Derivation

Our objective is to find out the following:-

1. Who wins?
2. Losses sustained by both sides.
3. Time in which the other army is defeated.

ASSUMPTIONS

There are some assumptions to be made to the question.

1. There are two armies- X-army and Y-army.
2. The number of soldiers in X-army is given by x.
3. The number of soldiers in Y-army is given by y.
4. The lethality factor of X-army is given by B and that of Y-army is given by A.
5. x > 0, y > 0, A > 0, B > 0 always.
6. A and B are constants.
7. The enemies are killed continuously(without any delay) by both sides of the army- i.e. time for reload, rest etc is not considered.

DERIVING LANCHESTER’S SQUARE LAW EQUATION

The equation of the two armies is given by:-

dx/dt = -Ay (1)

x(0) = Xo

The minus sign indicates decrease. The equation basically states that the rate of decrease of soldiers of X-army is equal to the product of the lethality factor of the Y-army and the number of soldiers in the Y-army. x(0) indicates the initial strength of the X-army which is assumed to be Xo.

Similarly,

dy/dt = -Bx (2)

y(0) = Yo

Dividing equation (2) by (1) we get,

(dy/dt)/(dx/dt) = Bx/Ay

Transposing, we get

Aydy = Bxdx

Integrating both sides,

A∫ydy = B∫xdx

(Ay^2)/2 = (Bx^2)/2 + C

Hence, Ay^2 – Bx^2 = constant

The above equation is the mathematical model for Lanchester’s square law. It tells us that the fighting effectiveness of an army is directly proportional to the lethality factor of an army and to the square of the number of soldiers in the army.

We can even include other factors and more number of armies in the equation. For example:-

x’= Ay + R(t) – Cx

Where R(t) is the rate of reinforcements coming to the army (note the plus sign), C(t) is the rate of decrease of soldiers of X-army due to other factors (e.g. disease etc). There can be more number of armies too in the battlefield model.