2.1. Description of the Initial Military Situation and the Decision Problem
A proxy war between a coalition of countries, BLUE, and a country, RED, is considered. RED wants to increase the size of the RED territory and rule more regions. BLUE wants to involve more regions in trade and other types of cooperation. GREEN is a small and independent nation that wants to become a member of BLUE. RED attacks GREEN and tries to take control of that country.
The map in
Figure 1 shows the territory of country GREEN. The X axis, with direction east, is used to determine the location of the war front, x, at time t, denoted x(t). In order to simplify the notation, we define the western border from the condition X = 0 and the eastern border from X = 10. In some parts of the analysis, we use X = K, as a more general definition of the eastern border.
The war front is illustrated as a dashed line from south to north.
The war starts this way: RED attacks GREEN from the east and rapidly sends combined armor and infantry units along the roads, in direction west. At time t, the RED units reach the frontline x(t). The area east of the frontline x(t) is not controlled by RED, since GREEN has several GREEN military units in the area. GREEN can attack RED east of the front line. GREEN army units have been positioned to secure the area west of x(t). BLUE supports GREEN with ammunition, combat service support and artillery. This way, GREEN can temporarily stop RED from going further west from x(t).
BLUE and RED both have large amounts of nuclear weapons and other weapons of mass destruction. BLUE wants to avoid using these in order not to start a world war that would completely destroy the territories of BLUE, RED and most other parts of the planet. BLUE is economically stronger than RED and has more advanced conventional weapons, artillery with longer shooting ranges, more efficient missiles and antitank weapons.
BLUE decides not to participate in the war with troops on the ground, in order not to make RED start using nuclear weapons. However, BLUE decides to give arms support to GREEN. This support can help GREEN in the war against RED and simultaneously reduce the military power of RED, which is valuable to BLUE, also outside this particular proxy war, since RED may confront BLUE also in other regions. BLUE demands that the arms support is only used within the territory of GREEN.
2.2. Briefing on the Determination of the Optimal Strategy
The analysis contains the following parts:
The optimal dynamic arms support problem, from the BLUE perspective, is defined in general form.
The objective function is a weighted sum of the present value of the free GREEN territory, west of the front line, and the present value obtained by BLUE, represented by the net loss of military resources in the RED army, during the war.
The net loss of RED at a particular point in time is a function of the location of the front line and the size of the mobile GREEN forces east of the front line.
First, it is assumed that the expected RED net loss is proportional to (a particular definition of) the force ratio east of x, the location of the front line. Then, it is proved that the net loss function is a strictly concave quadratic function of x. It is also proved that the unique maximum of the expected RED net loss function occurs at the same warfront location, x; this also occurs if the net loss function is proportional to the force ratio raised to some strictly positive exponent plus some constant. Neither the particular value of the exponent nor the value of the added constant influence the value of x that maximizes the RED net loss function.
The location of the front line is dynamically changing and determined by a differential equation, influenced by the level of attack from RED and the level of arms support from BLUE.
Since military analysis has already convinced BLUE that RED has too limited resources and competence to win this proxy war and to gain the GREEN territory, BLUE does not think that RED is optimizing the strategy in a logical way. Furthermore, the war clearly implies considerable costs of dead and injured soldiers and noncombatants, destroyed cities, infrastructure and military resources. These costs hurt all participants in the war, in particular GREEN and RED. Furthermore, these costs are in general nonlinear functions of the strategies of all parties. For these reasons, a zero-sum game theory approach is simply not relevant. Since the war will not only determine a modified location of the front line and the borders between GREEN and RED, a standard differential game model of the war cannot capture the true and relevant problem.
The observed level of attack from RED is not possible to interpret, by BLUE, as economically optimized by RED, in the interest of the people in country RED. The BLUE interpretation is that the RED command has other motives for the attack on GREEN. BLUE has, however, qualified intelligence resources that can give a reliable prediction of the time path of the military resources that RED can and will send to the front.
