In what time does Interstellar take place?

Question: Any ideas on the timeline of Interstellar's plot? How many years from now does the story take place? We know that it begins in near future, but any approximations or clues about "how many years from now?"
Comment: At a guess, because I'm not seeing anything in writing, I'm going to say 10-20 years in the future. The main reason for this is the drone. Doesn't he say something to the effect of "It must have been up there for 10 years or so" ... I'm fairly familiar with drone tech. I don't think the Indians have solar powered drones at this point.
Answer: This post from Bustle.com gives some good reasoning for Interstellar being set around the 2070s.
In summary: the descriptions of halcyon days gone by from Cooper's live-in father-in-law Donald sound very much like our present day. Assuming Donald was born between 1980s-2000s, he may have had Cooper's (now deceased) wife around 2030-2035. Cooper is in his late 30s, so assuming his wife was a roughly similar age this takes us up to 2070-ish.
Not sure this maths completely holds up under scrutiny, but I think it's clear from that logic that it's not as near-future as the 10-20 years from now other posts suggest.
EDIT: Thanks to Keybrd Basher for pointing out a silly mistake, now corrected.
Comment: I feel the reason he deliberately left out what year it takes place in is because he's also trying to invoke the story with his own message and musings on a possible future, one he feels could actually happen (from an earth dying standpoint anyway). By being deliberately vague as to when this happens, it allows for two things: 1) It expounds on the idea that this will happen and we have no idea when and where and how and why. 2) It keeps the movie from being dated, as no one can say, "Well, this didn't happen," when the year/date actually arrives in the event it actually doesn't.
Answer: Jonathan Deamer has done a great job at drawing the timeline. Much of what he says resounds with me. To add to it, I think we can gain a more precise idea of Donald's birth from his dialogue with Coop about "his time". A time when everyday seemed like Christmas.
Somewhere in there he mentions that the global population at that time was 6 billion. Now, according to this Wikipedia article that must be after 1999. That would indicate that Donald is Generation Z. Which indeed implies that Cooper would had been born in 2030s and the present day to be in 2070s.
This is inline with the fact that John Lithgow who plays Donald in the movie is pushing 70 and looks his age for the part. Somebody who is born in the late 90's or early 2000's would definitely be close to 70 by 2070. So that's where the present day is set.
Comment: The 6 billion population clue is a great point, of course!
Answer: At the end of the film, the space station doctor tells Cooper that he is the ripe age of 124 years old. Several minutes before this, Amelia says that Cooper doesn't look too bad for someone pushing 120 years old. So, at that point, Cooper is 119 / 118 years old. A five to six year gap lost inside the Tesseract. Cooper states, just moments before Amelia's comment, that they lost 51 years using Gargantua as a sling-shot, meaning Cooper was 68 / 67 years old when Mann died.
The next mention of time passage is when the crew returns to the Endurance after discovering Miller's planet was a bust. 23 years and change. Making Cooper 45 / 44 years old at that moment.
It was also mentioned that it would take two years for the crew to make the journey to Saturn from Mars, making Cooper 43 / 42 years old when he left Earth. This corresponds with Donald's comment about Cooper being born 40 years too early or too late. So, Cooper is in his 40's. The hard thing to understand is the statement of how Murph is 10 years old when Cooper leaves Earth, and when older Murph transmits her "birthday" message, it had only been 26 years, at best. Making Murph 35 / 36 years old, nearly six to seven years younger than Cooper at the time of her message. Hard to explain. (There is not a six to seven year leap mentioned. Though, it is possible that on Mann's planet there was still a time difference. That is not proven, but we will go with it, nonetheless.) So, we have to make Murph 43 / 42 years old at this point, by her statement.
Using the same logic, elderly Murph would be 98 / 97 years old at the end of the film; and had entered cryo-sleep around 96 / 95 years old. Understanding this, the mid 90's for a human in the 'future-future' is difficult to maintain for a older person. Meaning, an elderly person like Donald would not survive into his early 90's, especially in the state of how the Earth was and how the film portrayed medical care for children, let alone geriatrics. As of right now in 2014, the average life expectancy is 79 years old. So, to be fair, let's say Donald died at 87. We know that 25 years passed while Cooper was on the Endurance, and the like; giving proof that Donald had to be 62 when Cooper left Earth. That's not a bad sell when John Lithgow is 69 now, 68 when filming. Nearly, a six year gap in his physical appearance. For his physical appearance to make sense though, it would have to mean that Donald died at 93 / 94 years old. (Assuming the message from Tom about burying Grandpa is at the cusp of the 23 year gap.) Elderly Murph in the 'future-future' could barely hold out to reach her mid to late 90's without medical watch, and she wasn't on Earth; so I find it hard to believe that Donald died in his 90's.
Now, using this logic, Cooper is 43 / 42 years old and Donald is 62 years old. A 19 / 20 year gap. Meaning Donald was 19 or 20 years old when Cooper was born. Plausible. Now, Donald could have had children at any point at those ages. Making Cooper possibly one or two years younger than Donald's daughter. She would be 17 or 18 at the earliest for marriage, making Cooper 15 / 16 years old. Though it would be unlikely for them to marry right away, because Cooper was working for NASA and in college for awhile. But, if they did marry right away at 17(Cooper) and 19 / 18(Cooper) and 20, Murph would not have been born for 15 to 16 years. Making Tom 15 or 16 years old, at best.
Back to Donald, if he is 62 years old when he makes the comment about 6 billion people, and inventions happening each day when he was a kid, then we have to do some research here. October 1999, marked when the world hit 6 billion people. So, in order for Donald to be a kid in 1999, he would have had to have been born in the early 90's. Which would conflict with some of Cooper's statements and things of that nature. Donald must have been generalizing then, and the last year 6 billion people were on Earth was 2011. So, he couldn't have be born in 2011 to state he was a kid when the world held 6 billion people on it. So, guessing, we'll say Donald was born in 2003. Making the year Cooper left Earth, 2065. We could go even farther and whittle it down to being late May to mid June; due to how okra and corn are planted and harvested around the New York area. I say New York because of the baseball team seen. (Nearly 51 years into our future, or 50 years if we consider it to be 2015 now. Which rounds everything off nicely.) This would also make Cooper be born in 2022 - 2023. Tom was born in 2049 / 2050, meaning Cooper was 26 or 27 years old at that point. Murph was born in 2055, making Cooper 32 or 33 years old. Donald dies before 2090. And when Cooper finally returns from his journey it is the year, 2146.
I don't think Nolan intended for the film's timeline to be torn apart like this though, and the film is really more about a father and his daughter's relationship. I enjoy the fact that a specific year is not mentioned. I think that it takes away from the message of the film, and focuses on other "background noise". But us humans, we need to know the answers to our questions. Our curiosity will never die.
If anyone has better information, please share. My math could be bad, and I could have missed out on something. Feel free to help correct this.
Answer: There is one place where a specific year is mentioned.
In Kip Thorne's book, The Science of Interstellar, he says in the original treatment he wrote that Professor Brand discovered the wormhole in 2019. Almost every other significant event in the film, including the present day which is 48 years later, can be found using that one date though there are 7 years seemingly missing like the Will W. above pointed out.
I've done an extensive timeline here: http://joshbyers.com/blog/2015/4/what-year-does-interstellar-take-place
According the 2019 wormhole date, Cooper and Endurance leave earth April 18th, 2067.
Answer: I think the year is around 2040. Just a guess because Donald states that that there was 6 billion people when he was a kid. And that's around 2000. And Donald looks like he's about 40, probably 50. And Donald is not Cooper's dad. Any ways cooper was probably born in 2002 would be my guess. And he is 33 in the movie so that would make The year 2035. So my estimation was close. Correct if  wrong. So Donald would be 45, which is 12 years older than cooper because I'm estimating that he was born in 1990.
Comment: John Lithgow was born in 1945 ... which makes him 65 when Interstellar was filmed, and imho is about what he looked in the movie, not 40-50.  So if he were a kid around 2000, that makes it closer to 2055-2060... which aligns with the other answers.


Partition function of weakly interacting Bose gas

Question: In the Bogoliuobov theory of superfluidity, the hamiltonian of a system of weakly interacting bosons in second quantized notation is diagonalized with the following result:
$$\hat H = H_0  \hat1 + \sum_k E_k \hat\xi_k^+\hat \xi_k\ $$
Where $H_0$ is the ground state energy (a constant) and $E_k$ is the energy spectrum of the collective excitations:
$$ E_k = \sqrt{\epsilon_k (\epsilon_k + 2n\tilde U(\vec k))} $$
Now suppose I want to calculate the grand canonical partition function of this system. My initial approach was to imagine an ideal gas of these collective excitations (or quasiparticles) as the hamiltonian suggests, and just use :
$$Z_{GC} = Tr(e^{-\frac{\hat H - \mu \hat N}{k_BT}})$$
With N being the number operator of these quasi particles ($\hat N = \sum_k \hat\xi_k^+\hat \xi_k\ $). But then it occured to me that:
1) As the collective excitations are not really particles, their particle number doesn't need to be conserved and the lagrange multiplier $\mu$ for this constraint is not needed (like a gas of photons in blackbody radiation). So I thought about setting $\mu=0$ like the gas of photons. But this doesn't seem right either because in the limit of large momentum (k), the quasiparticles can essentially be replaced with our real bosons, and the number of these real bosons is undoubtedly conserved.
2) The derivation of the above hamiltonian includes an approximation  of taking the occupation number of the ground state to be large (macroscopic) like a Bose-Einstein condensate, and thereby taking the corresponding creation and annihilation operators of this particular state to be commuting. This means that our hamiltonian is not applicable to all temperatures.
So my question is : Is there any way to derive an approximate grand canonical partition function for the system using the above hamiltonian to derive its thermodynamic properties; and if yes, how do we take the non conserving nature of the quasiparticles into account? Can we just set $\mu = 0$ like an ideal photon gas? In general, how can we study the thermodynamics of such a system from first principles?
Answer:
Is there any way to derive an approximate grand canonical partition function for the system using the above hamiltonian to derive its thermodynamic properties; and if yes, how do we take the non conserving nature of the quasiparticles into account?

