An Node-Child Matrix-Based Algorithm to the Quickest Path Reliability Problem under Transmission Cost Constraints

1. Introduction

The quickest path problem involves identifying a path from a source node, 1, to a destination node, n, within a network. This path is used to efficiently transmit a specific flow quantity, d, from node 1 to node n while minimizing the transmission time [1,2]. In this problem, each network arc is characterized by two key attributes: a lead time value and a capacity value. The significance of this optimization problem is well-recognized by researchers due to its applicability across a broad spectrum of flow network scenarios [1,2,3,4,5,6,7,8,9,10,11]. This problem initially emerged in discovering the fastest route for convoy-type traffic within flow-rate-constrained networks [1]. Subsequently, it found application in communication networks, where nodes represent transmitters/receivers and arcs symbolize communication channels [2].

While deterministic (non-stochastic) flow networks have undeniably been instrumental in understanding and optimizing various systems, the practical reality is that many real-world systems exhibit dynamic characteristics, necessitating the adoption of a more nuanced approach. Multistate (stochastic) flow networks (MFNs) have gained prominence as a result of their ability to model complex systems where fluctuations, failures, maintenance, and other dynamic factors play a significant role [12,13,14,15,16,17,18,19,20]. Within an MFN, arcs and nodes can exist in various potential states, influenced by traffic conditions, maintenance activities, failures, or other underlying causes. Consequently, the network itself assumes multiple states, each reflecting the dynamic nature of the system. Numerous performance metrics have been introduced in the literature to evaluate the effectiveness of an MFN, with particular emphasis on network reliability, which stands out as a primary indicator. Network reliability is commonly defined as the system’s capacity to fulfill a predefined function within specified conditions and over a known time frame [21]. A well-known reliability indicator is the two-terminal reliability of an MFN. It is the probability of transmitting at least a given demand d units of flow/data/commodity from node 1 (source) to node n (destination). Numerous exact and approximation algorithms have been proposed in the literature to compute this indicator [8,13,22,23,24,25,26,27,28,29,30,31,32,33,34,35].

Due to the inherent random variability in arc capacities within MFNs, the transmission time likewise exhibits stochastic behavior. In light of this uncertainty, the classical quickest path problem has evolved into a more comprehensive challenge known as the quickest path reliability problem (QPRP) within the context of MFNs [4,5,27,36,37,38,39,40]. The primary objective of the QPRP is to ascertain the probability of successfully transmitting a minimum of d units of flow from source node 1 to the destination node n via a single path, all while adhering to a stipulated time constraint of T units. This extension of the problem accounts for the dynamic and unpredictable nature of network conditions, making it particularly relevant in scenarios where both speed and reliability are paramount, such as in telecommunications, transportation, and various other domains [4,27,36,41,42].

Lin [36] introduced an algorithm requiring all the MPs as input. It determines the minimum capacity required for each MP to meet the time constraints for transmitting d flow units. Subsequently, by systematically evaluating each MP, the algorithm derives the solutions to the problem. Yeh et al. [41] harnessed the kth shortest path approach to devise an algorithm for addressing the problem. In subsequent work [42], they further refined and enhanced their algorithm. The QPRP was expanded to encompass scenarios involving two disjoint MPs in [38,43], as well as situations with multiple disjoint MPs in [38]. In a different approach, the researchers in [40] considered both time and budget constraints, utilizing these constraints to reduce computational complexity efficiently. Their extensive numerical analysis underscored the effectiveness of the proposed algorithm. This work was subsequently improved upon in [44]. Furthermore, researchers in [5] recognized the computational limitations of the algorithm presented in [36], mainly when the network configuration involves over thirty relevant MPs. To address this challenge, they introduced an unbiased Monte Carlo estimator as an alternative to exact evaluation, offering a more scalable solution for large-scale scenarios. In a recent development, detailed in [4], the authors addressed integrating budget constraints into the QPRP. They introduced an innovative approach that capitalizes on budget and time constraints to streamline the process by eliminating redundant MPs before algorithm execution. The authors then conducted extensive numerical experiments to underscore the enhanced performance of their approach when compared to existing methods in the literature. However, their algorithm still needs all the MPs as input.

Recognizing the inherent computational challenge of determining all MPs, which is an $NP-$hard, this study introduces an efficient approach designed to address the QPRP under the cost constraints, all without prior knowledge of MPs. Based on the node-child matrix, our proposed algorithm offers a novel methodology for solving the problem. To underscore its efficiency, we provide complexity analyses and present a wealth of experimental results, establishing the algorithm’s superior performance compared to existing methods in the literature.

The subsequent sections of this paper are organized as follows. Section 2 states the required notations, nomenclature, and assumptions. Section 3 provides some preliminaries on the problem. The proposed algorithm is stated in Section 4. The complexity results and an illustrative example are given in Section 5. We conduct several experimental results on benchmarks and randomly generated test problems in Section 6. Finally, we conclude the work in Section 7.

