that this includes A MSB's = ....?; if A MSBs, prob that all N bits are 0 is ...?
Adding this up, prob that all N bits are 0 is ...?
P: 54
 Quote by haruspex N=8A+B, B<8, means there are A whole bytes and B odd bits, so either A or A+1 MSBs included. See what gaps you can fill in here: Prob that this includes A+1 MSB's = ....?; if A+1 MSBs, prob that all N bits are 0 is ...? Prob that this includes A MSB's = ....?; if A MSBs, prob that all N bits are 0 is ...? Adding this up, prob that all N bits are 0 is ...?
 Quote by haruspex if A+1 MSBs, prob that all N bits are 0 is ...?
That is not easy for me Sir.
Analysis: Since the N bits are drawn consecutively from ASCII, there is only 1 character (out of 2^256), which is are all 0. So only 1 MSB. Thus, the Prob =1/2^256.
Others seem to follow the conception , or I misunderstood your point?

Question:
1) Pr[0] in ASCII (assume each character appears with same ratio) equals = 1/8+1/2=5/8. Is it OK?
2) Successive 7 bits are drawn at random from ASCII bits (e.g. no bias of character distribution), what is Pr[0] in the 7 bits?
Successive 4 bits are drawn (same condition with above), what is Pr[0] in the 4 bits?
So, say, Successive N bits are drawn (same condition) , what is Pr[0] in the N bits?
Analysis:
Do you think it is same case? remember you explained that N/8*2^-N + (1-N/8)*2^-(N+1) . Does the formula apply to the case of 2).

Homework
HW Helper
Thanks ∞
P: 9,215
 Quote by Cylab Analysis: Since the N bits are drawn consecutively from ASCII, there is only 1 character (out of 2^256), which is are all 0. So only 1 MSB. Thus, the Prob =1/2^256.
The N bits might start in the middle of one byte, span several whole bytes, and finish part way through the last. For each whole byte, prob of all zeroes is 1/128 (since MSB always zero).
If the N bits include A MSBs then how many non-MSBs do they include?
What is the prob that the A MSBs are all 0?
What is the prob that the non-MSBs are all 0?
So what is the prob that all N bits are 0?
 Question: 1) Pr[0] in ASCII (assume each character appears with same ratio) equals = 1/8+1/2=5/8. Is it OK?
No. 1/8 that bit is MSB, so 7/8 that it is non-MSB. P[0] = 1/8 + 7/8*1/2 = 9/16. We went through that much earlier in the thread.
 2) Successive 7 bits are drawn at random from ASCII bits (e.g. no bias of character distribution), what is Pr[0] in the 7 bits? Successive 4 bits are drawn (same condition with above), what is Pr[0] in the 4 bits? So, say, Successive N bits are drawn (same condition) , what is Pr[0] in the N bits? Analysis: Do you think it is same case? remember you explained that N/8*2^-N + (1-N/8)*2^-(N+1) . Does the formula apply to the case of 2).
I believe you are misquoting the formula. In post #27 I explained that the prob of N consecutive 0 bits (N≤8) is 2-N(1+N/8).
P: 54
 Quote by haruspex I believe you are misquoting the formula. In post #27 I explained that the prob of N consecutive 0 bits (N≤8) is 2-N(1+N/8).
That is right and it was good explanation.
Now, say, two N consecutive bits are taken (or two groups) (N1=7, and N2=4).
So the prob is different in following cases in comparing of same amount of 0`s?
1st case: (in N1=7) Pr[0], Pr[00],..,Pr[0000] =?
2nd case: (in N2=4) Pr[0], Pr[00],..,Pr[0000] =?
Homework
HW Helper
Thanks ∞
P: 9,215
 Quote by Cylab 1st case: (in N1=7) Pr[0], Pr[00],..,Pr[0000] =? 2nd case: (in N2=4) Pr[0], Pr[00],..,Pr[0000] =?
I don't understand your question. What does Pr[0] mean in the context of N1=7? Is it the probability that the next bit is zero given the preceding 7 were?
P: 54
 Quote by haruspex I don't understand your question. What does Pr[0] mean in the context of N1=7? Is it the probability that the next bit is zero given the preceding 7 were?
Sorry!
I meant the prob of 0 within N1 =7 consecutively drawn from ASCII .
In other words, say, now we have a group of bits consists of many N1, each of which is consecutive 7 bits drawn from ASCII. what is Prob[0], Prob[00] in the group respectively?

