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Abstract While the utility of intelligence as force multiplier during warfare is widely accepted there have been few attempts to quantify its benefits. In this paper Lanchester combat models are developed to understand how superiority in intelligence can compensate for an inferior force ratio and how the time for one side to defeat the other is affected by the use of intelligence. It is found that intelligence does act as a force multiplier; however, its utility to compensate for inferior force ratio is less than commonly appreciated, proportional to the square root of the relative advantage in intelligence. Similarly, the time to defeat is proportional to the inverse of the square root of the relative advantage in intelligence, so that greatly increasing one side’s superiority in intelligence only produces a modest decrease in the time to defeat. The Lanchester combat models are extended to a hyperbolic system of partial differential equation (PDE) to investigate how intelligence influences manoeuvre warfare. These suggest that high tempo attacking operations are less sensitive to the effects of intelligence than slower operations. 1. Introduction There have been few attempts to model the benefits that intelligence contributes to a force engaged in combat. Despite a lack of quantitative evidence, intelligence is widely seen as an enabling capability that confers distinct advantages to a force during combat. Handel (1990) describes intelligence as a ‘force multiplier’ and Khan (2001) calls intelligence an ‘optimizer of resources’. Conversely, Keegan (2003) considers that intelligence may be of benefit to a commander but does not ‘point out unerringly the path to victory’. Here, we develop models to quantify how much of a force multiplier intelligence acts as during military engagements. In this paper, we are concerned with operational intelligence, the process by which information on the nature of the terrain, enemy forces and their intentions is collected and analysed to predict the enemy’s future actions. Rather than distinguish between processed intelligence disseminated by a military force’s intelligence staff and target identification from reconnaissance assets, we consider intelligence to consist of all timely information that can assist a military commander’s decision-making. Every decision made by a commander can be considered as a set of possible outcomes for a number of events. For example, one decision may be whether to attack or defend. Both of these events will have a probability of success associated with them. The outcome of the decision cannot be known with full certainty until after the event has occurred; only the probability of the outcome can be estimated prior to the event. Here, intelligence is considered to be any information that enables a commander to estimate or refine the probability of an outcome. Many factors influence how individual commanders actually make decisions such as their personal experience, training, orders from superiors and advice from subordinates. Intelligence received during or prior to actual combat has to compete with all of these factors in order to influence the decision. This approach will be considered in later work. Previous models to examine the value of intelligence have combined dynamic models of combat with fixed decision points (Cesar et al., 1992) and the application of inventory control models to consider the use of intelligence in resource allocation problems (Riordan, 2003). Models to quantify decision making in risk management have recently been developed (Yang & Qu, 2016) share many similarities with modelling intelligence (de Almeida et al., 2017). One approach to building a model of intelligence value would be to assume that the effects of intelligence can be represented by, say, a power-law of the ratio of the sizes of the opposing forces. The value of the indices of the power-law could be found by fitting historical data of the attrition of one force against another over the course of an engagement. The problem with this method is that it is rare to find case studies from military history that unambiguously allow the effects of intelligence to be identified. Handel (1990) recognized the problem in studying the utility of intelligence in military operations is that its precise effects are difficult to isolate and measure with accuracy. This lack of reliable data sets limits the utility of a data fitting or stochastic methods in understanding the benefits that intelligence provides. Instead of trying to fit actual combat data to a model, we adopt a deterministic approach and consider a highly abstract military engagement against to forces where the only difference between them is their ability to use intelligence to direct combat power. This approach avoids the need to isolate the effects of intelligence in actual combat, by exploring its utility