From the BLUE perspective, there is an optimal position of the front. This position is a function of the weights in the objective function and all other parameters.
The optimal control solution shows that the optimal arms support strategy for BLUE is to initially send an optimized volume of arms to GREEN, which rapidly makes it possible for GREEN to move the front to the optimal position. Then, the support should be almost constant during most of the war, keeping the war front location stationary. In the final part of the conflict, when RED has almost no military resources left and has to retire from the GREEN territory, BLUE should strongly increase the arms support and make sure that GREEN rapidly can regain the complete territory and end the war.
2.3. Derivation of the Optimal Net Profit Principle
Below, fundamental mathematical methods will be used. These are well presented by Chiang (1974). We consider a country, GREEN, with a rectangular land surface. Compare the illustration in
Figure 1. The coordinate in the west to east direction is denoted X. At the border to the west, X = 0. At the border to the east, X = K. At time t, the war front has the X coordinate x(t). The war front is a line in direction north, from the southern border to the northern border.
GREEN has complete control of the territory to the west of the front. RED attacks GREEN from the east.
The number of troops that can be supported by GREEN and can be active at the front and to the east of the front, behind the RED line, attacking RED logistics during transport to the front, is proportional to the area controlled by GREEN and denoted n
G.
The distance that the RED logistics support has to travel, at time t, from the RED border to the front, is K-x(t). RED has a fixed number of tanks that can be used to protect the RED logistics. Hence, if the distance from the eastern border to the front increases, and the amount of support needed at the front per time unit is constant, then n
R, the number of tanks per protected and transported unit, decreases.
In
Figure 2, the war front is located to the west of the war front in
Figure 1. Furthermore, in
Figure 2, the number of GREEN units east of the front is lower than in
Figure 1. This illustrates Equation (1). In
Figure 2, the RED logistics arrows are thinner than in
Figure 1. This illustrates Equation (2).
Figure 2.
The war map of the country GREEN at time t. Explanations are given in the main text. In this case, the frontline x(t) = 3.0. Compare
Figure 1, where the frontline at the same point in time has another location.
Figure 2.
The war map of the country GREEN at time t. Explanations are given in the main text. In this case, the frontline x(t) = 3.0. Compare
Figure 1, where the frontline at the same point in time has another location.
y is a particular military force ratio, defined in (3).
Clearly, as we see from Equations (4) and (5), y is a quadratic function of x.
Equations (6) and (7) show that y takes the value zero in case the front is identical to the western or eastern border. In all other war front locations, y is different from zero.
First, we assume that the expected net profit of BLUE caused by RED losses is proportional to y. Equations (8)–(10) show that y has one unique optimum and that this is a unique maximum. A star indicates an optimal value. This optimum occurs when the war front is located exactly in the middle of the country GREEN, when the war front has location K/2. The absolute values of the constants c
G and c
R do not affect this result, as long as they are both strictly positive.
2.6. The Simplified Stationary Problem with Optimal Solutions in Three Different Cases
In this paper, we determine the optimal solutions to the dynamic decision problems. The optimal solutions will be reported as explicit functions and as graphical solutions to alternative numerically specified cases. First, however, we determine the optimal locations of the war front via static problems. In the later analysis, these optimal static solutions are compared to the optimal dynamic solutions. We have to select x within the region defined in (24).
STATIC CASE A:
The value of the free GREEN region, to the west of the war front, is given in (25) and illustrated in
Figure 3.
The value of the expected net profit of BLUE, caused by expected RED losses, is given in (26) and illustrated in
Figure 4.
Now, we construct the total static objective function, (27), using (25) and (26). Since we are still only interested in the optimal static warfront solution, we do not have to specify the details of the arms support cost function yet. That will, however, be relevant and important in the later parts of this paper.
If we only care about the value of the free GREEN region, the optimal value of x is 10 = K, which means that all of the GREEN territory should be liberated from RED troops. This is also found via (28) and (29). The optimal objective function value is then found in (30). Compare
Figure 3.
