Wikipedia:Reference desk/Archives/Mathematics/2009 December 5

= December 5 =

optimisation tric??
this code has come as an outcome of heat conduction modelling of spot modelling process.After solving the general heat conduction in one dimension equation by divided differences method we are arriving at eqation (1).If we take initial heat (due to atmospheric conditions) in each point as unity.this in coded in initialisation section. Now dq is sent to ode45 for solving in a prescribed time domain and with initial condition y0=0. now my problem is that i want to optimise this process i.e. minimize dq.i need the welding to be cooled fastly.so what parameter should i consider for optimisation and what method should i adopt. SCI-hunter (talk) —Preceding undated comment added 02:43, 5 December 2009 (UTC).


 * See this duplicate inquiry at WP:RD/Science. Takes your pick but not both. You are in a little maze of twisting passages.
 * ( hydnjo (talk) 03:45, 5 December 2009 (UTC)

Probability of object being owned by person
Imagine there are 20 distinct objects and 4 people. Each person has 5 of the objects. The probability of a given object being owned by a given person is quite simple and equals one quarter.

However, if it is then said that object X is definitely owned by person Y, or definitely not owned by person Y, i'm having trouble adapting the probability equation such that it gives the correct answer for the remaining unknowns. I can work it out simply enough by just thinking about it, I just can't get a single equation representing the probability for any given object being owned by any given player.

To at least demonstrate I've given it some thought before some hero strolls in and instantly sees what i'm doing wrong, here's my working so far:

So for each person Y, i know:


 * objects_revealed(Y) - the number of objects Y has said they definitely own
 * objects_excluded(Y) - the number of objects Y has said they definitely do not own
 * objects_revealed_total - the total number of objects for whom an owner is definitely known

And for each object X:


 * people_excluded(X) - the number of people who said they definitely do not own X
 * people_possible(X) - the count of people who could possibly own X = 4 - people_excluded(X)

The probability of X being owned by Y for an unknown object is:

p(X, Y) = (5 - objects_revealed(Y)) / (20 - objects_revealed_total - objects_excluded(Y))

However this doesn't work if someone says they definitely do NOT own an object, so I know I must need to include some of the other variables somehow. I just can't see how...

Thanks --Iae (talk) 16:07, 5 December 2009 (UTC)
 * I don't know if the general solution can be written with a simple formula using only the variables you have defined. What does work in general is to consider that there are $$\binom{20}{5}\binom{15}{5}\binom{10}{5} = 11,732,745,024$$ equally probable ways to divide the objects between the people. To find the probability that X is owned by Y given some information, you need to calculate the total number of ways in which the information is true and Y owns X, divided by the total number of ways in which the information is true.
 * Example: You are told that object 1 is not owned by person 1. There are $$\binom{19}{5}\binom{15}{5}\binom{10}{5} = 8,799,558,768$$ ways this can happen. Among these, in $$\binom{18}{5}\binom{14}{4}\binom{10}{5} = 2,161,295,136$$ cases person 2 owns object 2. Therefore, the probability that person 2 owns object 2 given that person 1 does not own object 1 is $$\frac{2,161,295,136}{8,799,558,768} = \frac{14}{57}$$. -- Meni Rosenfeld (talk) 21:05, 5 December 2009 (UTC)


 * Thanks for the response. It makes a lot of sense; I think my approach had been clouded by specifics. I'll give this another go today. --Iae (talk) 11:13, 6 December 2009 (UTC)

Speed calculation
A local newspaper reports that "...from the length of the skid marks, the police decided that the van was travelling at 58mph." How is that calculated please?--88.109.243.127 (talk) 16:46, 5 December 2009 (UTC)


 * It could be estimated theoretically, but with all probability they have a detailed data base, which I guess comes from the crash tests made by the builder of the car. However, I suggest to redirect the question to the RD/Science. --pma (talk) 17:41, 5 December 2009 (UTC)


 * It won't be calculated, it will be looked up from experimental results. If the report was about a van hitting a pedestrian or something else that wouldn't slow it down much on impact then it is simply a matter of looking up the stopping distances for that model of car under those conditions at various speeds. The length of the skid marks will be equal to the stopping distance (assuming the car started to skid immediately, which is a pretty good assumption, I think). If the van hit something heavy or stuck down (another car, a wall, etc.) then you would need to estimate the collision speed (based on the amount of damage, I guess) and factor that in, which would be a little more complicated (since brakes don't have constant effect - the braking power will reduce as they heat up, I believe. I don't know how significant an effect that would have during one emergency stop.). --Tango (talk) 18:02, 5 December 2009 (UTC)


 * I'm thinking this is a matter for forensics experts rather than mathematicians. I could come up with some other theories of how they compute the speed, but not being an expert in forensics it would just be theory. It might be better to consult the Vehicular accident reconstruction article than to ask here.--RDBury (talk) 02:33, 6 December 2009 (UTC)

I seem to recall that there is a relatively simple formula in conventional use for this purpose, but of course it's not in any sense a mathematical theorem. Michael Hardy (talk) 02:51, 6 December 2009 (UTC)


 * Thanks for this, I'm sorry to have been a trouble to you.--88.109.243.127 (talk) 07:48, 6 December 2009 (UTC)


 * A cop once told me they get a similar car and actually try it out, because it depends on the angle of the ground, what the pavement is made of, weather conditions, tread remaining on the tire, weight of the car, braking force, and a lot more stuff. It's only done for the most severe cases since it's so complicated. Ariel. (talk) 09:25, 7 December 2009 (UTC)