Optimal Attack and Defense for a Number of Targets in the Case of Imperfect Interceptors

Summary

This article describes the optimal defense of a set of targets against an optimal attack, in the case of imperfect interceptors. This solution is obtained from a mathematical representation of strategic nuclear warfare as a two-sided resource-constrained optimization problem. The attacker and defender are assumed to know each other's total force sizes, and it is assumed that the attacker "moves last," i.e., the attacker allocates his weapons to targets after observing the defender's allocation of interceptors to targets. A one-on-one (or fixed salvo-to-one) firing doctrine is assumed. The attacker's goal is to maximize the total damage to the targets, and the defender's goal is to minimize this damage. It is assumed that damage on any single target can be adequately described by the "exponential" damage function defined below.

A. Mathematical Description of the Problem

1. Notation

Let us denote the total number of weapons by X and the total number of interceptors by Y. Further, let us denote the number of targets by t, and the number of weapons and interceptors allocated to the i-th target by x_i and y_i, respectively. The attacker's and defender's strategies are hence specified by vectors

x' = (x₁, x₂,..., x_t)

and

y' = (y₁, y₂,..., y_t)

where each x_i > 0, y_i > 0, and

Σ(i:1,t) x_i = X

and

Σ(i:1,t) y_i = Y

where the notation Σ(i:1,t) denotes the summation operator over the index i, ranging from 1 to t.

Using the exponential damage function, the expected damage at the i-th target is

D_i(z_i) = P_i(1 - exp(-β_iz_i))

where z_i is the number of weapons reaching the i-th target, P_i is the value of the i-th target, and β_i is the damage function parameter for the i-th target.

The interceptor kill probability, κ, is assumed to be the same for all targets.

The attacker is assumed to choose his strategy x after observing the strategy y selected by the defender. He will choose a strategy x*(y) such that

H(x*(y*),y) = max(x) H(x,y),

where H(x,y) denotes the total expected damage corresponding to the strategies x and y, and max(x) denotes maximization with respect to x (of the following function of x, in this case H(x,y)). The defender hence must choose his strategy y* such that

H(x*(y*),y*) = min(y) max(x) H(x,y) ,

and min(y) denotes minimization with respect to y.

2. An Expression for the Expected Damage

Using the exponential damage function, the expected damage on the i-th target is shown in Appendix A to be P_iQ_i where

1 - exp(-α_ix_i) , x_i < y_i

Q_i =

1 - exp(-β_i(x_i-y_i)) exp(-α_iy_i) , x_i > y_i ,

and α_i < β_i is defined by the equation

exp(-α_i) = (1 - κ)exp(-β_i) + κ .

The total expected damage is

H(x,y) = Σ(i:1,t) H_i(x_i,y_i)

= Σ(i:1,t) P_iQ_i .

We shall denote the expected damage corresponding to the optimal solutions x*, y* as H(X,Y).

3. Procedure for Finding the Solution

Our problem, thus, is to determine x*,y* such that

H(x*,y*) = min(y) max(x) H(x,y)

where

H(x,y) = Σ(i:1,t) P_iQ_i

subject to the constraints

Σ(i:1,t) x_i = X

Σ(i:1,t) y_i = Y

and x_i > 0, y_i > 0. The function H(x,y) is additively separable over the targets so the Everett-Pugh double Lagrange multiplier technique (References 2 and 3) can be used to find a solution. The problem becomes one of finding x_i, y_i corresponding to

min(y_i) max(x_i) (P_iQ_i - λx_i + μy_i)

where λ, μ are adjusted so that the constraints

Σ(i:1,t) x_i = X

and

Σ(i:1,t) y_i = Y

are satisfied. If there are several choices of (μ,λ) for which the constraints can be met (i.e., there are several solutions to the double Lagrange multiplier problem), the correct optimal solution to the problem is the solution corresponding the least damage.

For convenience we drop the subscript i, so that the problem is to find x, y corresponding to

min(y) max(x) (PQ - λx + μy) .

The solution is derived in Appendix A (Technical Details). The next section will describe the results.

B. Description of the Solution

The solution to the above two-sided optimization problem is complicated to describe analytically, and can be described best in geometrical terms. Appendix A contains such a description. In this section we shall present only a qualitative description of the optimal attack and defense allocations.

Note that the mathematical approach used to solve the present problem assumes that the x_i's and y_i's are continuous variables, rather than integers. This approximation introduces a negligible error whenever the number of interceptors and weapons allocated to targets are zero or large enough that the fractional parts are insignificant.

The optimal solution can conveniently be described by observing how the weapon and interceptor allocations to targets change for a fixed total defense level, Y, as we very the total offense level, X, from zero to a very large number. In all situations, weapons are allocated to targets so that the marginal payoff per weapon is the same on all targets. Note that for a specified X, there may be some targets that, even if undefended, yield so little payoff to the attacker that they are not attacked. Such targets are not defended, either. For a specified total defense level, Y, there are two total attack levels, X_min(Y) and X_max(Y), that are of special significance. We shall describe the nature of the solution in terms of these levels.