when it is contrived to be the only differentiator between combatants. A limitation with a deterministic approach is that it ignores the fact that intelligence collection and analysis are highly uncertain activities, where random variables often influence intelligence’s overall utility. While the use of deterministic modelling can provide an understanding of the overall effects of intelligence to explore the value of the different stages of acquiring and processing intelligence must include an element of uncertainty. One technique that may assist this understanding is to use of stochastic simulations of combat to model engagements between to forces. By repeating the same engagement but with varying levels of intelligence use for each force it should be possible to gain an understanding of how intelligence use effects the outcome of combat. This approach has been will be considered in later work. Here we develop a series of high-level Lanchester-type combat models to quantify the value of intelligence using the three traditional high-level measures of effect used to evaluate military operations: the probability of success, time take to achieve an effect and the number of casualties (attrition) (Rowland, 2006). Lanchester type equations have previously been used to model aimed and unaimed fire between two forces (Keane, 2011) and the value of reconnaissance in combat (Johnson, 1996); both these studies only considered attrition as a measure of effect. The aim of the Lanchester approach is to model the relative ability of each force to use intelligence to locate and manoeuvre to engage an opponent. We assume that the only difference in combat power between the opposing forces is their ability to use intelligence. Further, we will model the effects of using intelligence at the operational level, so that we are concerned with the overall ability of a force to correctly identify the course of action of the opposing force at the highest level. These two assumptions are consistent with the description of operational intelligence given above; however, one limitation with this approach is that it may not give sufficient weight to tactical differences between the forces, such as differences in speed of manoeuvre. As a starting point to understand this limitation, we introduce a system of partial differential equations to consider spatial effects in Section 3. The models describe how the use of intelligence by each force affects its ability to locate and target or manoeuvre to engage the other force at the operational rather than the tactical level. That is we are concerned with how intelligence aids the efficient deployment of large formations rather than the performance of individuals or small units. Hence the form of the equations (e.g. linear, squared, etc) relating the change in the size of each force is of greater interest than determining values for the parameters used in a given model. The structure of this paper is as follows, Section 2 develops Lanchester’s equations to model the combat multiplying effect of intelligence when both sides make use of intelligence, neither side uses intelligence, and where one side uses intelligence. While the form and the solutions of Lanchester’s equations presented here are well known, their application to determine the impact of intelligence on each of three measure of campaign effect is novel. Section 3 develops Lanchester type partial differential equations to take account of spatial and temporal changes during combat. Protopopescu et al. (1987; 1989) derived a reaction-diffusion equation form of Lanchester’s equations to model movement in space and time. This approach has been extended by Spradlin & Spradlin (2007) to take account of three-dimensional movement across the battlespace. The diffusion approach has been criticized (Keane, 2011; González & Villena, 2011) for introducing smearing, an unrealistic random-walk type motion, that is not representative of troop movements. To overcome these limitations, we propose an original system of linear, hyperbolic partial differential equations, similar in form to the transport equation. Section 4 describes some of the limitations to the simplified models presented here, using the time for defeat of one force by another as a representative measure of success, and discusses some extensions to overcome these limitations. 