2nd case. another group of bits with same condition, where N2=4. What is Prob[0], Prob[00] within the second group respectively?

Is (N1)Prob[0] = (N2)Prob[0] right, or should it be unequal?
Homework
HW Helper
Thanks ∞
P: 9,215
 Quote by Cylab Is (N1)Prob[0] = (N2)Prob[0] right,
Of course. The bits don't know how many others were chosen.
P: 54
 Quote by haruspex Of course. The bits don't know how many others were chosen.
So you are saying following are correct?

(N1=7)Prob[0] = (N2=4)Prob[0]
(N1=7)Prob[00] = (N2=4)Prob[00]
(N1=7)Prob[000] = (N2=4)Prob[000]
....
 Homework Sci Advisor HW Helper Thanks ∞ P: 9,215 To be completely clear: If you choose N consecutive bits, the probability that the first R of those bits are all zero (R <= N), depends only on R. It cannot depend on N. Further, if you choose N consecutive bits, then choose R consecutive bits from those N, the probability that the first R of those bits are all zero depends only on R. (This seems so obvious that I worry that I have not understood the question.)
P: 54
 Quote by haruspex To be completely clear: If you choose N consecutive bits, the probability that the first R of those bits are all zero (R <= N), depends only on R. It cannot depend on N. Further, if you choose N consecutive bits, then choose R consecutive bits from those N, the probability that the first R of those bits are all zero depends only on R. (This seems so obvious that I worry that I have not understood the question.)
case: N1=7 & N2=4 . Assume N is taken from X bits, which is ASCII.
take R=2 bits from N1 & N2 respectively, what is prob that they are two 0 bits.

1st. (N1 case) : {(9X/16)C2 * (7X/16)C5 } / xC7.
1st. (N2 case) : {(9X/16)C2 * (7X/16)C2 } / xC4.

Seems it depends on N too.
Homework
HW Helper
Thanks ∞
P: 9,215
 Quote by Cylab 1st. (N1 case) : {(9X/16)C2 * (7X/16)C5 } / xC7.
I have no idea what that notation means.
P: 54
 Quote by haruspex I have no idea what that notation means.
Hypergeometric Distribution.

X: number of ASCII bits , from which N is taken.
Pr[0] = 9/16.

xC7 : The number of combinations of x , taken 7 at a time.
Homework
HW Helper
Thanks ∞
P: 9,215
 Quote by Cylab Hypergeometric Distribution. X: number of ASCII bits , from which N is taken. Pr[0] = 9/16. xC7 : The number of combinations of x , taken 7 at a time.
So how do I read (9X/16)C2? If I plug in X=7, that gives (63/16)C2, which is meaningless.
P: 54
 Quote by haruspex So how do I read (9X/16)C2? If I plug in X=7, that gives (63/16)C2, which is meaningless.
C: combinations
Pr[0] = 9/16.
X: number of ASCII bits , from which N is taken.
case: N1=7 & N2=4 . Assume N is taken from X bits, which is ASCII.
Other definitions should be clear
Homework
HW Helper
Thanks ∞
P: 9,215
 Quote by Cylab C: combinations Pr[0] = 9/16. X: number of ASCII bits , from which N is taken. case: N1=7 & N2=4 . Assume N is taken from X bits, which is ASCII. Other definitions should be clear
You wrote (9X/16)C2, and you have still offered no reasonable explanation for that notation. Did you mean (9/16)XC2?
P: 54
 Quote by haruspex You wrote (9X/16)C2, and you have still offered no reasonable explanation for that notation. Did you mean (9/16)XC2?
X: The number of bits in ASCII.
9X/16: The number of 0 bits in the X that are classified as successes.
7 or 4: The number(s) of bits taken consecutively from X.
2: The number of 2 zeros in the 7 or 4 that are classified as successes.
(9X/16)C2 : The number of combinations of 9X/16, taken two 0 bits at a time.
Homework
HW Helper
Thanks ∞
P: 9,215
 Quote by Cylab X: The number of bits in ASCII. 9X/16: The number of 0 bits in the X that are classified as successes. 7 or 4: The number(s) of bits taken consecutively from X. 2: The number of 2 zeros in the 7 or 4 that are classified as successes. (9X/16)C2 : The number of combinations of 9X/16, taken two 0 bits at a time.
Now that you have explained that, thankyou, I can see where it is wrong.
For one thing, that analysis treats all bits as independently 0 or 1, regardless of their proximity to each other. Bits multiples of 8 positions apart will be positively correlated, and at other distances negatively correlated.
More significantly, let's look at what these represent:
1st. (N1 case) : {(9X/16)C2 * (7X/16)C5 } / xC7.
1st. (N2 case) : {(9X/16)C2 * (7X/16)C2 } / xC4.
The first is the probability of picking 7 bits that are exactly two 0 bits and 5 1 bits; the second is the prob of picking 4 bits that are exactly 2 and 2. No wonder they're different! In the problem I thought we were discussing, P[00] doesn't care what the remaining 2 or 5 bits are.

