Demand-driven publish/subscribe in mobile environments

Abstract

We propose a novel routing protocol, Self-Balancing Supply/Demand (SBSD), for Publish/Subscribe in mobile ad hoc environments. SBSD is a controlled flooding that reduces network congestion by constraining how far subscriptions replicate and how many times nodes broadcast them. SBSD ranks subscriptions by a utility function. This function matches the supply of publications with the recent demand for them; more popular subscriptions are replicated farther and their replicas are retained longer. SBSD is therefore demand-driven, as more popular subscriptions are more likely to receive their matching publications and receive them sooner. We show that SBSD is scalable; routing distance is independent of network size. SBSD’s performance is examined under random mobility, under challenging conditions including high node mobility and broadcast failure rates.

Appendices

Appendices

1.1 Appendix A: The SBSD algorithm

For a subscription s matching a publication p and two brokers B and B′, this algorithm describes processing at B′ for the broadcast of s from B to B′. In this presentation of the algorithm, we assume the only publication matching s is p but that i versions of p (p ₀, p ₁,…, p _i , with p _i being the most recent) have been published and may be present in the network. Details not associated with the broadcast itself, i.e., queuing operations, are omitted.

SBSD Algorithm at broker B′ :

if(p ∈ B′)//B′ publishes p

 if (s ∈ B′) update → B ′ .s (B.s);//note that here B′.s.M will already hold p _i

 else if (s ∉ B′) replicate → B′.s (B.s);

  B′.s.M = p _i );//s is replicated to B′ and matched to p _i at B′

else if (p ∉ B′)//B′ does not publish p

 if (B.s.M = ϕ)//dissemination phase for B.s

  if (s ∈ B′) update → B′.s (B.s);

  else if (s ∉ B′) replicate → B′.s (B.s);

 else if (B.s.M ≠ ϕ)//discovery phase for B.s

  if (s ∈ B′) update → B′.s (B.s);

  else if (s ∉ B′)//no action

End_SBSD.

*replicate → B′.s(B.s): B′.s = B.s; B′.s.h = B.s.h + 1;

*update → B′.s(B.s): if (B′.s.M = ϕ) B′.s.M = B.s.M;

else if ((B′.s.M = p _j and B.s.M = p _k ) and (j < k)) B′.s.M = p _k; //B′ takes the more recent version

if ((B′.s.D ∩ B.s.D) ≠ B.s.Dand(B.s.h > B′.s.h))

B′.s.h = B.s.h + 1 and B′.s.D = (B′.s.D ∪B.s.D);//cascading to B′.s.f, B′.s.a, and B′.s.u

1.2 Appendix B: Proof of theorem 1

Proof Consider a subscription s, matching only a publication p of which, as in Appendix A, i versions may exist. For any two brokers B and B′, there are six possible cases for the broadcast of B.s to B′. We first consider the two cases where B′ is a publisher for p:

[case 1]: B′ has s, which will already be matched to p _i . B.s and B′.s may have different known_demand sets; e.g., B may have recently posted an instance of s. B′.s.D is updated by inserting any instances of s in B.s.D it does not already contain. If B′.s.D changes, B′.s.f, B′.s.a, and B′.s.u are updated and if B.s.h < B′.s.h then B′.s.h = B.s.h + 1. B′ also updates its B′.s.M field to hold the most recent version of p (if any) in (B′.s.M ∪ B.s.M).

[case 2]: B′ does not have s. In this case, B.s is replicated to B′.s and p _i is inserted into B′.s.M. Note that B.s is replicated to B′ even if B.s is matched. This is an exception to our rule that matched subscriptions do not replicate; it is necessary to queue B′.s for broadcasting in case B′ has published a newer version of p (or will do so shortly).

If B′ is not a publisher for p, we must separately consider the dissemination and discovery phases of s. In dissemination, B.s.M = ϕ. There are two cases here:

[case 3]: B′ has s. B′.s.D is updated by inserting any instances of s in B.s.D it does not already contain. If B′.s.D changes, B′.s.f, B′.s.a, and B.s.u are updated and, if B.s.h < B′.s.h then B′.s.h = B.s.h + 1. B′ also updates its B′.s.M field (which has no effect since B.s is unmatched).

[case 4]: B′ does not have s; B.s is replicated to B′.

