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Research article Special Issues

Lifting mobility restrictions and the effect of superspreading events on the short-term dynamics of COVID-19

  • SARS-CoV-2 has now infected 15 million people and produced more than six hundred thousand deaths around the world. Due to high transmission levels, many governments implemented social distancing and confinement measures with different levels of required compliance to mitigate the COVID-19 epidemic. In several countries, these measures were effective, and it was possible to flatten the epidemic curve and control it. In others, this objective was not or has not been achieved. In far too many cities around the world, rebounds of the epidemic are occurring or, in others, plateaulike states have appeared, where high incidence rates remain constant for relatively long periods of time. Nonetheless, faced with the challenge of urgent social need to reactivate their economies, many countries have decided to lift mitigation measures at times of high incidence. In this paper, we use a mathematical model to characterize the impact of short duration transmission events within the confinement period previous but close to the epidemic peak. The model also describes the possible consequences on the disease dynamics after mitigation measures are lifted. We use Mexico City as a case study. The results show that events of high mobility may produce either a later higher peak, a long plateau with relatively constant but high incidence or the same peak as in the original baseline epidemic curve, but with a post-peak interval of slower decay. Finally, we also show the importance of carefully timing the lifting of mitigation measures. If this occurs during a period of high incidence, then the disease transmission will rapidly increase, unless the effective contact rate keeps decreasing, which will be very difficult to achieve once the population is released.

    Citation: Mario Santana-Cibrian, Manuel A. Acuña-Zegarra, Jorge X. Velasco-Hernandez. Lifting mobility restrictions and the effect of superspreading events on the short-term dynamics of COVID-19[J]. Mathematical Biosciences and Engineering, 2020, 17(5): 6240-6258. doi: 10.3934/mbe.2020330

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  • SARS-CoV-2 has now infected 15 million people and produced more than six hundred thousand deaths around the world. Due to high transmission levels, many governments implemented social distancing and confinement measures with different levels of required compliance to mitigate the COVID-19 epidemic. In several countries, these measures were effective, and it was possible to flatten the epidemic curve and control it. In others, this objective was not or has not been achieved. In far too many cities around the world, rebounds of the epidemic are occurring or, in others, plateaulike states have appeared, where high incidence rates remain constant for relatively long periods of time. Nonetheless, faced with the challenge of urgent social need to reactivate their economies, many countries have decided to lift mitigation measures at times of high incidence. In this paper, we use a mathematical model to characterize the impact of short duration transmission events within the confinement period previous but close to the epidemic peak. The model also describes the possible consequences on the disease dynamics after mitigation measures are lifted. We use Mexico City as a case study. The results show that events of high mobility may produce either a later higher peak, a long plateau with relatively constant but high incidence or the same peak as in the original baseline epidemic curve, but with a post-peak interval of slower decay. Finally, we also show the importance of carefully timing the lifting of mitigation measures. If this occurs during a period of high incidence, then the disease transmission will rapidly increase, unless the effective contact rate keeps decreasing, which will be very difficult to achieve once the population is released.


    In the past few decades, the research on fuzzy relation systems, including equation systems and inequality systems, has developed rapidly and comprehensively. The concept of fuzzy relation equation (FRE) was first proposed by E. Sanchez [1]. In [1], the composition operation was max-min. A sufficient and necessary condition for the solvable (having at least one solution) FRE system composed of max-min was displayed in [1]. Moreover, for the solvable system, the full solution set contains a finite set of lower solutions and a greatest solution. It was widely acknowledged that numerous issues related to body knowledge can be addressed as FRE problems [2]. In an FRE system, the initial composition max-min was soon later generalized to max-T, where T represents a triangular norm [3,4,5,6]. The structure of the solution set in a solvable max-T FRE system is the same as that in a solvable max-min one. The identification of solvability and the determination of the solution set are the central and foundational aspects of the research on FRE. Many scholars devoted their time to searching all the solutions to an FRE system, with the max-T composition [7,8,9,10,11].

