Speed of Performance of a Group of Controlled Objects

Abstract

The problem of the speed of performance (speed, performance) of a group of controlled objects is considered, the initial state of each of which is given, and the final state is selected from a fixed set of possible terminal states (targets). For each object, it is required to assign a target and find the control that ensures that the group simultaneously reaches all the selected targets in the shortest time. Since the speed-optimal controls of individual objects do not solve the problem of group speed, the so-called minimally delayed trajectories are used. A technique for solving the problem is developed, which includes an algorithm for solving the minimax assignment problem and a procedure for using the optimal and minimally delayed trajectories. The problem of the speed of a group of objects, whose planar motion is represented by Markov–Dubins trajectories, is solved.

Appendices

APPENDIX

ALGORITHM FOR SOLVING THE MINIMAX ASSIGNMENT PROBLEM

The minimax assignment problem or, equivalently, the linear bottleneck assignment problem (see Section 1) is reduced to the following integer optimization problem. For the given matrix $T = ({{T}_{{ij}}})$ of dimensions $M \times N$, whose elements are equal to the time required to reach the targets (${{T}_{{ij}}}$ is the time of the motion of the $i$th object to the jth target), we find the minimum value T* and optimal placement $(j_{1}^{*},...,j_{M}^{*})$ on which this value is reached:

$$T\text{*} = \mathop {\min }\limits_{({{j}_{1}},...,{{j}_{M}})} \mathop {\max }\limits_{i = 1,...,M} {{T}_{{i\,{{j}_{i}}}}}.$$

(A.1)

Minimization is carried out over all placements (j₁, …, j_M) by M of the possible N numbers.

To solve the problem, the branch and bound method is used, according to which, instead of a complete enumeration of all placements, a minimizing sequence of placements $\{ (j_{1}^{s},...,j_{M}^{s})\}$, $s = 1,2,...$, is constructed, leading to a monotonic decrease in the minimized function (A.1):

$${{T}^{1}} > {{T}^{2}} > ... > {{T}^{s}} > ...,$$

(A.2)

where ${{T}^{s}} = T(j_{1}^{s},...,j_{M}^{s})$. Sequence (A.2) is finite, since the number of placement variants is limited.

For the algorithm to work, apart from the matrix T of $M \times N$ dimensions, whose elements are equal to the time required to reach the targets (T_ij is the time of the motion of the ith object to the jth target), we will also need an $M \times N$-sized matrix J to record the placements. In the ith row of matrix J, we will write sequentially the numbers of targets available for the ith object. The remaining elements of this row are zeros. The first column of matrix J determines the placement $({{J}_{{11}}},...,{{J}_{{M1}}})$, i.e., the first object is assigned the target ${{J}_{{11}}}$; the second, ${{J}_{{21}}}$; etc. The remaining columns are auxiliary. For example, at the beginning, while there is not a single element of sequence (A.2), we write down the numbers of all targets in the first row; in the second, all targets, except for the target ${{J}_{{11}}}$, which is assigned to the first object; etc. If the value ${{T}^{s}}$ of sequence (A.2), has been obtained, then in the first row we enter the numbers of those targets that can be reached by the first object in less than ${{T}^{s}}$, i.e., ${{T}_{{1j}}} < {{T}^{s}}$, $j = 1,...,N$; in the second row, the numbers of the targets for which ${{T}_{{2j}}} < {{T}^{s}}$, $j = 1,...,N$, $j \ne {{J}_{{11}}}$, except for the one selected for the first object, etc. We emphasize again that only the first column of matrix J defines the real location. The rest of the columns are for generating variants. For example, when shifting all elements of the last row one position to the left in the first column, we get a new placement.

We proceed to the description of the algorithm. We assume matrix J is a zero matrix ($J = O$), $s = 0$, ${{T}^{0}} = + \infty$.

Step 1. If matrix $J = O$, then we assume k = 1 and go to Step 2. If matrix $J \ne O$ (in it only the first k rows are nonzero, $k < M$, and the nonzero elements of the first column form part of the destination $({{J}_{{11}}},...,{{J}_{{k1}}})$), then we assume $k = k + 1$ and go to Step 2.

Step 2. We complement the rows of matrix J, starting from the first zero row k:

in the kth line, we write the numbers of the columns of the elements of the kth row of matrix $T$, which are less than T^s, while columns with previously selected pairs are not taken into account $(i,{{J}_{{i1}}})$, $i = 1,...,k$;

in the (k + 1)th line, we write the column numbers of the elements of the (k + 1)th row of matrix T, which are less than T^s, while columns with previously selected pairs are not taken into account $(i,{{J}_{{i1}}})$, $i = 1,...,k + 1$, etc.

If we were able to complete all $M$ rows, a new assignment is received $({{J}_{{11}}},...,{{J}_{{M1}}})$. Then we calculate the new value

$${{T}^{{s + 1}}} = \mathop {\max }\limits_{i = 1,...,M} {{T}_{{i\,{{J}_{{i1}}}}}}.$$

We assume $J = O$, $s = s + 1$, and go to Step 1.

If it was not possible to fill in all the rows, we go to Step 3.