 Related Discussions Calculators 2 Programming & Computer Science 5 Programming & Computer Science 0 Programming & Computer Science 1 Computing & Technology 1


text:
Home > Pre-algebra > Fractions > Subtracting Fractions

## Subtracting Fractions

#### Introduction

In order to subtract fractions, we find the highest common factor of the denominator, and subtract one equivalent numerator from the other.

#### Terms

Fraction - Fractions are used when using numbers to express parts of whole.
Least/Lowest Common Multiple - The lowest multiple of two or more numbers.
Equivalent Fraction - Equivalent fractions are fractions that have the same value.

## Lesson

When subtracting fractions, we have to make sure the denominators are the same. So if we are given

in this case the denominators are the same, so we subtract 1 from 5

, which reduces to

When the denominators are different, we ‘make’ them the same by finding their least common multiple (LCM).

For example, in the problem , the LCM of 7 and 2 is 14

So to make the denominator of 6/7 = x/14, we have to multiply the denominator by 2 (don’t forget to multiply the numerator by 2 as well!). We then get 6*2 / 7*2 = 12/14. Similarly for 1/2, we multiply the numerator and denominator by 7 to get 1*7 / 2*7 = 7/14. The equivalent fractions of 6/7 (12/14) and 1/2 (7/14) now have the same denominators, so we simply subtract 7 from 12 to get 5/14.
When subtracting mixed fractions, we follow the same procedure but we need to convert the mixed fraction to an improper fraction first. So if we are given 1 2/3 – 1/4, we convert 1 2/3 to the improper fraction 5/3. Then we ‘make’ the denominators the same by finding the LCM, so we end up with 5*4/3*4 – 1*3/4*3 = 20/12 – 3/12. Since the denominators are now the same, we subtract 3 from 20 to get 17/12, or 1 5/12 (expressed as a mixed fraction).

## Examples

Since these fractions have the same denominator, we can just subtract the numerators
Since these fractions have different denominators, we need to find the least common multiple of the denominators
The least common multiple of 3 and 4 is 12, so we need to multiply to make each of the denominators = 12
Since these fractions have the same denominator, we can just subtract the numerators
Since these fractions have the same denominator, we can just subtract the numerators
can be reduced, since is a factor of both and :
The fraction is now in lowest terms


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# DPMO Calculator

Created by Hanna Pamuła, PhD
Reviewed by Dominik Czernia, PhD and Jack Bowater
Last updated: Feb 02, 2023

Welcome to the DPMO calculator (Defects per Million Opportunities) - a tool that will help you calculate this popular measure of process quality. Haven't you heard about this mysterious abbreviation before? No? Well, scroll down and you'll learn all about the DPMO meaning.

However, if you're wondering how to calculate DPMO and what's the formula behind it, you can also breathe a sigh of relief - you'll find all the explanations below.