In the discovery phase, B.s has been matched to p _j , j ≤ i. This phase also has two cases:

[case 5]: B′ has s. B′.s.D is updated by inserting any instances of s in B.s.D it does not already contain. If B′.s.D changes, B′.s.f, B′.s.a, and B.s.u are updated and if B.s.h < B′.s.h then B′.s.h = B.s.h + 1. B′.s.M takes the most recent version of p (if any) in (B′.s.M ∪ B.s.M).

[case 6]: B′ does not have s; B′ does nothing.

All possible cases of interaction between brokers B and B′ have been examined and are captured correctly by the SBSD algorithm. Therefore, the algorithm is complete.□

1.3 Appendix C: Changes to next

The following cases are based on a broker B broadcasting a subscription s to a broker B′. The decision to change next depends on the states of the received and [any] stored replicas of subscription s.

• [case 1]: B.s is matched; B′ does not hold s. B′ does not insert s into its queue.

• [case 2]: B.s is unmatched; B′ does not hold s. B′ inserts s into its queue with next = 2.

• [case 3]: B.s is matched; B′.s is unmatched. B′.s is updated and assigned next = 1.

• [case 4]: B.s is matched; B′.s is matched. B′.s is updated but no changes occur to its associated next field.

• [case 5]: B.s is unmatched; B′.s is matched. If B.s.D ≠ B′.s.D, B′.s has its associated next field set to 1; otherwise, it is left unchanged.

• [case 6]: B.s is unmatched; B′.s is unmatched. If B.s.D ≠ B′.s.D, B′.s has its associated next field set to 2; otherwise, it is left unchanged.

1.4 Appendix D: Proof of maximum replication proposition

Proposition: In a uniform network, h_max = 3q/2tλ, where q is the active queue size, t is the time per hop, and λ is the local arrival rate, the rate at which stored subscriptions are pushed out of a broker’s active queue by new ones.

Proof With t constant, the age of a new replica s (i.e., when s is received) is t·s.h. When hops = h_max, age = t·h_max; the maximum replica age, a_max, equals t·h_max. Recall our assumption that s.f is equal for all s. Then, at all brokers, the expected exponent of minimum utilityu_min is:

$$ \ln (u_{\min } ) = - {\frac{{t \cdot h_{\max }^{3/2} }}{{\sqrt[3]{f}}}} $$

(2)

All subscriptions posted within h_max of a broker B will be replicated to B and inserted in B′s active queue. Consider a new replica s reaching a broker B with age = a, hops = h, and frequency = f. B will hold s until B.s.u is less than u_min. Once B queues s, s.a will increase while s.h will not. Let Δa be the time B holds s before s becomes inactive. Since the ln(B.s.u) when B.s.age = (a + Δa) equals ln(u_min), Δa can be found by solving the equation:

$$ \ln (u_{\min } ) = - \left( {{\frac{\sqrt h \cdot (t \cdot h + \Updelta a)}{{\sqrt[3]{f}}}}} \right) = - {\frac{\sqrt h }{{\sqrt[3]{f}}}}\;\left( {t \cdot h + \Updelta a} \right) $$

(3)

$$ \Updelta a = - \ln (u_{\min } )\left( {{\frac{{\sqrt[3]{f}}}{\sqrt h }}} \right) - t \cdot h $$

(4)

Substituting for ln(u _min ):

$$ \Updelta a = {\frac{{a_{\max } \sqrt {h_{\max } } }}{{\sqrt[3]{f}}}}\left( {{\frac{{\sqrt[3]{f}}}{\sqrt h }}} \right) - t \cdot h $$

(5)

$$ \Updelta a = {\frac{{t \cdot h_{\max }^{{{\raise0.7ex\hbox{$3$} \!\mathord{\left/ {\vphantom {3 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$2$}}}} }}{\sqrt h }} - t \cdot h $$

(6)

$$ \Updelta a = t \cdot \left( {{\frac{{h_{\max }^{{{\raise0.7ex\hbox{$3$} \!\mathord{\left/ {\vphantom {3 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$2$}}}} }}{\sqrt h }} - h} \right) $$

(7)

