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\title{\ifnum\TeacherEdition=1 TEACHER VERSION \else STUDENT VERSION \fi\\M\&M Attrition Warfare}




\markboth{M\&M Attrition Warfare}{M\&M Attrition Warfare}

\author{Lukasz Grabarek\\
Matanuska-Susitna College\\
University of Alaska Anchorage\\
Palmer AK \teacher99645 USA
}

\makeStitlePDFLaTex


\begin{abstract}
Students will model attrition between two opposing forces using M\&M candies and discover a system of linear differential equations of order one, often called the Lanchester equations.  Through experiment and data collection, students will determine the parameters describing the fighting strength of each force, solve the system, arrive at Lanchester's N-square Law and use it to analyze a historical battle. 
\end{abstract}




\section*{SCENARIO DESCRIPTION}
In his 1916 work \cite{FWL}, Frederick William Lanchester proposed a model of how two opposing forces of aircraft would attrit one another in combat.  Lanchester's model came two years before the establishment of the first air force as an independent branch of any nation's armed forces. Since then, Lanchester's model continues to find applications in operations research (see \cite{Fox2010, NPG, RLJ}). In this scenario we will investigate mock military engagements between opposing forces of M\&M candies and derive Lanchester's model.


To proceed, consider two opposing forces. Suppose that the opposing units engage one another in repeated single combat rounds until one opponent is defeated.  Modern weaponry, capable of inflicting mass casualties, is not considered in this model, so for example, imagine a duel between two knights, two fighter airplanes in a dog fight, or two pirate ships attempting to sink one another.  In each situation one ultimately will defeat the other.  We will use M\&M's to simulate such combat all the while recognizing the severe constrains and heavy assumptions implicit in the model.  For a discussion of the obstructions to implementing Lanchester's model in practice see \cite[Chapter VI]{RLJ}.  

\subsection*{The experiment}
Begin with M\&M's of two different colors in equal numbers.  Say, 12 blue and 12 red pieces. Record the initial size of each force as Battle 0 in a table.  Subsequent rows (see Table \ref{table1}) track the number of casualties on each side and each row represents a single battle.

\subsection*{A typical round - of sequential battles}
\begin{itemize}
 \item Toss all the M\&M's onto a plate and correct any ambiguities by slightly jostling the plate until each piece is resting properly with either the M facing up, or facing down.
 \item Any piece with the M facing up is vulnerable to defeat, but defeat occurs only if a piece of the opposing color with the M facing down is available.  So, group  opposing colors in pairs with exactly one M facing up, then remove the piece with the M facing up from the plate -- it is a casualty -- and place it off to the side.   The other piece survives until the next battle, but until then it must be removed into the cup.
 \item Repeat the process until all pieces with M facing up are removed, or until no more pairs remain.  
 \item Record the casualties on each side in a table.
 \item Collect all pieces in the cup.  The round is over.
 \end{itemize}

We will return to this sequential battle modeling, but first let us consider just several simulations of a single battle, in each case starting with the same Red and Blue configuration.

\subsection*{A closer look at a single battle simulations, each with same initial forces}

Let us first consider just one battle simulation, rather many many such battle simulations, but in each case we always use the same starting values of Red forces and Blue forces.

 \begin{enumerate}
  \item Make a hypothesis on the attrition {\it rates\/} between two opposing forces of equal numbers for a single battle.  How do you expect the rate at which the number of Red units decreases is related to the number of Blue units?
  \vspace{1 ex} \\
  \noindent \textbf{Initial hypothesis:}
  \vspace{2 cm}
  
  
  \item Experiment and collect your data in  Table  \ref{table1}, Table  \ref{table2}, and Table  \ref{table3}.  You may divide the labor between groups. Each entry, e.g., ``Battle 1'' in Table  \ref{table1}, indicates the number of casualties suffered by respective forces at the hands of the other when the forces were at strength 12 for Red and 12 for Blue. 
  