Yes, you start with the grand canonical partition function for the original particles.  Doing this, is easiest using coherent states and working with a path integral.  This gives you
$
Z= \int D\phi D\phi^*e^{-S(\phi,\phi^*)}
$
With
$
S(\phi,\phi^*) = \sum_{k,w}\phi^*(\epsilon_k-iw)\phi+\sum_{k1,k2,k3,k4}\phi^*_{k1}\phi^*_{k2}\phi_{k3}\phi_{k4}V(k1,k2,k3,k4)
$
Here, $\epsilon$ contains the kinetic energy, potential energy and chemical potential of the real particles. $V$ decribes the interaction.
By doing a steepest decent approximation, you will derive a Gross Pitiveski equation for the classical solution and a non-diagonal second order action.  This is basically assuming the occupation of the classical solution is large.  After diagonalizing this action by the Bogolubov transform you get an energy as:
$
E_k = \sqrt{\epsilon_k (\epsilon_k + 2n\tilde U(\vec k))}
$
This approximate second-order action is that for a Bose gas with 0 chemical potential.
You can study thermodynamic properties by repeating this process with what ever coupling you need to generate your response functions.
The moral of the story is you should start with the partition function fo the bare particles and do the classical-mean-field approximation to get the partition function for the Bogolubov particles.
If you don't like path integrals or coherent states you can check out Bogolubov's original paper, which is some sense required reading for this subject:
https://ufn.ru/dates/pdf/j_phys_ussr/j_phys_ussr_1947_11_1/3_bogolubov_j_phys_ussr_1947_11_1_23.pdf
Comment: I think so.  That will definitely get you  Helmholtz free energy, heat capacity and entropy.  I think it will also get compressibility, but figuring out linear response like currents won't be as simple as directly coupling  Bogolubov particles to a voltage in the partition function, you'll have to start with the real particle theory and go through the approximation again.  I haven't done that calculation though so it might end up matching.
Comment: Thank you so much for your detailed answer. What you suggest is a bit beyond what we study in our graduate statistical mechanics course. I will definitely check the original paper by Bogolubov. But for the time being, I had another idea, can I just use the energy spectrum for $E_k$ and compute the canonical partition function with $Z=Tr(e^{-\hat H/k_BT})$ ? From that I can get the Helmholtz free energy, and subsequently other thermodynamic quantities.


How to disable event notifications for group I'm not in?

Question: I used to be a part of the "Tomb Raider Multiplayer Madness" Steam Group. They have regular weekly events. I became annoyed with the event notifications, so I thought leaving the group would cease them. I left the group. I'm still getting event notifications from the group. How do I stop the event notifications?
Answer: Apparently, with the new "Explore your Queue", when you choose to follow the game, it adds that game to your groups and it can only be removed through the community hub of that game.
Do not follow games if you do not want notifications from that game.
Answer: In your friends list, there's two tabs - one for friends, another for groups. Click the groups tab, and you'll likely see the tomb raider group there. You can remove yourself from there.
If that doesn't work, go to the Tomb Raider community hub. There should be an unfollow button on the page:

Between these two locations, you're guaranteed to never be notified of any Tomb Raider events ever again.
Comment: This is the correct answer, at least for me. This stopped games from sending me constant weekly/monthly 'event' notifications, games I don't even play or remember buying.
Comment: I am not following the group any longer on either of these tabs, yet I expect another event notification. It shows the green "Join Group" button for the group page. Is following different from joining?
Comment: Just updated my answer now that I'm at my desktop computer.
Comment: I'm having the same issue, I left the group but still getting notifications.
Answer: Go to your name on Steam and click "activity" in the dropdown menu. On the top of the next page, click "friend activity settings" and from there you should be able to turn off all the popup notifications you don't want.
Answer: Open FRIENDS > CLICK SETTINGS (gear icon) Then you'll see Group Events and Announcements at the bottom of the list:
Comment: I think you mean 'Friends & Chat' - there is another 'Friends' button at the top which doesn't give access to Settings or Notifications.
Comment: this solved my problem, thank u
Answer: You only need to open STEAM. Click "STEAM" in the top left corner, then click "Settings", this opens the setting window.  Now click "Friends" and near the bottom locate "For Group events and announcements" and uncheck "Display a notification".  This will stop the issue.
Comment: Can not find "friends" in settings.
Comment: Now you need to open up the Friends List and edit the settings there.
Answer: I found it! As of 2020 Steam. Open Steam, click your Name under the bar:
[Store Library Community Your Name] (Takes you to "activity")
Then on the bottom right you'll see View Groups, and even Upcoming Events including the ones you are sick of seeing.
Click View all X groups.
Find whatever group won't stop spamming you with events and click Leave Group.


Sidescrolling shooter where you rescue Vietnam POWs

Question: I'm trying to identify a game I played this weekend. The main game appeared to be a sidescrolling shooter, where you have to shoot open cages to rescue Vietnam POWs. The environment was destructible, and there was a mission select screen where you move around a little helicopter on a spherical map. The graphics are in a pixel-style, but the game appears to be very modern. It was optionally co-op, and I played it in a polycade arcade cabinet but I suspect it is not an "arcade" game. There was an online mode as well. Anyone have any clue what I played?
Comment: @chicks it was at a sort of art collective and the games weren't labeled. Also I'm wondering why mods deleted the comments from users reminiscing about that Atari game that shared many similarities? Seemed useful
Comment: Where did you play this that you wouldn't have been able to find out what the name of game is?
Answer: Looks like Broforce or The Expendabros. The screenshot and description match perfectly.
Comment: It's Broforce. The character(s) playing in this screenshot are [The Boondock Bros](https://broforce.fandom.com/wiki/The_Boondock_Bros) who are not in Expendabros.
Comment: Looks like those games were a sort of homage to (and/or ripoff of) Metal Slug?
Answer: That's level two of Broforce.
Here is a random Youtuber playing that level:


how to send only one request instead of multiple request in angularjs?

Question: I have a demo in which user type anything in input field and request goes to server. Currently whenever user type it fire the request. I want only one request will fire. Example if i type "abc" it fire three request. Is this possible user type anything without stop, after one sec stop I will fire a request.
i know  Inputs can be debounced with the ng-model-options directive: but it fire after time given, but i want user type as long as without stop ,but fire request after stop
Here is my code:
http://plnkr.co/edit/npiA2abAo5SEQFMMpKZO?p=preview
var app = angular.module('plunker', []);

app.controller('MainCtrl', function($scope,$http) {
   $scope.name = 'World';

    $scope.keyupevt = function(){
       console.log('xx')
       $http.get("data.json")
            .then(function(response) {
               console.log(response)
            });
    }
});

Comment: you could have it fire if you lose focus and input > 0. You could also have a minimum character limit and wait until you at least hit that before sending a request.
Answer: Implement your own debouncing using setTimeout/clearTimeout, something like this will do:
app.controller('MainCtrl', function($scope,$http) {
    $scope.name = 'World';

    var timeout = null;                                // the timeout handle
    $scope.keyupevt = function() {
       clearTimeout(timeout);                          // clear any previous timeout (if null, nothing will happen)

       timeout = setTimeout(function() {               // set a new timeout
           console.log('xx');                          // that will make the request after 1 second
           $http.get("data.json")
             .then(function(response) {
               console.log(response);
           });
           timeout = null;                             // reset timeout (not really necessary but it won't do any harm)
       }, 1000);                                       // the code inside will be called after exactly 1000 ms
    }
});

Every time a key is down, a request will be set to happen after 1 second from that keydown event. And if the previous request didn't happen yet, the request will be cancelled and a new one will be set to happen after 1 second. etc


How to write for cartoons?

Question: How do the techniques for writing for animated cartoons differ from the techniques for other script-writing?  Is there a guide for writing for cartoons?  I suddenly find myself writing for a flash Youtube cartoon series.
Comment: I don't think it should be any different than writing a screenplay, but since I've never done either, I might be wrong. Or writing for comics.
Comment: I've made some edits to your question to focus it on cartoons in particular, since asking about screen-writing in general would be too broad (and I think not what you meant).  Please review my edit and feel free to [edit] further if I've misunderstood your intent.  Thanks for the interesting question; I'll be there *are* differences when you're not constrained by "we have to actually be able to stage this", but I don't know what.
Comment: There are differences indeed, since you need to worry about the number of pages and the size of what characters are saying. I found some interesting material some time ago but they were in Portuguese and won't help here. In any case, maybe this will help you [Short cartoon writing tips](http://bobcampcartoonist.blogspot.pt/2009/08/short-cartoon-writing-tips.html)
Answer: Have your target audience in mind, but don't stereotype and don't dumb it down. I mean, if that's a cartoon for kids, make it a nice tale for kids, but get the characters a secondary, non-obvious set of characteristics; have them play it through some behaviors, things that kid will simply shrug away as some insignificant discrepancy but an adult will notice as a nice nod - example, a benevolent wise master is secretly a prankster, half of the trouble - the more harmless half - is not random accidents or omissions but caused by them secretly. Make the story layered - the simple, obvious layer for kids, and a second, hidden layer, an implied play behind the scenes for adults that watch it. That truly will make a difference between that being "just another toon" and something excellent.
Also: Learn your animation crew skill level. There are things that will be difficult to animate and draw. Full views of streets of lively, busy city may be impressive but take ten times as much time to animate as a sequence of talking heads. A sideways walk cycle is a standard pulled out of library, three minutes of animator work, but a rich, dynamic fighting scene will need to be drawn from scratch, a week of work for the whole team for a minute of animation. Talk to them and if that's a series, balance their workload by picking enough "rich animation" and enough "filler" so that making an episode takes roughly the same effort and cost.
You can afford some more flair for pilot, season finale, and 3-4 episodes, but prepare 2-3 episodes of "filler" with roughly half of the core crew missing each - voice actors want a little time off too, and this gives them a week or two of holidays.
In essence, if you write a scene, try imagining how it will look like, when done - and then imagine the workload needed to make it. Make it too hard and expect it will be cut by the storyboard artists at request of the director.
Comment: Actually this is going to be an adult cartoon. A comedy about alcoholism. It's also a parody of a children's cartoon.
Answer: Well, a lot of big animation studios (Disney, Pixar, etc) work with storyboards rather than scripts. Try that.
Comment: Actually, a storyboard comes next after the script. The script is less detailed; the storyboard is closer to a shooting script, except in graphical form (all the key frames), and is THE document upon which the actual animation is based - but still it's made basing on a script; script is the phase when all the major edits and changes to the plot are made, such editing of the storyboard would be extremely cumbersome.
Comment: Can you say more about that?  Maybe describe the essential differences, since the OP is coming from a screenwriting perspective?  Thanks.