Step 3. If the matrix $J = O$, then the solution is complete and we go to step 4. If the matrix $J \ne O$ (in it only the first k rows are nonzero, $k < M$). We move all elements of the kth line one position to the left ${{J}_{{kj}}} = {{J}_{{kj + 1}}}$, $j = 1,N - 1$, and we add the last zero element ${{J}_{{kN}}} = 0$ to this row. If the kth row turned out to be zero, then we shift the elements in the ($k - 1$)th row, etc. If there are nonzero rows in matrix J, go to step 2. Otherwise (when J = O), the solution is completed and we go to Step 4.

Step 4. The last assignment $({{J}_{{11}}},...,{{J}_{{M1}}})$ and value ${{T}^{{s + 1}}}$ of the function to be minimized, found in Step 2, are the optimal assignment and the minimax value of the time required to reach all targets, respectively.

The number of calculations can be slightly reduced by replacing the initial value ${{T}^{0}} = + \infty$, for example, by the value

$${{T}^{1}} = \mathop {\max }\limits_{i = 1,...,M} {{T}_{{ii}}},$$

which corresponds to the placement $(1,2,...,M)$. The first placement can be obtained by choosing for the first object the closest target to it, which corresponds to the smallest element in the first row of matrix T; for the second, the nearest target to it, which does not coincide with the target chosen for the first; etc. Of course, such a choice of the first placement does not change the estimate of the number of computations in the general case, but it can reduce them in a particular example. Note also that the algorithm finds one of the possible solutions if there are several of them.

At M = N, it is possible to set the problem that certain targets must be reached by several objects. For such purposes, it is necessary to write several equal columns in matrix T. For example, if the first three columns of matrix T coincide, then three moving objects will be directed to one target.

Example 5. The time required for each of the four moving objects to reach each of the five targets is indicated in the matrix

$$T = \left( {\begin{array}{*{20}{c}} 5&1&5&2&4 \\ 4&3&3&5&2 \\ 1&3&5&5&2 \\ 2&2&1&4&1 \end{array}} \right).$$

It is required to find the minimum value T* of the maximum duration of the motion,

$$T\text{*} = \mathop {\min }\limits_{({{j}_{1}},{{j}_{2}},{{j}_{3}},{{j}_{4}},{{j}_{5}})} \max \{ {{T}_{{1\,{{j}_{1}}}}},{{T}_{{2\,{{j}_{2}}}}},{{T}_{{3\,{{j}_{3}}}}},{{T}_{{4\,{{j}_{4}}}}},{{T}_{{5\,{{j}_{5}}}}}\} ,$$

as well as the optimal assignment $(j_{1}^{*},j_{2}^{*},j_{3}^{*},j_{4}^{*},j_{5}^{*})$ on which this value is reached.

Solution. Applying the algorithm, we successively obtain assignments and the corresponding values of the minimized function $T({{j}_{1}},{{j}_{2}},{{j}_{3}},{{j}_{4}},{{j}_{5}})$ $= \max \{ {{T}_{{1\,{{j}_{1}}}}},{{T}_{{2\,{{j}_{2}}}}},{{T}_{{3\,{{j}_{3}}}}},{{T}_{{4\,{{j}_{4}}}}},{{T}_{{5\,{{j}_{5}}}}}\}$:

$$T(1,2,3,4) = 5;\quad T(2,1,5,3) = 4;\quad T(2,3,1,5) = 3;\quad T(2,5,1,3) = 2.$$

Four assignment had to be built; i.e., the minimizing sequence (A.2) has four elements. A complete search of all 120 placements shows that there are two more solutions, in addition to the one obtained, (4, 5, 1, 2) and (4, 5, 1, 3), which, of course, give the same minimum.

Example 6. The time required for each of the four moving objects to reach each of the four targets is indicated in the matrix

$$T = \left( {\begin{array}{*{20}{c}} 1&1&2&2 \\ 3&3&1&1 \\ 2&2&5&5 \\ 2&2&3&3 \end{array}} \right).$$

Since the first and second, as well as the third and fourth columns, are the same, there are actually two targets; and two objects should be directed to each of them. It is required to find the minimum value T* of the maximum duration of the motion,

$$T\text{*} = \mathop {\min }\limits_{({{j}_{1}},{{j}_{2}},{{j}_{3}},{{j}_{4}})} \max \{ {{T}_{{1\,{{j}_{1}}}}},{{T}_{{2\,{{j}_{2}}}}},{{T}_{{3\,{{j}_{3}}}}},{{T}_{{4\,{{j}_{4}}}}}\} ,$$

as well as the optimal assignment $(j_{1}^{*},j_{2}^{*},j_{3}^{*},j_{4}^{*})$ on which this value is reached.

Solution. Applying the algorithm, we successively obtain the assignments and the corresponding values of the minimized function $T({{j}_{1}},{{j}_{2}},{{j}_{3}},{{j}_{4}}) = \max \{ {{T}_{{1\,{{j}_{1}}}}},{{T}_{{2\,{{j}_{2}}}}},{{T}_{{3\,{{j}_{3}}}}},{{T}_{{4\,{{j}_{4}}}}}\}$:

$$T(1,2,3,4) = 5;\quad T(1,3,2,4) = 3;\quad T(3,4,1,2) = 2.$$

Three destinations were required to be built; i.e., the minimizing sequence (A.2) has three elements. A complete search of all 16 permutations shows that there are three more solutions, in addition to the one obtained, namely, (4, 3, 2, 1), (4, 3, 1, 2), and (3, 4, 2, 1), which, of course, give the same minimum.