2. Notations, nomenclature, and assumptions

G	$G(N,A,M,L,C)$ represents a Multistate Flow Network (MFN), with $N=1,2,\cdots ,n$ as the node-set, where n signifies the total number of nodes. The set of arcs is denoted by $A=a_1,a_2,\cdots ,a_m$, where m corresponds to the number of arcs. The MFN is further characterized by: (1) $M=(M_1,\cdots ,M_m)$, a maximum capacity vector, where $M_i$ signifies the maximum capacity of arc $a_i$ for $i=1,\cdots ,m$. (2) $L=(l_1,\cdots ,l_m)$, a lead time vector, with each $l_i$ representing the lead time of arc $a_i$ for $i=1,\cdots ,m$. (3) $C=(c_1,\cdots ,c_m)$, a cost vector in which $c_i$ designates the transmission cost of arc $a_i$ for transmitting each unit of flow, for $i=1,\cdots ,m$. One also notes that nodes 1 and n are source and destination nodes, respectively. To illustrate, Figure 1 depicts an MFN defined by $N=1,2,3,4,5$ and $A=a_1,a_2,\cdots ,a_9$. As an example, the network has maximum capacity, lead time, and transmission cost vectors respectively as follows: $M=($4, 2, 5, 4, 3, 3, 4, 5), $L=($3, 1, 2, 3, 4, 3, 2, 3), and $C=($2, 3, 4, 3, 2, 3, 2, 1). Consequently, for instance, the values within these vectors indicate that at most four units of flow can be transmitted concurrently, at any time, through $a_1$, $a_4$, or $a_7$ due to $M_1=M_4=M_7=4$. Likewise, for example, $l_8=3$ denotes that passing up to $M_8=5$ units of flow through $a_8$ lasts three units of time. Furthermore, $c_2=3$ signifies that transmitting any flow unit on $a_2$ incurs a cost of three currency units.
X	$(x_1$, $x_2,\cdots ,x_m)$ is the current system state vector (SSV) in which $0\le x_i\le M_i$ is an integer-valued number and denotes the current capacity of arc $a_i$, for $i=1,\cdots ,m$. For instance, $X=($3, 2, 3, 4, 3, 3, 3, 5) can be considered as a current SSV for Figure 1.
$I_i$	shows the number of incoming arcs to node i, for $i=1,2,\cdots ,n$. We call this number the in-degree of the respective node. One notes that $I_1=0$ because all flows originate from the source node 1, and no flow goes to the source node. For instance, we have $I_1=0$, $I_2=I_3=I_5=3$, and $I_4=4$ in the network depicted in Figure 1.
$O_i$	shows the number of outgoing arcs from node i, for $i=1,2,\cdots ,n$. We call it the out-degree of the respective node. One notes that $O_n=0$ because all flows go to the destination node, and no flow goes out from this node. For instance, we have $O_1=O_4=3$, $O_2=O_3=4$, and $O_5=0$ in the network depicted in Figure 1.
$P_j$	is the jth minimal path (MP), for $j=1,\cdots ,h$. So, h is the number of MPs in the network. For instance, $P_1=\{a_1,a_4,a_5,a_8\}$ is an MP for the given network in Figure 1.
$K{P}_j\left(X\right)$	$K{P}_j\left(X\right)=min\left\{x_i\right|a_i\in P_j\}$ is the capacity of $P_j$ under SSV X, for $j=1,2,\cdots ,h$. For instance, in Figure 1, the capacity of $P_1=\{a_1,a_4,a_5,a_8\}$ under $X=($3, 2, 3, 4, 3, 3, 3, 5) is equal to $K{P}_1\left(X\right)=min\{$3, 4, 3, 5$\}=$3.
$L{P}_j$	$L{P}_j={\sum }_{i:a_i\in P_j}{l}_i$ is the lead time of the MP $P_j$, for $j=$1, 2, $\cdots ,h$. For instance, considering $L=($3, 1, 2, 3, 4, 3, 2, 3), we have $L{P}_1=l_1+l_4+l_5+l_8=$13 for $P_1=\{a_1,a_4,a_5,a_8\}$.
$C{P}_j$	$C{P}_j={\sum }_{i:a_i\in P_j}{c}_i$ is the transmission cost to send one unit of flow through $P_j$, for $j=$1, 2, $\cdots ,h$. For instance, considering $C=($2, 3, 4, 3, 2, 3, 2, 1), we have $C{P}_1=c_1+c_4+c_5+c_8=$8 for $P_1=\{a_1,a_4,a_5,a_8\}$.
d	a non-negative integer number that shows the demand value, the required flow to be transmitted from node 1 to node n.
$b,T$	b and T are the budget and time limits, respectively.
$R_{d,T,b}$	is the network’s reliability, which is the probability of successful transmission of at least d units of flow within T units of time through a single MP while incurring a cost of no more than b currency units.

Figure 1. A benchmark network example.

Figure 1. A benchmark network example.