How to write for cartoons?


Comparison of Board Shape between Arduino MEGA and Arduino DUE

Question: I'm using KiCAD to design boards, and while it has a lot of Arduino templates, it doesn't have one for the Arduino Due. Now, I've did some searching, and it looks like the Arduino MEGA is very similar to the Arduino DUE, not the specifications, but the number of I/O pins and the overall shape of the board. I couldn't find a straight answer, so I decided to ask here: Can the template of the Arduino MEGA be used as a template for the Arduino DUE when designing custom 'shields' and such based on the number of I/O pins and overall layout?
Answer:
The Due has the same footprint as the Mega 2560.

source


Why CCl4 does not dissolve in water while SiCl4 does?

Question: Why $\ce{CCl4}$ does not dissolve in water while $\ce{SiCl4}$ does?
Does it have to do with d-orbitals? If yes, then how are they related?
Answer: $\ce{SiCl_4}$ does not quite dissolve in water; rather, it reacts with water. So does $\ce{CCl_4}$, albeit extremely slowly, so for most purposes one may safely assume it doesn't. The difference is mostly due to the atomic size of C and Si. Smaller С atom is completely blocked off by four bulky chlorine atoms, so the water molecule can't reach it. Larger Si is easier to attack.
Another factor is that silicon is a less electronegative element than carbon, so that the Si-Cl bonds will have more polar character.
Whether or not the d-orbitals of silicon contribute significantly to the transition state, is (IMHO) a question of secondary importance.
Comment: Similarly, while methane will burn in air, silane is pyrophoric, that is it will spontaneously burst in to flame (and then there is disilane!).
Answer: A silicon tetrachloride reacts with water, while carbon tetrachloride does not. This is due to the fact that the carbon does not have d-orbitals to accept lone pair of electron from water, while silicon has vacant d-orbitals to accept lone pair of electron from water.


$\mathcal S(\mathbb R^n) \hat \otimes_\pi \mathcal S(\mathbb R^m) \simeq \mathcal S(\mathbb R^{n+m})$?

Question: If $S(\mathbb R^n)$ is the Scwartz space of smooth rapidly decaying functions equipped with the topology generated by the family of semi-norms
$$\mathcal N_p (\varphi)= \sum_{|\alpha|, |\beta| \leq p} \sup_{x\in \mathbb R^n} |x^\alpha \partial^\beta \varphi (x) |\, ,$$
Is it true that $\mathcal S(\mathbb R^n) \hat \otimes_\pi \mathcal S(\mathbb R^m)\simeq \mathcal S(\mathbb R^{n+m})$, where $\mathcal S(\mathbb R^n) \hat \otimes_\pi \mathcal S(\mathbb R^m)$ is the completion of $\mathcal S(\mathbb R^n) \otimes \mathcal S(\mathbb R^m)$ for the topology $\pi$ defined by the family of semi-norms
$$\mathcal N_{p,q} (\varphi\otimes \psi)=\mathcal N_p(\varphi)\mathcal N_q (\psi)\, .$$
Comment: As in @PeterMichor's answer, the nuclearity means that _genuine_ tensor products exist, as opposed to the more typical situation that one can have one half or the other, but not both, of the characterizing properties of a tensor product. Some of the notes on my functional analysis page http://www.math.umn.edu/~garrett/m/fun/ talk about the impossibility of having a "genuine" tensor product of Hilbert spaces, and also about nuclear Frechet spaces occuring as suitable (proj) limits of Hilbert spaces.
Answer: Yes, if you understand the tensor product topology in the right way.
Since the spaces are nuclear, inductive and projective tensor products coincide.
The result is theorem 51.6 of Treves: Topological vector spaces, distributions, and kernels.
Added later:
Attention: The description of  seminorms on the tensor-product given in the question is not sufficient to specify a locally convex topology on the tensor product. There are many satisfying this description; all between the the projective one and the the inductive one and even more. See the source I have given, or many other books.
Comment: Grothendieck called the $\varepsilon$ tensor topology *injective*. The *inductive* topology is still finer and even for nuclear spaces it may be different from the injective one.
Comment: Thank you very much! The topology on the tensor product is then generated by the family of semi norms $\mathcal N_p$ defined above?


statistical test to compare a distribution of values versus a fixed number

Question: I have over a 100 observations of human decisions. If my hypothesis is correct, they should all be a fixed number, or around that fixed number (allowing for small errors around that fixed point). How can I test that the real distribution is different from a distribution that is essentially just that number, with small mistakes or deviations around it?
Comment: You cannot, because you haven't made any quantitative characterization of what you're trying to test.  If you could state what "small" means, then you might be able to make some progress.
Comment: Sounds like tests/estimates of location (errors "around that fixed point" & scale ("small" errors) are what you want; but to justify a specific test you need to explain what assumptions you can make about those errors.
Answer: Is there some reason you can't do a t-test comparing the mean value of your decisions to a fixed value. In other words, test the null hypothesis that $H_0: \mu = c$, where c is whatever your predicted value is? You would have to make sure you have sufficient power to reject that null hypothesis if it is false.
Alternatively -- perhaps better -- you could use a Bayesian test. This will allow you to determine a probability that the null hypothesis is actually true. See Rouder, J. N., Speckman, P. L., Sun, D., Morey, R. D., & Iverson, G. (2009). Bayesian t tests for accepting and rejecting the null hypothesis. Psychonomic Bulletin & Review, 16(2), 225–237. https://doi.org/10.3758/PBR.16.2.225
Comment: I have read the question several times without increased enlightenment. (+1)
Comment: Interesting alternative interpretation of a somewhat vague question: I took "fixed number" and "that number" to be the same population _means,_ and thought (I guess along with commenters) that the issue is about a possible difference in population variances. Maybe OP can clarify.
Comment: I guess it is somewhat vague. If your interpretation is correct - and rereading it now I suspect it probably is - then I think your first two solutions (which will, essentially, give the same result) are the way to go. If it were me, I'd prefer the first solution (confidence interval) because of the asymmetry of null hypothesis testing.
Answer: Just to get started with some specifics, here are three of approaches
that might be helpful. Somehow (as in @whuber's Comment), you need to develop a 'standard' of what you
mean by a 'small' variance. I hope your situation is similar enough to one
of the three scenarios below that this Answer will help you to do that.
CI for a single variance. First, you might make a confidence interval for the variance $\sigma^2$ or the standard deviation $\sigma$ of the population from which your $n = 100$ observations were sampled at random. I assume that the population is normal.
Consider $n = 100$ observations with $S^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i - \bar X)^2 = 9.943.$ For normal data, one has $Q=
\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(df=n-1).$ Thus one can find
values $L$ and $U$ such that $P\left(L < \frac{(n-1)S^2}{\sigma^2} < U\right) = 0.95,$
so that a 95% CI  for $\sigma^2$ is of the form
$\left(\frac{(n-1)S^2}{U}, \frac{(n-1)S^2}{L}\right).$
The figure below shows $L = 73.36$ and $U = 128.4.$

In our example, a 95% CI for $\sigma^2$ is $(7.664, 13.417)$ and, taking
square roots of endpoints, a 95% CI for $\sigma$ is $(2.77, 3.67).$ [Computations below are from R, but you could also use printed tables of chi-squared distributions.]
99*9.942/qchisq(c(.975,.025), 99)
[1]  7.664248 13.416624
sqrt(99*9.942/qchisq(c(.975,.025), 99))
[1] 2.768438 3.662871

With these results, you can ponder whether the variability
is greater than you anticipated, "allowing for small errors around a fixed point."
Test of a sample variance against a known population variance. Second, you may have done previous research on this kind of human decision
making and you may believe that 'small' means
a standard deviation around $\sigma = 2.5.$ Looking at the CI for the population standard deviation just above,
you should already be suspicious that the variability in my fake data is
not 'small' because $\sigma = 2.5$ is not 'small'. But if you want to do
a formal test of the hypothesis $H_0: \sigma^2 \le 6.25$ against the one-sided
alternative $H_1: \sigma^2 > 6.26.$
Assuming that $H_0$ is true the test statistic $Q = \frac{(n-1)S^2}{6.25}$
has $Q \sim \mathsf{Chisq(99)}$ and the critical value for a test at the 5% level
is $c = 125.0.$  Because $Q = 157.6 > 125.0,$ you would reject $H_0$ in favor
of the right-sided alternative at the 5% level of significance. (The P-value of
this test is about 0.00017.)
qchisq(.96, 99)
[1] 124.955
1 - pchisq(157.5, 99)
[1] 0.0001680777

F-test to compare two sample variances. Third, if you have a control group in which you suspect variability is 'small' and a treatment group in which you suspect variability may be larger, then you can do a standard F-test to see if variances differ. This kind of test comparing two sample variances has notoriously bad power, especially for small sample sizes. But if you have sample sizes around 100, you should be able to make
meaningful distinctions.
My fake data x above were sampled from a normal
population with $\sigma_X^2 = 9.$ Suppose you had a control group with $n = 100$ observations y from a population with $\sigma_Y^2 = 6.25.$ Sample variances are
$S_X^2 = 9.942,\, S_Y^2 = 6.364.$
var(x); var(y)
[1] 9.942942
[1] 6.36421

Then in R, the
F-test of $H_0: \sigma_X^2/\sigma_Y^2= 1$ against $H_1: \sigma_X^2/\sigma_Y^2 > 1,$ might look like this:
var.test(x, y, alte="greater")

        F test to compare two variances

data:  x and y
F = 1.5623, num df = 99, denom df = 99, p-value = 0.01373
alternative hypothesis: true ratio of variances is greater than 1
95 percent confidence interval:
 1.120698      Inf
sample estimates:
ratio of variances
          1.562321

The figure below show the density function of $\mathsf{F}(99,99)$ with critical
value $c = 1.394$ for a one-sided test at the 5% level (dotted red line). The area under the density curve to the right of the observed value $F = S_X^2/S_Y^2 = 1.562$ is $0.0137.$

Notice that comparison of a sample variance with a known population variance (constant) is different than comparing two sample variances to assess whether
the corresponding population variances differ.