    J. Drewniak mentioned the fuzzy relation inequality (FRI) with max-min for the first time [12]. The max-min FRIs were completely solved in [13], using the so-called conservative path approach. In [13], the authors also studied a kind of optimization problem, in which the objective function was expressed in a latticized linear form and the constraint was the max-min FRIs. F. Guo et al. also proposed the FRI path approach for solving the minimal solutions in a max-min FRIs system [14].

    In 2012, J.-X. Li et al. [15] employed a new composition operator, i.e., addition-min, in a system of FRIs. The addition-min of FRIs

    1jnaijxjbi,i=1,2,,m, (1.1)

    were introduced for describing the peer-to-peer (P2P) educational information sharing system. In such a P2P network system, n terminals are represented by the notations T1,,Tn as shown in Figure 1.

    Figure 1.  Peer-to-Peer educational information sharing system.

    In system (1.1), the variable xj represents the quality level on which the terminal Tj sends out (shares) its local data. The parameter aij represents the bandwidth. The download traffic requirement of Ti is requested to be no less than bi. Obviously, the total download traffic of the ith terminal Ti is satisfied if and only if it holds that

    1jn(aijxj)bi. (1.2)

    Satisfying the download traffic requirements of all the terminals, the above system (1.1) is naturally generated and established.

    Some detailed properties, particularly the number of minimal solutions and the convexity of the solution set, were extensively discussed in [16]. It was highlighted in [16] that the addition-min FRIs system often has infinitely many minimal solutions. Thus, it is difficult to compute its complete solution set. Of course, when the scale of the problem is small enough, it is still possible to find out all the minimal solutions, as well as the complete solution set [17]. The min-max programming and the leximax programming were formulated and studied subject to the addition-min FRIs system [18,19,20]. Considering the random line fault [21] or the distinguishing quality levels on which the terminals send out their local files [22,23], some variants of the addition-min FRIs were also investigated.

    As pointed out above, characterizing the P2P network system by the addition-min of the FRIs system (1.1), the authors only considered the total download traffic of the ith terminal. However, when considering the highest download traffic, the above system (1.1) is ineffective. In fact, the highest download traffic of the ith terminal Ti should be

    1jn(aijxj)bi. (1.3)

    As a consequence, adopting the highest download traffic, the P2P educational information sharing system was characterized by the max-min FRIs below [24,25,26,27],

    1jn(aijxj)bi,i=1,2,,m. (1.4)

    G. Xiao et al. [25] discussed the classification of the solution set of (1.4), while X. Yang [26] attempted to find the approximate solution for the unsolvable system (1.4). Considering the rigid requirement, exactly equal to bi, system (1.4) turns out to be

    1jn(aijxj)=bi,i=1,2,,m. (1.5)

    Different resolution methods were proposed for solving the approximate solutions of the above system (1.5) [28,29] when it was unsolvable. In the solvable case, three types of geometric programming problems subject to system (1.5) were solved [30,31,32]. Considering the flexible requirement, [33,34,35] assumed the highest download traffic of Ti to be no less than ci and no more than di, and established the following max-min FRIs with bidirectional constraints.

    ci(bi1y1)(bi2y2)(binyn)di,i=1,2,,m. (1.6)

    After standardization, it is assumed that ci,bij,yj,di[0,1], iI,jJ, where

    I={1,,m},J={1,,n}.

    Similar to system (1.1) for describing the P2P educational information sharing system, any solution of system (1.6) stands for a feasible flow control scheme for the P2P network system. A given solution is indeed a feasible scheme prepared in advance. However, in the actual implementation process, the values of components in the given solution might encounter some temporary adjustment or uncontrollable variation. Thus, we consider the tolerable variation for a given solution in this work. The widest symmetrical interval solution will be defined and investigated. It embodies the tolerable variation for a given solution. Moreover, we will further define the centralized solution for system (1.6). The centralized solution is indeed the solution with the biggest tolerable variation. As is well known, a feasible scheme is considered to be more stable if it is able to bear a bigger tolerable variation. As a result, the centralized solution would be considered to be the most stable feasible scheme in the P2P network system.