2.1. Nomenclature

• Vector X = (x₁, x₂, ⋯, x_m) is considered less than or equal to vector Y = (y₁, y₂, ⋯ , y_m), denoted as X ≤ Y, if x_i ≤ y_i holds for all i = 1, 2, ⋯ , m. If, in addition to X ≤ Y, there exists at least one j such that x_j < y_j, we express it as X < Y. For instance, if we take X = (4, 2, 1), Y = (3, 1, 1), and Z = (2, 2, 2), we can observe that Y < X, Z ≮ X, X ≮ Z, Y ≮ Z, and Z ≮ Y.
• We define a vector X ∊ Ψ as a minimal vector when there is no other Y ∊ Ψ such that Y < X. For example, every vector in the set {(4, 3, 1), (2, 1, 3), (3, 4, 1), (1, 2, 2)} is a minimal vector. It is worth noting that a vector does not need to be less than or equal to all other vectors in the set to be considered minimal
• Noting that a path is a set of adjacent arcs enabling data transmission from source node 1 to destination node n, we say path P₁ is a subset of path P₂, denoted by P₁ ⊂ P₂ when P₂ encompasses all the arcs present in path P₁.

2.2. Assumptions

• The capacity of each arc $a_i\in A$ is a random integer ranging from 0 to $M_i$ for $i=1,2,\cdots ,m$, following a predefined probability distribution function. It is important to emphasize that $M_i$ is a known integer value, representing the maximum capacity of arc $a_i$.
• The arcs’ capacities are statistically independent.
• The network adheres to the flow conservation law, which means that no other node generates or accumulates flow apart from the source and destination nodes.
• All the required flow is sent through a solitary path from node 1 to node n.
• Each node is perfectly reliable.

It is worth highlighting that in cases where an unreliable node exists within the network, it can be represented as a pair of reliable nodes connected by an arc [45]. As a result, the final assumption does not impose any artificial constraints on the problem.

3. Background

We have two constraints in the problem: the budget and the time constraints, and we are looking to calculate the probability of successfully transmitting at least d units of flow from node 1 to node n such that the constraints are satisfied. One notes that the cost for transmitting d units of flow through a minimal path (MP) $P_j$ under system state vector (SSV) X is equal to

$$g(d,P_j)=d\times C{P}_j,$$

(1)

provided that $K{P}_j\left(X\right)>0$. In fact, as we are considering the time parameter, as far as the capacity of the MP is nonzero, its amount does not play any role in computing the transmission cost. However, the capacity of an MP directly affects the transmission time.

To illustrate it, consider the network of Figure 1 with current SSV, $X=($3, 2, 3, 4, 3, 3, 3, 5) and lead time vector $L=($3, 1, 2, 3, 4, 3, 2, 3). Consider the scenario where we aim to transmit a flow of $d=7$ units from node 1 to node n through path $P_1=\{a_1,a_4,a_5,a_8\}$. Observing that $K{P}_1\left(X\right)=min\{$3, 4, 3, 5$\}=3\ge 1$, it is feasible, and the transmission cost is calculated as $g(7,P_1)=7\times C{P}_1=$ 21. Additionally, we find that $L{P}_1=l_1+l_4+l_5+l_8=$13. Since $K{P}_1\left(X\right)=$3, the transmitted flow is limited to 3 units of flow at a time, and since $L{P}_1=$13, no flow can arrive at node n during the first 13 time units. Following this initial period, the flow is steadily pumped through, three units at a time, until the entire $d=7$ units of flow have successfully traversed path $P_1$. Consequently, it takes a total of 13 + $\lceil 7/3\rceil$=16 time units to send $d=7$ units of flow from node 1 to node n through $P_1$. Generally, the required time to transmit d units of flow from node 1 to node n through MP $P_j$ under SSV, X,, provided that $K{P}_j\left(X\right)>0$, is equal to

$$f(d,X,P_j)=L{P}_j+\lceil \frac{d}{K{P}_j\left(X\right)}\rceil ,$$

(2)

where $\lceil x\rceil$ is the smallest integer number not less than x. One notes that if $K{P}_j\left(X\right)=0$, it is impossible to transmit any flow through $P_j$, and one can define $f(d,X,P_j)=\infty$ for such a case.

To compute the reliability, one needs to find all the SSVs under which d units of flow can be sent through the network within the time T and budget b. The following result from [4] shows that it is sufficient to determine at least the minimal vectors with this property and not all of them.

Lemma 1.

[4] Assume that X and Y are two SSVs for the network G. If $X\le Y$, then for any MP $P_j$ with $K{P}_j\left(X\right)>0$, we have $f(d,X,P_j)\ge f(d,Y,P_j)$.

We now define the following function to take care of the time and budget limits simultaneously.

$$F(d,X,b)=min\left\{f(d,X,P_j)\right|K{P}_j\left(X\right)>0\&g(d,P_j)\le b,j=1,2,\cdots ,h\}$$

(3)

This way, $F(d,X,b)\le T$ signifies that one can transmit at least d units of flow from node 1 to node n through some MP in the network while adhering to the time and budget constraints. To elaborate, assuming ${\mathsf{\Psi }}_{d,T,b}=\{0\le X\le M|F(d,X,b)\le T\}$, it is evident that $R_{d,T,b}$$=Pr\left\{X\right|X\in {\mathsf{\Psi }}_{d,T,b}\}$. Moving forward, let

$${\mathsf{\Psi }}_{d,T,b}^{min}=\{X^1,X^2,\cdots ,X^{\sigma }\}$$

represent the collection of minimal vectors within ${\mathsf{\Psi }}_{d,T,b}$, and define $E_r=\left\{X\right|X\ge X^r\}$ for $r=1,2,\cdots ,\sigma$. By forming sets $B_1=E_1$, $B_2=E_2-E_1,\cdots$, $B_{\sigma }=E_{\sigma }-{\cup }_{r=1}^{\sigma -1}{E}_r$, it becomes apparent that ${\cup }_{r=1}^{\sigma }{E}_r={\cup }_{r=1}^{\sigma }{B}_r$. As a result, the computation of reliability, denoted as $R_{d,T,b}$, can be determined using the sum of disjoint products [46,47,48] as follows.