If $p^k|n!$ then $p!^k|n!$

Question: Is there a direct proof of the result that if $p$ is a prime and if $p^k|n!$ then $p!^k|n!$? The proof that I know is rather circuitous (take $k$ to be given by de Polignac's formula and construct a subgroup of $S_n$ of order $p!^k$). The result seems too elegant to not admit a slick combinatorial proof.
Answer: Induction on $n$: If $n=mp+r$ with $0\le r<p$, there are  $\frac{n!}{p!^mr!m!}$ ways to partitoin $n$ objects into $m$ indistinguishable subsets of size $p$ each and a rest of size $r$.
Note that $p^{k-m}\mid m!$ (because the multiples of $p$ among $1,2,\ldots, n$ are precisely $p,2p,,\ldots, mp$) and by induction hypothesis $p!^{k-m}\mid m!$.
Hence $p!^n\mid p!^mm!\mid n!$.
Comment: If you pull out one power of $p$ from each of the $m$ numbers $p,2p,\ldots,mp$, then you get a factor of $p^m$ times the powers of $p$ dividing $1,2,\ldots,m$. This shows that the power of $p$ dividing $(mp+r)!$ is exactly $p^m$ times the power of $p$ dividing $m!$.
Comment: I have a little doubt.  What if k>m>p. Then the no. dividing $m!$ will be less than $p^{k-m}$
Comment: This looks correct (although I think the $p!^n$ should be $p!^k$ in the last line). It would be nice if there was a non-inductive proof though.
Comment: I don't follow the reasoning in "*because the multiples of $p$ among $1,2,\ldots,n$ are precisely $p,2p,,\ldots,mp$*", but there's an alternative way of arguing it: the $p$-adic valuation of $p!^m$ is clearly $m$, and the $p$-adic valuation of $r < p$ is clearly $0$.
Answer: I have not been able to show this in general.  But in special case
when $k<p$,  which will imply $pk \le n$,
So, the no. of of ways of dividing $n$ objects into $k+1$ different boxes such that $k$ boxes contain $p$ objects, while 1 box contain $n-pk$ object is
$\frac{n!}{(p!)^k\times(n-pk)!}$
From here we get that $(p!)^k|n!$
Comment: Unfortunately, the result only starts to get interesting once $k\geq p$ and extra factors of $p$ start showing up. You can extend this method up to $k<p^2$ if you partition the boxes into blocks of size $p$ and divide out by rearrangement within the blocks. Repeating this construction ends up being equivalent to the group-theoretic proof.
Answer: I was talking to some friends about this problem and we came up with a direct proof that (barely) avoids using de Polignac's formula.
For each factor of $p$ in $n!$, we wish to group that factor of $p$ with something divisible by $(p-1)!$. You group the first factor of $p$ dividing a number $x\in\{1,\ldots,n\}$ with the numbers $x-1,x-2,\ldots,x-(p-1)$. You group the second factor of $p$ dividing a number $px\in\{1,\ldots,n\}$ with the numbers $x-1,x-2,\ldots,x-(p-1)$. These are precisely what is leftover from the numbers $px-p,px-2p,\ldots,px-(p-1)p$ after the first step. Repeating this process, you group the $(k+1)$-st factor of $p$ dividing a number $p^kx\in\{1,\ldots,n\}$ with the numbers $x-1,x-2,\ldots,x-(p-1)$. These are precisely what is leftover from the numbers $p^kx-p^k,p^kx-2p^k,\ldots,p^kx-(p-1)p^k$ after $k$th step. However, the binomial coefficient
$$\binom{x-1}{p-1}=(x-1)(x-2)\ldots(x-(p-1))/(p-1)!$$
is an integer. This shows that each group is divisible by $p!$. Also, each factor of $p$ in $n!$ is contained in some group and each group contains exactly one factor of $p$. This shows that if $p^k|n!$ then $p!^k|n!$.
This is still not quite the slick combinatorial proof that I was looking for but it's pretty close.

If $p^k|n!$ then $p!^k|n!$


Can we somehow restrict automatic bumping of old questions to the front page?

Question: It has become a quite frequent situation for me that I start typing an answer to a question on the front page only to realize after the first few lines (if I'm attentive enough) that it is a bump from some year in the distant past asked by a person who is, most likely, no longer interested, if still staying here at all. Checking the timestamp on every post is, of course, not too hard, but it is not what I would call my "knee reflex" yet. So I wonder if this bumping as it is now really seems to make sense to other people, and if not whether we can make it happen less often or in some more reasonable fashion.
Comment: See https://meta.stackexchange.com/questions/213468/why-do-people-answer-old-questions.
Comment: Also related, on Meta SE: [Which sites impose limits on Community bumps, and what are those limits?](https://meta.stackexchange.com/q/350958)
Answer: At the top, do you have newest or active selected (or something else)? I have newest, and this doesn't happen to me. You can look at the 'filter' to get more of a sense of how this works.
[While playing with this setting, I left it on active by accident, and did almost exactly what you do. I was reading a question from 5 years ago, thinking it was new!]


Could medieval European merchants travel through the Islamic world?

Question: I've heard that the Islamic world was closed to European merchants in the Middle Ages.  Is this true?  If so, when did it close and when did it open again?
Comment: [The Polos](http://en.wikipedia.org/wiki/File:Travels_of_Marco_Polo.png) were all European merchants who travelled through the Islamic world. Presumably this means that any "reopening" occurred before the 13th century.
Comment: .. and the other way round muslim traders were active along the lower dnjepr during (and I think before) the time of the Kiever Rus. An interesting question would be how permeable western europe was to muslim traders in the middle ages.
Answer: It's difficult to give a proper answer, because during any century of Middle Ages there could be many reasons for closing trading routes for European merchants as outsiders in various parts of Islamic world (which is huge). And it didn't need to have anything in common with religion - it could be the level of civilization (early medieval Europe), an economic matter in the particular case, or just a result of war.
But let's try to clarify few things.
First of all, Islamic world, in fact, wasn't that closed to Europeans as we think. Jews and Christians could live at Muslim lands if they paid special tax. The same with merchants from Europe. What's more, some Christians could even serve to Muslim rulers at important positions, such as revenue collectors or administrators of various types.
It's nicely described in the book "Daily Life In The Medieval Islamic World" by James E. Lindsay. You can find it's full text in the link, look for pages 120-121.
Also there's a good source for that in "Europe and the Islamic World: A History" by John Tolan, Henry Laurens and Gilles Veinstein, at pages 80-81. Unfortunately in both cases I'm not able to copy-paste it.
I believe that a bit about the overall history of the trade between Christians and Muslims during Middle Ages could help you solve the problem. I'll quote Jihoon Ko and his article "Economic Impact the Islamic World Had on Christian Europe (11th ~14th century)".

The conquest of Muslims of various regions of Europe had actually
  ceased the major long-distance trade of the Christian Europeans
  overseas for a while and the commerce of Christian Europe had been
  limited to the local ones. But the regions which were under Islamic
  rules were included in the commercial sphere of the Muslims—who were
  the most active adventurers and entrepreneurs at the time—saw the rise
  of international and complicated commercial activities. The
  international trades in Christian regions were late to appear. The
  crusade created a large influx of Eastern goods and luxuries into the
  Christian Europe. This created a large interest among rich Europeans
  for the Eastern goods, and the Christian merchants saw this to be an
  opportunity for a dramatic profit. As the merchants began to actively
  engage in the trade with the Islamic world, the trade of the Christian
  states overseas began to flourish. The leading states of the trade
  with the East were Italian city-states, most notably Genoa and Venice.
  The extent of the international trades before the 11th century,
  however, was nothing compared with those of the Muslim world. The case
  of Al-Andalus indicates how well established was the Andalusi
  merchants' trades compared with the Europeans in the same era.

This way first, European (Christian) international trade wasn't well developed until 11th century, mainly due to the fall of ancient culture. It took European civilization a lot of time to get up after that.
But there were some smart merchants who made a fine business by initiating contacts between both worlds. They were, of course, Jews.

There were some global commercial activities going on in the Christian
  world, however, before the Christians began to trade over long
  distance, by the Jewish merchants. The Jews were also the first
  merchants in the Middle Ages to introduce goods from the East to the
  Christian Europeans. They supplied goods from the Islamic world as
  soon as the Carolingian era. The Jews could travel to Europe with
  relative ease. Some of them settled in Italy, but majority of them
  were those who based their commerce in the Muslim countries. The Jews
  from the Muslim world reached Western and Northern Europe through
  Spain. At first, they exclusively traded spices, pepper, manufactured
  goods, and other precious stuffs from Egypt, Syria, and Maghrib. They
  provided the goods to the churches which used the goods to make
  incense for religious activities, and to the selected nobility.

When European trade market became more international, the crusades started. One could imagine that it made the trade more difficult, from the most obvious reason which is "you don't go trading with your enemy if you don't want to be killed".
Somehow, in fact it helped to open both markets on each other.

The crusades, in turn, helped to change the economic activity of
  Christian Europe. The crusades "enabled Western Europe to monopolize
  the whole trade from the Bosphorus and Syria to the Straits of
  Gibraltar."[10] Even if they failed to take the holy place from the
  hands of Muslims, they politically weakened the powers of the Islamic
  states and Byzantine Empire. The Byzantine Empire's maritime power was
  weakened seriously and their merchants now could not compete with the
  Italian rivals, which now controlled the Byzantine Empire's export and
  import trades. The crusades also caused the large influx of Eastern
  luxuries into Christian Europe, thus stimulating the interest of
  people for the trade with the Islamic world. So, it is no exaggeration
  to say that the crusades were crucial factor in the revival of
  extensive economic activities of Medieval Christian Europe.

