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| Not WW1, but air combat was the basis of Lanchester's Theory of Battles. |
In the previous instalment, I discussed a theory of battles suggested by one F.W. Lanchester. What he had in mind was the random melee of the type imagined of the aerial dogfights of World War One. It was in 1916 that Lanchester published his ideas.
The theory was that in such a random environment, the outcome was determined by the squares of the numbers involved on each side. One of the assumptions behind this was that all combatants were in it with an equal chance of killing or being killed. If one side has more battle prowess (man for man) than the other, it might affect the scale of the outcome, but, if outnumbered two to one, they had to have four times the battle prowess (however that might be measured) than their opponents, and that just to come out even (mutual annihilation).
An article by Graham Jones in Wargames illustrated back in 1991 pointed out the limitations of this theory in the absence of randomness. However, upon reading that article, it occurred to me that some randomness did exist in the scenario he used to illustrate the problem. This was a force of 1200 people against 600, in which both sides had 200 in the firing line.
Extending the Theory
...It seemed to me that Lanchester's theory could be extended to cope with this, and even more general cased, i.e. to include firefights. True, the randomness of elimination of individuals does not extend over the entire forces, but it is reasonable to suppose that it will apply to the embattled front ranks - the forces actually engaged. This will be our first assumption. Our second will be that the unengaged ranks will act as a reserve from which losses will be replaced as they occur.A generalised battle along these lines might have as many as three 'phases'.
PHASE I:
Both sides have engaged in the actual fighting some number less than their overall numerical strength. The remainder on each side forms a reserve pool of replacements. The number engaged on each side need nit be equal.
PHASE II:
One side has been so reduced in numbers as to have exhausted its reserve pool. Its losses among those engaged can no longer be made good; and the force's fighting strength now declines. A battle could end in this phase.
PHASE III:
Both sides have used up their reserve pools. This is the straightforward Lanchester battle.
Our extended battles might pass through any or all of these 'phases'.
General Postulate:
A more general postulate than Lanchester's might beThe outcome of a battle depends upon the product of overall numerical strength and the numbers engaged at any given moment. This, of course, becomes the square of either if they are equal.
Symbolic Expression of the Generalised Theory:
What follows is in symbolic form the outcomes of the battles of the types just described. For reasons of (a) space and (b) I'd have otherwise to fiddle around with more symbols than indices and square roots, I'll omit the mathematical process I took to reach these conclusions. They weren't hard.
In what follows:
R = numerical strength of RED
B = numerical strength of BLUE
r = maximum number of RED individuals that can engage at any time
b = maximum number of BLUE that can engage at any time
p = relative effectiveness of RED soldiers, compared with the basis of...
q = 1 = relative effectiveness of BLUE soldiers.
Then:
RED wins in PHASE II if :
prR - pr² > bB - b²/2 or R > bB/pr - b²/2pr + r [A1]
with survivors
s(RED) = R - bB/pr + b²/2pr
BLUE wins in PHASE II if:
bB - b² > prR - pr²/2 or B > prR/b - pr²/2b + b [A2]
with survivors
s(BLUE) = B - prR/b + pr²/2b
RED wins in PHASE III if:
prR - pr²/2 > bB - b²/2 or R > bB/pr - b²/2pr + r/2 [B1]
with survivors:
____________________
s(RED) = √ {(r(2R-r) - (b/p)(2B-b)}
BLUE wins in PHASE III if:
bB - b²/2 > prR - pr²/2 or B > prR/b - pr²/2b + b/2 [B2]
with survivors
_____________________
s(BLUE) = √ {(b(2B-b) - (pr)(2R-r)}
Can you see at what point mutual annihilation will occur?
Back to Outremer...
Now, imagine you are one Yakub Salah ed-Din, surnamed 'the BLUE', a distant relative of Saladin, charged with the safekeeping of the vital caravan mentioned in the previous article. You are particularly concerned about Bedouin raids. Suddenly a scout comes in from the west, with word that that rapacious Frank, Reynald of Chatillon, has violated the truce and is, even now, in full march to attack the caravan. Seventy-one of his followers, according to the ever-reliable scout, Ali Muhammad the Far Seeing, knights and sergeants all, are riding hither filled with malice aforethought.
Realising that a melee in the plain will lose retainers, caravan and credibility with Saladin, you recall that Reynald will have to pass through a defile not far west of here. Ten men abreast is the maximum the defile can accommodate. But just as they debouche from the pass into the open, they will have to face twenty Saracens at a time - not more will be in reach - to their ten. This promises your men a fighting chance.
