Using Linear Programming with Interval Coefficients in Optimization of a Military Action

In military field is more realistic to use models of linear programming with interval coefficients method (LPIC) when we want to find the optimum solution.

In paper 1 we present the LPIC problem:

“Suppose we are given the following LPIC problem: (1) minZ=∑j=1n[cj_,cj¯]xj∑j=1n[aij_,aij¯]xj≥[bi_,bi¯],i=1,…,mxj≥0∀j \matrix{{\min \,Z = \sum\limits_{j = 1}^n {[\underline {{c_j}} ,\overline {{c_j}} ]{x_j}}} \hfill \cr {\sum\limits_{j = 1}^n {[\underline {{a_{ij}}} ,\overline {{a_{ij}}} ]{x_j} \ge [\underline {{b_i}} ,\overline {{b_i}} ],\,\,\,i = 1, \ldots ,m}} \hfill \cr {{x_j} \ge 0\,\forall j} \hfill \cr}

The best optimum and worst optimum are computed as follows: (I)

For the best optimum: (2) minz_=∑j=1ncj′xj∑j=1naij′xj≥bi_,∀ixj≥0∀j \matrix{{\min \,\underline z = \sum\limits_{j = 1}^n {c_j^{'}{x_j}}} \hfill \cr {\sum\limits_{j = 1}^n {a_{ij}^{'}{x_j} \ge \underline {{b_i}}} ,\,\,\,\forall i} \hfill \cr {{x_j} \ge 0\,\forall j} \hfill \cr} where cj′=cj_ c_j^{'} = \underline {{c_j}} and aij′=aij¯ a_{ij}^{'} = \overline {{a_{ij}}}

So the best optimum solution will be ẕ, at (x^′₁, x^′₂, … , x^′_n) and with coefficient settings c^′_j, a^′_ij and ḇ_i, ∀i and j and also the worst optimum solution z̄ with (x^″₁, x^″₂, … , x^″_n) and with coefficient settings c^″_j, a^″_ij ∀i și j.

The general solution of LPIC will be Z = [ẕ, z̄] i.e. optimal value of the problem will be between ẕ (the best case scenario) and z̄ (the worst case scenario).“

In the military field, there are a multitude of situations that can be “optimized” using various mathematical models.

Problems of distribution of forces and means on objectives refers to: optimal implementation of defense against tanks, optimal implementation of defense against air attacks, optimal choice of combinations of means of combat against different objectives, distribution of resources on different types of weapons and others.

Inspired by Căruțașu (2014) who presented and found an optimal solution to a problem of distribution of firepower on objectives, we create a similar problem, but we used numerical intervals instead of exact real numbers:

The own unit has 5 combat means that use ammunition of the M₁, M₂, M₃, M₄ and M₅ types, and the enemy has 7 types of objectives O₁, O₂, O₃, O₄, O₅, O₆ and O₇ (meaning both equipment and personnel).

Based on the probabilities of hitting each of the 7 enemy targets, the average ammunition consumption of each type necessary to destroy them was established, because of the uncertainty the values are written as intervals (the data is given in the table below).

Each destroyed objective causes the enemy certain percentage losses given in the table below, also these values are intervals because we can't be 100% sure.

Also in the table are given the quantities of ammunition of each type that the own unit has at its disposal as well as the number of objectives of each type that the enemy has.

The problem is to determine the number of enemy objectives of each type that must be destroyed so that the losses caused to the enemy are maximum, taking into account the quantities of ammunition of each type that the own unit has at its disposal.

Let x_i be number of O_i type objectives selected for destruction, i=1,7¯ i = \overline {1,7} .

The function to be maximized has the form:

Max Z=[0.5,1]x₁ + [0.25,0.5]x₂ + [1.25,1.5]x₃ + [1.75,2.5]x₄ + [0.5,0.75]x₅ + [0.75,1.25]x₆ + [2.5,3]x₇

For the constrains: ➢

no more ammunition can be used than the one the unit has at its disposal

[10,12]x₁+[8,10]x₂+[6,8]x₃+[7,9]x₄+ [9,11]x₅+[7,9]x₆+[8,10]x₇ ≤ [690,720]

Sign restrictions

-are the sign conditions imposed on decision variables:

x₁ ≥ 0

We will convert the problem in to a minimization problem and change all “≤” inequalities into “≥” inequalities.

The new model is:

Min(−Z)=[−1,−0.5]x₁+[−0.5,−0.25]x₂+ [−1.5,−1.25]x₃+[−2.5,−1.75]x₄+ [−0.75,−0.5]x₅+[−1.25,−0.75]x₆+[−3,−2.5]x₇

We obtain: ➢

The best optimum:

min −ẕ =−x₁−0.5x₂−1.5x₃−2.5x₄−0.75x₅− 1.25x₆−3x₇

We use LINDO (a linear programming solver) to find the both optimums:

Figure no. 1:

LINDO best optimum

(Source: Author's compilation)

Figure no. 2:

LINDO solution for best optimum

(Source: Author's compilation)

Figure no. 3:

LINDO worst optimum

(Source: Author's compilation)

Figure no. 4:

LINDO solution for worst optimum

(Source: Author's compilation)

So the best optimum is −ẕ=−87.25 (Figure no. 2) for the minimal problem, so for our maximum problem ẕ=87.25, and also in Figure no. 2 we have the variables. Similar the worst optimum is z̄=43.75 for the maximum problem with the fixed variables used to get this solution.

So we can conclude that optimum solution Z of this problem lies between 43.75 and 87.25, so the enemy's losses will be between 43.75% and 87.25%.

In recent decades, mathematical modeling of combat operations has developed as an indispensable tool in decision-making at all levels. These models provide the possibility of analyzing different variants of the use of forces and combat means, as well as their dynamics during the conduct of military operations.

Many operational research optimization models, such as linear programming, assume that the data used to build a model is accurate, but in the real world most of these assumptions are only approximately true. So it is more realistic to use models of linear programming with interval coefficients (LPIC).