## What is DPMO, and what does it stand for? DPMO meaning

DPMO stands for defects per million opportunities. It's a measure of process performance used to assess the quality of a process - e.g., the quality of a service or production. The lower the value of DPMO, the better, as it is tied with the probability of the presence of a defect.

The concept of DPMO may look similar to reporting defective parts per million (PPM). However, in DPMO it's possible to have in one unit:

• Multiple types of defects; or
• Multiple defects of the same type.

while PPM describes defective units per million units. Another related concept is DPU, which stands for defects per unit. DPU describes the average number of defects per unit of product/service.

## How to calculate DPMO? DPMO formula & example

The DPMO formula is pretty straightforward:

$\footnotesize \text{DPMO} = \frac{1,\hspace{-0.04cm}000,\hspace{-0.04cm}000\times \text{defects}}{\text{units}\times\text{defect opportunities}},$

where:

• $\small\text{defects}$ - number of defects;
• $\small\text{units}$ - number of units; and
• $\small\text{defect opportunites}$ - number of defect opportunities per unit.

Let's show how we can calculate DPMO with an example.

Assume that you're the owner of a cloth factory. You'd like to calculate the DPMO for the production of jeans 👖. You've gathered some information from your company managers, and you found out that every pair of jeans can be defective in five different ways:

• Missing button 🔘
• Dingy material;
• Faulty seams 🧵
• Inaccurate size; and
• Broken zip 🤐

Moreover, when 10,000 pairs of jeans were sampled, we found 11 defects. So how to calculate DPMO? Use the DPMO formula defined above!

$\small\text{defects}$ is the number of defects found in a sample, so it's 11;

$\small\text{units}$ - for this example, it's 10,000 pairs of jeans; and

$\small\text{defect opportunites}$ - the number of defects opportunities per unit, so in our case, it's 5.

\footnotesize \begin{align*} \text{DPMO} &= \frac{1,\hspace{-0.04cm}000,\hspace{-0.04cm}000\times \text{defects}}{\text{units}\times\text{defect opportunities}}\\[1.0em] &= \frac{1,\hspace{-0.04cm}000,\hspace{-0.04cm}000\times 11}{10,\hspace{-0.04cm}000\times 5}\\[1.0em] &= 220 \end{align*}

How good or bad is that result? The best possible DPMO is 0, while the DPMO will get a value of 1 million in the worst case. According to the six sigma methodology, highly capable processes experience less than 3.4 defects per million opportunities - an extremely unlikely event.

## DPMO & six sigma relationship

You may have heard about the three sigmas (more of that in our empirical rule calculator). How about the six sigmas? The six sigmas are a set of methods, metrics, and tools that help improve any kind of process by eliminating waste and defects, reducing variation, and ensuring quality control. DPMO is one metric that can be found in the six sigma methodology.

So what does six sigma mean, and what does it tell us?

The standard deviation - sigma σ - measures the dispersion of a set of values. (See our standard deviation calculator for an in-depth tour about standard deviation.)
The number of standard deviations between the mean of the process and the nearest specification limit describes the sigma level (does z-score ring a bell?🔔). If you want to express the amount of defect-free products created during your process, you can use the sigma levels instead of standard percentages:

1 Sigma = 30.85% defect-free;

2 Sigma = 69.146% defect-free;

3 Sigma = 93.319% defect-free;

4 Sigma = 99.379% defect-free;

5 Sigma = 99.977% defect-free; and

6 Sigma = 99.99966% defect-free.

The above values comply with the six sigma industry standard, where the empirically found value of the is used. Sigma shift is used to allow for the increase in process variation over time, which occurs in real-life situations.
The sigma level measure is useful when you want to compare different processes or find the ones that need improvement.

For the DPMO-sigma level conversion, this table may come in handy:

Sigma level (with 1.5σ shift)

DPMO

Yield, defect-free

691,462

30.850%

308,538

69.146%

66,807

93.319%

6,210

99.380%

233

99.977%

3.4

99.99966%

0.019

99.999998%

Alternatively, our DPMO calculator has a in-built feature that calculates both the DPMO and the sigma level using a 1.5σ shift. Check it out!