But you can't send off your whole hundred retainers: there's still the Bedouin to consider. So, what is the fewest number to send against the Franks? Maybe fifty will do? How about matching the Frankish numbers and send 71?
To determine the numbers required, you want to send the minimum that will stop the Franks, whilst retaining the maximum possible protection for the caravan. Even if you lose the whole column, just so long as Reynald is stopped. That will suffice. We'll use relation [B2] and state that your strength B must be at least -
B = prR/b - pr²/2b + b/2
where R=71, p=2 (RED fighting ability twice as effective as BLUE), r=10 (RED front line) , b=20 (BLUE front line)
B = (2x10x71)/20 - (2x10x10)/(2x20) + 20/2
= 71 - 5 + 10
= 76.
So you must send at least 76 - although if you do send that many, don't expect to see anyone - Frank or Saracen - to come back from the west. You might, withal, send 77 instead, and leave just 23 men to defend the caravan. According to my calculations, you'll see 6 men come back.
Applications... Sort of.
At this point I added a short paragraph indicating what I thought might be applications for war games. Of course, I didn't want mutual annihilation, but, having come this far, it was not that hard to envisage what might be going on if one side or the other was reduced by, say, one-third. Take that as the bench mark, and calculate (this again is where the calculus comes in) the losses on the other side. |
I was going to suggest the Lanchester model for a Space Opera game based on what I've heard and read of the Waddington board game 4000A.D. I gather that a RED fleet of 20 ships would defeat with equal ease an opposing BLUE fleet of 19 or a squadron of 5 - and at zero cost. I thought it would be nice to have a table that offered standard results according to the Lanchester model. According to that model, the BLUE 5-ship squadron would be scarcely more than a speed bump for the 20-ship fleet - losing all five and taking out, with luck, maybe one RED ship. The 19-vessel BLUE fleet would inflict a deal more damage before going under, and leave just 6 RED ships out of 20 to limp homeward.
I once put together a Table of Outcomes for fleets up to 20 or so ships a side, the losers being wiped out, of course, but it showed the numbers of survivors from the winning side. I would have produced it here, but the thing seems to have vanished the way of all good and useful things...
Limitations:
Note that in all this, we are concerned with the overall result as a statistical expectation: who wins, and by how much. What is going on internally is another matter altogether. I will admit that, upon working out that a force outnumbered 2 to 1 had to fight, man for man, with 4 times the efficiency of the enemy just to break even, I concluded that, without that skill superiority, the odds of their winning was 4 to 1 against.It turns out that was wrong. In a forum discussion (long since defunct) I was informed that a study concluded that the odds against the smaller side's winning was 8:92 against - 1:11.5. Odd looking numbers, that, and I'm not 100% sure what they really mean. Something for discussion another time, perhaps...
Many decades ago I used to play 4000AD, and at the time we thought the combat system was.... eccentric. Iirc we introduced a differential CRT. The larger side generally still won, but suffered some losses in doing so. I have to confess I remain deeply unconvinced by Lanchester as a model of combat for all sorts of reasons, Dupuy and Biddle have much to say on force ratios and combat outcomes, in fact Biddles model has relative combat competence as the primary determinant of success, based on operational experience in GW1, Iraq and Afghanistan. However some of my pals in the professional wargaming community tell me the Chinese are basing their autonomous drone combat tactics on Lanchester. So who knows.
ReplyDeleteMartin -
DeleteRelative combat competence - whatever that means - is surely a factor. But the 'p' factor, as I called it, can mean all sorts of things., and may be a very large number. A sniper with a rifle will cope with many times his numbers of people armed with sling shots.
I'm not sure that many insights will be available from enumerating the successes of the US and its friends in Iraq or Afghanistan - the technological edge was overwhelming, and how do you assess the numbers? Now, whatever failures the US and its allies experienced - that might well be very enlightening.
All belligerents in a fight will look to randomise - or to de-randomise - the situation in their favour. That is why flank attacks are so effective. Consider the methods adopted by Hamas and Hezbollah in recent times - in effect drawing a more powerful opponent upon a battlefield that neutralises the disparity in strength. Such were the methods of British at Agincourt, Lee and Jackson at Chancellorsville, Maori against British Imperialists and colonist militia. Whatever ruses, subterfuges, contrivances or manoeuvres; whatever the superiority in weaponry, training, self-belief, the idea comes down to 'God fights on the side of the Big Battalions'.
Someone once put it to, I think, Napoleon. He explained that although he often went into battle with the smaller army, he always contrived to establish a local superiority where it mattered most. So, he agreed, the bigger numbers ensured his victory.
These two articles are what I wrote 30 years ago. I may add a third. I'll think about it.