(*) Now conduct a single battle simulation, by tossing the M\&M's representing the initial forces of Red and Blue as represented by M\&M's   onto a plate.  Recall, any piece with the M facing up is vulnerable to defeat, but defeat occurs only if a piece of the opposing color with the M facing down is available.  So, after grouping  opposing colors in pairs with exactly one M facing up,  remove the piece with the M facing up from the plate -- it is a casualty -- and place it off to the side.   The other piece survives until the next battle, but until then it must be removed into the cup. So for example, if after one battle (a single toss of your initial number of Red and Blue M\&M's) you obtain 7 Red casualties and  5 Blue casualties then enter 7 under Battle 1 in column denoted Red and under Battle 1 enter 5 in column denoted Blue. We have thus recorded the outcome of a single battle in which we had   7 casualties for Red and 5 casualties for Blue.

Repeat the simulation of a single battle, by first restoring the forces to full strength, in the case of Table \ref{table1},12 Red and 12 Blue, and then return to (*) above and conduct another single battle simulation. Continue this process reporting casualties for Red and Blue for additional battles until you have conducted 10 battles from initial value (each) of 12 Red and 12 Blue.


\begin{figure}[h]   
\centering
\includegraphics[width=.95\textwidth]{Example.png} 
\caption{Illustration of casualties of ``Battle 1'' in Table \ref{table1}}
\end{figure}


Table  \ref{table2} and Table  \ref{table3} suggest doing the same thing as done in generating entries in  Table  \ref{table1}, only with different starting values for each battle, namely 16 Red and 16 Blue in Table \ref{table2} with 10 Red and 20 Blue in Table \ref{table3}.

  
  \teacher{\it We provide results of our experiments and data collection on 30 battles in  Table  \ref{table1}, Table  \ref{table2}, and Table  \ref{table3}  in case time is limited and the teacher wishes to use this pre-collected data, because collecting battle data can take time.}
  \vspace{1 ex}
\end{enumerate}
 