As it's described later, the main European ports that were trading with Islamic World were Genoa and Venice. And even if there were some difficulties, a business is still a business.

The trades with the Islamic world at first time were limited because
  of religious reason, and because of technical reasons, but gradually
  the Italian merchants began to seek profits and the economic
  incentives were large enough to overcome religious obstacles for
  greater incomes. As a result, the Italian city-states rose as major
  maritime commercial centers from 11th century, and the scope of their
  trades even exceeded that of the Muslim ports.

I'll quote only two paragraphs more and I'll leave the rest to you, if you're more interested in the trade with Islamic World in overall, not only from the point of it's connection to religious problematic.

Christians and Muslims have been enemies by principles. Moreover, as
  the Middle Age was the time when the religion affected the lives of
  Christians most significantly, the Muslims were considered evil by
  Christians. This aversion of Christians toward Muslims, however, could
  not stop the economic interest of the Christian merchants, and the new
  pattern of global trade appeared.
The Muslim world indeed acted as an important trade partner of Europe
  in the Late Middle Age. They facilitated the European trade and thus
  strengthened its economy. Also, they provided the goods that were not
  available in Europe, and this fact made Europe a favorite place for
  European merchants. The religious principle could not separate the
  two. The trades between Europe and the Muslim World sometimes stopped
  because of political situations, and were greatly reduced when the
  Turkish soldiers occupied Constantinople and destroyed Byzantine
  Empire. However, it is evident that the two had long-time economic
  connections, and such connectivity helped the European economy a lot.
Answer: As a terminus post quem for such closure (if it ever existed at all) one might suggest the early 9th century, as evinced by the following classic story:

In 828, relics believed to be the body of Saint Mark were stolen from
  Alexandria (at the time controlled by the Abbasid Caliphate) by two
  Venetian merchants with the help of two Greek monks and taken to
  Venice. A mosaic in St Mark's Basilica depicts sailors covering
  the relics with a layer of pork and cabbage leaves. Since Muslims are
  not permitted to touch pork, this was done to prevent the guards from
  inspecting the ship's cargo too closely.


Is Testing a necessary part of Agile methodology?

Question: I have been on numerous teams that try to practice Agile methodologies and often these teams are test centric. Is testing a necessary part of practicing the Agile methodology or is it just a XP practice that has been latched on over the years?
Comment: Testing is a necessary part of any quality software development.
Comment: I disagree with the duplicate. Testing is broader than TDD and the broader question has different answers.
Comment: possible duplicate of [Can you be Agile without doing TDD (test driven development)?](http://programmers.stackexchange.com/questions/21870/can-you-be-agile-without-doing-tdd-test-driven-development) - if you disagree let me know...
Comment: I agree with @BartvanIngenSchenau. Not only is testing a far broader activity than just doing TDD, but TDD and *unit testing* are not the same. I'm surprised so many people confuse the last two.
Comment: It may have been folded into the concept of "Definition of Done", which in Agile/Scrum means it is context-dependent. When Agile is applied to marketing, sales etc., they don't have similar concepts as software testing, but still have a "definition of done". For software, "definition of done" should consider the quality (both visible and internal e.g. code quality) of the end result, as well as the level of testing that has been performed.
Answer: Testing is absolutely essential to agile, primarily because agile is based around incremental improvements: the difficulty is that it can sometimes be hard to see how the current changes will effect your old code. The best way to be confident  that you haven't broken something is to test it, and to know HOW to test it. That way you find the bug immediately, not down the road when you have forgotten exactly what you did when you were writing the code that broke some old feature.
The reason this is different from more traditional, top-down design type programming is that in that environment its a) very difficult to test until you have the finished product b) in theory you are considering all the design criteria at the same time, and so you are less likely to make a design decision that breaks previous design decisions.
Comment: It is also important that future incremental changes do not break previous functionality, and unit tests are an easy way to check for that.
Comment: I'd say testing is absolutely essential for *software development*.
Comment: @Snowman Testing != unit testing. Also, unit testing != TDD (just in case...).
Comment: @AndresF. true, I normally think of _unit testing_ being an integral part of Agile for the reasons I outlined above. Reading failure.
Answer: If you do not have tests, how do you know your code works?
Edit: the assertion that tests cannot prove that the code works fails to define one crucial term, namely works. What does it mean for a program to work? If you keep this term vague, then there is no way at all to prove or be sure that any program works. Ever.
On the other hand, you can define works as "behaves according to a specification". Now you can not only use tests to show that code works, but the tests themselves can serve as an executable specification of your code's behavior. In other words, a well written test suite defines what works means.
This way of thinking also forces you to re-examine the meaning of a bug. If your code passes all the tests, then there are no bugs in the code. If, despite that, the system does not behave as it should, then its behavior is not correctly specified. I. e. the bug is in the spec, defined by tests.
This approach to software development decouples the functional specification of a system from its implementation, which, according to every software engineering book in the world, is a very good thing. At the same time, this approach ensures that your implementation always corresponds to the functional spec.
Comment: when customers stop filing bug reports? :-)
Comment: You could prove it correct. Clean Room Software Engineering is very similar to TDD, actually, but it *only* has automatically generated statistical tests that are generated from the specs and proofs.
Comment: -1 - "Testing shows the presence, not the absence of bugs."
Comment: @Telastyn, Not having tests shows the presence of bugs with near certainty. On the other hand, a well-written set of tests provides an executable specification of what your code does.
Comment: I don't disagree, but no amount of testing _proves_ that your code works.
Comment: @Telastyn, I've updated the answer to address your comment.
Comment: Testing always deals only with a limited number of test cases, which makes it insufficient to act as a full specification for the code. I would consider "the input string is converted to lowercase and returned" to be a reasonable specification, but not something like "`"Foo"` should return `"foo"`, `"FOO"` should return `"foo"` and `"foo"` should return `"foo"`".
Answer: You are probably talking about automated testing, unit tests, integration tests etc. These are more important to agile than manual testing (with testers and such) because they are too slow, thus its not possible to test every small change that you make. Since agile is about fast small iterations, having tests that verify correctness in seconds or minutes, rather than hours or days, is very useful.
Answer: No, it's not "necessary"
The principles of Agile say nothing directly about testing.
But It's Highly Advisable
Given Agile's commitment to a sustainable process, continuous/incremental delivery, and software quality, automated testing is the best solution currently available for most projects
Exceptions (as noted by Jörg W Mittag) include provably-correct development tools, meta-programming systems that generate code, et al. But these kinds of systems are rare.
Answer: Both Agile and XP try to avoid Big Design Up Front. In BDUF, requirements are gathered, a formal specification is created, then coding is done, then testing is done. This makes sense for well-defined, mission- and life-critical systems like medical equipment, space probes, etc.
Agile avoids this flow because it doesn't work well for problems which aren't well-defined, for example "whatever changes the client asks for this week". We still need a formal specification, so we know what to do and when we're done, but rather than some kind of written document we use code in the form of automated tests.
Automated unit tests are quick to write, quick to run and very modular/decoupled. This makes them a fast way to formally specify and check the requirements.


How keep uninstall files inside uninstaller?

Question: Has a any way to make the uninstall files (JPEG, DLL, PNG...) stay inside of the unins000.exe? If so, please show a code.
Answer: There is not any way to do this.  The Inno uninstall engine cannot contain embedded files.
But on the other hand there's no particular need for it to do so either -- you can just install any files that it requires (into a subfolder of the app folder, if you want to keep things tidy).
The only real reason that the setup files are embedded into the installer is that it makes downloads easier.  The uninstaller doesn't have that excuse.
Comment: Actually with some creativity, there's a way. Se [my answer](http://stackoverflow.com/a/38336841/850848).
Comment: @MartinPrikryl True, but the effort to do so typically outweighs the benefit.  It's simpler to just install the files.  Still, points for creativity. :)
Answer: Inno Setup has no native support for embedding files into the uninstaller.
But with some creativity, you can embed the files into the code (in a form of some constant).
Unfortunately the Unicode Inno Setup is quite limited when dealing with binary data as it tends to try to convert everything to UTF-8. But after numerous tries I've ended up with some working code (with use of a PowerShell code invoked from Inno Setup preprocessor).

Add this code somewhere to the front of your [Code] section:
function CryptStringToBinary(
  sz: string; cch: LongWord; flags: LongWord; binary: string; var size: LongWord;
  skip: LongWord; flagsused: LongWord): Integer;
  external 'firstname.lastname@example.com stdcall';

const
  CRYPT_STRING_HEX = $04;

procedure WriteBinaryStringToStream(S: string; Stream: TStream);
var
  Buffer: string;
  Size: LongWord;
  L: Integer;
begin
  L := Length(S);
  SetLength(Buffer, (L div 4) + 1);
  Size := L div 2;
  if (CryptStringToBinary(S, L, CRYPT_STRING_HEX, Buffer, Size, 0, 0) = 0) or
     (Size <> L div 2) then
  begin
    RaiseException('Error decoding binary string');
  end;

  Stream.WriteBuffer(Buffer, Size);
end;

function StreamFromBinaryString(S: string): TStream;
begin
  Result := TStringStream.Create('');
  WriteBinaryStringToStream(S, Result);
  Result.Position := 0;
end;

procedure LoadBitmapFromBinaryString(Bitmap: TBitmap; S: string);
var
  Stream: TStream;
begin
  Stream := StreamFromBinaryString(S);
  try
    Bitmap.LoadFromStream(Stream);
  finally
    Stream.Free;
  end;
end;

procedure SaveBinaryStringToFile(FileName: string; S: string);
var
  Stream: TStream;
begin
  Stream := TFileStream.Create(FileName, fmCreate);
  try
    WriteBinaryStringToStream(S, Stream);
  finally
    Stream.Free;
  end;
end;

#define FileToBinaryString(str FileName) \
  Local[4] = ExtractFileName(FileName), \
  Local[0] = AddBackslash(GetEnv("TEMP")) + Local[4] + ".pas", \
  Local[1] = \
    "-ExecutionPolicy Bypass -Command """ + \
    "Write-Host 'Generating code for " + Local[4] + "'; " + \
    "$bytes = [System.IO.File]::ReadAllBytes('" + FileName + "'); " + \
    "$s = '''' + (($bytes | foreach { $_.ToString('X2') }) -join '') + ''''; " + \
    "Set-Content -Path '" + Local[0] + "' -Value $s;" + \
    """", \
  Exec("powershell.exe", Local[1], SourcePath, , SW_HIDE), \
  Local[2] = FileOpen(Local[0]), \
  Local[3] = FileRead(Local[2]), \
  FileClose(Local[2]), \
  DeleteFileNow(Local[0]), \
  Local[3]

And then you can use the FileToBinaryString preprocessor macro to convert a file on compile-time (or more precisely, when pre-processing) to a hex string like:
'4D5A50000200000004000F00FFFF0000B8000....'