2. Lanchester modelling 2.1. Both sides use intelligence Consider two opposing forces x and y with time varying populations, x = x(t) and y = y(t) with initial populations x(0) = x0 and y(0) = y0, respectively. Lanchester (1916) modelled combat between the two forces as a coupled pair of differential equations, with rate of loss (i.e. casualty rate) on each side proportional to the size of the opposing force. So that at time t, \begin{align} \begin{aligned} &\frac{dx}{dt}=-ay,\qquad a>0,\\ &\frac{dy}{dt}=-bx,\qquad b>0. \end{aligned} \end{align} (2.1) The factors a and b, assumed to be positive constant in (2.1) represent the weighting factors between the effectiveness of fire for x and y. In general, the values of these constants will be a combination of different factors including firepower, target acquisition and manoeuvrability. Lanchester considered (2.1) to represent two forces engaging each other with aimed (i.e. targeted) fire. Rather than represent the effectiveness of fire, the models introduced in this paper consider the relative ability of each force to use intelligence to locate and manoeuvre to engage its opponent. To model the effects of intelligence we assume that both forces have identical weapons and capabilities and the only difference between them is their capability to collect and process operational intelligence. Since we are only interested in the relative values of constants a and b we can rescale the problem by normalizing with respect to a and write (2.1) as a Cauchy problem \begin{align} \dot{\boldsymbol{x}} = A{\boldsymbol{x}},\quad{\boldsymbol{x}}\left(0\right)={{\boldsymbol{x}}}_{{\mathbf 0}} = \left( \begin{array}{@{}c@{}} x_{0} \\ y_{0} \end{array} \right)\! , \end{align} (2.2) where A is a (2 × 2) matrix with eigenvalues |$\pm \sqrt {b}$| and corresponding eigenvectors |$\underline {v}_{1}=\left ({1\atop{-\sqrt {b}}}\right )$| for |$\sqrt {b}$| and |$\underline {v}_{2}=\left ({1\atop{\sqrt {b}}}\right )$| for |$-\sqrt {b}$|. So that the rate of decrease in the size of each force is proportional to the square root of its relative capability to use intelligence. For realistic combat modelling we are only concerned with the region x ≥ 0 and y ≥ 0. The solution to (2.2) is \begin{align} {\boldsymbol{x}}\left(t\right)={{\boldsymbol{e}}}^{{\boldsymbol{At}}}{{\boldsymbol{x}}}_{0,} \end{align} (2.3) where A in (2.2) has two distinct, independent eigenvectors |${{\boldsymbol{e}}}^{{\boldsymbol{At}}}=Ve^{Dt}V^{-1} $|, where V is the matrix of eigenvectors, |$V = \left ( {\boldsymbol{v}}_{1} \quad {\boldsymbol{v}}_{2} \right )$| and D is the diagonal matrix of the eigenvalues of A. \begin{align*} {\boldsymbol{e}}^{\boldsymbol{At}}=\left( \begin{array}{@{}cc@{}} \cosh\sqrt{b}t & -\frac{1}{\sqrt{b}}\sinh\sqrt{b}t \\ -\sqrt{b}\sinh\sqrt{b}t & \cosh\sqrt{b}t \end{array} \right)\!. \end{align*} So, \begin{align*} x(t)=x_{0} \cosh\sqrt{b}t-\frac{y_{0}}{\sqrt{b}}\sinh\sqrt{b}t \end{align*} and \begin{align*} y(t)=y_{0} \cosh\sqrt{b}t-x_{0}\sqrt{b}\sinh\sqrt{b}t. \end{align*} Eliminating t from (2.2) gives the well-known Lanchester square law \begin{align} {y_{0}^{2}}-y^{2} =b\left({x_{0}^{2}}-x^{2}\right)\!. \end{align} (2.4) If the opposing forces are of equal size at the start of the conflict, the force with the superior target acquisition intelligence wins: so b > 1 implies that force x wins and b < 1 implies that force y wins. If b > 1, (2.4) implies that force y can still win the engagement if its initial strength is greater than the multiple of the square root of force x’s relative advantage in intelligence capability, i.e. |$y_{0}>\sqrt {b}\ x_{0}$|. Similarly, if b < 1, force x can win if |$x_{0}>\frac {y_{0}}{\sqrt {b}}$|. Hence, applying Lanchester’s model of direct fire implies that the capability of intelligence in terms of compensating for inferior numbers is proportional to the square of the relative advantage in intelligence. So to defeat a force twice its size an army would need an intelligence capability/advantage four times greater than its numerically superior foe. The value of intelligence to one side is proportional to the square root of their advantage, or disadvantage, in intelligence capability over their opponent. There may be other limitations that prevent a force out-thinking its rival in this manner. Historical analysis of actual military combat (Dupuy, 1979) suggests that if one force outnumbers the other by greater than 6:1 it has a probability of victory of >95%. If x0 > y0 and b > 1, the time taken for force x to defeat force y, τ, is \begin{align} \tau=\frac{1}{2\sqrt{b}}\ln\left(\frac{\sqrt{b}x_{0}+y_{0}}{\sqrt{b}x_{0}-y_{0}}\right)\!. \end{align} (2.5) The time to defeat is not particularly sensitive to the initial force ratio |$\frac {x_{0}}{y_{0}}$|; however, doubling b, force x’s intelligence capability more than halves the time for x to defeat y. Hence, (2.5) suggests that for a superior force, increasing its intelligence capability increases its operational tempo by a greater amount than the increase in capacity. The third measure of effectiveness is the number of casualties suffered by one side after a given time. As above, if x0 > y0 and b > 1 after a time |$\frac {1}{\sqrt {b}}$|, force x is reduced by a factor of \begin{align*} \left\{ e^{1} \left(1-\left(\sqrt{b} x_{0} /y_{0} \right)^{-1}\right)+e^{-1} \left(1+\left(\sqrt{b} x_{0} /y_{0}\right)^{-1}\right) \right\} \Big/2. \end{align*} As with (2.5), the number of casualties suffered is not very sensitive to the initial force