Reminds me of an article I once wrote on the Lanchester model: https://snv-ttm.blogspot.com/search?q=lanchester
ReplyDeleteSee also https://arxiv.org/abs/math/0606300
Phil -
DeleteI've just looked at your article (July 2011). Your comment on 'Tactical Consequences' is pretty much what I discuss with more detail in my 'Extending the theory'. The numbers you gave (10 vs 6 with only 2 available on either side to fight at any time) I ran through my own scheme ... and ran into a problem!
For a wild moment I thought my maths must be wrong. I came to the same conclusion using one method, but something different using my equation for s(RED) (PHASE II) above. The latter gave me 5 survivors, not 4. Then I realised: it was not for nothing I mentioned the phases. In your example, the sides reduce 2 by 2 until they are 6 to 2. But suppose the losses accrue 1 x 1 until they are 6 to 2. Then, both sides lose 1 more, and it's 5 to 1, and 2 to 1 in the front line. Does the single man have the same chance of knocking over one of the two as the two have of knocking him over?
Now, we do have a problem with the fact that we are dealing with discrete bodies, which is why 0.7071 of a man doesn't mean a whole bunch. It is especially acute when we look at small numbers. Suppose we increase the example across the board by a factor of 100: 1000 RED vs 600 BLUE; 200 on both sides in the fighting line. Losses are equal until BLUE is reduced to 200. Now its 600 to 200. Whole new ball-game.
From first principles, there are indeed 500 RED survivors.
Using my equation for PHASE II (because RED can never run out of reserves)
s(RED) = 1000 - (200x600)/200 + (200x200)/(2x200)
= 1000 - 600 + 100
= 500.
As an application if this idea, I had a look at an action in a Napoleonic battle in which an 8000-man column (French) faced a 3000 man line (British). I think we're talking Albuera here (I'm going by memory). Let us say that the column comprised ranks of 600, that is to say, 1200 French muskets can be brought to bear. All 3000 British muskets are in action. Again, this is a PHASE II battle, assuming the French don't run out of reserves (there is a quick way of figuring out the likelihood).
The French win, wiping out the British, but at a cost of 3750 of their own number.
Now, in fact I looked into the actual losses, which I think were 1500 British TO 3000 French. So, what does the calculus indicate? A quick calculus suggested that in inflicting 1500 casualties, the French 'should' have taken just over 3300. Pretty close to the 3000 estimated earlier.
Now, I am aware there were other factors involved, such as artillery (on both sides). But overall, I think there remains something to be said for this type of analysis.
Cheers,
Ion
All true, but as you also have said, this is due to the discretization of the time-steps. The Lanchester mathematical model assumes continuous functions modeled on infinitesemal timesteps and force strengths modeled by real numbers (as opposed to integers). Once we start to discretize that, discretization errors occur. So that's why Lanchester should rather be seen as a crude approximation to what could be happening in a real situation (with discrete time events and discrete force strengths).
DeletePhil -
DeleteI think we are saying the same thing to each other - just in different ways. To be sure, the method is idealised, but I found it very useful in figuring out war games mechanics that seem to me plausible.
Mind you, my applications were to my 'Old School' war games. These I don't play so often these days.
Cheers,
Ion
In the end, the Lanchester model is rather simple: your losses in each "timestep" are proportional to my strength and vice versa, and this applies throughout the entire duration of the conflict. LAnchester is then simply the solution of that differential equation.
ReplyDeleteYes. My time step example is very crude, but can be made less crude if you wanted by making them shorter, with smaller decrements. Lanchester is really the limit as the decreasing decrements approach zero.
DeleteHaving said that, if we are dealing with individuals, the decrement should strictly speaking be not less than 1 (individual). This rather favours the weaker side incidentally. I am, therefore, conscious of the rather idealised nature of this analysis. In the Napoleonic 'encounter' of my previous comment, a 'step approach' of say 1% decrements would seem to result in rather fewer than 3000 losses to the French - a little more than 2800, by the look - against the British loss of 1500.
Cheers,
Ion
The modeling of discrete events by continuous approximations (which can then be solved by calculus) is of course typical of the pre-digital age, when calculus was the tool of choice (and often the only tool available). These days such an analysis would simply be run brute-force on computer using stochastic Monte Carlo or Markov chain models, and averages and standard deviations would be reported.
DeleteOh, but it is so much more fun this way!
DeleteWell, I think running simulations using MCMC is fun too. The continuous v discrete debate is one reason to consider Poisson which I suggested in your earlier post.