 \begin{table}
 \begin{minipage}{\linewidth}
 \begin{minipage}{.3\linewidth}
  \begin{tabular}{|c|c|c|}
 \hline
 Battle & Red & Blue \\
 \hline
 0     & 12  & 12 \\
 \hline\hline
 1     & \ifnum\TeacherEdition=1 5 \else \fi    & \ifnum\TeacherEdition=1 5 \else \fi   \\
 \hline
 2     & \ifnum\TeacherEdition=1 8 \else \fi   & \ifnum\TeacherEdition=1 4 \else \fi   \\
 \hline
 3     & \ifnum\TeacherEdition=1 5 \else \fi    & \ifnum\TeacherEdition=1 7 \else \fi   \\
 \hline
 4     & \ifnum\TeacherEdition=1 5 \else \fi    & \ifnum\TeacherEdition=1 6 \else \fi   \\
 \hline
 5     & \ifnum\TeacherEdition=1 5 \else \fi    & \ifnum\TeacherEdition=1 4 \else \fi   \\
 \hline
 6     & \ifnum\TeacherEdition=1 8 \else \fi    & \ifnum\TeacherEdition=1 4 \else \fi   \\
 \hline
 7     & \ifnum\TeacherEdition=1 7 \else \fi    & \ifnum\TeacherEdition=1 3 \else \fi   \\
 \hline
 8     & \ifnum\TeacherEdition=1 5 \else \fi    & \ifnum\TeacherEdition=1 7 \else \fi   \\
 \hline
 9     & \ifnum\TeacherEdition=1 4 \else \fi    & \ifnum\TeacherEdition=1 5 \else \fi   \\
 \hline
 10     & \ifnum\TeacherEdition=1 7 \else \fi    & \ifnum\TeacherEdition=1 5 \else \fi   \\
 \hline
 \hline
 Mean & \ifnum\TeacherEdition=1 5.9 \else \fi  & \ifnum\TeacherEdition=1 4.9 \else \fi   \\
 \hline
 \hline
 & \ifnum\TeacherEdition=1 $\rho = 0.41$ \else $\rho = \hspace{.5cm}$ \fi  & \ifnum\TeacherEdition=1 $\beta = 0.49$ \else $\beta =\hspace{.5cm} $\fi   \\
 \hline
\end{tabular}
\centerline{\parbox{4,2 true cm}{\caption{\label{table1} Starting with 12 R and 12 B forces.}}}
 \end{minipage}
 \hspace{.025\linewidth}
 \begin{minipage}{.3\linewidth}
  \begin{tabular}{|c|c|c|}
 \hline
 Battle & Red & Blue \\
 \hline
 0     & 16  & 16 \\
 \hline\hline
 1     & \ifnum\TeacherEdition=1 8 \else \fi    & \ifnum\TeacherEdition=1 5  \else \fi \\
 \hline
 2     & \ifnum\TeacherEdition=1 9 \else \fi    & \ifnum\TeacherEdition=1 5 \else \fi   \\
 \hline
 3     & \ifnum\TeacherEdition=1 9 \else \fi   & \ifnum\TeacherEdition=1 7 \else \fi   \\
 \hline
 4     & \ifnum\TeacherEdition=1 9 \else \fi    & \ifnum\TeacherEdition=1 7 \else \fi   \\
 \hline
 5     & \ifnum\TeacherEdition=1 6 \else \fi    & \ifnum\TeacherEdition=1 7 \else \fi   \\
 \hline
 6     & \ifnum\TeacherEdition=1 4 \else \fi    &  \ifnum\TeacherEdition=1 6 \else \fi  \\
 \hline
 7     & \ifnum\TeacherEdition=1 9 \else \fi    & \ifnum\TeacherEdition=1 5 \else \fi   \\
 \hline
 8     & \ifnum\TeacherEdition=1 8 \else \fi    & \ifnum\TeacherEdition=1 8 \else \fi   \\
 \hline
 9     & \ifnum\TeacherEdition=1 8 \else \fi    & \ifnum\TeacherEdition=1 8 \else \fi   \\
 \hline
 10     & \ifnum\TeacherEdition=1 7 \else \fi    & \ifnum\TeacherEdition=1 9 \else \fi   \\
 \hline
 \hline
 Mean & \ifnum\TeacherEdition=1 7.7 \else \fi  & \ifnum\TeacherEdition=1 6.7 \else \fi   \\
 \hline
 \hline
    & \ifnum\TeacherEdition=1 $\rho = 0.42$ \else $\rho = \hspace{.5cm}$\fi  & \ifnum\TeacherEdition=1 $\beta = 0.48$ \else $\beta = \hspace{.5cm}$\fi   \\
 \hline
\end{tabular}
\centerline{\parbox{4.2 true cm}{\caption{\label{table2} Starting with 16 R and 16 B forces.}}}
 \end{minipage}
 \hspace{.025\linewidth}
 \begin{minipage}{.3\linewidth}
  \begin{tabular}{|c|c|c|}
 \hline
 Round & Red & Blue \\
 \hline
 0     & 20  & 20 \\
 \hline\hline
 1     & \ifnum\TeacherEdition=1 8 \else \fi    & \ifnum\TeacherEdition=1 5 \else \fi   \\
 \hline
 2     & \ifnum\TeacherEdition=1 7 \else \fi    & \ifnum\TeacherEdition=1 10 \else \fi   \\
 \hline
 3     & \ifnum\TeacherEdition=1 9 \else \fi    & \ifnum\TeacherEdition=1 8 \else \fi   \\
 \hline
 4     & \ifnum\TeacherEdition=1 8 \else \fi    & \ifnum\TeacherEdition=1 11 \else \fi   \\
 \hline
 5     & \ifnum\TeacherEdition=1 9 \else \fi    & \ifnum\TeacherEdition=1 11 \else \fi   \\
 \hline
 6     & \ifnum\TeacherEdition=1 6 \else \fi    & \ifnum\TeacherEdition=1 11 \else \fi   \\
 \hline
 7     & \ifnum\TeacherEdition=1 10 \else \fi    & \ifnum\TeacherEdition=1 9 \else \fi   \\
 \hline
 8     & \ifnum\TeacherEdition=1 8 \else \fi    & \ifnum\TeacherEdition=1 11 \else \fi   \\
 \hline
 9     & \ifnum\TeacherEdition=1 11 \else \fi    & \ifnum\TeacherEdition=1 9 \else \fi   \\
 \hline
 10     & \ifnum\TeacherEdition=1 8 \else \fi    & \ifnum\TeacherEdition=1 8 \else \fi   \\
 \hline
 \hline
 Mean & \ifnum\TeacherEdition=1 8.4 \else \fi  & \ifnum\TeacherEdition=1 9.3 \else \fi   \\
 \hline
 \hline
    & \ifnum\TeacherEdition=1 $\rho = 0.46$ \else $\rho =\hspace{.5cm}$\fi  & \ifnum\TeacherEdition=1 $\beta = 0.42$ \else $\beta = \hspace{.5cm}$\fi   \\
 \hline 
\end{tabular}
\centerline{\parbox{4.2 true cm}{\caption{\label{table3}  Starting with 20 R and 20 B forces.}}}
 