$$R_{(d,T,b)}=Pr\left({\cup }_{r=1}^{\sigma }{B}_r\right)=\sum _{r=1}^{\sigma }Pr\left(B_r\right),$$

(4)

where $Pr\left(B_r\right)={\sum }_{X\in B_r}Pr\left(X\right)$, and $Pr\left(X\right)={\prod }_{i=1}^{m}Pr\left(x_i\right)$. Therefore, the essential task is to determine the set ${\mathsf{\Psi }}_{d,T,b}^{min}=\{X^1,X^2,\cdots ,X^{\sigma }\}$.

Definition 1.

A system state vector X is called a ($d,T,b$)-$MP$ candidate if there exists an MP $P_j$ such that $K{P}_j\left(X\right)>0$, $g(d,P_j)\le b$, and $f(d,X,P_j)\le T$.

Proposition 1.

The set ${\mathsf{\Psi }}_{d,T,b}=\{0\le X\le M|F(d,X,b)\le T\}$ is the set of all the ($d,T,b$)-$MP$ candidates.

Definition 2.

A system state vector X is a ($d,T,b$)-$MP$, if and only if it is a ($d,T,b$)-$MP$ candidate and any $Y<X$ is not a ($d,T,b$)-$MP$ candidate.

Proposition 2.

The set ${\mathsf{\Psi }}_{d,T,b}^{min}$ is the set of all the (real) ($d,T,b$)-$MP$s.

In the next section, we provide an efficient algorithm to search for all the ($d,T,b$)-$MP$s.

4. The NCM-based algorithm

As all the flow must pass through a single MP, and no MP is a subset of another MP, it is possible to determine the minimum required capacity for each MP to facilitate the transmission of d units of flow within T units of time. Let $P_j$ be a designated MP with $C{P}_j\le b/d$. Now, if one creates an SSV, X, by setting the capacity of all the arcs within $P_j$ to an arbitrary positive value ${\alpha }_j$ and the capacity of all other arcs to zero, several observations can be made: (1) The capacity of $P_j$ under X is ${\alpha }_j$, that is, $K{P}_j\left(X\right)={\alpha }_j$. (2) The vector X is the minimal SSV under which the capacity of $P_j$ equals ${\alpha }_j$. (3) The capacity of all other MPs under X is zero, as each of them contains at least one arc not belonging to $P_j$.

Hence, one needs to determine the value ${\alpha }_j$ for $P_j$ such that d units of flow can be transmitted through it within T units of time. From Eq. (2), one sees that the required time to transmit d units of flow through $P_j$ under an arbitrary SSV, X, is equal to $L{P}_j+\lceil \frac{d}{K{P}_j\left(X\right)}\rceil$. Assume that $K{P}_j\left(X\right)={\alpha }_j>0$. As d and ${\alpha }_j$ are positive numbers, then $\lceil \frac{d}{{\alpha }_j}\rceil \ge 1$, and thus $L{P}_j$ should be less than T. Now, for the MP, $P_j$, that satisfies $C{P}_j\le b/d$ and $L{P}_j<T$, we have

$$L{P}_j+\lceil \frac{d}{{\alpha }_j}\rceil \le T\to \lceil \frac{d}{{\alpha }_j}\rceil \le T-L{P}_j\to {\alpha }_j\ge \lceil \frac{d}{T-L{P}_j}\rceil .$$

(5)

As a result, ${\alpha }_j=\lceil \frac{d}{T-L{P}_j}\rceil$ is the minimum possible capacity for $P_j$ such that d units of time can be transmitted through it within T units of time. If ${\alpha }_j\le K{P}_j\left(M\right)$, it is possible to create the corresponding SSV, X, described above, which is the corresponding ($d,T,b$)-$MP$ to $P_j$. Otherwise, it is impossible to have such a ($d,T,b$)-$MP$.

This forms the fundamental concept behind several algorithms presented in the literature, which rely on having access to all the MPs and inspecting each one individually to assess the feasibility of conducting the necessary transmission [4,36]. Nonetheless, the primary drawback of such algorithms lies in their dependence on the complete set of MPs. It is noteworthy that determining all the MPs is intrinsically an $NP$-hard problem, as established in [37,49]. Here, we use the idea of the node-child matrix utilized in [50] to propose an efficient algorithm that does not need any MP in advance.