## An example on how to use the DPMO calculator

We've created this DPMO calculator to help you do the calculations in the blink of an eye!

1. Enter the number of defects that occurred in your sample. Let's say that you've found 17 errors in the spreadsheets.
2. Input the number of units in your sample. Let's assume that these 17 defects appeared in 20,000 spreadsheets that were checked (not by hand we hope!).
3. Fill in the defect opportunities field. Every spreadsheet has 75 fields, so there are 75 opportunities for error.
4. The calculator shows the result: DPMO = 11.333! You don't need to rack your brain anymore on how to calculate the DPMO, yay!

Additionally, the tool found out the sigma level of this process: 5.74. Experiment with the values and check out how many defects you could have to fulfill the six sigma rule.

Don't forget to look through our Advanced mode, where you'll find DPU and PPM calculated for you.

Hanna Pamuła, PhD
# of Defects
# of Units
# Defect opportunities
/ unit
Defects per Million Opportunities
DPMO
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###### Question:

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# What about all those zeroes? Measuring performance of models on intermittent demand

In one of the previous posts, we have discussed how to measure the accuracy of forecasting methods on the continuous data. All these MAE, RMSE, MASE, rMAE, rRMSE and other measures can give you an information about the mean or median performance of forecasting methods. We have also discussed how to measure the performance of models in terms of prediction intervals, and should now be comfortable with such measures as Range, Coverage, pinball, MIS. But all of this might become irrelevant when we face the intermittent demand – it can really make a bright day dark and screw with you and your measures. So, care is needed if you have data with randomly occurring zeroes.

We have already discussed aspects of intermittent demand in a post about the intermittent exponential smoothing model some time ago, and we have seen some time series examples in one of the previous posts, so I will not repeat myself, but I still think that there are a couple of important notes worth making about it.

Intermittent time series is the series that has non-zero values occurring at irregular intervals. This implies that there might be some periods of time, when we observe zeroes (e.g. no one buys our product). An example of such time series is shown below:

Intermittent Demand

As you can see, there are two sources of randomness in this data: there is a randomness about the non-zero value itself (e.g. demand size) and there is also a randomness about the occurrence of the non-zero value (e.g. demand occurrence). So, a very general model for intermittent demand can be represented as:
\label{eq:general}
y_t = o_t z_t ,

where $$y_t$$ is the observed value at time $$t$$, $$o_t$$ is the binary occurrence variable and $$z_t$$ is the demand sizes variable. While there are some statistical models that don’t do this distinction, the intermittent demand forecasting methods that are most popular in practice do it either directly or indirectly.

Although we can create very fancy forecasting models (there are even tries to use neural networks for this), we usually cannot accurately predict, when specifically the product will be bought and how many units we will sell. All we usually can do is to produce the mean value and / or prediction intervals for the demand itself. And even if we do that, the next thing to ask ourselves is: “What do we do with that?”

In a supply chain context, the typical decision would be how many units of product to order or to produce, given the amount that we already have. In this case we are usually talking about the safety stock – how many units of the product we should hold, so that we satisfy the demand and don’t have empty shelves. Typically, this is determined based on calculation of quantiles of a distribution. In many cases, in practice, the Normal distribution is used, and I cannot describe on how many levels this is wrong (the most obvious thing to point out is that usually we cannot have negative amounts of product). But let’s not get too much distracted.

What does the safety stock typically depend on? In many cases this is not just the demand of a product, but specifically the demand over the lead time – for example, how many products we will sell over the next week. Why? Because the deliveries tend to happen once over a time period, e.g. once a week, and we tend to order products once in a period of time – it does not make sense to order the product each time we run out of it. So, we do periodic reviews of the inventory and decide, how much we should order, so that we don’t run out of stock until the next replenishment.

Good. So, how do we get the information about the demand over the lead time? Here come our forecasting model! But all of this implies that we should focus not just on the point and interval forecasts for each specific horizon, but on the values aggregated over the lead time. This is the first important difference between conventional forecasting and forecasting for inventory purposes (see for example, Kourentzes et al., 2019).