ratio, but doubling force x’s intelligence capability reduces its attrition by slightly less than a factor of 2. Operational level Lanchester modelling of two forces both using intelligence, but where one side has a relative advantage in intelligence capability over the other, suggests that if the dominant force doubles its intelligence capability, it increases its probability of success by only a factor of |$\sqrt {2}$|. The operational tempo; however, is slightly more than doubled and the attrition is reduced by around a half. The value of intelligence appears to be greatest in reducing operational tempo, followed by reduction of casualties; while the increase in probability of success is only proportional to the square root in the increase in intelligence capability. 2.2. Neither side uses intelligence Lanchester also considered two opposing forces engaging each other with unaimed or area fire. In this case, the rate of decease of one side is proportional to its own population (i.e. the size of the target) as well as the population of the opposing force (rate of fire). The equations for area fire are \begin{align} \begin{aligned} \frac{dx}{dt}=-axy,\qquad a>0 \;\\ \frac{dy}{dt}=-bxy,\qquad a<0. \end{aligned} \end{align} (2.6) In an intelligence context, (2.6) can be said to represent the case where neither side uses intelligence to locate the enemy. Solving (2.6) gives a linear relationship between the populations of both forces \begin{align} a\left(\,y_{0}-y(t)\right)=b\left(\,x_{0}-x(t)\right)\!. \end{align} (2.7) When a = 1, if the opposing forces are of equal size at the start of the conflict, neither side uses intelligence, b > 1 implies that force x wins and b < 1 implies that force y wins. If b > 1, (2.7) implies that force y can still win the engagement if its initial strength is greater than the multiple of force x’s non-intelligence capability (e.g. firepower or training), i.e. y0 > bx0. Similarly, if b < 1, force x can win if |$x_{0}>\frac {y_{0}}{b}$|. Compared with the square-root dependency of (2.4). With a = 1, and bx0 > y0, eliminating y from system (2.6) produces a second order, non‐linear differential equation in x \begin{align} x\ddot{x}-\dot{x}^{2}+b\dot{x}x^{2}=0. \end{align} (2.8) For |$x\neq 0$|, using the substitution |$u=\frac {\dot {x}}{x}$|, this reduces to a linear equation with the solution \begin{align} x(t)=\frac{x_{0}(bx_{0}-y_{0})}{bx_{0} - y_{0}e^{-(bx_{0}+y_{0})t}},\quad x\neq0. \end{align} (2.9) From (2.7), the dependency of y with time is \begin{align*} y(t)=y_{0} -b\left(x_{0} -\frac{x_{0} \left(bx_{0} -y_{0} \right)}{bx_{0} -y_{0} e^{-(bx_{0} +y_{0} )t} } \right)\!. \end{align*} As |$t\to \infty $|, |$x\to x_{0} -\frac {y_{0} }{b} $|, and the time taken for y to be defeated is \begin{align*} \tau =\frac{1}{\left(bx_{0} +y_{0} \right)} \ln \left(\frac{y_{0} }{x_{0} \left(b-1\right)} \right)\!, \end{align*} so that in the absence of intelligence to direct combat, the side that has a combat advantage multiplied by its initial force size greater than the opposition is victorious. 2.3. One side uses intelligence A limiting case for (2.2) is when one force, x say, has an intelligence capability that can be described by a constant coefficient b and the opposing force, y, has no intelligence capability to locate the enemy and instead must rely on area fire. In this situation, the Lanchester equations become \begin{align} \begin{aligned} &\frac{dx}{dt}=-xy,\\ &\frac{dy}{dt}=-bx,\qquad b>0. \end{aligned} \end{align} (2.10) Eliminating t and solving gives a quadratic in y \begin{align} {y_{0}^{2}} -y^{2} \left(t\right)=2b\left(x_{0} -x(t)\right)\!. \end{align} (2.11) So that for b > 1, for force y to win, |$y_{0}>\sqrt {2bx_{o}}$| . Equation (2.11) implies that a force unable to use intelligence to locate an enemy, must outnumber the opposition by the square root of twice that force’s initial population multiplied by the square root of twice that force’s intelligence capability. Eliminating x from system (2.11) produces a second order, non-linear differential equation in y \begin{align} \ddot{y}+\dot{y}y=0. \end{align} (2.12) (2.12) can be solved using the substitution |$\nu =\dot {y}$| gives \begin{align} y(t)=A\left[(\textrm{tanh}(At/2)+y_{0}/A)/(1+(y_{0}/A))\left.