The node-child matrix of an MFN is structured as an $n\times q$ matrix, where q represents the maximum out-degree of all the nodes within the network, determined as $q=max\left\{O_i\right|i=1,2,\cdots ,n-1\}$. In this matrix, each row corresponds to a specific node in the network and indicates its child nodes. For instance, the following is the node-child matrix related to the network depicted in Figure 1.

$$\mathbf{B}=\left[\begin{array}{ccc}2& 3& 4\\ 4& 5& 0\\ 4& 5& 0\\ 2& 3& 5\\ 0& 0& 0\end{array}\right]$$

One notes that the out-degrees of different nodes in the network may not be uniform. Consequently, when the out-degree of a specific node is less than the maximum out-degree q, we add “0” to the node-child matrix. For instance, in Figure 1, where we have $q=O_1=O_4=3$, we assign “0” in the last column of the respective rows for nodes 2 and 3, as $O_2=O_3=2$. Additionally, the last row in the node-child matrix always consists of zeros, as $O_n=0$. With this matrix in hand, a backtracking procedure can be employed to identify all the MPs [50]. We utilize this approach to discover all the ($d,T,b$)-$MP$s. We enhance the procedure by including two conditions for checking the lead time and transmission cost of the in-progress MPs. When either the lead time equals T, or the transmission cost exceeds b, we terminate the construction and move on to construct the next MP. It is worth noting that as the path is built and new arcs are added, the lead time and transmission cost of the in-progress path increases incrementally. Therefore, the algorithm continuously evaluates these two conditions after incorporating each new arc into the path. Significantly, once the lead time matches T or the transmission cost surpasses b for the in-progress path, the algorithm discontinues checking paths leading from that point to the destination node and instead reverts to building other paths.

One also notes that in the algorithm below, P is a vector that shows the ordered nodes in the under-construction MP, and $Lt$ and $cap$ are, respectively, its lead time and transmission cost.

The proposed NCM-based algorithm

• Input: $G(N,A,M,L,C)$, demand level d, budget limit b, and time limit T.
• Output: The set $\mathsf{\Theta }$ of all the ($d,T,b$)-$MP$s.

Step 0. Let $f={(1,\cdots ,1)}_{1\times n}$, $P=\left(1\right)$, $i=s=1$, $Lt=0$, $cap=\infty$, and $\mathsf{\Theta }=\left\{\right\}$.

Step 1. Determine the NCM, B.

Step 2. If $B(s,f(s\left)\right)\in P$, then let $f\left(s\right)=f\left(s\right)+1$ and repeat this step. Otherwise, let $t=B(s,f(s\left)\right)$.

Step 3. If $t\ne 0$, then go to Step 7.

Step 4. If $s=1$, then stop. Otherwise, if $s=n$, then go to Step 5; else, go to Step 6.

Step 5. Calculate the corresponding SSV with P and add it to $\mathsf{\Theta }$. If $i=2$, then stop. Otherwise, let

$$\begin{array}{cc}\hfill & f\left(P\right(i-1\left)\right)=1,\hfill \\ \hfill & c=c-C\left(P(i-1),P\left(i\right)\right)-C\left(P(i-2),P(i-1)\right),\hfill \\ \hfill & lt=lt-L\left(P(i-1),P\left(i\right)\right)-L\left(P(i-2),P(i-1)\right),\hfill \end{array}$$

remove the last two components from P, let $s=P\left(end\right)$ and $i=i-2$, and update $cap$. Go to Step 2.

Step 6. Let

$$\begin{array}{cc}\hfill & f\left(s\right)=1,\hfill \\ \hfill & c=c-C\left(P(i-1),P\left(i\right)\right),\mathrm{and}\hfill \\ \hfill & lt=lt-L\left(P(i-1),P\left(i\right)\right).\hfill \end{array}$$

Remove the last component from P, and let $s=P(i-1)$ and $i=i-1$. Update $cap$ and go to Step 2.

Step 7. If $lt+L(s,t)<T$, then let $\eta =\lceil \frac{d}{T-lt-L(s,t)}\rceil$. If $lt\ge T-L(s,t)$, $(c+C(s,t\left)\right)\times d>b$, or $\eta >min\{cap,M(s,t\left)\right\}$, then let $f\left(s\right)=f\left(s\right)+1$, else let $c=c+C(s,t)$, $lt=lt+L(s,t)$, $f\left(s\right)=f\left(s\right)+1$, $i=i+1$, $P\left(i\right)=t$, $s=t$, and $cap=min\{cap,M(s,t\left)\right\}$. Go to Step 2.

One notes that the last two nodes of the MP, P, are removed in Step 5 of the algorithm, and accordingly, the $cap$ is updated as follows. After removing these nodes, if P includes only node 1, then we have $cap=\infty$. Otherwise, $cap$ is equal to the minimum capacity of the arcs in P. To have a better understanding of the proposed algorithm, its flowchart is provided in Figure 2.

As our proposed algorithm is based on the node-child matrix in constructing the new MPs and correctly checks the lead time and budget constraints after adding any arc to the in-progress path, it is seen that the algorithm calculates all the ($d,T,b$)-$MP$s in the given MFN correctly and with no duplicates. Hence, the following Theorem is at hand.

Theorem 1.

The proposed algorithm above calculates all the ($d,T,b$)-$MP$s with no duplicates.