On runtime, you use the hex string with some of the functions WriteBinaryStringToStream, StreamFromBinaryString, LoadBitmapFromBinaryString or SaveBinaryStringToFile.

So for example to load a bitmap to TBitmapImage, use:
LoadBitmapFromBinaryString(
  BitmapImage.Bitmap, {#FileToBinaryString("C:\path\WizModernSmallImage.bmp")});

On compile-time, this gets converted to a code like:
LoadBitmapFromBinaryString(
  BitmapImage.Bitmap, '4D5A50000200000004000F00FFFF0000B8000....');

To extract a DLL, use:
SaveBinaryStringToFile(
  ExpandConstant('{tmp}\InnoCallback.dll'),
  {#FileToBinaryString("InnoCallback.dll")});

To use a DLL extracted on runtime, you have to use the delayload flag:
function WrapTimerProc(callback:TTimerProc; paramcount:integer):longword;
  external 'wrapcallback@{tmp}\innocallback.dll stdcall delayload';

The pre-processor/Pascal compiler has a limit of about 100M characters for a string. Though you will actually first hit a [compile-time] memory limit of the PowerShell script for files larger than about 20-30 MB.  Though even for smaller sizes (larger than few MBs), the compile-time time performance of the PowerShell script is bad. The script can be optimized significantly though.
Due to the hex encoding, the size of the installer increases twice as much. This could be improved by using some more efficient encoding, like Base64 (CRYPT_STRING_BASE64). The code section is not even compressed too, comparing to files included with the [Files] section (not a problem for images as these are compressed already, but makes a difference with DLLs for example).

This solution was initially developed for the question Inno Setup: Reading a file from installer during uninstallation.

The code requires the Unicode version of Inno Setup. With latest Inno Setup 6, it's the only version available anyway. Even if you use Inno Setup 5, you should not use the Ansi version anyway, in the 21st century. Though ironically, implementing this in the Ansi version would be way easier. See my answer to Writing binary file in Inno Setup for a use of the CryptStringToBinary that's compatible with both Ansi and Unicode version of Inno Setup. Though in the Ansi version you can actually do with a binary string, instead of a hex string.


What are my public transport options in Tbilisi and further afield in Georgia

Question: What are the different transport options for getting around the city of Tbilisi, and travelling to the Svaneti region.
Comment: There's a metro, taxis, and marshrutkas in the city, there's marshrutkas and a couple of train lines only for going long distance. But you need to ask less broad questions, we're not going to go into the details of every kind of transport for the whole country.
Comment: Any particular locations you would like to visit?  Without it your question is too broad.  In addition please review existing question on Tbilisi as well.
Answer: That's a lot of questions!
Should I hire a car or are there tours available to tourist areas?
You probably don't need a car, the public transport in Tbilisi is pretty efficient, and when it's not it's still interesting :-)
Are there regular bus services?
Yes, there is a metropolitan bus service. Most locals use the minibuses that you have to wave down. If your Georgian is rusty, this isn't a great option, but a surprising number of drivers do speak English. You have to request a stop, and request it before you get on, otherwise you may get an unexpected tour of the city before you get off!
and what is the rail network like?
Clean and efficient. There's tons of kids acting like hoodlums, but in my experience they're harmless. Watch your stuff though!
კეთილ მგზავრობას გისურვებთ!


Can I increase my credit limit by transferring money to my Credit Card account?

Question: I have a credit card with a credit limit of $1000, but I would like to make a purchase of $1500. If I transfer $500 into my credit card account, would that increase my credit limit to $1500?
Comment: "Credit" (the amount of debt you are allowed to take on) will still be limited at $1000. But if you add *extra* cash, then that may increase your purchase power with your card, because you are starting out at -$500. (`$1000 - (-$500) = $1500`).
Comment: @MrChrister What I do know is that if you overpay, you will simply achieve a negative balance.  You can, from that point, purchase `limit + X` before reaching the limit, but I *don't* know that you can do it all once (which is why I didn't post an answer, but a comment).
Comment: @Renesis - got any documentation?  You seem sure, but I don't see banks being so logical.  Just an opinion and I am curious now.  From the few anecdotes I read online (at Metafilter) it doesn't work that way.
Comment: For reference:  http://ask.metafilter.com/143018/Can-I-prepay-my-credit-card-and-use-it-like-a-debit-card-to-help-me-spend-more-responsibly#2047093
Comment: @MrChrister I'm sorry if I didn't make it clear, but that's why I said "may".  My main point is that *regardless*, it doesn't increase your *limit* -- instead, it increases the difference between the balance (negative) and your limit (constant).
Answer: When in doubt, call (the card issuer) and ask. Ask if you overpay your current bill if the overpayment becomes available credit and tell them why you are asking. It can go either way.

For what it's worth, I made a payment that was well over what was due, and this is a snapshot of my balance. My credit line is $25,000, but since I overpaid, I can now charge $26,600 if I wish. This doesn't "answer" the question, as each card issuer has his own rules, and OP needs to ask. But, it proves that what he proposes is possible.
As the comment below notes, trying to charge $26,600 may (most likely) will not go through, a fraud alert would block it. But. I doubt if one does this with a $1000 credit limit card, pushing the available credit to $2000, that it would trip, or trip as fast.
To close this one - personal anecdotes only can suggest what 'might' be. To actually do this, call the card issuer, they'll be happy to answer the question.
Comment: +1 for practical example. Being pedantic, this doesn't _guarantee_ that all the "extended credit" can be spent in one transaction ("_available for purchase**s**_"), but I can't see any real reason why they _wouldn't_ let you do so (although you may have to confirm the transaction as one of this size would probably trigger their anti-fraud "spending pattern" alarms).
Answer: I live in Canada and have a BMO mastercard. I called them and asked them and their answer was "of course!". I have put thousands of dollars on my mastercard from my bank account to pay for rare, large purchases. The money I put on appears differently on my online mastercard account though.
EX:
account balance: $6,000.00 CR
available credit: $3,000.00
This confused me at first, but when I called and asked them, they said my available credit doesn't  change (ie: how much BMO lends me), but when I add my available credit + what I've put on my card (my account balance, which is CR (meaning my balance has a surplus of money)), then my spending limit is $9,000.00
So, I don't increase my "credit" limit, but I do increase my spending limit. It just comes down to terminology. I assume it is like this for other credit cards, but I would recommend calling and asking, just to be on the safe side.
Heath
Answer: No. You will need to call your credit card issuer and ask for a credit limit increase.  If you plan to pay off your card then I see no problem in this.
You could also ask the seller if they will put $1000 on your card, and $500 in cash.  (Ask for a cash discount too.)
Found some anecdotes!
http://ask.metafilter.com/143018/Can-I-prepay-my-credit-card-and-use-it-like-a-debit-card-to-help-me-spend-more-responsibly#2047093
http://ficoforums.myfico.com/t5/Credit-Cards/Can-I-prepay-to-increase-limit/m-p/670376#M214222
Comment: I've had cases where for one reason or another, I've had a credit on my account. Why do you think this wouldn't work?
Comment: Well, having a credit on your account does not increase your credit *limit*.  (At least I have never read as much.  Every bank is different).  I guess I do assume that electronic confirmation of funds goes against the credit limit and only subtracts, never subtracting a negative.
Comment: I understand. I was thinking that most cards will let you go over the limit but charge you a penalty (as compared to being rejected at point of sale). In that case, the charge goes through and the credit keeps the balance from being over his $1000 limit. If the limit is checked for each charge, I agree it might not clear.
Comment: +1 for the link, that seems to answer the question pretty well (some work that way, some won't, but there are concrete stories both ways).
Comment: This definitely depends on the card. (My issuer even encourages positive balances on card accounts by paying interest on them!)
Answer: My card keeps a separate 'cash advance' limit, that's lower than the regular rate. I believe balance transfers also trigger that limit and (much higher) interest rate.
Answer:
I have an upcoming purchase that will be just over $2000, and recalled this question. I tried to pay $2000 more than my balance on my card, and this is the note I got back via email. Now, I did try to pay it directly to the card issuing bank. I don't (yet) know what they'd do if I made a payment via ACH through my own bank.
So for this type of situation, I'd suggest calling customer service and ask how they handle these. (Yes, the Joe was edited in. The rest is factual) I know this contradicts my older answer, but it serves as a counter-example. Not all banks handle the same.
Comment: The part about "plus 7.5%" is interesting.
Comment: It does. If one runs $1000/mo, they'll accept an extra $75. Not helpful to the OP, in this case.
Answer: I had a macys card which only had $75.00 credit limit... I accidently paid over the limit so the card had $100.00 in it. I left it that way for a month.. My credit limit turned into 100.. So I do think its possible to increase your credit limit that way.. I've tried many times requesting for a credit limit increase.. I was denied many times.. The only thing I have is to add money but the tricky thing is that you'll have to add money and spend the whole amount and then pay it off at once for the credit limit to stick. But since you have great credit assuming because your limit is 1000, you should request for an increase of your credit limit.
Comment: Your limit may not actually have been increased.  You had an effective limit of $25 because you had a credit on your account.  Once you use that $25 in purchases I would expect your limit to return to $75
Comment: interesting situation. I'd caution that a store card may have different rules than a bank issued visa/MasterCard.
Comment: @Chad I'm almost certain that you are correct. About 10 years ago, I overpaid my VISA card by $200. I called customer service. They sent me a check for the over paid amount of $200, very promptly. My credit limit temporarily was X + 200. It returned to X, as of the next billing cycle though. OP should directly request a credit limit increase. If granted, it may not be for an increase of $500, but whatever the credit issuer decides is appropriate given usual criteria.
Answer: Overpaying a credit card to create a large positive balance may cause a bank to red flag your account.  This is a technique used in fraud for check kiting (write or deposit a fraudulent check to overpay your credit card, then demand a refund on the balance overpaid before the check bounces.)
Every bank is different: Talk to your bank first before you try this.  For a small balance ($5-20) overpayment isn't a big deal, it happens regularly... just spend down the balance.  Past that, you might be harming your credit record or risk closing the account if the bank disagrees with how you are using an overpaid balance for a larger purchase, or you risk unwanted law enforcement attention aimed at your finances.
If you are trying to do this to build your creditworthiness, a secured card is better for this purpose.  Disagree with your credit limit? Deposit more in the holding account.