(\textrm{tanh}(At/2))\right)\right]\!, \end{align} (2.13) where |$A=\sqrt {{y_{0}^{2}} -2bx_{0} } $|. If |$y_{0}>\sqrt {2bx_{0}}$|, this implies that A ∈ |$ \mathbb{R} $| and that force y will beat force x with the size of force y reduced to A. From (2.11), the dependency of x with time is \begin{align} x(t)=(2b)^{-1}\left[2bx_{0}-{y^{2}_{0}}+A^{2}((\tanh({At}/{2})+({y_{0}}/{A}))\big(1+(y_{0}/{A})\textrm{tanh}({At}/{2}))^{-1}\big)^{2}\right]\!. \end{align} (2.14) Hence (2.14) is a solution to the non-linear, second order differential equation \begin{align} x\ddot{x}-\dot{x}^{2}-bx^{3}=0. \end{align} (2.15) If |$y_{0} <\sqrt {2bx_{0} } $|, then (2.12) can be integrated to give \begin{align} y(t)=B\left[\left((\,y_{0}/{B})-\tan({Bt}/{2})\right)\left(1+(\,y_{0}/{B})\tan({Bt}/{2})\right)^{-1}\right]\!, \end{align} (2.16) where |$B=\sqrt {2bx_{0} -{y_{0}^{2}}}$|. From (2.11), the dependency of x with time is \begin{align} x(t)=(2b)^{-1}\left[2bx_{0}-{y_{2}^{0}}+B^{2}\left(\left((\,y_{0}/B)-\tan(Bt/2)\right)\left(1+(\,y_{0}/B)\tan(Bt/2)\right)^{-1}\right)^{2}\right]\!. \end{align} (2.17) When |$y_{0} <\sqrt {2bx_{0} } $|, x, the force using intelligence defeats force y leaving force x reduced to |$\frac {2bx_{0} -{y_{0}^{2}}}{b}$|. Similar equations were derived by Deitchman (1962) to model ambush attacks in guerilla warfare. (2.16) implies that the time to defeat y by x is |$\frac {2}{B} \arctan \left (\frac {y_{0} }{B} \right )$| when |$y_{0} <\sqrt {2bx_{0} } $| and when |$y_{0}>\sqrt {2bx_{0} } $|, from (2.13), the time for x to be defeated by y is |$\frac {2}{A} \arctan h\left (\frac {y_{0} }{A} \right )$|. In the combat model considered in Section 2.1 where both sides use intelligence, the parameter b represents the relative advantage in intelligence that one side (force x) has over the other. Typical values for b in Section 2.1 range from 1–10. When only force x uses intelligence, the parameter b is the absolute advantage in intelligence that x has over y. From (2.11), |$b\ ~ \ y_{0} $|, this suggests a clear advantage in force ratios and hence probability of success for a force exploiting intelligence against a force that does not. The Lanchester equations consider only the temporal effects of attrition warfare and not the spatial effects of the battlefield. In the next section we consider how intelligence value compares with intelligence capability using battlefield models that take account of both spatial and temporal factors. 3. Hyperbolic systems and relative intelligence To include spatial effects in Lanchester equations, consider two opposing forces with populations f and g, such that |$\mathbb{R} ^{2}\otimes \mathbb{R} ^{+}\rightarrow \mathbb{R} ,\,f=f(x,y,t)$| and g = g(x, y, t), where x and y are orthogonal directions in a plane. Suppose that both sides have an intelligence capability to exploit intelligence to locate each other’s forces in the battle space, then from (2.2), the Lanchester system of equations is \begin{align} \begin{aligned} &\frac{df(x,y,t)}{dt}=-G(x,y,t,f,g)\\ &\frac{dg(x,y,t)}{dt}=-bF(x,y,t,f,g)\qquad b>0. \end{aligned} \end{align} (3.1) Then by the chain rule \begin{align*} \frac{df(x,y,t)}{dt}=\frac{\partial f}{\partial t}+\underline{u}\cdot\nabla f, \end{align*} where |$\underline {u}=\frac {\partial x}{\partial t}+\frac {\partial y}{\partial t}$|, can be considered to be the velocity of the force. Similarly, \begin{align*} \frac{dg(x,y,t)}{dt}=\frac{\partial g}{\partial t}+\underline{v}\cdot\nabla f, \end{align*} where |$\underline {v}=\frac {\partial x}{\partial t}+\frac {\partial y}{\partial t}$|. So that (3.1) can be written as a system of partial differential equations, a system of transport equations \begin{align} U_{t}+AU_{x}=-B, \end{align} (3.2) where |$U=\left ({{f(x,y,t)}\atop{g(x,y,t)}}\right )$| and |$B=\left ({{G(x,y,t,f,g)}\atop {F(x,y,t,f,g)}}\right )$|. If |$\underline {u}=\underline {v}$|, then (3.2) is parabolic; otherwise, (3.2) is a hyperbolic system. Initial conditions f(x, y, 0) = fo and g(x, y, 0) = go, along with suitable boundary conditions are required for solutions to exist to (3.2). 3.1. Both sides use intelligence The partial differential equation equivalent to the Lanchester equations when both sides use intelligence is \begin{align} \begin{aligned} &\frac{\partial f}{\partial t}+u\frac{\partial f}{\partial x}=-g\\ &\frac{\partial g}{\partial t}+v\frac{\partial g}{\partial x}=-b f,\qquad b>0\\ &t\geq0,\quad x\in[0,1] \end{aligned} \end{align} (3.3) with initial conditions \begin{align*} f(\,x_{t},0)=f_{0} \end{align*} and \begin{align*} g(\,x,0)=g_{0}. \end{align*} These initial conditions correspond to all of force f being initially located at some point x0, and all of force g being initially located at the origin. In this example, we consider the following Robin boundary conditions \begin{align*} f(\,x_{1},t)+\left.\frac{\partial f}{dt}\right|_{x=x_{1}}=0 \end{align*} and \begin{align*} g(\,x_{1},t)+\left.