1. Low Attack Level (X < X_min(Y))

When the attack level is less than the value X_min(Y), every attacked target is defended by a number of interceptors that exceeds the number of weapons assigned to it. Thus damage to targets can occur only through "leakage", i.e., failure of the interceptors to kill all incoming weapons. The defense allocation is arbitrary, so long as there are more interceptors than weapons on each attacked target. The attack allocation for specified X is unique. The total payoff to the attacker is an increasing concave ("diminishing returns") function of the total attack level, X.

2. Intermediate Attack Level (X_min(Y) < X < X_max(Y))

For all attack levels lying between the values X_min(Y) and X_max(Y), the defense is unique. Such a defense is called "stable." In this region of attack levels, the attacker can choose between two different attack levels on each target. The lower level may be zero on some targets. The attacker must attack each target at either the lower level or the higher level, and can choose to attack at any combination of these permissible levels that meets his total attack level, X. The payoff is a linear function of the total attack level in this region.

The situation in the case of the low and intermediate attack levels is referred to as "defense dominant".

3. High Attack Level (X > X_max(Y))

As the total number of weapons is increased above the level X_max(Y), the defender must change his allocation. He does so by defending fewer targets, but defending the defended targets at a higher level. As the attack size increases, eventually only one target would be attacked. Also, for a sufficiently large attack size all targets would be attacked. Since the defense allocation changes as the attack size changes in this region, the defense is called "unstable". Both the attack and defense allocations corresponding to specified X are unique.

The situation in the case of a high attack level is called "attack dominant".

4. Qualification Concerning the Lagrangian Solution

The solutions to the present problem are derived by the method of generalized Lagrange multipliers. Using this technique, we can find a solution for every total attack level X < X_min(Y). Over the range X_min(Y) < X < X_max(Y), however, there are only a finite number of Lagrangian solutions, corresponding to all possible combinations of low level or high level attacks on each attacked target. For a large number of weapons, interceptors, and targets, there is a very large number of such solutions, and there would be a Lagrangian solution corresponding to a total attack size quite close to any specified total attack level. Hence in practice we need only concern ourselves with the Lagrangian solutions. (Solutions corresponding to other total attack levels differ very little from Lagrangian solutions corresponding to nearby total attack levels. These other solutions are less "efficient" than the Lagrangian solutions, since the payoff function for them lies below the marginal payoff for the additional weapons used (beyond those required by the closest Lagrangian solution using fewer weapons) is less than the total marginal payoff achieved by going to the next higher Lagrangian solution. This latter marginal payoff is defined by the slope of linear payoff function defined by the Lagrangian solutions.)

Over the range X > X_max(Y), there can be found a Lagrangian solution for any specified total attack level, X, but not for every specified total defense level, Y. The reason for this difficulty lies with the fact that, in the Lagrangian solution, a target can be defended at only two levels, one of which is zero (no defense). (The situation is analogous to the intermediate attack situation in which targets could be attacked at only two levels.) As before, this problem is not of practical significance, since we can always find a Lagrangian solution corresponding to a total defense level close to the specified total defense level.

C. Comparison of Solution in the Case of Imperfect Interceptors to That for Perfect Interceptors

It is of interest to compare how the nature of the solution to the allocation problem in the case of imperfect interceptors differs from the solution in the case of perfect interceptors. Two questions of interest are:

1. For what range of the kill probability, κ, are the allocations and total damage essentially the same as for the perfect interceptor case, κ = 1?

2. For specified interceptor kill probability, κ, and total defense level, Y, is the solution to the problem "similar" (in allocations and total damage) to the perfect interceptor problem with total defense level κY? (Note that κY is the expected number of interceptors that do not fail, and represents a possible "effective" number of perfect interceptors.)

The questions were answered by examining the nature of the solution for a "reasonable" set of hypothetical targets, described in Table 1.

Table 1. Hypothetical Target Set

Target Value Exponential Damage

Target Number (Population in millions) Function Parameter (β)

1 10 .03

2 6 .04

3 4 .05

4 3 .06

5 2 .08

6 1.5 .10

7 1.0 .20

8 1.0 .30

9 1.0 .40

10 .5 .50

The total expected damage, H(X,Y), corresponding to the optimal attack and defense allocations is plotted in Figures 1 and 2 as a function of total attack level, X, for fixed total defense level, Y.

Figure 1 corresponds to Y₁ = 400, and Figure 2 corresponds to Y₂ = 800. On each figure are three curves:

1. H(X,Y_i) for κ = 1

2. H(X,Y_i) for κ = .75

3. H(X,.75Y_i) for κ = 1.

The figures indicate first that there are substantial differences in total damage between the κ = 1 and κ = .75 cases, and that the imperfect interceptor case is not always closely approximated by the perfect interceptor case for total defense level κY. However, the difference between these latter two cases is not so great as to preclude using the "equivalent perfect interceptor" approximation for some purposes.