5. The complexity results and an illustrative example

5.1. The complexity results

To compute the time complexity of the proposed algorithm, we recall that n and m are the number of nodes and arcs in the network, respectively. Moreover, as the considered network is assumed to be connected, we have $O\left(n\right)\le O\left(m\right)\le O\left(n^2\right)$. Step 0 includes some simple considerations and is of the order of $O\left(1\right)$. To determine the node-child matrix, one needs to check all the outgoing arcs from each node, which takes at most $O\left(n\right)$ for each node, and hence $O\left(n^2\right)$ in total. Thus, the time complexity of Step 2 is $O\left(n^2\right)$. Steps 3 and 4 are of the order of $O\left(1\right)$. The corresponding SSV with the obtained MP is an $m-$tuple vector, and hence its calculation in Step 5 is at most of the order of $O\left(m\right)$. Updating $cap$ in Step 5 may need to find the minimum of $i-1$ numbers, and as i is bounded by n, the time complexity of calculating $cap$ is at most $O\left(n\right)$. The other calculations in Step 5 are simple and of the order of $O\left(1\right)$. Therefore, the time complexity of Step 5 is $O\left(m\right)$, reminding that $O\left(n\right)\le O\left(m\right)$. The updating $cap$ in Step 6 is of the order of $O\left(n\right)$ in the worst case, and the other calculations in this step are of the order of $O\left(1\right)$. Hence, Step 6 is of the order of $O\left(n\right)$. Step 7 includes some simple calculations and is of the order of $O\left(1\right)$.

One notes that Step 5 is run when we have a new solution to save, and one of steps 6 or 7 is run during the verification of each new node to determine a new solution. On the other hand, any MP has at most n nodes. Hence, the time complexity of steps 2 to 7 for each MP is at most $O\left(n^2\right)$, reminding that $O\left(m\right)\le O\left(n^2\right)$. As a result, recalling that h is the number of MPs in the network, the time complexity of steps 2 to 7 is at most $O\left(h{n}^2\right)$. As steps 0 and 1 are run parallel to other steps, the time complexity of the proposed NCM-based algorithm is $O\left(h{n}^2\right)$, and the following theorem is at hand.

Theorem 2.

The time complexity of the proposed node-child matrix-based algorithm to address the quickest path reliability problem under budget constraint is $O\left(h{n}^2\right)$.

One notes that the number of solutions to this problem is far less than the number of MPs in practice, and accordingly the time complexity of the proposed algorithm in practice is far less than the computed one for the worst case.

5.2. An illustrative example

Consider the provided flow network in Figure 3 as the communication infrastructure for a smart grid. In this network, each communication line comprises multiple dedicated fiber cables. These cables are exclusive to their respective lines, susceptible to failures, and possess distinct transmission capacities. Additionally, each cable requires a specific duration for data transmission and incurs a corresponding cost. Consequently, based on the type and quantity of available fiber cables, each arc in the network exhibits a probability distribution for capacity, lead time, and transmission cost, as detailed in Table 2. The objective is for the administrator to ascertain the likelihood of successfully transmitting a data volume $d=7$ units from node 1 to node seven within a time frame $T=8$ time units and a budget $b=213$ currency units using this network. We employ the proposed NCM-based algorithm to achieve this objective.

• Solution: There are $n=7$ nodes and $m=12$ arcs in the given network. We have $M=$ (3, 3, 3, 3, 5, 4, 4, 5, 3, 5, 5, 4), $L=$ (1, 4, 2, 3, 2, 4, 2, 3, 1, 1, 1, 3), and $C=$ (8, 8, 9, 8, 7, 8, 6, 6, 7, 8, 4, 3) according to Table 2, and $T=8$, $b=213$, and $d=7$ are given.
• Step 0. We let $f=(1,1,1,1,1,1,1)$, $P=\left(1\right)$, $i=s=1$, $Lt=0$, $cap=\infty$, $R=0$, and $\mathsf{\Theta }=\left\{\right\}$.
• Step 1. The NC matrix is equal to

  $$\begin{array}{c}\hfill B=\left[\begin{array}{cccc}2& 3& 4& 0\\ 5& 7& 0& 0\\ 4& 5& 6& 0\\ 3& 6& 0& 0\\ 2& 3& 6& 7\\ 3& 4& 5& 7\\ 0& 0& 0& 0\end{array}\right]\end{array}$$
• Step 2. $B(1,f(1\left)\right)=2\notin P$, so we let $t=2$.
• Step 3. $t\ne 0$, the transfer is made to Step 7.
• Step 7. $Lt+L(1,2)=1<8$, so $\eta =\lceil \frac{7}{8-1}\rceil =1$. As $Lt<8-1$, $7\times 8<213$, and $\eta =1\le 3=min\{\infty ,M(1,2\left)\right\}$, we let $c=8$, $Lt=1$, $f\left(1\right)=2$, $i=2$, $P=(1,2)$, $s=2$, $cap=3$, and go to Step 2.
• Step 2. $B(2,f(2\left)\right)=5\notin P$, so we let $t=5$.
• Step 3. $t\ne 0$, the transfer is made to Step 7.
• Step 7. $Lt+L(2,5)=4<8$, so $\eta =\lceil \frac{7}{8-4}\rceil =2$. As $Lt<8-3$, $7\times 16<213$, and $\eta =2\le 3=min\{3,M(2,5\left)\right\}$, we let $c=16$, $Lt=4$, $f\left(2\right)=2$, $i=3$, $P=(1,2,5)$, $s=5$, $cap=3$, and go to Step 2.
• Step 2. $B(5,f(5\left)\right)=2\in P$, we let $f\left(5\right)=1+1=2$ and repeat this step.
• Step 2. $B(5,f(5\left)\right)=3\notin P$, so we let $t=3$.
• Step 3. $t\ne 0$, the transfer is made to Step 7.