\frac{\partial g}{dt}\right|_{x=x_{1}}=0. \end{align*} These absorbing boundary conditions correspond to the removal of forces out of the battlespace, so forces crossing the boundary no longer have any military effect (Protopopescu et al., 1987). Writing this (2 × 2) hyperbolic system gives an inhomogeneous, linear equation in the form of \begin{align} u_{t}+Au_{x}=Bu,\quad x\in \left[0,1\right]\!,\quad t\in [0,+\infty ), \end{align} (3.4) where |$u:[0,1]\times [0,\infty )\rightarrow \mathbb{R} ^{2},uC^{2}$| and A is the diagonal matrix |$\left ({{u}\atop {0}}\quad {{0}\atop {v}}\right )$| and B is the anti-diagonal matrix|$\left ({{0}\atop{-1}}\quad {{-b}\atop{0}}\right )$|. Differentiating with respect to t and combining the two equations in (3.3) eliminates g(x, t) leaving a second‐order equation \begin{align} f_{tt}+(u+v)f_{tx}+uvf_{xx}=bf. \end{align} (3.5) The characteristic equations of (3.5) are |$\frac {dx}{dt}=u$| and |$\frac {dx}{dt}=v$|. Consider two functions μ(x, t) and η(x, t), that are constant along the characteristic directions, \begin{align} \mu(x, t)=x-ut \end{align} (3.6) and \begin{align} \eta(x, t)=x-vt. \end{align} (3.7) Using the chain rule to transform |$f(x,t)\rightarrow f(\mu (x,t),\eta (x,t))$| reduces (3.5) to canonical form \begin{align} f_{\mu\eta}+\frac{b}{(u-v)^{2}}f=0. \end{align} (3.8) Assuming that f(μ(x, t), η(x, t)) can be written as the product of two functions \begin{align} f(x,t)=X(x)T(t). \end{align} (3.9) Equation (3.9) can be solved using separation of variables technique to give two differential equations, \begin{align*} \frac{X^{\prime}}{X}=-c\frac{T}{T^{\prime}}={\lambda}, \end{align*} where |$c=\frac {b}{(u-v)^{2}}$|. From the boundary conditions \begin{align} f(\,x,t)=Ae^{\lambda_{+}(x-ut)}e^{-\frac{c}{\lambda_{+}}(x-vt)}+Be^{\lambda_{-}(\,x-ut)}e^{-\frac{c}{\lambda_{-}}(x-vt)}, \end{align} (3.10) where the two eigenvalues are given by \begin{align} \lambda_{\pm}=\frac{1}{2(u-v)}\left\{(v-u)\pm\sqrt{(u-v)^{2}+4b}\right\} \end{align} (3.11) so that (3.10) can be simplified to \begin{align} f(x,t)=Ae^{x}e^{\left(\frac{cv}{\lambda_{+}}-\lambda_{+}u\right)t}+Be^{-x}e^{\left(\frac{cv}{\lambda_{-}}-\lambda_{-}u\right)t}. \end{align} (3.12) Using \begin{align*} \frac{\partial f}{\partial t}+u\frac{\partial f}{\partial x}=-g \end{align*} and the initial conditions, the constants can be found such that \begin{align} A=A(u,v,x_{1},b,f_{0},g_{0}) \end{align} (3.13) and \begin{align} B=B(u,v,x_{1},b,f_{0},g_{0}). \end{align} (3.14) Using the same method to write g as a decoupled second-order equation \begin{align} g(x,t)=Ce^{x}e^{\left(\frac{cv}{\lambda_{+}}-\lambda_{+}u\right)t}+De^{-x}e^{\left(\frac{cv}{\lambda_{-}}-\lambda_{-}u\right)t}. \end{align} (3.15) Applying the boundary conditions implies that \begin{align} \lambda^{\prime}=\lambda=\frac{1}{2(u-v)}\left\{(v-u)\pm\sqrt{(u-v)^{2}+4b}\right\}\!. \end{align} (3.16) From \begin{align*} \frac{\partial g}{\partial t}+v\frac{\partial g}{\partial x}=-b\ f\quad b>0 \end{align*} and the initial conditions the determine the values of constants C and D \begin{align} C=C(u,v,x_{1},b,f_{0},g_{0}) \end{align} (3.17) and \begin{align} D=g_{0}-C. \end{align} (3.18) When both forces, use intelligence to direct their firepower, the eigenvalue equations (3.11) and (3.13) indicate that the combat advantage scales as the square root of the relative advantage in intelligence between opposing forces. Similar to the dependency found in the classical Lanchester model derived in Section 2. Modelling combat when both sides use intelligence as a series of partial differential equations introduces a dependency on the relative speed |u − v| between the combatants. In the case where the relative speed of the engagement is small compared with the relative intelligence superiority we recover the |$\sqrt {b}$| dependency found in section 2. As the expressions obtained for the constants in (3.12) and (3.15) are not very amenable to analysis, a simplified example is helpful to see how the partial differential approach differs from the Lanchester equations when both sides use intelligence. 3.2. Both sides use intelligence; one side mounts a static defence Assume that force f has a superior advantage in intelligence relative to force g and moves with a speed u. Force g adopts a static defence, holding ground at the point x = 0, then v = 0 and (3.15) reduces to \begin{align} g(x,t)=Ce^{-\lambda_{+}ut}+(g_{0}-C)Ce^{-\lambda_{-}ut}. \end{align} (3.19) The time taken for force f to defeat force g is \begin{align} \tau =\frac{1}{\sqrt{u^{2} +4b}}\ln(C-g_{0}), \end{align} (3.20) where C is the constant given in (3.17), with v = 0. This compares with the time to defeat for the classical Lanchester engagement derived in Section 2 where the time to defeat is proportional to |$\frac {1}{2\sqrt {b}} $|. Figure 1 compares the classical Lanchester time to defeat with (3.20) for varying superiority in intelligence of the attacking force. For an attacker with superior intelligence and an advantageous force ratio of 4:3, the predicted time to defeat is shorter for the classical Lanchester combat model than that predicted by (3.20) over the range 1 < b < 10. Fig. 1. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model (dashed line) using (3.20) with the speed of attack equal to 1.0. Fig. 1. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model (dashed line) using (3.20) with the speed of attack equal to 1.0. Figures 2 and 3 compare the time to defeat for different attacking speeds over a range of superiority for the attacking force (1 < b < 10). When the attacking speed is less than the superiority in intelligence, the time to defeat decreases as intelligence superiority increases as in Fig 1. Conversely, if the attacking speed is much greater than the superiority in intelligence, then the time to defeat remains roughly constant as b increases (Fig. 2). This suggests that high tempo attacking operations are less sensitive to the effects of intelligence than slower operations. Fig. 2. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model (dashed line) using (3.20) with the speed of attack equal to 50. Fig. 2. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model (dashed line) using (3.20) with the speed of attack equal to 50. Fig. 3. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model (dashed line) using (3.20) with the speed of attack u = 2.5. Fig. 3. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model (dashed line) using (3.20) with the speed of attack u = 2.5. When the speed of attack is approximately equal to the superiority in intelligence the relationship between time to defeat and superiority in intelligence becomes more complicated. The time to defeat attains both a local minimum and a local maxima in the neighbourhood of b ≈ u (Fig. 3). A possible reason for this relationship is consider in the next section. The solution to the system of partial differential equations in (3.3) is dependent on the initial and boundary conditions, indicating that changes to both the tempo of operations, the initial position of combatants and the battlespace conditions (i.e. the boundary conditions) can all affect the utility of intelligence. 4. Discussion In contrast to previous models of intelligence value (e.g. Cesar et al., 1992) which considered intelligence as flow of information through forces in combat, in this paper the use (or absence) of intelligence is used to develop the structure of the equations in each combat model. The simplified Lanchester type models developed assume that with the use of intelligence to identify and locate enemy forces, the rate of attrition of the enemy force is proportional to its size. Without the use of intelligence to locate the enemy, the rate of attrition is proportional to the rate of fire which is proportional to both the size of the force and the size of the enemy. This section reviews some of the features of the models presented here and describes further work to investigate limitations with current approach. The intelligence parameter The combat models developed here used a single parameter b to represent the relative advantage in intelligence of force x over force y. No attempt was made to place a value on b. In these, very idealized, models the only difference between the forces except for their sizes was in their ability to process intelligence. All other factors such as types of weapons, rates of fire, combat effectiveness, etc are assumed to be the same. So that in Section 3.2, where no side used intelligence, b should be equal to unity. The parameter b is an aggregation of all types of intelligence ranging from tactical reconnaissance to all-source intelligence analysis. In reality, the process of collecting information and turning it into intelligence that can inform a commander’s decision-making operate at different timescales depending on the environment and the source of intelligence. Typically, high-level, all-source intelligence takes longer than tactical intelligence to process. There has been no attempt here to quantify the relationship between the time taken to conduct intelligence assessment and the accuracy of the final product. Further work will investigate the differences in intelligence value between rapid compared with in-depth intelligence analysis and assessment. To account for these different features of different intelligence sources b in (3.1) can be written as a summation of the functional dependence of each intelligence source \begin{align*} b(t)=\sum b_{i} (t), \end{align*} where each bi is an independent source of intelligence, now a function of time. Giving b an explicit dependence on time alters the solutions to (3.1) and will be the subject of further work. Modifications The transport equations derived in Section 3.1 can be modified to take account of specific dependencies on the collection and processing of intelligence by one or both sides. For example, historical analysis of intelligence operations conducted in the Middle East in 1917 (Syk, 2009) indicate that for the British force attacking fixed Turkish positions, a situation analogous to the model developed in Section 3.2, the level of intelligence received was inversely proportional to the distance of the attacker from the defender. So that in (3.1), the intelligence function \begin{align*} G \propto \frac{b}{|\,x-x_{1}|}. \end{align*} Other, refinements can be made to take account of functional dependencies on time and can be extended to include different sources of intelligence. Speed of attack versus speed of decision-making In Section 3.2 the time to defeat was dependent on the speed of the attacking force. Fig. 4 summarizes the variance on time to defeat for different values of u, the attacking speed. Fig. 4. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model using (3.20) with the speed of attack u = 0.8, 2.5, 5, 10 and 40. The force ratio of attackers to defenders is 4:3. Fig. 4. View largeDownload slide Time to defeat versus relative advantage in intelligence for the Lanchester combat model using (2.5) and the transport equation model using (3.20) with the speed of attack u = 0.8, 2.5, 5, 10 and 40. The force ratio of attackers to defenders is 4:3. In the traditional Lanchester model, the time to defeat increases as b increases from 0 to 1, i.e. when the attacking force has an inferior intelligence capacity relative to the defender. When the attacking force has superior intelligence, the time to defeat decreases as b increases. From Fig. 4, it can be seen that the greatest reduction in time to defeat occurs as the attacker’s intelligence advantage increases from 1 to 2, after that the effects of intelligence become more marginal. Using the transport equation to model take account of the speed of the attacker, the time to defeat is greater than the Lanchester engagement for 0 < b < 10, unless the attacking speed is much greater than the relative advantage in intelligence. This reflects the time taken to close with the static defender. When |$u\ ~\ b$|, the relationship between the relationship between increasing the relative advantage in intelligence and the time to defeat becomes more complicated, as discussed in Section 3.2, with the curve attaining both local maxima and minima values over the range of b. The distance between the maxima and minima increases with the speed of approach. Further work will be conducted to investigate if this relationship is real or an artefact of the model. The consequence of this relationship is that for some speeds of approach superiority in intelligence acts as an advantage to the attacking force, but at other speeds, any advantage in intelligence does not assist the attacker and the defending force wins the engagement (i.e. the time to defeat is negative). This complexity may explain why different authors have come to different conclusions as to whether intelligence favours an attacker or a defender. Khan (1978) argued that intelligence is essential to victory only in defence; however, Herman’s (1996) review of a wider range of case studies than Khan, was inconclusive noting only that intelligence can favour the attacker in some cases and the defender in others. Decoupling the influence of intelligence received by a commander during combat from prior information may help in reconciling these differing conclusions. A possible explanation for the action of intelligence during manoeuvre warfare comes from considering the dimensions of the parameter b. In the transport equations (3.5) b has dimensions of |$\left (time\right )^{-1} $|, hence we can write \begin{align*} b\equiv b^{\prime}\omega, \end{align*} where b′ is the (dimensionless) relative superiority in intelligence and ω represents the speed at which information is collected, processed and analysed to produce intelligence and disseminated to decision-makers. This suggests that when the speed at which intelligence is produced is less than the attacking speed, intelligence has little effect on the outcome of combat. Boundary conditions Solutions to the transport equations in Section 3 are highly dependent on the initial distribution of the two forces and the boundary conditions. Later work will investigate the range of boundary conditions for which unique solutions to (3.1) are possible. This dependency on initial conditions suggests that the influence of intelligence will be very different when both forces are concentrated and initially separate from each other as occurs during conventional warfare, compared with a situation when one force has initially infiltrated the other as is in the case during insurgency or terrorist conflicts. Conclusions Applying classical Lanchester models to the use of intelligence in conflict indicates that the intelligence does act as a force multiplier; however, its utility to compensate for inferior force ratio is less than commonly appreciated, proportional to the square root of the relative advantage in intelligence. Similarly, the time to defeat is proportional to |$\frac {1}{\sqrt {b} } $|, so that greatly increasing one side’s superiority in intelligence only produces a modest decrease in the time to defeat. This dependency is confirmed by the application of transport equations to model the use of intelligence in combat. 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IMA Journal of Management Mathematics – Oxford University Press
Published: Feb 1, 2019
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