Optimization of Information Exchange in a Network of Autonomous Participants

Abstract

Suboptimal algorithms for planning a sequence of information messages in the communication network of mobile autonomous participants are proposed. The algorithms use as optimality criteria the upper bounds of either the total delivery time of all messages from the transmitted sequence or the average delivery time over the messages. Several formulations of the problem are considered, which differ in the presence or absence of preliminary ordering for the set of messages.

APPENDIX

Proof of Statement 1. We transform expression (2.1):

$${{\hat {\Delta }}_{s}} = \sum\limits_{k = 1}^n {e_{{[k]}}^{w}} + \sum\limits_{k = 1}^n {e_{{[k]}}^{t}} + \sum\limits_{k = 1}^n {({{r}_{{[k]}}} - 1)\bar {n}E} + \sum\limits_{k = 1}^n {\sum\limits_{i = 2}^{{{r}_{{[k]}}}} {e_{{[k],i}}^{t}} } .$$

It is easy to see that in this expression only the first sum depends on the ordering of the messages, which means that only it is subject to minimization:

$$\tilde {\Delta } = \sum\limits_{k = 1}^n {e_{{[k]}}^{w}} .$$

Let us expand this expression

$$\tilde {\Delta } = \sum\limits_{k = 1}^n {\sum\limits_{j = 1}^{k - 1} {e_{{[j]}}^{{}}} } = ({{e}_{{[1]}}} + {{e}_{{[1]}}} + {{e}_{{[2]}}} + {{e}_{{[1]}}} + {{e}_{{[2]}}} + {{e}_{{[3]}}} + ... + {{e}_{{[1]}}} + ... + {{e}_{{[n - 1]}}}) = \sum\limits_{k = 1}^n {(n - k){{e}_{{[k]}}}} .$$

It is known that the sum of pairwise products of members of two numerical sequences has a minimum value if one of these sequences increases and the other decreases [18]. Since the coefficients (n – k) decrease with increasing k, then the minimum average delivery time is reached only when ${{e}_{{[k]}}}$ increase with increasing k or at least do not decrease.

Proof of Statement 2. We assume the opposite, namely that the order of the queue π minimizes the upper bound $\hat {\Delta }_{s}^{w}$, but the statement condition is not satisfied for it, i.e., there is such a position l that

$$\frac{{{{e}_{{[l]}}}}}{{{{w}_{l}}}} > \frac{{{{e}_{{[l + 1]}}}}}{{{{w}_{{l + 1}}}}}.$$

(A.1)

Let L and $L{\kern 1pt} '$ be the numbers of messages in the queue π in positions l and l + 1. Let us form a queue $\pi {\kern 1pt} '$, swapping around messages L and $L{\kern 1pt} '$ in the queue π. Delivery of the first l – 1 and last ${{n}_{i}} - l - 1$ messages are completed in both queues at the same time. As a result, these messages contribute equally to the weighted sum in (2.3). Thus, the difference in the value of criterion (2.3) for two variants of queues π and $\pi {\kern 1pt} '$ determined by the terms for messages L and $L{\kern 1pt} '$. Let

$$t = \sum\limits_{j = 1}^{l - 1} {{{e}_{{[j]}}}} .$$

Then for π

$${{w}_{L}}\Delta _{L}^{{}} = {{w}_{{[l]}}}\Delta _{{[l]}}^{{}} = {{w}_{L}}(t + {{e}_{L}}),$$

$${{w}_{{L{\kern 1pt} '}}}\Delta _{{L{\kern 1pt} '}}^{{}} = {{w}_{{[l + 1]}}}\Delta _{{[l + 1]}}^{{}} = {{w}_{{L{\kern 1pt} '}}}(t + {{e}_{L}} + {{e}_{{L{\kern 1pt} '}}}),$$

$${{w}_{L}}\Delta _{L}^{{}} + {{w}_{{L{\kern 1pt} '}}}\Delta _{{L{\kern 1pt} '}}^{{}} = {{w}_{L}}t + {{w}_{L}}{{e}_{L}} + {{w}_{{L{\kern 1pt} '}}}t + {{w}_{{L{\kern 1pt} '}}}{{e}_{L}} + {{w}_{{L{\kern 1pt} '}}}{{e}_{{L{\kern 1pt} '}}}.$$