    The remaining sections are arranged as follows. Section 2 presents the basic foundations on the max-min FRIs system (1.6). In Section 3, we propose an effective approach for solving the widest symmetrical interval solution regarding a provided solution. In Section 4, we further provide a resolution approach for searching the centralized solution for system (1.6). Some numerical examples are enumerated for illustrating our presented resolution approaches. Section 5 concludes the work.

    For convenience, we abbreviate

    (bi1y1)(bi2y2)(binyn)=(bi1,,bin)(y1,,yn),

    for any iI. Furthermore, the above max-min FRIs system (1.6) could be rewritten in its abbreviation form as

    cByd, (2.1)

    where c=(c1,,cm), B=(bij)m×n, y=(y1,,yn) and d=(d1,,dm). Accordingly, the set of all solutions to system (1.6) or system (2.1) is exactly

    S(B,c,d)={yVcByd}. (2.2)

    where V=[0,1]n.

    Definition 1. (Consistent)[33,34,35] System (1.6) is called consistent if it has a solution, i.e., S(B,c,d). Conversely, S(B,c,d)=; we call it inconsistent.

    Definition 2. [33,34,35] ˆvV is a maximum solution if ˆvS(B,c,d) and ˆvv hold for any vS(B,c,d). vmV is a minimal solution if vmS(B,c,d) and vvmv=vm holds for any vS(B,c,d).

    Define the operator "@" as follows:

    bij@di={1,ifbijdi,di,ifbij>di. (2.3)

    Let ˆv=(ˆv1,ˆv2,,ˆvn), where

    ˆvj=iIbij@di. (2.4)

    Theorem 1. [33,34,35] S(B,c,d) if and only if ˆvS(B,c,d).

    Proposition 1. [33,34,35] If S(B,c,d) and vS(B,c,d) is an arbitrary solution, then we have vˆv.

    According to Theorem 1 and Proposition 1, when (1.6) is a consistent system, ˆv is always its maximum solution.

    Proposition 2. [33,34,35] Let u,vS(B,c,d) be two solutions of system (1.6), with uv. Then for any x[u,v], it holds xS(B,c,d), i.e., x is also a solution of (1.6).

    Theorem 2. [33,34,35] Let system (1.6) be consistent. Then the solution set is

    S(B,c,d)=vmSm(B,c,d)[vm,ˆv], (2.5)

    where Sm(B,c,d) is the collection of all minimal solutions, while ˆv is the maximum one.

    In this section, we posit the hypothesis that

    v=(v1,v2,,vn)V

    is one of the given solutions in the max-min FRIs system (1.6), i.e., vS(B,c,d). We will define the concept of the widest symmetrical interval solution regarding the provided solution v. Moreover, some effective procedures will be proposed for obtaining the widest symmetrical interval solution.

    Definition 3. (Intervalsolutionwidth) Let u,v be two vectors in V, with uv. If

    [u,v]S(B,c,d),

    then we say [u,v] an interval solution of (1.6). Moreover, the following non-negative number, denoted by w[u,v], i.e.,

    w[u,v]=(v1u1)(v2u2)(vnun) (3.1)

    is said to be the width of the interval solution [u,v].

    Remark 1. Let [u,v],[x,y]S(B,c,d) be two interval solutions of (1.6). If [u,v][x,y], then it holds w[u,v]w[x,y]. However, this conclusion does not hold in reverse. For example, suppose [u,v]=([0.2,0.8],[0.3,0.4],[0.4,0.9]),[x,y]=([0.3,0.5],[0.4,0.6],[0.4,0.7]). It is clear that w[u,v]=0.10.2=w[x,y]. However, the inclusion relation [u,v][x,y] doesn't hold.

    Definition 4. (Symmetricalintervalsolution(SIS)regardingv) Let ε=(ε1,ε2,,εn)V be an arbitrary vector. Denote the following vectors v+ε and vε, according to ε and v,

    {v+ε=(v1+ε1,,vn+εn),vε=(v1ε1,,vnεn). (3.2)

    If [vε,v+ε] is an interval solution of system (1.6), we say [vε,v+ε] a symmetricalintervalsolution regarding the provided solution v.