\end{minipage}

 \end{minipage}

\end{table}

\begin{enumerate}
\setcounter{enumi}{2}
\item Average each column.  Do the results appear significant?  Offer an explanation for any anomalous results.  Alter your initial hypothesis if necessary.

\item For each Red and each Blue column, compute 
\[ \rho = \frac{\mbox{Mean Blue casualties}}{\mbox{\# Red units}} \mbox{ and } \beta = \frac{\mbox{Mean Red  casualties}}{\mbox{\# Blue units}} \]


\item Incorporate the constants $\rho$ and $\beta$ into your initial hypothesis to make an updated hypothesis.

  \vspace{1 ex}
  \noindent \textbf{Updated hypothesis:} 
  \vspace{1cm}


\item Average the values of $\rho$ and $\beta$.
\begin{itemize}
 \item The parameter \ifnum\TeacherEdition=1 $\rho = 0.43$ \else $\rho = $ \hspace{1cm} \fi represents the rate at which the number of Blue units is decreasing per unit of Red.  It is the {\it fighting strength\/} of a typical \ifnum\TeacherEdition=1 Red \else \underline{\hspace{1cm}} \fi unit.
 \item The parameter \ifnum\TeacherEdition=1 $\beta = 0.46$ \else $\beta = $ \hspace{1cm} \fi represents the rate at which the number of Red units is decreasing per unit of Blue.  It is the {\it fighting strength\/}  of a typical \ifnum\TeacherEdition=1 Blue \else \underline{\hspace{1cm}} \fi unit.
 \item The rate of change of the number of Red units is proportional to the number of the Blue units, or in the language of mathematics
 \[ \frac{dB}{dt} = \ifnum\TeacherEdition=1 -\rho R \else \fi \]
 \item The rate of change of the number of Blue units is proportional to the number of the Blue units, or in the language of mathematics
 \[ \frac{dR}{dt} = \ifnum\TeacherEdition=1 -\beta B \else \fi\]
\end{itemize}
Together, these are called the {\it Lanchester equations\/}.
\end{enumerate}


\subsection*{Solving the system}
Let $R(0) = R_0$ and $B(0) = B_0$ be the initial strengths of each force, respectively .  Solve the system 
\[ \left\{ \begin{tabular}{c}
            $\frac{dR}{dt} = -\beta B$ \\
            $\frac{dB}{dt} = -\rho R$             
           \end{tabular} 
           \right.\]
 by observing that the opposing forces would be equally matched if the attrition rates were equal on both sides; i.e.  set $\frac{dR}{dt} = \frac{dB}{dt}$ and integrate both sides. The solution,
 \[ \rho\left(R^2 - R_0^2\right) = \beta\left(B^2 - B_0^2\right) \]
 gives rise to Lanchester's $N$-square Law:
 
\begin{center}
 \textit{The fighting strengths of two forces are equally matched when the \ifnum\TeacherEdition=1 square \else \underline{\hspace{1cm}} \fi of the \ifnum\TeacherEdition=1 numerical \else \underline{\hspace{1.5cm}} \fi strength multiplied by the fighting value of the \ifnum\TeacherEdition=1 individual \else \underline{\hspace{1.5cm}} \fi for the respective forces  are equal.} \cite[p.~2145]{FWL2}
\end{center}


\subsection*{A Lanchester model}
We will analyze the battle of Trafalgar in terms of Lanchester's equations. For this exercise, assume that $\rho = \beta = 1$.