For $\pi {\kern 1pt} '$

$${{w}_{L}}\Delta _{L}^{{}} = {{w}_{{[l + 1]}}}\Delta _{{[l + 1]}}^{{}} = {{w}_{L}}(t + {{e}_{{L{\kern 1pt} '}}} + {{e}_{L}}),$$

$${{w}_{{L{\kern 1pt} '}}}\Delta _{{L{\kern 1pt} '}}^{{}} = {{w}_{{[l]}}}\Delta _{{[l]}}^{{}} = {{w}_{{L{\kern 1pt} '}}}(t + {{e}_{{L{\kern 1pt} '}}}),$$

$${{w}_{L}}\Delta _{L}^{{}} + {{w}_{{L{\kern 1pt} '}}}\Delta _{{L{\kern 1pt} '}}^{{}} = {{w}_{L}}t + {{w}_{L}}{{e}_{{L{\kern 1pt} '}}} + {{w}_{L}}{{e}_{L}} + {{w}_{{L{\kern 1pt} '}}}t + {{w}_{{L{\kern 1pt} '}}}{{e}_{{L{\kern 1pt} '}}}.$$

We compare for queues π and $\pi {\kern 1pt} '$ the values of the total contributions of the criteria of the studied messages L and $L{\kern 1pt} '$. It is easy to see that for π the contribution contains ${{w}_{{L{\kern 1pt} '}}}{{e}_{L}}$ but does not contain ${{w}_{L}}{{e}_{{L{\kern 1pt} '}}}$ and vice versa. However, from (A.1) it follows that ${{w}_{{L{\kern 1pt} '}}}{{e}_{L}} \geqslant {{w}_{L}}{{e}_{{L{\kern 1pt} '}}}$. This implies that the value of criterion (2.3) for π is more than for $\pi {\kern 1pt} '$ and that the assumption that this ordering minimizes the upper bound $\hat {\Delta }_{s}^{w}$ is incorrect. Thus, the statement being proved is true.

Proof of Statement 3. We denote by ${{\hat {\Delta }}_{{g[k]}}}$ the upper bound of the delivery time for a group of messages that is among the groups at the k-th position. Then the upper bound of the total delivery time of all groups is determined by the expression

$${{\hat {\Delta }}_{g}} = \sum\limits_{k = 1}^p {{{{\hat {\Delta }}}_{{g[k]}}}} .$$

(A.2)

Let us rewrite it by analogy with (2.1)

$${{\hat {\Delta }}_{g}} = \sum\limits_{k = 1}^p {{{{\hat {\Delta }}}_{{g[k]}}}} = \sum\limits_{k = 1}^p {\left( {\sum\limits_{i = 1}^{k - 1} {e_{{[i]}}^{'}} + e_{{[k]}}^{'} + {{{\bar {\delta }}}_{{g[k]}}}} \right)} = \sum\limits_{k = 1}^p {\sum\limits_{i = 1}^{k - 1} {e_{{[i]}}^{'}} + \sum\limits_{k = 1}^p {e_{{[k]}}^{'}} + \sum\limits_{k = 1}^p {{{{\bar {\delta }}}_{{g[k]}}}} } ,$$

where ${{\bar {\delta }}_{{g[k]}}}$ is the upper bound for the time interval from the moment emission is completed to the moment of delivery of all messages from the group located at the k-th position. In this expression, the second and third sums representing the total duration of the group and the interval ${{\bar {\delta }}_{{g[k]}}}$ are constant. Moreover, the first (double) sum depends on the order of the groups. Applying an analysis similar to that used in the proof of Statement 1, we conclude that the groups in the plan must be ordered in accordance with (2.4).

Proof of Statement 4. Let us write an expression for the upper bound on the delivery time of the j-th message of the group located at the k-th position:

$$\hat {\Delta }_{{[k],j}}^{{}} = {{\hat {\Delta }}_{{[k]}}} - {{\bar {\delta }}_{{[k],j}}},$$

(A.3)

where ${{\bar {\delta }}_{{[k],j}}}$ is the upper bound of the time interval from the moment of completion of emission of the j-th message of the group located at the k-th position, until the delivery of all messages in this group is completed.