Table 2. The arcs data for Figure 3.

Arcs	Lead time	Cost	Capacities/Probabilities
			0	1	2	3	4	5
$a_1$	1	8	0.01	0.04	0.05	0.9	0	0
$a_2$	4	8	0.01	0.02	0.03	0.94	0	0
$a_3$	2	9	0.01	0.09	0.1	0.8	0	0
$a_4$	3	8	0.01	0.04	0.1	0.85	0	0
$a_5$	2	7	0.01	0.02	0.02	0.02	0.03	0.9
$a_6$	4	8	0.01	0.02	0.05	0.1	0.82	0
$a_7$	2	6	0.01	0.05	0.1	0.1	0.74	0
$a_8$	3	6	0.01	0.01	0.05	0.02	0.01	0.9
$a_9$	1	7	0.01	0.02	0.02	0.95	0	0
$a_{10}$	1	8	0.01	0.02	0.04	0.02	0.06	0.85
$a_{11}$	1	4	0.01	0.03	0.03	0.03	0.05	0.85
$a_{12}$	3	3	0.01	0.05	0.05	0.05	0.84	0

Table 2. The arcs data for Figure 3.

Arcs	Lead time	Cost	Capacities/Probabilities
			0	1	2	3	4	5
$a_1$	1	8	0.01	0.04	0.05	0.9	0	0
$a_2$	4	8	0.01	0.02	0.03	0.94	0	0
$a_3$	2	9	0.01	0.09	0.1	0.8	0	0
$a_4$	3	8	0.01	0.04	0.1	0.85	0	0
$a_5$	2	7	0.01	0.02	0.02	0.02	0.03	0.9
$a_6$	4	8	0.01	0.02	0.05	0.1	0.82	0
$a_7$	2	6	0.01	0.05	0.1	0.1	0.74	0
$a_8$	3	6	0.01	0.01	0.05	0.02	0.01	0.9
$a_9$	1	7	0.01	0.02	0.02	0.95	0	0
$a_{10}$	1	8	0.01	0.02	0.04	0.02	0.06	0.85
$a_{11}$	1	4	0.01	0.03	0.03	0.03	0.05	0.85
$a_{12}$	3	3	0.01	0.05	0.05	0.05	0.84	0

• Step 7. $Lt+L(5,3)=6<8$, so $\eta =\lceil \frac{7}{8-6}\rceil =4$. As $\eta =4>3=min\{3,M(5,3\left)\right\}$, we let $f\left(5\right)=3$ and go to Step 2.
• Step 2. $B(5,f(5\left)\right)=6\notin P$, so we let $t=6$.
• Step 3. $t\ne 0$, the transfer is made to Step 7.

   ⋮

• The final set of solutions is obtained { (3, 0, 0, 3, 0, 0, 0, 0, 0, 0, 3, 0), (2, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0), (0, 0, 3, 0, 0, 0, 0, 0, 3, 3, 3, 0)}.

Notably, 368,640 potential state vectors exist for this modestly-sized network. Consequently, directly validating all these vectors is an exceedingly time-consuming endeavor. Furthermore, the presence of 25 MPs in this network underscores the inefficiency of algorithms that utilize all MPs as input and systematically check them individually to identify solutions. The next section discusses our proposed algorithm’s efficiency in more detail.

6. Experimental results

Recently, the authors in [4] demonstrated the superiority of their proposed algorithm over other exact algorithms in the literature by evaluating complexity results and conducting numerous numerical experiments. In light of this, we compare their algorithm with ours to showcase the practical effectiveness of our approach in contrast to existing literature. Both algorithms are implemented in the MATLAB programming environment and compared on the Arpanet topology—a rather large-sized benchmark with 20 nodes, 32 arcs, and 1,610 MPs (depicted in Figure 4). Additionally, we employ one thousand randomly generated large-sized test problems for a comprehensive assessment of algorithmic efficiency. The computations are carried out on a computer with an Intel(R) Core(TM) i5-12500 Duo CPU clocked at 3.00 GHz and 32.0 GB of RAM.

The capacities, lead times, and transmission costs of the arcs in both cases, the Arpanet and the randomly generated test problems, are assigned random integer values within the intervals [5, 20], [3, 10], and [5, 15], respectively. It is essential to note that with sufficiently large time and budget limits, any SSV can be a solution, rendering the algorithms redundant. To ensure meaningful constraints, we define a specific time limit $T=\overline{LP}$ and budget limit $b=d\times \overline{CP}$ for each test problem, where $\overline{LP}$ and $\overline{CP}$ are the arithmetic means of paths’ lead times and paths’ costs, respectively.