In a naval battle in 1805 off of Cape Trafalgar, Spain, 40 British vessels faced a combined French and Spanish force of 46 vessels.  The British force was outgunned, yet under the brilliant command of Admiral Horatio Nelson the British devastated the opposing force.  Nelson's plan was to split the enemy line into two halves by sending a vanguard of 8 vessels (see \cite[pp.~2154-2157]{FWL2} or \cite[pp.~25-27]{Fox2010}).

\begin{itemize}
 \item According to the N-square law, what is the expected outcome of 8 units fighting against 23?
 \item Answer: \ifnum\TeacherEdition=1 With $\rho = \beta = 1$, $R_0 = 8$, $B_0=23$, and $R=0$, we have $-8^2 = B^2 -23^2$ or $B = 465$.  We expect that the combined French and Spanish force wins the engagement with $\sqrt{465}$ or approximately 21 vessels remaining.  \else \vspace{2 cm}  \fi
 \item What advantage does a force of 32 have against 23?
 \item Answer: \ifnum\TeacherEdition=1 We have $32^2 - 23^2 = 1024-529 = 495$ so the larger force has nearly a 2 to 1 fighting advantage equal to $\sqrt{495}$ or approximately 22 vessels.   \else \vspace{2 cm}  \fi
 \item With this strategy, the British fighting strength of \ifnum\TeacherEdition=1 $32^2 + 8^2$ \else \underline{\hspace{2 cm}} \fi slightly tipped the French and Spanish strength of \ifnum\TeacherEdition=1 $23^2+23^2$ \else \underline{\hspace{2 cm}} \fi.  Research the Battle of Trafalgar to find out whether adopting this strategy paid off.
\end{itemize}

\subsection*{M\&M wargame scenario}
Using the Lanchester model with one of the sets of parameters $\rho$ and $\beta$ you obtained from the averages of entries in Table \ref{table1}, Table \ref{table2}, or Table \ref{table3}, analyze a scenario in which 20 Red M\&M's face a stronger army of 26 Blue M\&M's.  
\begin{itemize}
 \item Determine how Blue should engage Red to at least in part enjoy a 2 to 1 fighting strength advantage.
 
 \item Is it possible for Red to engage Blue in such a way to at least neutralize Blue's advantage?  Explain in detail the optimal attack strategy for the Red force.
\end{itemize}


\subsection*{Duration of battle}
The Lanchester equations may be solved for $B = B(t)$ and $R = R(t)$ through various methods.  Applying methods of linear algebra, the solution to the matrix differential equation
\[ 
 \left( \begin{tabular}{c} $R'$ \\ 
                            $B'$
        \end{tabular}
 \right)
 = \left( \begin{tabular}{rr} 0 & $-\beta$ \\
                            $-\rho$ & 0
          \end{tabular}
          \right) 
 \left( \begin{tabular}{c} $R$ \\ 
                            $B$
        \end{tabular}
 \right)
\]
may be obtained through the fundamental matrix.  Thus, we have
\[ 
 \left( \begin{tabular}{c} $R$ \\ 
                            $B$
        \end{tabular}
 \right)
 = \left( \begin{tabular}{rr} $\cosh \sqrt{\beta \rho} \ t$ & $  -\frac{\beta}{\sqrt{\beta \rho}} \sinh \sqrt{\beta \rho} \ t$ \\
                             $  -\frac{\beta}{\sqrt{\beta \rho}} \sinh \sqrt{\beta \rho} \ t$ & $\cosh \sqrt{\beta \rho} \ t$
          \end{tabular}
          \right) 
 \left( \begin{tabular}{c} $R_0$ \\ 
                            $B_0$
        \end{tabular}
 \right).
\]
It follows that the solution to the Lanchester equations is 
\[
 \left\{ \begin{tabular}{l}
          $R(t) = \cosh\left(\sqrt{\beta \rho} \ t \right)R_0 - \frac{\beta}{\sqrt{\beta \rho}} \sinh\left( \sqrt{\beta \rho} \ t \right)B_0$\\
          $B(t) = - \frac{\beta}{\sqrt{\beta \rho}} \sinh\left( \sqrt{\beta \rho} \ t \right)R_0 + \cosh\left(\sqrt{\beta \rho} \ t \right)B_0$\\
         \end{tabular}
         \right. 
\]
Then, if it is determined that, for example, Red will prevail over Blue, the duration of battle is the solution to the equation $B(t) = 0$, because the number of Blue units will be reduced to zero.