As you can see, the relation between the actual demand forecast and the decision of how much to order is complex. And this means that all those nice error measures, discussed in the previous posts, might not give us the important information of how our model performs in terms of orders. The model can produce very accurate forecasts, but this does not necessarily translate directly to correct ordering policy.

In order to align the model with the specific decisions, one can revert to simulations, so that it becomes more apparent how the model performs in terms of achieved service level, lost sales and excess inventory (which all are sort of proxies to the object of real interest in practice – the cost). This is one of the common approaches in the modern supply chain and inventory literature. Unfortunately, this approach is computationally expensive and may vary from one situation to another, so you would need to set up the simulation for each separate company and potentially for different products.

An alternative approach would be to at least asses the performance of models over the lead time, not on each observation. This means that we need to refer to cumulative values over the period of time:
Y_{t+h} = \sum_{j=1}^h {y}_{t+j} .

Then we can measure, for example, how our models perform in terms of working stock (aka “cycle stock”). In this case we are only interested in the mean demand over the lead time, which for the additive models is relatively easy to deal with because the expectation of the sum of the demands is equal to the sum of the expectations:
\begin{multline} \label{eq:workingStock}
\text{E} \left(\sum_{j=1}^h {y}_{t+j} \right) = \text{E}\left(\sum_{j=1}^h (\hat{y}_{t+j}+e_{t+j})\right) = \sum_{j=1}^h \text{E}(\hat{y}_{t+j}) + \sum_{j=1}^h \text{E}(e_{t+j}) = \\
\sum_{j=1}^h \text{E}(\hat{y}_{t+j}).
\end{multline}
where $$\hat{y}_{t+j}$$ is the point forecast of a model. This means that we can produce the $$h$$ steps ahead forecast and aggregate it over the lead time $$h$$ and then compare it with the actual values for the same period. However, if we deal with the additive models in this context, then we are assuming that the demand can be negative – which is usually unrealistic, especially for intermittent demand. This means that a different model is needed, and \eqref{eq:workingStock} might not hold any more. So, for example, in case of ETS(M,N,N) simulations are necessary in order to produce the correct cumulative conditional expectation over the lead time.

Now, let’s say that we have managed to produce meaningful cumulative point forecast. What’s the next step? We need to measure its accuracy. The good news is, RMSE-based measures can be used in order to assess the performance of models in terms of the working stock (I hope you still remember, why MAE- and MAPE-based measures are not appropriate for intermittent demand). So, we can evaluate the Squared Cumulative Error and compare models based on it:
\label{eq:workingStockRMSCE}
\text{SCE} = \left( \sum_{j=1}^h y_{t+j} -\sum_{j=1}^h \hat{y}_{t+j} \right)^2 .

We can use relative or scaled measures, if we want to compare performance of models across products – all the things discussed in the previous post are applicable here. The main limitation here is that, given that we deal with intermittent demand, Naive method might cause problems if used as a benchmark. An example is a situation, when it produces zero forecasts for the holdout sample (because the last observation was zero), and the holdout contains only zeroes. This might happen by chance, but it will mess your beautiful relative error measures. A simple average of the whole series might be more appropriate as a benchmark for the intermittent demand.

There is a measure, which is ideologically close to the SCE. It is called “Periods-In-Stock” – PIS (Walstrom, 2010):
\label{eq:workingStockPIS}
\text{PIS} = \sum_{j=1}^h \hat{y}_{t+j} -\sum_{j=1}^h y_{t+j} .

Have you noticed that it is calculated as forecast minus the actual values? This was done by the author intentionally, so that the sign of the measure aligns with the inventory decisions. If the value is negative, then this means that we missed sales. If it is positive, then this means that we had too much stock and did not sell as much as we should have. Petropoulos & Kourentzes (2015) have proposed several modifications of PIS, which allow aggregating this measure over several products. So, have a look at the paper if you are interested, but I will not expand here.

