One of the objectives of this paper is to understand the impact of link addition and deletion on storage availability for those agents who are involved in the link addition/deletion as well as those who are not. The storage availability is determined by the distances between them and their closeness (from Equation (
2)). Therefore, first we study how addition and deletion of a link impacts the shortest distances between pairs of agents and, therefore, their closeness. This analysis provides a base for understanding the effect of link-addition/deletion on agents’ storage availability in
.
3.1. Effect of Link Alteration on Closeness
Lemma 1. Suppose . Then, .
Proof. Clearly, . As , we have, . Also, . Thus, and increase by at least in . ☐
Lemma 2. Suppose . Then, .
Proof. Let us assume there is no path between i and j in , then , thus, and decrease by 1 in .
Now, let us assume there exists a path between i and j in , the distance between i and j in being at least 1 more than that in . Thus, and decrease by at least in . ☐
Lemmas 1 and 2 show that, with respect to closeness, every link benefits agents on either side of the link. An action of link addition or deletion between a pair of agents not only impacts their closeness, but also that of other agents. Now, we study the impact of link addition or deletion between a pair of agents (say, i and j) on the closeness of the other agents .
Lemma 3. Suppose and . Then, if and only if for all .
Proof. If
for all
, then by Equation (
1),
.
Conversely, suppose .
It is easy to see that, if for some , if , then . (Paths in exist in too).
We have for all and, if there exists x such that , then , a contradiction. ☐
Lemma 4. Suppose and . Then, if and only if for all .
Proof. As for all l, the proof follows in lines similar to that of Lemma 3. ☐
We now show necessary and sufficient conditions for increase in the closeness of agents who are not involved in link addition or deletion.
Theorem 1. Suppose , and let k be an agent distinct from i and j. Then, if and only if there exists at least one agent such that and all shortest paths from k to l in contain .
Proof. Let . Then, by Lemma 3, there must be at least one agent, say l, such that .
Suppose and l are all distinct. Note that j may be the same as l.
If possible, let for all . Then, for all . From Lemma 1, , a contradiction. Therefore, there exists an such that .
As , . As , and . Hence, .
Now,
It follows that every shortest path between k and l in contains . (Note that if there exists a shortest path from k to l in that does not contain , then this shortest path exists in too).
Conversely, let such that and all shortest paths from k to l in contain .
Clearly, .
If possible, let . This means for every l in there exists a shortest path from k to l in that does not contain , a contradiction. Therefore, . ☐
Theorem 2. Suppose , and let k be an agent distinct from i and j. Then, if and only if there exists at least one agent such that and all shortest paths from k to l in contain .
We skip the proof as it is similar to the proof of Theorem 1.
In subsequent sections, we present our results due to link addition. We present our results on link deletion in
Appendix B.
3.2. Effect of Closeness on Distances of Agents Not Involved in Link Alteration
In this section, we classify agents whose mutual distances from each other remain the same after link alteration. We use the same to analyze the effect of closeness on distances between agents who are not involved in the link addition or deletion.
Given k such that , we use to denote the set of all such that all shortest paths from k to l in contain . We use to denote an agent in .
Proposition 1. Suppose i, j, and k are distinct agents in . Suppose l is another agent, distinct from i and k, and suppose . If , then .
Proof. We have . Then, from Theorem 1, there exists such that all shortest paths from k to l in contain .
We consider the two cases and .
Suppose . As , k observes i before j on all shortest paths . This implies .
Suppose . As , k observes i before j, and j before l, on all shortest paths . This implies . ☐
Definition 3. Suppose and k is an agent such that . A -shortest-path-network, , is a subnetwork of that consists of all shortest paths from k to in , which contain , for all .
Definition 4. An (all -shortest-path-network, , is , the smallest network consisting of all -shortest-path-networks.
Definition 5. A sub--network, of , is the induced subnetwork of consisting of all agents such that . Similarly, we define the sub--network, of , as the induced subnetwork of consisting of all agents such that .
Refer to
Appendix A for an illustration of the above definitions.
Proposition 2. For all , .
Proof. If then, from Definition 5, and . As as well, there exists l and such that and .
It is sufficient to show that, given , can never be k.
If possible, let . Then, from Definition 5, implies observes i first, and subsequently j to reach k, on all shortest paths from to k in . Then, . This is because, if , . Therefore, , which is a contradiction. Now, if , then k must first visit j, and later i, to reach on all shortest paths from to k. This implies and hence, , again, a contradiction. Thus, . ☐
We discuss our results on shortest distances due to link deletion in
Appendix B.1.
3.3. Effect of Link Alteration on Storage Availability
Our aim here is to analyze under what conditions agents’ chance of obtaining storage space in the network increases or decreases by adding a new link. We present our results in the case of link deletion in
Appendix B.2.
Lemma 5. Suppose agent i and j add a direct link in and let . Then, and .
Proof. If agent
then
. Thus,
. Therefore, from Equation (
2),
. A similar proof holds for
j too. ☐
Lemma 6. Suppose agents i, j, k, and l are such that , , , and . (Agents i and k may be the same, and agents j and l may be the same). Suppose , , and . Then,
, and
implies that . Similarly, if , then .
Lemma 7. Let k and be agents in . Then, and .
Proof. The proof follows from Proposition 2. ☐
Theorem 3. Suppose agents i and j are such , and . Then, if and only if .
Additionally, if and only if .
Proof. The proof follows from Lemmas 5, 6, and 7. ☐
3.4. Externalities
In this section, we study externalities, that is, how a link that is added between a pair of agents affects the utility of others. (Refer to Definition 6). The particular form of externalities (positive, negative, or none) is crucial in determining which network is likely to evolve and the conditions under which it will lead to a stable and efficient network.
Definition 6. [32] Consider a network, , with agents such that and . Suppose agents i and j form a direct link . Then, agent experiences Positive externalities if ;
Negative externalities if ;
No externalities if .
We now show that the type of externalities an agent experiences, can be determined using conditions on the storage availability, independent of the data loss rate and the value that agents associate with their data.
Proposition 3. In an SSSC , an agent experiences
Positive externalities if ;
Negative externalities if ;
No externalities if .
Proof. By Definition 6,
.
As agent k does not pay the cost for link , we have .
Thus, .
For Cases 2 and 3, the proof is similar to that of Case 1. ☐
The following results provide a necessary and sufficient condition under which an agent experiences positive or negative externalities.
Lemma 8. Let i, j, and k be distinct agents in . Suppose . Then, k experiences only negative externalities.
Proof. If agents
i and
j add a direct link in
, then, from Lemma 1,
. If
, then, from Theorem 1,
, thus,
for all
. Therefore,
, by Equation (
2). Now, for all
,
. If
, then
and, if
, then
. Thus,
. ☐
Theorem 4. Suppose agents i, j, k, , and l are such that , , , , , and . (Agents i and may be the same, and agents j and l may be the same). Suppose , and . Then, agent k experiences positive externalities if and only if and , otherwise k experiences negative externalities.
Proof. From Lemma 8, it is required to increment in agent k’s closeness. It is straightforward to observe that , then . Thus, k experiences positive externalities.
Conversely, let , then either from Proposition 8, , for all or . ☐
Lemma 8 and Theorem 4 show that an increase in the closeness of an agent (who is not involved in the link formation) is necessary in order for that agent to experience positive externalities. Although we have provided a necessary and sufficient condition for positive and negative externalities by performing a microscopic analysis of externalities, it is hard to obtain a general characterization of networks where agents experience only positive externalities. This leads us to the following question. At least for specific network structures, can we show positive (or negative) externalities? For instance, we can argue that in a two diameter network, agents never experience positive externalities.