\subsection*{Returning to a round of sequential battles until one side loses}
\begin{enumerate}
\item Use one of the sets of parameters $\rho$ and $\beta$ you obtained from the averages of entries in Table \ref{table1}, Table \ref{table2}, or Table \ref{table3} and the corresponding set of differential equations to model one round of a set of battles until the size of one side, Red or Blue, is reduced to 0. How many battles did it take until the size of one side, Red or Blue, is reduced to 0? 

\item Fix $\rho$ from one of the average sets of parameters in Table \ref{table1}, Table \ref{table2}, or Table \ref{table3} and let the value of $\beta$ vary ($\rho < \beta$, $\rho = \beta$, or $\rho > \beta$. Using your differential equations, model one round of a set of battles until the size of one side, Red or Blue, is reduced to 0. How many battles did it take until the size of one side, Red or Blue, is reduced to 0. How does the duration of the round of battles relate to the value of  $\beta$? 
Recalling the meaning of $\beta$ as the fighting strength of a typical Blue unit, do the results you have obtained make sense?  Explain.

\item For fixed $\rho$ and given initial sizes for Red and Blue forces can you determine the value(s) of $\beta$ for which neither side wins, or that both sides annihilate themselves. 

\begin{quote}
The concept of parity in combat modeling is important. We define {\it parity\/} as a fight to finish that ends in a draw—neither side wins. We assume that under the conditions of parity that both sides annihilate themselves (mathematically they both go to zero). No side wins and both sides lose.'' \cite[p.~21]{Fox2010}
\end{quote}


\item In the explicit formulas $R = R(t)$ and $B = B(t)$ given above, what are the units of $t$?  Does $t$ represent time, the number of battles, or another parameter?  Explain.
\end{enumerate}



\begin{thebibliography}{98}
\bibitem{Fox2010} Fox, W. P. 2010. Discrete combat models: investigating the solutions
to discrete forms of Lanchester's combat models. {\it International Journal of Operations Research and Information Systems\/}. 1(1): 16-34.

\bibitem{NPG} Gleditch, Nils Petter. 1999. \textit{Lewis F Richardson - An Exceptional Scholar} in PRIO Blogs. Available at \href{https://blogs.prio.org/2019/12/lewis-f-richardson-an-exceptional-scholar/}{https://blogs.prio.org/2019/12/lewis-f-richardson-an-exceptional-scholar/} Accessed 29 January 2020.

\bibitem{RLJ} Johnson, Ronald L. 1989. \textit{Lanchester's Square Law in Theory and Practice}. Forth Leavenworth, Kansas: School of Advanced Military Studies.

\bibitem{FWL} Lanchester, Frederick W. 1916.  \textit{Aircraft in Warfare: The Dawn of the Fourth Arm}. London: Constable and Company.

\bibitem{FWL2} Lanchester, Frederick W. 1956.  \textit{Mathematics in Warfare} in \textit{The World of Mathematics, Vol. 4}, Newman, J. R., ed. New York: Simon and Schuster. pp.~2138-2157.
\end{thebibliography}



\teacher{
\section*{NOTES FOR TEACHERS}
\begin{itemize}
 \item It may be worthwhile to have a discussion of the assumptions implicit in the the Lanchester model.  Effects of terrain, fire suppression, and range are ignored.  Moreover, the model assumes that the opposing forces have direct line of sight of each other.  See \cite{RLJ} for other considerations and discussion of how the Lanchester model stands up to empirical evidence from the modern battlefield. 

 \item Another historical battle worthy of analysis through the Lanchester model is the Battle of the Alamo (see \cite[pp.~23-25]{Fox2010}).
 
 \item The last two sections address the duration of battle and should be considered at the same time.  They may be dropped from the scenario without affecting the experiment and basic modeling with Lanchester equations.
 
 \item You may wish to discuss the stability of the matrix differential equation appearing on p.7.  Since $p = 0$, $q=-\beta \rho$, and $p^2-4q >0$, it follows that the critical point at the origin corresponds to an unstable saddle.  In the stability diagram, parity is represented by the line $\beta B = \rho R$ while the remaining curves intersect the coordinate axes at right angles.
\end{itemize}


}





\end{document}
