CYBERNETICS AND CONTROL THEORY

B. S. VERKHOVSKII

MULTIDIMENSIONAL LINEAR PROGRAMMING PROBLEMS OF THE TRANSPORTATION TYPE

(Presented by Academician A. I. Berg, 31 X 1962)

Elements of an \(s\)-dimensional matrix are considered, each cell of which is determined by \(s\) coordinates (indices) \(i_1,\ldots,i_s\), where the \(k\)-th coordinate takes the values \(i_k = 1,\ldots,n_k\); all \(n_k \ge 2\).

Let \(M\) be the set of numbers of all coordinates; \(R\) a family of subsets of the set \(M\); \(M_j \in R\) the \(j\)-th subset, and \(|M_j| = m(j)\) the number of its elements; \(j=1,\ldots,t\).

Consider an arbitrary cell. Denote by \(j(i_1,\ldots,i_s)\) those coordinates of the cell whose numbers are not elements of \(M_j\). Let, further,

\[ \sum_{k \in M_j} \]

denote summation over all values of all those coordinates whose numbers belong to \(M_j\). All elements of the matrix \(\{p_{i_1\ldots i_s}\}\) are given.

We consider the problem of finding all elements of the matrix \(\{x_{i_1\ldots i_s}\}\) satisfying the conditions

\[ \sum_{k \in M_j} x_{i_1\ldots i_s} = a_{j(i_1\ldots i_s)}, \qquad x_{i_1\ldots i_s} \ge 0 \tag{1} \]

and giving a minimum to the functional

\[ L = \sum_{k \in M} p_{i_1\ldots i_s} x_{i_1\ldots i_s}. \tag{2} \]

All \(a_{j(i_1,\ldots,i_s)} \ge 0\) are prescribed.

Remark. In the case when \(1 \le m(1)=\cdots=m(t)\le s-1\); \(t=C_s^{m(j)}\), we obtain the symmetric multi-index transportation problem of type \(T_m\) \((^3)\).

Theorem 1. In order that the problem have a feasible solution, it is necessary that the following compatibility conditions be satisfied:

\[ \sum_{k \in (M_{j_1} \cup M_{j_2}) \setminus M_{j_1}} a_{j_1(i_1,\ldots,i_s)} = \sum_{k \in (M_{j_1} \cup M_{j_2}) \setminus M_{j_2}} a_{j_2(i_1,\ldots,i_s)}, \qquad \text{where } 1 \le j_1 < j_2 \le t. \tag{3} \]

Proof.

\[ \sum_{k \in (M_{j_1} \cup M_{j_2}) \setminus M_{j_1}} a_{j_1(i_1,\ldots,i_s)} = \sum_{k \in (M_{j_1} \cup M_{j_2}) \setminus M_{j_1}} \sum_{k \in M_{j_1}} x_{i_1\ldots i_s} = \]

\[ = \sum_{k \in M_{j_1} \cup M_{j_2}} x_{i_1\ldots i_s} = \sum_{k \in (M_{j_1} \cup M_{j_2}) \setminus M_{j_2}} a_{j_2(i_1,\ldots,i_s)}, \]

which was required to be proved.

Further, in Theorems 2–5, cases are indicated in which problem (1)—(2) reduces to one or several simpler ones, and constructive proofs are given.

Theorem 2. If there exist \(M_{j_1}\) and \(M_{j_2}\) such that \(M_{j_1} \subset M_{j_2}\) \((j_1 \ne j_2)\), then the conditions

\[ \sum_{k \in M_{j_2}} x_{i_1\ldots i_s} = a_{j_2(i_1,\ldots,i_s)} \]

are dependent.

Proof. All the conditions

\[ \sum_{k\in M_{j_2}} x_{i_1\ldots i_s}=a_{j_2(i_1,\ldots,i_s)} \]

can be obtained from the conditions

\[ \sum_{k\in M_{j_1}} x_{i_1\ldots i_s}=a_{j_1(i_1,\ldots,i_s)}, \]

if the latter are summed over the indices whose numbers belong to the set \(M_{j_2}\setminus M_{j_1}\).