Let us analyze the expression for the upper bound of the average message delivery time over the entire queue:

$${{\bar {\hat {\Delta }}}_{m}} = \frac{{\sum\limits_{k = 1}^p {\sum\limits_{j = 1}^{{{n}_{{[k]}}}} {\hat {\Delta }_{{[k],j}}^{{}}} } }}{n}.$$

We transform this expression using the relation (A.3)

$${{\bar {\hat {\Delta }}}_{m}} = \frac{{\sum\limits_{k = 1}^p {\sum\limits_{j = 1}^{{{n}_{{[k]}}}} {(\hat {\Delta }_{{[k]}}^{{}}} } - {{{\bar {\delta }}}_{{[k],j}}})}}{n} = \frac{{\sum\limits_{k = 1}^p {\sum\limits_{j = 1}^{{{n}_{{[k]}}}} {\hat {\Delta }_{{[k]}}^{{}}} } }}{n} - \frac{{\sum\limits_{k = 1}^p {\sum\limits_{j = 1}^{{{n}_{{[k]}}}} {{{{\bar {\delta }}}_{{[k],j}}}} } }}{n}.$$

In this expression, the second term is a constant value that does not depend on the order of the groups. The first term does not depend on j and therefore transforms to the form

$${{\bar {\hat {\Delta }}}_{m}} = \frac{{\sum\limits_{k = 1}^p {{{n}_{{[k]}}}\hat {\Delta }_{{[k]}}^{{}}} }}{n}.$$

From a comparison of this expression, for example, with expression (2.3), the validity of this statement follows.

Proof of Statement 5. The idea of the proof is based on the fact that the distinguished subgroups can be considered as the groups from the previous statement, i.e., although they can be interrupted, it is proved that this interruption does not improve the plan.

We consider the subgroup b of some group formed in accordance with the statement algorithm and containing n_b messages. Let

$${{\bar {\tilde {\Delta }}}_{b}} = \frac{{\sum\limits_{j = 1}^{{{n}_{b}}} {{{{\hat {\Delta }}}_{{b,j}}}} }}{{{{n}_{b}}}}.$$

If you designate a message in the subgroup b of the pair (b, j), where j is the sequence number of the message in the subgroup, then the subgroup is given by the sequence (b, 1), …, (b, ${{n}_{b}}$). Breaking a subgroup means splitting it into two parts $b{\kern 1pt} '$ and $b{\kern 1pt} ''$:

$$b{\kern 1pt} ' = (b,1),(b,2),...,(b,i),$$

$$b{\kern 1pt} '' = (b,i + 1),(b,i + 2),...,(b,{{n}_{b}}).$$

Each of these new parts of the subgroup has its own conditional upper bound on the delivery time

$${{\bar {\tilde {\Delta }}}_{{b{\kern 1pt} '}}} = \frac{{\sum\limits_{j = 1}^i {{{{\hat {\Delta }}}_{{b,j}}}} }}{i},\quad {{\bar {\tilde {\Delta }}}_{{b{\kern 1pt} ''}}} = \frac{{\sum\limits_{j = i + 1}^{{{n}_{b}}} {{{{\hat {\Delta }}}_{{b,j}}}} }}{{{{n}_{b}} - i}},$$

which, in accordance with Statement 4, would determine the place of the parts of the subgroup in the plan.

We need to prove that such partitioning of the subgroup b is inappropriate because part $b{\kern 1pt} '$ must immediately precede $b{\kern 1pt} ''$, and the structure remains unchanged.

