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On Lanchester’s Differential Equations and WWII: Modeling the Iwo Jima Battle

Almost at the end of World War II, a battle to the death between US and Japanese forces took place off Japan in the island of Iwo Jima.  We can definitely apply a Lanchesterian analysis to this encounter; in fact, it has already been done by Martin Braun in Differential Equations and their Applications, 2nd ed. (New York: Springer and Verlag, 1975).  If $x$ represents the number of US soldiers and $y$ represents the number of Japanese troops, Braun estimates that $a = 0.05$ and $b = 0.01$ approximately.  Assuming then that the initial strength of the US forces was hovering around 54,000, and that of the Japanese was 21,500, we'd be interested in knowing the trajectory of the battle and see who wins.  I've used the method described in my previous post to obtain estimates for the proportion of  soldiers alive in $X = US$ and in $Y = Japan$, namely the Markovization or discretization (or Eulerization) method, and this is what we come up with.  We assume the battle is fought to the death and there are no reinforcements from either side:

$L = \left[ \begin{array}{ccc} 1 - 0.05 \cdot \frac{p_{alive,y}}{p_{alive,x}} & 0 & 0.05 \cdot \frac{p_{alive,y}}{p_{alive,x}} \\ 0 & 1 - 0.01 \cdot \frac{p_{alive,x}}{p_{alive,y}} & 0.01 \cdot \frac{p_{alive,x}}{p_{alive,y}} \\ 0 & 0 & 1 \end{array} \right]$

The attrition constants are within the range so that all entries are less than one as required by a Markov transition matrix provided the excess ratio of one army population to another is not too large (also, each row sums to 1).  The matrix is time-dependent, however, because each time the proportion of alives in any army changes according to the strength of the other army (proportional to both numerical superiority and technology, as can be seen in the matrix above, thus, from the Markov chain vantage point, the probability of death at each time step changes), as modeled by Lanchester's continuous differential equations.  Here are the proportion trajectories I came up with, computing at each time step the above matrix and pulling the original vector through:

Iwo Jima battle

It's pretty clear that, despite having the technology advantage, the Japanese lose because of numerical inferiority after about 59 time-steps.  The approximate proportion of US soldiers at the end of the battle is 0.35 (of the total soldier population), or about 26,276 soldiers, a HUGE decline from the initial 54,000 (roughly 0.72 of the total soldier population).  The Japanese army essentially obliterated half the soldier population of the US, a testament to their ferocity in battle.

My Calculus book (Deborah Hughes-Hallet, Andrew M. Gleason, et al., New York, John Wiley & Sons, Inc., 1994, p.551) suggests that the US in fact would have had 19,000 soldiers that could have stepped in as reinforcements: this does not matter really if the reinforcements come at the end of the battle, the battle has already been won with 54,000 soldiers, but it does alter the outcome significantly if they are added to the original soldiers at the beginning of battle (namely the US has 73,000 troops, or 0.78 of the total soldier population): the Japanese army would lose in roughly 34 time-steps (to the death), and the number of US soldiers left standing would be about 55,420, or 0.59 of the total soldier population: the Japanese obliterate only about 25% of the US troops.

In my next post, I want to discuss Dan's (from Giant Battling Robots) pdf document that he provided, essentially looking at a different stochastic model related to warfare and battle.  It's pretty neat, although it counts each potential outcome as a Markovian state (and each state is linked "downward" by a transition probability so that a given probability governs the death of one soldier in either army).  Thanks Dan!

1. August 21st, 2010 at 09:00 | #1

You are most welcome, and thank for the link!

I should have some references to other analyses of the Iwo Jima data, and probably the day-to-day troops levels too. Give me a few days (busy weekend!) and I'll try to post something relevant.

I've been getting some "500" server errors when I try to load your blog. I'm not sure what that means, but I thought I should let you know.

---Dan

1. April 20th, 2012 at 08:24 | #1

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