If we want to maximize the expected net profit of BLUE, caused by expected RED losses, given in (26) and illustrated in
Figure 4, we should use Equations (31)–(33). Hence, as we already know from Equation (9), the optimal value of x would be K/2 = 5.
Now, we optimize the location of the war front based on the total objective function (27). Equations (34)–(36) show how this is achieved. The optimal war front is now located between the different solutions that were optimal with consideration of the objective functions f
1(x) and f
2(x). This is shown in (37). These results are also illustrated in
Figure 5.
Observe that we have now analyzed a function of the type (38). Then, the maximum value of (38) is less than or equal to the maximum values of the two components f
1(x) and f
2(x). The reader can check that this also is true in
Figure 5.
STATIC CASE B:
Now, we see how the optimal location of the war front is affected if the value per area unit of the free GREEN territory increases, by 40%, as illustrated in
Figure 6. Then, in
Figure 7, we observe that the optimal location of the war front moves to the east.
STATIC CASE C:
In
Figure 8, the net profit of BLUE, caused by expected net losses of RED, increases by 100%, for every possible level of x. Then,
Figure 9 illustrates how the optimal location of the war front moves west.
2.7. Observations and Conclusions from the Static Optimizations
We may consider the decision problem as a multi objective optimization problem. We make the following observations concerning the optimal static solutions:
Let us consider the total objective functions (20) and (23). We regard them as weighted objective functions, where, in STATIC CASE A, the weight of component f
1(x) is 1, and the weight of component f
2(x) is 1. Then, the optimal location of the war front is 7.5, as shown in Equation (36) and
Figure 5. The optimal total objective function value is 562.5. Compare Equation (36) and
Figure 5.
In STATIC CASE B, we have increased the weight of f
1(x), by 40%, to the new value 1.4. Then, the optimal static location of the front moves to the east, and the total objective function value increases, compared to STATIC CASE A. Compare
Figure 6 and
Figure 7.
In STATIC CASE C, we have increased the weight of f
2(x) by 100% to the new value 2.0. Then, the optimal static location of the front moves to the west and the total objective function value increases, compared to STATIC CASE A. Compare
Figure 8 and
Figure 9.
We conclude that, in a cooperate strategy negotiation between GREEN and BLUE, it is natural that GREEN is more interested in a high value of the weight of f1(x), since the inhabitants of the GREEN territory want to have a large free territory, and that BLUE is more interested in a high value of the weight of f2(x), since the expected net value of the war is expressed as that function.
Hence, depending on the relative negotiation powers of the parties GREEN and BLUE, the optimal static solution of x is found in some location in the interval between 5 and 10, or more generally, between K/2 and K.
2.10. The Limiting Value of the Complementary Solution
As time goes to infinity, the complementary function converges to infinity, zero or minus infinity, in case A2 is strictly positive, zero or strictly negative.
The Particular Solution:
Let the particular solution have this functional form:
Finally, we conclude that the particular solution is:
Observation:
The particular solution is always a linear function of time.
The solution:
Since the complete solution is the sum of the complementary solution and the particular solution, we have:
This can be explicitly stated as:
The time derivative of x is:
We remember the expression of the adjoint variable from Equation (47). Now, thanks to the explicit form of the time derivative of x, we can express the adjoint variable as an explicit function of time:
We note that x and the adjoint variable are both explicit functions of time. These functions contain a number of parameters that are already known. These functions, however, contain two more parameters, that have not yet been determined, namely A1 and A2.
We can now determine A1 and A2 from two boundary conditions. These two boundary conditions have strictly logical motivations:
We can observe the initial value of x, at time t = 0, denoted x
0.
At time T, we want x to take the value x
T.
In some cases, we may know that the “shadow price”, the marginal capacity value or the adjoint variable, at the time horizon, T, has to be zero. In such cases, the following equations are relevant:
So, if the adjoint variable, at the time horizon, T, has to be zero, the following equation would have to be included in the linear equation system that should determine A
1 and A
2.
