Transcript lanchester equations for MAT 5932

Dynamical Systems
MAT 5932
The Lanchester Equations
of Warfare Explained
Larry L. Southard
Thursday, July 16, 2015
Agenda
• History of the Lanchester
Equation Models
• Lanchester Attrition Model
• Deficiencies of the
equations
History
• The British engineer F.W. Lanchester
(1914) developed this theory based
on World War I aircraft engagements
to explain why concentration of
forces was useful in modern warfare.
• Lanchester equations are taught and
used at every major military college
in the world.
Two Types of
Models
• Both models work on the
basis of attrition
• Homogeneous
• a single scalar represents a unit’s
combat power
• Both sides are considered to have the
same weapon effectiveness
• Heterogeneous
• attrition is assessed by weapon type and
target type and other variability factors
The Homogeneous Model
• An “academic” model
• Useful for the review of
ancient battles
• Not proper model for
modern warfare
Heterogeneous Models
• CONCEPT: describe each type of
system's strength as a function
(usually sum of attritions) of all
types of systems which kill it
• ASSUME: additivity, i.e., no
synergism; can be relaxed with
complex enhancements; and
proportionality, i.e., loss rate of Xi
is proportional to number of Yj
which engage it.
• No closed solutions, but can be
solved numerically
The Heterogeneous
model
• More appropriate for “modern”
battlefield.
• The following battlefield functions are
sometimes combined and sometimes
modeled by separate algorithms:
• direct fire
• indirect fire
• air-to-ground fire
• ground-to-air fire
• air-to-air fire
• minefield attrition
The Heterogeneous
model
• The following processes are directly or indirectly
measured in the heterogeneous model:
• Opposing force strengths
• FEBA (forward edge of the battle area) movement
• Decision-making (including breakpoints)
• Additional Areas of consideration to be
applied:
• Training
• Morale
• Terrain (topographically quantifiable)
• Weapon Strength
• Armor capabilities
Decision Processing in
Combat Modeling
Attrition
Target Acquisition
Sensing
Engagement Decision
Target Selection
Physical Attrition Process
Command
and Control
Accuracy Assessment
Damage Perception by Firer
Damage Assessment
Movement
Lanchester Attrition Model

CONCEPT: describe the rate at which a force loses systems as a function of the
size of the force and the size of the enemy force. This results in a system of
differential equations in force sizes x and y.
dx
dt
 f 1 x, y,...
dy
dt
 f 2  x, y,...
The solution to these equations as functions of x(t) and y(t) provide insights
about battle outcome.

This model underlies many low-resolution and medium-resolution combat models.
Similar forms also apply to models of biological populations in ecology.
The Lanchester Equation
Mathematically it looks
simple:
dx
dt
  ay and
dy
dt
  bx
Lanchester Attrition Model - Square Law
Integrating the equations which describe
modern warfare
dx
  ay
dt
and
dy
  bx
dt
we get the following state equation, called
Lanchester's "Square Law":
b( x 2 
0
x 2)  a( y2  y2)
0
These equations have also been postulated
to describe "aimed fire".
ab
a
b

measures battle
intensity

measures
relative
effectiveness
Questions Addressed by
Square Law State Equation
 Who will win?
 What force ratio is required to gain victory?
 How many survivors will the winner have?
 Basic assumption is that other side is
annihilated (not usually true in real world
battles)
 How long will the battle last?
 How do force levels change over time?
 How do changes in parameters x0, y0, a, and b
affect the outcome of battle?
 Is concentration of forces a good tactic?
Lanchester Square Law - Force Levels Over
Time
After extensive derivation, the following expression for the X force
level is derived as a function of time (the Y force level is
equivalent):

x(t ) 1   x
2 
0

a
b

y  e
0

abt   a
 x0
b


y  e
0

abt 


Square Law - Force Levels
Over Time
Example:
X=30, Y=60, a=.04, b=.04
70
60
60
50
50
30
20
0
40
30
20
10
4
Time
Force Level
40
0
8
-10
12
X Force
x(t) becomes zero at about t = 14 hours.
Surviving Y force is about y(14) = 50.
10
0
0
Y Force
-10
1
2
3
4
5
Square Law - Force Levels
Over Time
How do kill rates affect outcome?
Reduce a to .01, Increase b to .1
60
50
40
30
20
10
25
20
-10
15
Time
10
5
0
0
X Force
Now y(t) becomes zero at about t =24 hrs.
Surviving X force is about x(24) = 20.
Y Force
Force Levels
Square Law - Force Levels
Over
Time
Can
Y overcome
this disadvantage by adding forces?
Increase Y by 30
90
60
50
40
30
20
60
48
36
12
Time
24
0
10
0
-10
X Force
Not by adding 30 (the initial size of X's whole force).
Y Force
Force Level
80
70
Square Law - Force Levels
Over Time
What will it do to add a little more to Y?
Increase Y by another 10
120
100
100
80
60
40
20
0
19
60
40
0
Y Force
0
0
X Force
-20
This is enough to turn the tide decidedly in Y's favor.
20
57
20
15
-20
10
38
5
Time
Force Level
80
Square Law - Who Wins a
Fight-to-the-Finish?


To determine who will win, each side must have victory
conditions, i.e., we must have a "battle termination model".
Assume both sides fight to annihilation.
One of three outcomes at time tf, the end time of the battle:

X wins, i.e., x(tf) > 0 and y(tf) = 0

Y wins, i.e., y(tf) > 0 and x(tf) = 0
Draw, i.e., x(tf) = 0 and y(tf) = 0
It can be shown that a Square-Law battle will be won by X
if and only if:


x 
y
0
0
a
b
Lanchester Square Law - Other Answers
How many survivors
are there when X
wins a fight-to-thefinish?
When X wins, how
long does it take?
x
f
a 2

2
  x 0  y0

b 
 y0
 1
1
x0
t x  
ln
2 ab  y0
 1
 x0
f
a

b
a
b 
Square Law - Breakpoint Battle Termination




t y   




How long does
it take if X wins?
(Assume battle
termination at
x(t) = xBP or
y(t) = yBP)
In what case does
X win?
If and only if:
BP
1
x0
x0 a
ln
if

y0 b
ab x BP

2
2 b 2
y
y
y
 BP  0  x 0 
1  BP
a  otherwise
ln
b
ab 

y0  x 0




a
 y2 
 1 2 
x  a y 
y
b  x2 
 1 2 
 x 
BP
0
0
0
BP
0