For the Arpanet topology, we have compared the algorithms in ten cases by assigning the demand $d=\alpha \times \lceil \overline{KP}\rceil$, where $\alpha =16,17,\cdots ,25$, and $\overline{KP}$ is the arithmetic mean of paths’ capacities, and $\lceil x\rceil$ is the smallest integer number not less than x. One notes that our tests showed very few solutions or no solutions for larger demand values in this benchmark example. One also notes that for every case, the arcs’ capacities, lead times, and transmission costs have been generated randomly, as explained above. That is, we compared the algorithms on this benchmark for ten different scenarios. Table 3 presents the final results, with columns detailing the demand level, the number of solutions, the runtime of our proposed algorithm, the runtime of the algorithm proposed in [4], and the time ratio $t_2/t_1$, respectively. The last column in the table illustrates that our proposed algorithm outperforms the other algorithm by solving all cases at least 3.5 times faster, with some instances exceeding eight times faster and averaging over five times faster. These outcomes unequivocally highlight the superiority of our algorithm in comparison to the alternative in this benchmark network example.

For a more meaningful comparative analysis of the algorithms, we leverage a dataset comprising one thousand randomly generated test problems. To construct this dataset, we vary the number of nodes, denoted as n, across the range of 31 to 40. For each value of n, we generate 100 distinct random networks, totaling 1000 test problems. To maintain a balanced distribution, avoiding overly dense networks with an abundance of MPs or extremely sparse networks with few MPs, we use the utilized limits in [4] for the number of arcs in each random network. Subsequently, the number of arcs in each randomly generated network is assigned random integer values within the interval $[3\times (\lceil n/2\rceil -1),2\times (\lceil n/2\rceil +10\left)\right]$. The specifics of the arcs, including data values, as well as time and budget constraints, are determined randomly, following a similar methodology as applied to the Arpanet network. Additionally, in each test problem, the demand level is established as the arithmetic mean of the capacities of the MPs.

This way, we created ten sets of randomly generated networks, each comprising one hundred test problems. Table 4 illustrates the average data for each set. The table columns present respectively the number of nodes, the average count of MPs, the average number of solutions, the average runtime of our proposed algorithm, the average runtime of the algorithm proposed in [4], and the average time ratio of the runtimes. The last column in this table also shows that our proposed algorithm has solved the random test problems on average six times faster than the other algorithm and notably indicates the superiority of our proposed algorithm.

Furthermore, for a meaningful comparison across the set of $N_p=1000$ test problems, we assess the runtimes of both algorithms by creating a performance profile following the framework established by Dolan and Moré [52]. This performance profile assesses the ratio of computation times for these algorithms concerning the best time achieved by any algorithm. In essence, let $t_{i,1}$ and $t_{i,2}$ denote the computation times for our proposed algorithm and the one proposed in [4], respectively, for $i=1,2,\cdots ,1000$. The performance ratios are then determined as $r_{i,j}=\frac{t_{i,j}}{min{t}_{i,j}:j=1,2}$, where $j=$ 1, 2 [52]. The performance of each algorithm is characterized by $P{r}_j\left(\tau \right)=\frac{1}{N_p}size\left\{i\right|r_{i,j}\le \tau \}$ for $j=1,2$. Here, the size represents the number of problems for which the respective algorithm achieves a performance ratio within a factor $\tau \in \mathbb{R}$ of the best possible ratio. Consequently, $P{r}_j\left(\tau \right)$ quantifies the probability that an algorithm’s performance ratio $r_{i,j}$ falls within the factor $\tau$. According to this profile, one algorithm is deemed superior to another when its performance chart surpasses that of the other [52].

Figure 5 provides the resulting performance profiles for both algorithms. This figure shows clearly that our proposed algorithm has solved almost all the test problems faster than the other algorithm. One also can see that almost 90% of the test problems have been solved by our proposed algorithm almost four times faster. Moreover, it shows that our algorithm has solved some of the test problems more than nine times faster than the other algorithm. In line with the previous experimental results, Figure 5 unequivocally highlights the efficiency of our proposed algorithm compared to the other one.

7. Conclusions

Typical algorithms proposed in the literature to tackle the quickest path problem in multistate flow networks (MFNs) often encompass three fundamental stages: (1) identifying all the minimal paths (MPs) of the network, (2) scrutinizing each MP to ascertain if it meets the necessary conditions for validity, and (3) computing the corresponding system state vectors associated with these validated MPs. It is worth noting, however, that the initial step of determining all MPs of an MFN belongs to the family of the $NP$-hard problems. Moreover, as the number of MPs increases exponentially with the network size, the second stage turns out to be very time-consuming for large-sized MFNs. To address this complexity and consider the cost constraints that are crucial for real-world systems, this study proposed an improved approach, capitalizing on the network’s node-child matrix structure, to resolve the problem without the prerequisite of acquiring MPs beforehand. We demonstrated the algorithm’s correctness, computed its time complexity, and substantiated it by a benchmark example. Moreover, several numerical results on known benchmarks and randomly generated test problems were provided to show the efficiency of our proposed algorithm in comparison with the existing ones in the literature.