However, in this particular analysis, we will not make the assumption that the marginal capacity value has to be zero at time T. On the other hand, we will demand that x takes the value xT at time T. This way, we have a linear equation system with two equations that will be used to determine the relevant values of A1 and A2.
Here is the linear simultaneous equation system with two equations and two unknowns:
Thanks to Cramer’s rule, we instantly obtain the solutions:
Observations concerning the values A1 and A2:
We observe that the expressions of A
1 and A
2, (89) and (90), have real value nominators, divided by the denominator:
We have already determined that s1 and s2 are strictly different and that they are both real. We also know that s2 > s1. Hence, for all values of T, equal to or greater than 0, the denominator is strictly positive. Hence, A1 and A2 both exist, are determined via the expressions, and are real.
The signs of A1 and A2 are the same as the nominators in the corresponding expressions.
As a result, we now know these two functions:
We also already know how u is linked to the adjoint variable. Compare Equation (45).
We can obtain the optimal control as a function time, via the adjoint function:
Clearly, this can also be expressed as an explicit function of time:
Observation:
The optimal control function values can also be obtained in another way, as seen below. The two alternative ways to calculate u(t) can be used to confirm the correctness of the calculations.
The reader may confirm that Equation (95) corresponds to Equation (96).
2.11. Results Based on the General Dynamic Optimal Control Problem
The optimal and explicit time dependent functions of the arms support level, the war front location and the adjoint variable, are found in Equations (78), (80) and (95).
Now, we use the optimal general dynamic results, expressed in the forms of equations, to derive some optimal dynamic results for numerically specified cases. These optimal dynamic results are compared to the optimal statics results derived in the earlier parts of this paper.
Below, six different dynamic cases will be investigated in detail. DYNAMIC CASE 0, represents the standard case and may be viewed as a dynamic version of STATIC CASE A.
Parameter values in DYNAMIC CASE 0: a = 150, b = 10, g = 1, h = 0.1, r = 0.05, v0 = 1, v1 = −0.1, x0 = 5, T = 1 and xT = 10.
In each of the dynamic cases 1 to 6, some parameter has been changed from the value according to DYNAMIC CASE 0. All other parameters are the same as in DYNAMIC CASE 0. This way, it is possible to investigate how sensitive the optimal dynamic solutions are to different parameter values.
The dynamic cases contain many more parameters and parameter values than the static cases, since these are needed to define and handle several things that were not present in the static problem. In (40), we have the objective function in the dynamic problem, and in (41), we have the differential equation the war front. Hence, several parameters in the dynamic optimal control problem can be found in (40) and (41).
More parameters are needed in the dynamic problem than in the static problem: the parameters of the cost function of the control u, namely g and h, the rate of interest in the capital market, r, the parameters of the RED attack level function, v0 and v1, the total time of the proxy war, T, and the initial and final locations of the war front, x0 and xT.
DYNAMIC CASE 0
This case may be viewed as a dynamic version of STATIC CASE A, since the values of a and b are the same. Compare Equation (27). In
Figure 10, we observe that the location of the war front starts at the initial location, x
0 = 5, and ends at the final value x
T = 10, which means that GREEN controls the complete territory at the end of the war. Most of the time, during the war, the war front is very close to location 7.5, which also is the optimal value in STATIC CASE A.
In
Figure 11, we see the time path of the optimal arms support from BLUE to GREEN, as a function of time. This level is very high in the beginning (t < 0.2), since the initial location of the war front, 5, is far below the optimal value of the war front, 7.5, according to the STATIC CASE A. Hence, it is important to rapidly move the front to a location close to the optimal location. That can be accomplished with massive arms support, a high value of u, for low values of t. The logic behind that is clear from (41). During most of the war, the war front should be close to the optimal static value, which means that the optimal level of arms support, u, should be almost the same as the level of attack from RED, which only changes very slowly. For that reason, the level of u is low and almost constant, for 0.2 < t < 0.8. When we reach the end of the war, it is important for BLUE to send large amounts of arms support to GREEN to rapidly move the war front to the original border between GREEN and RED, namely 10. This way, GREEN regains control of the complete GREEN territory exactly when the war ends.
