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Title: 10,000 - 5,000 = Post by rloginunix on Jul 11th, 2015, 12:02pm 10,000 - 5,000 = Initially an army A has A0 = 10,000 soldiers while an army B has B0 = 5,000 soldiers. Both armies are positioned directly across from each other in the open, they fire directly at each other with an optimum firing strategy with no place to hide for either army, the (constant) efficiency of the soldiers of both armies is the same and is a unity (for simplicity). The above parameters do not change over time during the battle, assume an infinite supply of ammunition, B will not run, etc. Basically both armies are equal in every respect except for the number of soldiers and they are committed to duking it out till the bitter end. 1). How long will it take for A to annihilate B? 2). How many soldiers of A will still be alive when 1) occurs? tniH: 1805 + 210 = 2015 [e] Disambiguation: assume that the attrition is continuous in time, efficiency of a soldier is the number of soldiers of the opposing army he/she kills in a unit of time (see remarks below). [/e] |
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Title: Re: 10,000 - 5,000 = Post by towr on Jul 12th, 2015, 12:08am [hide]In the first salvo A annihilates B, and B kills half of A Unless A aligned in two rows, and the guns are powerful enough to shoot through multiple people, then they're all dead. Or maybe both armies are aligned in single file, and everyone shoots the person in front, and then in the second salvo the last person from A and B kill each other.[/hide] |
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Title: Re: 10,000 - 5,000 = Post by rmsgrey on Jul 12th, 2015, 8:50am It depends a lot on your assumptions about how firing works - at one extreme, if every shot fired kills someone, and everyone fires simultaneously, then one salvo leaves only 5,000 of A standing. At the other extreme, where everyone has infinite ammo, people fire continuously and it takes a fixed amount of accumulated man-time to kill someone, the optimal strategy is to focus your entire army's fire on a single target until that target drops then switch (instantaneously!) to the next target - in that situation, B takes a shade over twice as long to kill the first member of A as A does of B because B's firepower drops a little as it takes casualties, while A's firepower stays at full strength until the first casualty. On the other hand, there's also the suppressing effect of taking fire - if you're under fire, your return fire is likely to be less well aimed than if you're not having to dodge bullets yourself - so the optimum firing strategy may be to disperse your fire to keep the other guys' heads down (even if there's no actual cover to start with, once the corpses start piling up, there's going to be plenty - and even before that, lying prone is going to make you a smaller target than standing up). Of course, if a mortally wounded soldier has enough time to mortally wound an enemy between being wounded and ceasing to be effective, then both forces end up dead. |
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Title: Re: 10,000 - 5,000 = Post by towr on Jul 12th, 2015, 11:23am Additional considerations: - if they fire randomly, someone from B has twice the chance of hitting someone from A (where there's twice as many targets) - twice as many guns are not twice as effective (if two people fire at a third with probability p of getting a kill-shot, then the chance of a kill is 1-(1-p)^2 < 2p) |
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Title: Re: 10,000 - 5,000 = Post by rloginunix on Jul 12th, 2015, 11:32am Very interesting. Don't want to affect your thinking too much but here are some pieces of how I was nudged towards the intended answer. I asked the person who gave me this problem towr's second paragraph question - does the geometry matter? I was told I am taking things too literally, so overall - no, as long as it is understood that no one takes cover. rmsgrey's third paragraph question - even when dead bodies/objects start to pile up? Too literal, no. Think an air battle - once the planes are shut down out of the sky they are no longer an obstacle. On to rmsgrey's second paragraph ideas. Can I transform them into some [hide]solid equations[/hide] to model this rather idealistic situation? Especially in terms of [hide]"it takes a fixed amount of accumulated man-time to kill someone"[/hide]? What would the [hide]rate of this diminishing of the opposing fighting objects/items[/hide] depend on? |
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Title: Re: 10,000 - 5,000 = Post by rloginunix on Jul 12th, 2015, 11:56am Sorry, towr. I have completely missed your reply number 3 - while I was typing my reply number 4. Apparently in this particular situation (direct fire in the open) randomness/probability does not play a crucial role (it is packed into the (constant) effectiveness of the firing objects and the optimality of the firing strategy). It does, however, when a partisan/guerilla/ambush-under-cover style fighting takes place. Then the square area of the cover comes into play as probability of a hit is much reduced. |
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Title: Re: 10,000 - 5,000 = Post by towr on Jul 12th, 2015, 1:35pm I feel a bit like the puzzle becomes "guess what model the puzzle-maker wants you to use". So I'll guess: attackA,B(t) = C1 * hitpointsA,B(t) and hitpointsA,B(t) = hitpointsA,B(t-1) - C2 attackB,A(t) with hitpointsA(0)= 10000 and hitpointsB(0)= 5000 (And obviously we can fold C1 and C2 into one constant.) If so desired we can move from discrete to continuous by taking the limit for a timestep from 1 to 0. |
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Title: Re: 10,000 - 5,000 = Post by towr on Jul 12th, 2015, 10:52pm [hide] Under above assumptions with C = 1 and solving the differential equation (for the continuous case): A(t) = 7500 e-t + 2500 et B(t) = 7500 e-t - 2500 et tend = log(3)/2 A(tend) = 5000 sqrt(3) [/hide] [edit] For clarification, in this case log is the natural logarithm ( https://en.wikipedia.org/wiki/Logarithm#Particular_bases ). [/edit] |
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Title: Re: 10,000 - 5,000 = Post by rmsgrey on Jul 13th, 2015, 6:27am on 07/12/15 at 11:32:01, rloginunix wrote:
In an air battle, the limited maneuverability of planes means the geometry of the engagement is critical :P And unless you're dealing with fixed gun-emplacements, you still have a trade-off between evasion and accuracy even in the absence of cover - a target in unpredictable motion is harder to hit, but also has more difficulty aiming... |
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Title: Re: 10,000 - 5,000 = Post by rloginunix on Jul 13th, 2015, 10:40am on 07/12/15 at 13:35:39, towr wrote:
Agreed - I had the same feeling and my initial response was [hide]5,000[/hide] for a discrete salvo-at-a-time model. Though for a continuous model one has to think - the [hide] "extra" or unmatched soldiers of A are also firing - they do not wait for their turn[/hide] ... Anyway, rmsgrey have said all the right words in his first reply. I hinted at that in mine but didn't want to make it too obvious. And towr - you have nailed it in reply #7 - that is the intended answer. Also agree with rmsgrey about the air battle - may be not a very good analogy overall but I gave it there to show that it is possible to have a battle scenario without the disabled units piling up as obstacles. I was told this result is called [hide]Lanchester's Law of Squares[/hide] - for continuous models. The intermediate calculations are long and boring (as a web page set in TeX it took me about 60 lines) but the result is sobering. In my hint 1805 is the [hide]year of the Trafalgar Battle[/hide] - apparently used as an example for this types of problems. I also applied this law to Wiki's "Battle of Kursk" - a tank battle of WW2 - and though theory and reality are different, still, the idea of numeric superiority in this types of engagements seems to hold (despite the fact that the Third Reich had by far superior weaponry). PS. Extra for experts. It may be of interest to consider the case when, in towr's notation, C1 and C2 are different. [e] I've relayed the problem as it was given to me but in an attempt to improve its quality I will add the attrition model and soldier's efficiency definition. [/e] |
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