How to provide custom software for easy installation on pi?

Question: I've seen several useful scripts programs that someone created for Raspberry Pi. Installation, however often requires to install dependent packages, manually create files and directories, copying and adjusting files etc. What's the easiest way to provide such software in a way it can simply be installed, updated, and removed with one command? I suppose one could create a Debian package but I doubt that everyone contributing to Raspberry Pi software is willing to learn details of Debian package management. Maybe there is some easier way, a simple tutorial, or sample software for Raspberry Pi that one could reuse?
Answer: Since there are only a select few operating systems used on the Raspberry Pi I would suggest detecting which distribution is being used and handle software installation using that distros default package manager.
This way you don't have to overcome any cross-distribution differences and software can be packaged as simple tar files.


Data Recovery using EasyUS Data Recovery

Question: I have accidentally deleted a 60gb folder from my 1TB HDD. The hdd is not corrupted so i decided to use the deep scan method of easyus data recovery which i have used in the past and successfully found my files. However, it is taking so much time to finish. When i deleted my files i turned off the computer, removed the hdd and put it on another pc using a USB cable. My questions are: Is their a chance of burning my hdd since it will be scanning for over 10 hours? If i stop the scan will i reduce my chances of finding the data (since it is only scanning not writing)? What do you suggest me since i only deleted them and the hdd is just fine?
Comment: Wait until the scan finishes.
Answer: There is no problem with the disk running for 10 hours or more, since it is not a problematic disk.
If you stop it, it will not reduce your chance of finding your data, unless you write something on the disk afterwards.
I don't know why you wish to stop it but if you do, when you scan the disk again you should first try without the deep scan feature, since your disk has not been formatted or heavily used after your accidental deletion. This will save you a lot of time.
Comment: It is producing a bit high temperature but if you say that it is okay i will leave it as is. To be honest i didn't look the files carefully in the quick scan so i am thinking of stopping it and go again to the quick scan.. i will see how it goes...


What is the difference between baud rate and bit rate?

Question: I am really having hard time understanding the difference. Some say they are same, while others say there is a slight difference. What's the difference, exactly? I would like it if you explained with some analogy.
Answer: Bits per second is straightforward. It is exactly what it sounds like. If I have 1000 bits and am sending them at 1000 bps, it will take exactly one second to transmit them.
Baud is symbols per second. If these symbols — the indivisible elements of your data encoding — are not bits, the baud rate will be lower than the bit rate by the factor of bits per symbol. That is, if there are 4 bits per symbol, the baud rate will be ¼ that of the bit rate.
This confusion arose because the early analog telephone modems weren't very complicated, so bps was equal to baud. That is, each symbol encoded one bit. Later, to make modems faster, communications engineers invented increasingly clever ways to send more bits per symbol.¹
Analogy
System 1, bits: Imagine a communication system with a telescope on the near side of a valley and a guy on the far side holding up one hand or the other. Call his left hand "0" and his right hand "1," and you have a system for communicating one binary digit — one bit — at a time.
System 2, baud: Now imagine that the guy on the far side of the valley is holding up playing cards instead of his bare hands. He is using a subset of the cards, ace through 8 in each suit, for a total of 32 cards. Each card — each symbol — encodes 5 bits: 00000 through 11111 in binary.²
Analysis
The System 2 guy can convey 5 bits of information per card in the same time it takes the System 1 guy to convey one bit by revealing one of his bare hands.
You see how the analogy seems to break down: finding a particular card in a deck and showing it takes longer than simply deciding to show your left or right hand. But, that just provides an opportunity to extend the analogy profitably.
A communications system with many bits per symbol faces a similar difficulty, because the encoding schemes required to send multiple bits per symbol are much more complicated than those that send only one bit at a time. To extend the analogy, then, the guy showing playing cards could have several people behind him sharing the work of finding the next card in the deck, handing him cards as fast as he can show them. The helpers are analogous to the more powerful processors required to produce the many-bits-per-baud encoding schemes.
That is to say, by using more processing power, System 2 can send data 5 times faster than the more primitive System 1.
Historical Vignette
What shall we do with our 5-bit code? It seems natural to an English speaker to use 26 of the 32 available code points for the English alphabet. We can use the remaining 6 code points for a space character and a small set of control codes and symbols.
Or, we could just use Baudot code, a 5-bit code invented by Émile Baudot, after whom the unit "baud" was coined.³

Footnotes and Digressions:

For example, the V.34 standard defined a 3,429 baud mode at 8.4 bits per symbol to achieve 28.8 kbit/sec throughput.
That standard only talks about the POTS side of the modem. The RS-232 side remains a 1 bit per symbol system, so you could also correctly call it a 28.8k baud modem. Confusing, but technically correct.
I've purposely kept things simple here.
One thing you might think about is whether the absence of a playing card conveys information. If it does, that implies the existence of some clock or latch signal, so that you can tell the information-carrying absence of a card from the gap between the display of two cards.
Also, what do you do with the cards left over in a poker deck, 9 through King, and the Jokers? One idea would be to use them as special flags to carry metadata. For example, you'll need a way to indicate a short trailing block. If you need to send 128 bits of information, you're going to need to show 26 cards. The first 25 cards convey 5×25=125 bits, with the 26th card conveying the trailing 3 bits. You need some way to signal that the last two bits in the symbol should be disregarded.
This is why the early analog telephone modems were specified in terms of baud instead of bps: communications engineers had been using that terminology since the telegraph days. They weren't trying to confuse bps and baud; it was simply a fact, in their minds, that these modems were transmitting one bit per symbol.
Answer: I don't understand why everyone is making this complicated (answers).
I'll just leave this here.

So above would be:

Signal Unit: 4 bits
Baud Rate [Signal Units per second]: 1000 Bd (baud)
Bit Rate [Baud Rate * Signal Unit]: 4000 bps (bits per second)

Bit rate and Baud rate, these two terms are often used in data
  communication. Bit rate is simply the number of bits (i.e., 0's and
  1's) transmitted per unit time. While Baud rate is the number of
  signal units transmitted per unit time that is needed to represent
  those bits.
Comment: Baud rate is the number of signal units sent per second. And bit rate is number of bits sent per second. Right?
Comment: @RishabhGupta I understand why you are uncertain. Updated :). But yes you are correct! :)
Comment: To complement this answer, it's recommended to read the articule:  http://www.tldp.org/HOWTO/Modem-HOWTO-23.html
Answer: Bit rate:-
      Bit rate is nothing but number of bits transmitted per second.For example if Bit rate is 1000bps then 1000 bits are i.e. 0s or 1s transmitted per second.
Baud rate:-
      It means number of time signal changes its state.When the signal is binary then baud rate and bit rate are same.
Comment: I like your concise answer. And when the signal is not binary, the baud rate is lower than bit rate. That doesn't mean the transmission speed is slowed down. Actually it means more bits can be transferred with the same time cost.
Comment: When you say the signal is binary does that mean the signal is a digital unlike that of an analog signal?
Answer: First something I think necessary to know:
It is symbol that is transferred on a physical channel. Not bit. Symbol is the physical signals that is transferred over the physical medium to convey the data bits. A symbol can be one of several voltage, frequency, or phase changes. Symbol is decided by the physical nature of the medium. While bit is a logical concept.
If you want to transfer data bits, you must do it by sending symbols over the medium. Baud rate describes how fast symbols change over a medium. I.e. it describes the rate of physical state changes over the medium.
If we use only 2 symbols to transfer binary data, which means one symbol for 0 and another symbol for 1, that will lead to baud rate = bit rate. And this is how it works in the old days.
If we are lucky enough to find a way to encode more bits into a symbol, we can achieve higher bit rate with the same baud rate. And this is when the baud rate < bit rate. This doesn't mean the transfer speed is slowed down. It actually means the transfer efficiency/speed is increased.
And the communicating parties have to agree on how bits are represented by each physical symbol. This is where the modulation protocols come in.
But the ability of sending multiple bits per symbol doesn't come free. The transmitter and receiver will be complex depending on the modulation methods. And more processing power is required.
Finally, I'd like to make an analogy:
Suppose I stand on the roof of my house and you stand on your roof. There's a rope between you and me. I want to send some apples to you through a basket down the rope.
The basket is the symbol. The apple is the data bits.
If the basket is small (a physical limitation of the symbol), I may only send one apple per basket. This is when baud/basket rate = bit/apple rate.
If the basket is big, I can send more apples per basket. This is when baud rate < bit rate. I can send all the apples with less baskets. But it takes me more effort (processing power) to put more apples into the basket than put just one apple. If the basket rate remains the same, the more apples I put in one basket, the less time it takes.
Here are some related threads:
How can I be sure that a multi-bit-per-symbol encoding schema exists?
What is difference between the terms bit rate,baud rate and data rate?
Answer: According to What's The Difference Between Bit Rate And Baud Rate?:

Bit Rate
The speed of the data is expressed in bits per second (bits/s or bps).
  The data rate R is a function of the duration of the bit or bit time
  (TB) (Fig. 1, again):
R = 1/TB
Rate is also called channel capacity C. If the bit time is 10 ns, the
  data rate equals:
R = 1/10 x 10–9 = 100 million bits/s
This is usually expressed as 100 Mbits/s.
Baud Rate
The term "baud" originates from the French engineer Emile Baudot, who
  invented the 5-bit teletype code. Baud rate refers to the number of
  signal or symbol changes that occur per second. A symbol is one of
  several voltage, frequency, or phase changes.
NRZ binary has two symbols, one for each bit 0 or 1, that represent
  voltage levels. In this case, the baud or symbol rate is the same as
  the bit rate. However, it's possible to have more than two symbols per
  transmission interval, whereby each symbol represents multiple bits.
  With more than two symbols, data is transmitted using modulation
  techniques.
When the transmission medium can't handle the baseband data,
  modulation enters the picture. Of course, this is true of wireless.
  Baseband binary signals can't be transmitted directly; rather, the
  data is modulated on to a radio carrier for transmission. Some cable
  connections even use modulation to increase the data rate, which is
  referred to as "broadband transmission."
By using multiple symbols, multiple bits can be transmitted per
  symbol. For example, if the symbol rate is 4800 baud and each symbol
  represents two bits, that translates into an overall bit rate of 9600
  bits/s. Normally the number of symbols is some power of two. If N is
  the number of bits per symbol, then the number of required symbols is
  S = 2^N. Thus, the gross bit rate is:
R = baud rate x log2S = baud rate x 3.32 log10S
If the baud rate is 4800 and there are two bits per symbol, the number
  of symbols is 2^2 = 4. The bit rate is:
R = 4800 x 3.32 log(4) = 4800 x 2 = 9600 bits/s
If there's only one bit per symbol, as is the case with binary NRZ,
  the bit and baud rates remain the same.
Comment: but why In Manchester encoding, the bit rate is half of the baud rate?
Comment: Great question, you should post ... it ... oh
https://stackoverflow.com/questions/25834577/why-in-manchester-encoding-the-bit-rate-is-half-of-the-baud-rate
Answer: bit rate : no of bits(0 or 1 for binary signal) transmitted per second.
baud rate  : no of symbols per second.
A symbol consists of 'n' number of bits.
Baud rate = (bit rate)/n
So baud rate is always less than or equal to bit rate.It is equal when signal is binary.
Answer: Baud rate is mostly used in telecommunication and electronics, representing symbol per second or pulses per second, whereas bit rate is  simply bit per second. To be simple, the major difference is that symbol may contain more than 1 bit, say n bits, which makes baud rate n times smaller than bit rate.
Suppose a situation where we need to represent a serial-communication signal, we will use 8-bit as one symbol to represent the info. If the symbol rate is 4800 baud, then that translates into an overall bit rate of 38400 bits/s. This could also be true for wireless communication area where you will need multiple bits for purpose of modulation to achieve broadband transmission, instead of simple baseline transmission.
Hope this helps.
Comment: Regarding your second paragraph: Not really. Usually in serial communication by cable (eg. RS232) symbol represents one bit. Baud means "state changes of the line per second" Look at this: [Baud Rate](http://www.pccompci.com/Baud_Rate.html)
Answer: Bit per second is what is means - rate of data transmission of ones and zeros per second are used.This is called bit per second(bit/s. However, it should not be confused with bytes per second, abbreviated as bytes/s, Bps, or B/s.
Raw throughput values are normally given in bits per second, but many software applications report transfer rates in bytes per second.
So, the standard unit for bit throughput is the bit per second, which is commonly abbreviated bit/s, bps, or b/s.
Baud is a unit of measure of changes , or transitions , that occurs in a signal in each second.
For example if the signal changes from one value to a zero value(or vice versa) one hundred times per second, that is a rate of 100 baud.
The other one measures data(the throughput of channel), and the other ones measures transitions(called signalling rates).
For example if you look at modern modems they use advanced modulation techniques that encoded more than one bit of data into each transition.
Thanks.
Answer: The bit rate is a measure of the number of bits that are transmitted per unit of time.
The baud rate, which is also known as symbol rate, measures the number of symbols that are transmitted per unit of time.
A symbol typically consists of a fixed number of bits depending on what the symbol is defined as(for example 8bit or 9bit data). The baud rate is measured in symbols per second.
Take an example, where an ascii character 'R' is transmitted over a serial channel every one second.
The binary equivalent is 01010010.
So in this case, the baud rate is 1(one symbol transmitted per second) and the bit rate is 8 (eight bits are transmitted per second).
Comment: Your example is wrong because serial lines send one bit at a time, by definition. Many bits per symbol generally isn't a matter of software-level encodings, but of hardware-level modulation schemes. This is why the bit to baud ratio need not be divisible by 8, as with most other data sizes in modern computers.
Answer: This topic is confusing because there are 3 terms in use when people think there are just 2, namely:
"bit rate": units are bits per second
"baud": units are symbols per second
"Baud rate": units are bits per second
"Baud rate" is really a marketing term rather than an engineering term. "Baud rate" was used by modem manufactures in a similar way to megapixels is used for digital cameras. So the higher the "Baud rate" the better the modem was perceived to be.
The engineering unit "baud" is already a rate (symbols per second) which distinguishes it from the "Baud rate" term. However, you can see from the answers that people are confusing these 2 terms together such as baud/sec which is wrong.
From an engineering point of view, I recommend people use the term "bit rate" for "RS-232" and consign to history the term "Baud rate". Use the term "baud" for modulation schemes but avoid it for "RS-232".
In other words, "bit rate" and "Baud rate" are the same thing which means how many bits are transmitted along a wire in one second. Note that bits per second (bps) is the low-level line rate and not the information data rate because asynchronous "RS-232" has start and stop bits that frame the 8 data bits of information so bps includes all bits transmitted.
Answer: Bit rate is a measure of the number of data bits (that's 0's and 1's) transmitted in one second. A figure of 2400 bits per second means 2400 zeros or ones can be transmitted in one second, hence the abbreviation 'bps'.
Baud rate by definition means the number of times a signal in a communications channel changes state. For example, a 2400 baud rate means that the channel can change states up to 2400 times per second. When I say 'change state' I mean that it can change from 0 to 1 up to 2400 times per second. If you think about this, it's pretty much similar to the bit rate, which in the above example was 2400 bps.
Whether you can transmit 2400 zeros or ones in one second (bit rate), or change the state of a digital signal up to 2400 times per second (baud rate), it the same thing.
Answer: Serial Data Speed:
Data rate (bps) = 1/Tb
Tb is the time duration of 1 bit
If the bit duration is 2ms then data rate is 1/2x10-3 , which is about 500 bps.
Baud rate:
Baud rate is defined as no. of signalling elements(symbols) in a given unit of time (say 1 sec) or it means number of time signal changes its state.When the signal is binary then baud rate and bit rate are same.
Bit rate:- Bit rate is nothing but number of bits transmitted per second.For example if Bit rate is 1000 bps then 1000 bits are i.e. 0s or 1s transmitted per second.
There are few other terms similar to this (i.e serial speed, bit rate, baud rate, USB transfer rate),and i guess(?) the values that are printed on serial monitor relates to serial speed, baud rate and USB transfer rate. Bit rate isn't an another term, please correct me if i am wrong, because serial monitor prints some values at an interval of time and value is definitely a set of bits. so if one value is printed i can say no of bits present in the respective value which gets printed on serial monitor per unit time will be the bit rate.
Answer: Replies here are misleading. Saying true, but no one tell that for UART a symbol is not a single character but a single bit and this way the question was tagged.
For example 115200/8n1 is 11520 bytes per second as a single ASCII character is a 1 start bit plus 8 data bit plus 1 stop bit.
Answer: As correctly pointed out in the other replies, the bitrate is the amount of logical (or "abstract high level") information transferred in a given time, while baud rate is the number of symbols (more or less "signal changes") in the physical line in a given time.
While it is easy to understand that if a transmitted symbol carries 4 bits of information, then the bitrate is four time the baud rate, things get blurred in case of, for example, a RS-232 serial line.
The classic serial line works on bytes (well, "frames"), not bits. There is no way to transmit fewer that 8 bits (i.e. a byte), because the serial line defines a "frame" (I assume frames with 8 data bits, no parity, 1 start bit and 1 stop bit); and this is usually OK, because devices (computers) work probably on bytes, not single bits.
Given that, when a device sends a byte, i.e. 8 bits, the physical lines transmits 10 symbols, because to the original data composed of 8 bits, 2 more are added (start and stop bits, they are needed for synchronization). Some confusion can arise because the symbols transmitted on the physical line are also called "bits", but they are really symbols (MARK and SPACE, actually).
So in that classic RS-232 (in case of "8N1" frame) the bitrate is actually 8/10 of the baudrate. If we add the parity bit, the ratio lowers further and becomes 8/11.
The number of bits or symbols per second translates directly to the duration of them (bits or symbols). What does it mean for an engineer designing a system? It means that if he is designing a line filter to protect the line or reduce the noise, he should take the duration (or frequency) of the symbols transmitted on that line. For a baudrate of 1000 baud, he knows that the frequency of the signal is 1 KHz, and that a symbol has a duration of 1 ms. Fine. But if he has to calculate how much time is needed to transfer a file from a device to another, say a file of 1000 bytes, he must consider the bitrate, not the baudrate! Because the devices, at higher level, do not even see the start and stop bits, they are only a burden which slow down the communication (but they are useful for error checking).
To take it to the extreme case, imagine that a serial frame is just a bit long. For every bit transmitted by a device, three symbols would travel in the physical line. And if a parity were added, then four symbol would travel: the bitrate would be 1/4 of the baudrate. And if we add a second stop bit, the bitrate goes down to 1/5 of the baudrate!