DYNAMIC CASE 1
DYNAMIC CASE 1 is identical to the DYNAMIC CASE 0, with respect to all parameters except for the fact that x
0 = 3. This means that the initial war front is located to the west of the corresponding war front in DYNAMIC CASE 0.
Figure 12 shows how the time path of the war front develops over time. We see that the front is rapidly moved east, until t = 0.4, when it reaches almost the same level as we had in DYNAMIC CASE 0 and in STATIC CASE A. In the end of the conflict, the DYNAMIC CASES 0 and 1 have almost identical war front developments.
In order to initially move the war front more to the east, in DYNAMIC CASE 1, than in DYNAMIC CASE 0, a higher level of arms support is needed in the early period of the war. This is also graphically seen in
Figure 13, before t = 0.4.
DYNAMIC CASE 2
DYNAMIC CASE 2 is identical to the DYNAMIC CASE 0, with respect to all parameters except for the fact that a = 170. This corresponds to the STATIC CASE B. One possible interpretation is that the relative weight in the objective function of the value of the free GREEN territory increases. Compare
Figure 6 and
Figure 7.
Figure 14 shows how the time path of the war front develops over time. We see that the front is rapidly moved further east, than in DYNAMIC CASE 0, until t = 0.4, when it reaches almost the same level as we had in STATIC CASE B. In the end of the conflict, the front moves to 10, and the complete GREEN territory is liberated from RED.
In order to initially move the war front more to the east, in DYNAMIC CASE 2, than in DYNAMIC CASE 0, a higher level of arms support is needed in the early period of the war. In the end of the war, less arms support is needed than in DYNAMIC CASE 0, since the front does not have to move very far during the final period. Compare
Figure 15.
DYNAMIC CASE 3
DYNAMIC CASE 3 is identical to the DYNAMIC CASE 0, with respect to all parameters except for the fact that b = 250. This corresponds to the STATIC CASE C. One possible interpretation is that the relative weight in the objective function of the value of the net profit of BLUE, caused by RED losses, increases. Compare
Figure 8 and
Figure 9.
Figure 16 shows how the time path of the war front develops over time. We see that the front is initially moved east to a position to the west of the corresponding position in DYNAMIC CASE 0. This position is almost the same as we had in STATIC CASE C. In the end of the conflict, the front moves to 10, and the complete GREEN territory is liberated from RED.
In order to initially move the war front less to the east, in DYNAMIC CASE 3, than in DYNAMIC CASE 0, a lower level of arms support is needed in the early period of the war. In the end of the war, more arms support is needed than in DYNAMIC CASE 0, since the front must move very far during this period. Compare
Figure 17.
DYNAMIC CASE 4
DYNAMIC CASE 4 is identical to the DYNAMIC CASE 0, with respect to all parameters except for the fact that v0 = 5. This means that the RED attack level is considerably higher than in DYNAMIC CASE 0.
Figure 18 shows how the time path of the war front develops over time. We see that the front develops exactly as in DYNAMIC CASE 0. In order to handle the increased level of RED attack, keeping the front line at in the same location as in DYNAMIC CASE 0, the level of arms support from BLUE to GREEN must be higher, during every time interval. This is clearly illustrated in
Figure 19.
DYNAMIC CASE 5
DYNAMIC CASE 5 is identical to the DYNAMIC CASE 0, with respect to all parameters except for the fact that v
1 = 1. This means that the level of RED attack is an increasing function of time. In DYNAMIC CASE 0, the RED level of attack was a slowly decreasing function of time.
Figure 20 shows how the time path of the war front develops over time. We see that the front develops exactly as in DYNAMIC CASE 0. In order to handle the increasing level of RED attack, keeping the front line at in the same location as in DYNAMIC CASE 0, the level of arms support from BLUE to GREEN must be a higher than in DYNAMIC CASE 0, particularly in the later part of the war. This is also shown in
Figure 21.