Theorem 3. If the union of all \(M_j\) does not form a cover of \(M\), then problem (1)—(2) decomposes into several problems, the number of indices in each of which is equal to

\[ \left|\bigcup_{j=1}^{t} M_j\right|. \]

Proof. The union of all \(M_j\) means the totality of the numbers of all those indices with respect to each of which summation is performed at least once in the conditions (1). If

\[ \left|\bigcup_{j=1}^{t} M_j\right|<s, \]

then there exist indices with respect to which no summation is performed. Let

\[ M\setminus \bigcup_{j=1}^{t} M_j=\{1,2,\ldots,r\}. \]

Then the original problem decomposes into

\[ \prod_{k=1}^{r} n_k \]

problems, each of which does not depend on the first \(r\) coordinates.

Theorem 4. If there exists a subset \(\widetilde M\) such that \(|\widetilde M|=r\geqslant 2\) and \(|M_j\cap \widetilde M|=r\), for all \(j=1,\ldots,t\), but

\[ \left|\bigcap_{j=1}^{t} M_j\right|=0, \]

then problem (1)—(2) reduces to a problem whose number of indices is \(s-r+1\).

Proof. Let \(\widetilde M=\{1,2,\ldots,r\}\). Then, instead of the indices \(i_1,\ldots,i_r\), introduce the unfolding index

\[ l=i_1+n_1(i_2-1)+n_1n_2(i_3-1)+\cdots+n_1n_2\cdots n_{r-1}(i_r-1). \]

The values \(i_1,\ldots,i_r\) corresponding to a given value of \(l\) are found by the following algorithm:

Operation 1, \(\lambda\):

\[ \frac{u_\lambda-u_{\lambda-1}}{n_1\cdots n_{\lambda-1}} =d_\lambda \to \text{ if }u_{\lambda-1}= \begin{cases} 0, & \text{then } i_\lambda=d_\lambda,\ i_1=n_1,\ldots,\ i_{\lambda-1}=n_{\lambda-1},\\ >0, & \text{then } i_\lambda=d_\lambda+1,\\ & \text{go to operation }2,\lambda . \end{cases} \]

Operation 2, \(\lambda\):

compare \(\lambda\) and \(2\) \(\to\) if

\[ \lambda= \begin{cases} 2, & \text{then } i_1=u_1,\ \text{all }i_k\text{ have been found},\\ >2, & \text{then go to operation }1,\ \lambda-1 . \end{cases} \]

\[ \lambda=r,\ldots,1,\qquad \text{with } u_r=l. \]

Theorem 5. If the intersection of all \(M_j\) is a nonempty set, then problem (1)—(2) reduces to another problem whose number of indices is

\[ \left|M\setminus \bigcap_{j=1}^{t} M_j\right|. \]

Proof. Let

\[ \bigcap_{j=1}^{t} M_j=\{s-r+1,\ s-r+2,\ldots,s\}=\overline M. \]

Then conditions (1) and (2) can be written in a somewhat different form:

\[ \sum_{k\in M_j\setminus \overline M} \left(\sum_{k\in \overline M} x_{i_1\ldots i_s}\right) =a_{j(i_1\ldots i_s)}; \tag{1′} \]

\[ L= \sum_{k\in M\setminus \overline M} \left(\sum_{k\in \overline M} p_{i_1\ldots i_s}x_{i_1\ldots i_s}\right). \tag{2′} \]

Denote

\[ \sum_{k\in \overline M} x_{i_1\ldots i_s}=y_{i_1\ldots i_{s-r}}; \tag{4} \]

\[ q_{i_1\ldots i_{s-r}} = \min_{\substack{1\leq i_k\leq n_k,\ s-r+1\leq k\leq s}} p_{i_1\ldots i_s}. \tag{5} \]

Then from the optimality principle it is not difficult to understand that problem \((1')\)—\((2')\) reduces to the following:

Find the minimum of the functional

\[ L=\sum_{k\in M\setminus \bar M} q_{i_1\ldots i_{s-r}} y_{i_1\ldots i_{s-r}}, \quad y_{i_1\ldots i_{s-r}}\geq 0 \tag{6} \]

under the condition that

\[ \sum_{k\in M_i\setminus \bar M} y_{i_1,\ldots,i_{s-r}}=a_{j(i_1,\ldots,i_s)} . \tag{7} \]

Let \(y^{(0)}_{i_1\ldots i_{s-r}}\) satisfy all conditions (7) and deliver the minimum of (6). Then all \(x^{(0)}_{i_1\ldots i_s}\) whose coordinates coincide with the coordinates

\[ \min_{1\leq i_k\leq n_k,\;s-r+1\leq k\leq s} p_{i_1\ldots i_s}, \]

are equal to \(y^{(0)}_{i_1\ldots i_{s-r}}\), and the remaining \(x^{(0)}_{i_1\ldots i_s}\) are equal to zero. Thus, in the last case it is possible not only to reduce the problem to a simpler one, but also to reduce the number of unknowns.

Definition. Problems that do not satisfy the conditions of Theorems 2—5 will be called irreducible. All symmetric problems of type \(T_m\) are irreducible.

For \(s=4\) there are 8 irreducible nonsymmetric problems. Schematically, each of these problems can be represented as follows: let each index be denoted by a point; arrows between indices denote summation over those indices that lie at the arrowheads and inside them; a small circle around a point means that summation is carried out over a single index. All schemes were constructed symmetrically (see Fig. 1).

Fig. 1

Let \(x^{(0)}_{i_1\ldots i_s}\) be an optimal solution of the problem with conditions (1), (2).

Theorem 6. Multiply all \(a_{j(i_1,\ldots,i_s)}\) by one and the same number \(c>0\). Then the new optimal solution has the form \(c x^{(0)}_{i_1\ldots i_s}\).

Proof. The validity of the assertion is easy to understand if one takes into account that multiplying all \(a_{j(i_1,\ldots,i_s)}\) by \(c\) is equivalent to choosing a new unit of measurement, which should not affect the optimum.

Theorem 7. Multiply all \(p_{i_1\ldots i_s}\) by one and the same number \(c>0\). Then the optimal solution of the new problem will be the same as that of the original one, while the value of the linear form will increase by a factor of \(c\).

Proof is analogous to Theorem 6.

Theorem 8. Consider a new problem in which all \(p_{i_1\ldots i_s}\) corresponding to \(x_{i_1\ldots i_s}\) that enter into some condition are either increased or decreased by one and the same number \(c\). Then the optimal solution of the new problem will be the same as that of the original one, while the value of the linear form will increase or decrease by a constant quantity.

Proof. Let \(M_1=\{1,2,\ldots,r\}\). Add to all \(p_{i_1\ldots i_r1\ldots1}\) corresponding to the condition

\[ \sum_{k\in M_1} x_{i_1\ldots i_r1\ldots1}=a_{1(1,\ldots,1)} \]

a constant

number \(c\). Then the functional takes the form

\[ L_1 = \sum_{k \in M} p_{i_1 \ldots i_s} x_{i_1 \ldots i_s} - \sum_{k \in M_1} p_{i_1 \ldots i_r 1 \ldots 1} x_{i_1 \ldots i_r 1 \ldots 1} + \]

\[ + \sum_{k \in M_1} \left(p_{i_1 \ldots i_r 1 \ldots 1}+c\right) x_{i_1 \ldots i_r 1 \ldots 1} = \sum_{k \in M} p_{i_1 \ldots i_s} x_{i_1 \ldots i_s} + c a_{1(1,\ldots,1)} . \]

Thus, \(L_1 = L + c a_{1(1,\ldots,1)}\), which is what was required to prove.

The last three properties can be used essentially in finding optimal solutions.

Institute of Complex Transport Problems
of Gosplan of the USSR

Received
27 X 1962

REFERENCES

¹ E. Shell, Proc. 2-d Symposium in linear Programming, Washington, 1, 1955.
² Progress in the Operation Research, 1961.
³ B. S. Verkhovskii, On one algorithm for solving a multi-index transportation problem on an electronic computer, Novosibirsk, 1962.