Subscriptions which reach B are randomly posted within a circle of h _max centered on B. The average time queued for all subscriptions reaching B is the integral of Eq. 7 between 0 and h _max, divided by the area of the circle of radius h _max. For any h in [0, h _max], the fraction of all subscriptions posted within h _max, is πh divided by the constant π(h _max ) ² and their time queued is Δa (a function of h) from Eq. 7. As all active queues are normally full, the average time queued must also equal q/λ (e.g., from Little’s Formula, λE(T) = E(n)). Because Δa decreases as h increases, we take the definite integral from h _max to 0. We can solve for h _max from the equation:

$$ {\raise0.7ex\hbox{$q$} \!\mathord{\left/ {\vphantom {q \lambda }}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$\lambda $}} = \left[ {\int\limits_{{h_{\max } }}^{0} {\pi \cdot h \cdot \Updelta a \, \cdot }\, dh} \right]/\left[ {\pi \cdot h_{\max }^{2} } \right] $$

(8)

Substituting for Δa and moving constants to the left of the integral symbol gives:

$$ {\raise0.7ex\hbox{$q$} \!\mathord{\left/ {\vphantom {q \lambda }}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$\lambda $}} = \left[ {\pi \cdot t\int\limits_{{r_{\max } }}^{0} {h\left( {h_{\max }^{{{\raise0.7ex\hbox{$3$} \!\mathord{\left/ {\vphantom {3 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$2$}}}} h^{{{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$2$}}}} - h} \right)} dh} \right]/\left[ {\pi \cdot h_{\max }^{2} } \right] $$

(9)

This becomes:

$$ {\raise0.7ex\hbox{$q$} \!\mathord{\left/ {\vphantom {q \lambda }}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$\lambda $}} = \left[ {\pi \cdot t\left| {_{{h_{\max } }}^{0} h\left( {h_{\max }^{{{\raise0.7ex\hbox{$3$} \!\mathord{\left/ {\vphantom {3 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$2$}}}} h^{{{\raise0.7ex\hbox{${ - 1}$} \!\mathord{\left/ {\vphantom {{ - 1} 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$2$}}}} - h} \right)} \right.dh} \right]/\left[ {\pi \cdot h_{\max }^{2} } \right] $$

(10)

Which expands to:

$$ {\raise0.7ex\hbox{$q$} \!\mathord{\left/ {\vphantom {q \lambda }}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$\lambda $}} = \left[ {\pi \cdot t\left[ {\left( {0 - 0} \right) - \left( { - \frac{1}{3}h_{\max }^{3} - \frac{1}{3}h_{\max }^{3} } \right)} \right]} \right]/\left[ {\pi \cdot h_{\max }^{2} } \right] $$

(11)

And reduces to:

$$ {\raise0.7ex\hbox{$q$} \!\mathord{\left/ {\vphantom {q \lambda }}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$\lambda $}} = \left[ {t \cdot \left( {\frac{2}{3}} \right)h_{\max } } \right] $$

(12)

Rearranging to find r _max yields:

$$ h_{\max } = {\raise0.7ex\hbox{${3q}$} \!\mathord{\left/ {\vphantom {{3q} {2t\lambda }}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{${2t\lambda }$}}. $$

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1.5 Appendix E: Proof of broadcast rate proposition

Proposition: For a subscription with one subscriber and a utility function ln(u) = -G(a)H(h), where G(a) = a and H(h) = √h, the number of broadcasts decreases at the rate H′(h).

Proof Let us set G(a) = a and H(h) = √h. For a subscription with one subscriber, the age (as well as the frequency) of all its replicas is the same. So, since ln(u) = -G(a)H(h), ln(u) decreases at the rate H′(h).

Active time approximately follows the function 1/G′(a)H(h), which reduces to 1/H(h) (5.3.3). As active time falls by a factor k, broadcasts fall by a factor between k and log ₂ k (5.3.4). To find how broadcasts change with respect to hops, we integrate the functions 1/H(h) and log ₂ (1/H(h)), average their integrals, then take the derivative of the average. The average of the integrals is:

$$ \frac{1}{2}\int {\left[ \log_{2} \left( h^{ - 1/2} \right) + h^{1/2} \right]} $$

(14)

$$ \frac{1}{2}{\left[-( \log_{2}e)/2h +2 h^{1/2} \right]} $$

(15)

$$ {\frac{{ - \log_{2} e}}{4h}} + h^{1/2} $$

(16)

As h increases, the left term rapidly becomes small and the average will approach √h, which equals H(h). Their derivatives are also the same, so the change in ln(u) matches the change in the number of broadcasts. Thus, when G(a) = a and H(h) = √h, our utility function yields a ln(u) directly corresponding to how many times a replica is expected to be broadcast.