By looking at those error measures, we can evaluate the performance in terms of working stock. But, as it was mentioned before, this does not necessarily translate to safety stock performance. So we need to figure out how to align our error measures with it. It seems that the closest we can get to it is by producing the specific quantiles for the values accumulated over the lead time. We typically don’t care about the lower bound, because in case of intermittent demand it is usually just zero, plus, the lower bound is not helpful for inventory decisions (e.g., knowing what will be the level of sales in 2.5% of cases is not helpful). So, we are more interested in the upper bound. In a realistic situation we would need to simulate the data from our model (something like a 1000 runs for the possible future outcomes), sum it up over the lead time and then take the upper quantile (for example, 95% or 99%). We have to do it this way, because we might not have expressions for the quantiles for the values accumulated over the lead time. But this way we will get quite close to the safety stock.

How do we measure performance of models in terms of this upper bound? I would be inclined to recommend pinball function, keeping in mind its limitations, discussed in the previous post. We can also analyse the coverage and range, and if we really need to use something like Mean Interval Score (MIS), then it makes sense to use Quantile Score instead (Gneiting, 2007), because it only takes the upper bound into account. However it needs to be modified in order to reflect the idea of cumulative values:
\label{QS}
\text{QS} = \left(\sum_{j=1}^h y_{t+j} -\sum_{j=1}^h u_{t+j} \right) \left(\mathbb{1} \left\{ \sum_{j=1}^h y_{t+j} \leq \sum_{j=1}^h u_{t+j} \right\} -\alpha \right) ,

where $$\alpha$$ is the confidence level (in our example 95% or 99%), $$u_{t+j}$$ is the upper bound of the prediction interval and $$\mathbb{1}(\cdot)$$ is the indicator function, returning one, when the condition is true and zero otherwise. The QS has similar interpretation and working principle as MIS: if the values lie below the specified quantile, then the measure is penalised less than in the case, when they lie above: in the former case the indicator function returns 1, so the weight becomes $$1-0.95=0.05$$, while in the latter case it is equal to zero, so the weight is $$0 -0.05 = -0.05$$. Note also that in that latter case the difference between the actual value and the bound is negative. This means that QS will always be positive, and in the ideal situation, when all the future values lie on the bound, the QS is equal to zero. The difference between QS and pinball function is in what specifically they measure. The QS measures the performance in terms of coverage and range for the bound, while the pinball tries to measure, how accurately we hit the specific quantile. So, arguably, QS is closer to what we would be interested in real life: we want to have the coverage closer to nominal (so that we achieve the nominal service level), but with the smallest range possible (so that we do not order too much and do not have huge holding costs). Based on QS, we can calculate scaled or relative measures in order to compare performance of forecasting models across the products.

So, what about the other intermittent demand contexts? In some of them the situation might be quite different and in some sense easier than in supply chain. For example, in case of the forecasting the patients flow in a hospital, we often do not need to deal with values over the lead time, which simplifies things substantially. In addition, we might not care much about the average number of patients for a specific day, but be more interested in the prediction interval. This way we can assess how many patients to expect in the “worst” / “best” case scenarios for the selected confidence level. Based on these values, we then decide, how many nurses, doctors and beds we should have for a specific day in order to service, let’s say, 95% of the patients in the next 4 hours. Note that yet again the prediction interval is not directly related to this goal, but at least it can be considered as an approximation to it, if we know how many nurse and doctor hours we typically need in order to process a specific type of patient. This also implies that we should evaluate the accuracy of the prediction intervals, not the accuracy of the mean. And, when doing all of that in case of intermittent demand, once again, we usually don’t care about the lower bound, because we can typically say even without any model that in the worst (best?) case we won’t have any patients at all. Instead, we should concentrate on the upper bound of the prediction interval, and we can use the measures, discussed in the previous context in this post.

Having said all this, the general recommendation for dealing with intermittent demand – avoid it at all costs. Why? Because intermittent demand is messy to identify and to deal with. Besides, are you sure that you really need to produce forecasts on that level of sales? What specific decisions do you make based on those forecasts? Maybe you don’t need to know how many watermellons you will sell in each separate shop every day. Maybe you should switch to weekly data, where the intermittency disappears, or aggregate sales of products over larger volumes and over several shops in the chain… So, before diving into the problem and trying to evaluate models, think about the decisions you usually make based on those models.