DeleteJonathan -
DeleteI think that is a whole different study, less focused upon simple outcomes, but more the way the incoming is distributed. For instance, my 'OS' combat mechanic supposes that, for a certain volume of incoming (bullets, say) there is a certain average of strikes upon a unit. The unit is made up of individuals. Now every hit might mean a man casualty, OR every hit may strike some unlucky individual only. The first is very unlikely; the second vanishingly unlikely.
There you might be looking at a Poisson (though, as I think I mentioned, when I thought through this I was thinking more Binomial). What I came up with was this:
Let s = number of strikes upon a target
Let t = number of individuals with the target
Let K = expected number of individuals struck
Then -
K = t - {(t-1)^s}/{t^(s-1)}
So if we have, say a target of 10 individuals, and 4 shots strike them, how many do we expect to get hit?
K = 10 - (9^4)/(10^3)
K = 10 - 6501/1000
K := 3.5 approximately
I interpret that as 4 people are as likely to be hit as 3 or fewer. All four shots might have hit Private Bloggs.
Unfortunately the expression really does need a number cruncher to be useful (supposing it is correct: it seems to work for small numbers). Imagine a 1000 man unit facing an equal opponent. Say that opponent manages to put 20% of its shots on target (a reasonable average). how are the 200 hits distributed?
K = 1000 - (999^200)/(1000^199)
K = 1000 - 999{(999/1000)^199}
It works out as close to 1000-820 = 180 individuals struck.
Unfortunately, I never did figure out the variance.
But this exercise (several years earlier than my look into Lanchester) was a whole different gig.
Cheers,
Ion
Jonathan -
DeleteYou can see from my examples that the distributions are skewed. The range of outcomes in the first is 1 to 4, with the high numbers most likely.
In the second, the range is 1-200, with the mean at 180.
I am reminded of the distribution of dart holes on a dartboard. Have you noticed that (apart from the heavy traffic around trip 20 and double tops), the holes form a sort of ring around the centre? Are we looking at a Gamma distribution maybe?
Cheers,
Ion
When I did a study on musket effectiveness more than a dozen years ago, I settled upon a Beta distribution.
DeleteI looked up the Beta function. Far out. The maths is beyond me. I'd probably have to do a whole undergraduate course in probability and statistics to understand what the hell they're talking about!
Deleteaargh!
Ion
"...four times the battle prowess (however that might be measured) ...".
ReplyDeleteFor Lanchester it is measured as each element 'killing' at four times the rate that the other side's elements are 'killing'. The 'randomisation' (IIRC simply that every element on each side is at risk of every element on the other) or fighting context does not matter provided the differential in killing rate per element is maintained. So, if you only consider those elements in combat (i.e. excluing those not in the firing line in Graham Jones' WI article who are effectively not in the combat until they enter the firing line) then the theory can still be applied.
Rob -
DeleteI think what I called battle prowess (proficiency would have been a better choice of word) Lanchester might have used 'efficiency' or 'effectiveness'. The question I didn't so much raise as acknowledge was how this might be measured. How do you compare, say, mounted knights in armour against unarmoured peasants armed with pitchforks? How about these same knights, on foot and floundering on the banks of a muddy rill, against burly but nimble hairy-chested types wielding bally great clubs with spikes driven through them? How do we measure well-trained veterans against raw conscripts? Acknowledging that such questions are reasonable to ask (if not so easy to answer), I think setting them aside helps with focus.
A couple of years back I soloed a campaign based upon an imaginary world, in which one side was armed with magazine rifles and machine guns, against an adversary armed with muskets less than half of which were rifled, and a very small proportion, single-shot breech loaders. The latter force began about 80,000 strong; the Union, something like 22,000 - reinforced about half-way through by about half as many again.
The author of the world this was based upon and I both knew full well the smaller army could not take on the larger in one go. It had to be a piece at a time, and even then the thing could have gone either way. If you like a good story, you might find this entertaining:
https://archdukepiccolo.blogspot.com/search/label/Woodscrew%20Miniature%20Army%20-%20Table%20of%20Posts
This is a 'Table of Posts' - 19 chapters in all.
At any rate, such issues I have set aside for the purposes of this article. I mentioned them in passing as an acknowledgement of this being an 'analysis in vacuo' so to speak.
Cheers,
Ion
I’m not sure I understand much of the maths, but your cartoon artwork is absolutely great 🚀🛸
ReplyDeleteCheers,
Geoff
Geoff -
DeleteI admit I haven't really explained the maths - at least not the method used to reach the my conclusions. But thanks for liking my cartoon. It was the 4000A.D. game thing that set that off...
I see I forgot to sign it...
Cheers,
Ion