Let us show that

$${{\bar {\tilde {\Delta }}}_{{b{\kern 1pt} '}}} > {{\bar {\tilde {\Delta }}}_{b}} > {{\bar {\tilde {\Delta }}}_{{b{\kern 1pt} ''}}}.$$

Inequality ${{\bar {\tilde {\Delta }}}_{{b{\kern 1pt} '}}} > {{\bar {\tilde {\Delta }}}_{b}}$ follows from the construction in accordance with the assertion algorithm:

$${{\bar {\tilde {\Delta }}}_{{b{\kern 1pt} '}}} = \frac{{\sum\limits_{j = 1}^i {{{{\hat {\Delta }}}_{{b,j}}}} }}{i} > {{\bar {\tilde {\Delta }}}_{b}} = \frac{{\sum\limits_{j = 1}^i {{{{\hat {\Delta }}}_{{b,j}}}} + \sum\limits_{j = i + 1}^{{{n}_{b}}} {{{{\hat {\Delta }}}_{{b,j}}}} }}{{i + ({{n}_{b}} - i)}}.$$

Multiplying both sides of the inequality by $i(i + ({{n}_{b}} - i)$, we get

$$(i + ({{n}_{b}} - i))\sum\limits_{j = 1}^i {{{{\hat {\Delta }}}_{{b,j}}}} > i\left( {\sum\limits_{j = 1}^i {{{{\hat {\Delta }}}_{{b,j}}}} + \sum\limits_{j = i + 1}^{{{n}_{b}}} {{{{\hat {\Delta }}}_{{b,j}}}} } \right).$$

By adding to both parts of the last inequality the quantity

$$({{n}_{b}} - i)\sum\limits_{j = i + 1}^{{{n}_{b}}} {{{{\hat {\Delta }}}_{{b,j}}}} - i\sum\limits_{j = 1}^i {{{{\hat {\Delta }}}_{{b,j}}}} ,$$

we find

$$({{n}_{b}} - i)\sum\limits_{j = 1}^{{{n}_{b}}} {{{{\hat {\Delta }}}_{{b,j}}}} > {{n}_{b}}\sum\limits_{j = i + 1}^{{{n}_{b}}} {{{{\hat {\Delta }}}_{{b,j}}}} ,$$

$${{\bar {\tilde {\Delta }}}_{b}} = \frac{{\sum\limits_{j = 1}^{{{n}_{b}}} {{{{\hat {\Delta }}}_{{b,j}}}} }}{{{{n}_{b}}}} > \frac{{\sum\limits_{j = i + 1}^{{{n}_{b}}} {{{{\hat {\Delta }}}_{{b,j}}}} }}{{{{n}_{b}} - i}} = {{\bar {\tilde {\Delta }}}_{{b{\kern 1pt} ''}}}.$$

It follows that the splitting of any group formed by the statement algorithm into two subgroups leads to the fact that the resulting subgroups do not satisfy the minimum order relation $\bar {\tilde {\Delta }}$, but they cannot be swapped because of the relation of precedence in the group.

Let us now show that the displacement $b{\kern 1pt} '$ forward or $b{\kern 1pt} ''$ back is impractical. Let in the original description a and c be subgroups, the first of which precedes b, and the second one follows b, i.e.

$${{\bar {\tilde {\Delta }}}_{a}} \leqslant {{\bar {\tilde {\Delta }}}_{b}} \leqslant {{\bar {\tilde {\Delta }}}_{c}}.$$

Subgroup $b{\kern 1pt} '$ cannot be placed before a because

$${{\bar {\tilde {\Delta }}}_{a}} \leqslant {{\bar {\tilde {\Delta }}}_{b}} \leqslant {{\bar {\tilde {\Delta }}}_{{b{\kern 1pt} '}}}.$$

It cannot be carried further forward because for any subgroup preceding a the value $\bar {\tilde {\Delta }}$ does not exceed ${{\bar {\tilde {\Delta }}}_{a}}$. Subgroup $b{\kern 1pt} '$ cannot be placed inside a, dividing a into two parts $a{\kern 1pt} '$ and $a{\kern 1pt} ''$ because

$${{\bar {\tilde {\Delta }}}_{{a{\kern 1pt} ''}}} < {{\bar {\tilde {\Delta }}}_{a}} \leqslant {{\bar {\tilde {\Delta }}}_{b}} < {{\bar {\tilde {\Delta }}}_{{b{\kern 1pt} '}}},$$

and $b{\kern 1pt} '$ cannot be placed before $a{\kern 1pt} ''$. Similarly, it is proved that $b{\kern 1pt} ''$ cannot be placed inside or after c.