    Obviously, if [vε,v+ε], represented by (3.2), is a symmetrical interval solution (SIS) regarding v, then by (3.1), its width is indeed

    w[vε,v+ε]=2(ε1ε2εn). (3.3)

    Definition 5. (Widest symmetrical interval solution (WSIS) regarding v) Let [vεv,v+εv] be an SIS regarding the provided solution v, where εvV. [vεv,v+εv] is said to be a widest symmetrical interval solution regarding v, if

    w[vεv,v+εv]w[vε,v+ε] (3.4)

    holds for any SIS [vε,v+ε] regarding v, where εV. Moreover, the width of [vεv,v+εv], i.e., w[vεv,v+εv], is said to be the (biggest) symmetrical width regarding v.

    According to our provided solution v, define m indicator sets as

    Jvi={jJ|bijvjci}, (3.5)

    where iI. Furthermore, define m indices as

    jvi=argmax{vj|jJvi}, (3.6)

    where iI. Now we find the indices jv1,jv2,,jvm. These indices enable us to further define

    Ivj={iI|jvi=j}, (3.7)

    where jJ. Next, we could construct a vector, denoted by ˘v=(˘v1,˘v2,,˘vn), in which

    ˘vj={iIvjci,ifIvj,0,ifIvj=,jJ. (3.8)

    Proposition 3. Assume that vS(B,c,d). Then there is Jvi for any iI.

    Proof. Since vS(B,c,d), according to system (1.6), we have

    ci(bi1v1)(bi2v2)(binvn)di,iI.

    Therefore, for any iI, there is jiJ satisfying

    bijiyjici.

    Observing Eq (3.5), we have jiJvi, i.e., Jvi.

    Proposition 4. For the solution vS(B,c,d) of (1.6) and the vector ˘v defined by (3.8), we have ˘vv.

    Proof. Let jJ be an arbitrary indicator.

    Case 1. If Ivj=, it follows from (3.8) that ˘vj=0vj.

    Case 2. If Ivj, it follows from (3.8) that ˘vj=iIvjci. Take arbitrarily iIvj. By (3.7), we have

    jvi=j.

    According to (3.6), it is clear j=jviJvi. It follows from (3.5) that

    vjbijvjci.

    In view of the arbitrariness of i in Ivj, there is vjiIvjci=˘vj.

    Cases 1 and 2 indicate ˘vjvj, jJ. Thus, we obtain ˘vv

    Theorem 3. For the solution vS(B,c,d) of (1.6) and the vector ˘v defined by (3.8), there is ˘vS(B,c,d). That is to say, ˘v serves as a solution of (1.6).

    Proof. Take arbitrarily iI. It is clear jviJvi by (3.6). According to (3.5), we have

    bijvibijvivjvici. (3.9)

    Denote j=jvi. Then we have iIvj by (3.7). According to (3.8), we have

    ˘vj=kIvjckci. (3.10)

    Considering j=jvi, by (3.9) and (3.10) we further have

    (bi1˘v1)(bi2˘v2)(bin˘vn)bij˘vjci,iI. (3.11)

    Since vS(B,c,d), according to system (1.6), we have

    ci(bi1v1)(bi2v2)(binvn)di,iI.

    Thus,

    bijvjdi,iI,jJ.

    Following Proposition 4, it holds ˘vjvj, jJ. Thus we get

    bij˘vjbijvjdi,iI,jJ,

    i.e.,

    (bi1˘v1)(bi2˘v2)(bin˘vn)di,iI. (3.12)

    Combining Inequalities (3.11) and (3.12), it is evident to have ˘vS(B,c,d).

    In the previous section, we have obtained the vector ˘v based on the provided solution v. Moreover, ˘v is found to be a solution of system (1.6), no more than v, i.e., ˘vv. In this section, we further construct a symmetrical interval solution regarding v, based on the solutions ˘v,v,ˆv.

    Remind that ˆv is the maximum solution, while v is a provided solution of (1.6). Besides, ˘v is as obtained following Eqs (3.5)–(3.8). Now we denote the vector εv=(εv1,εv2,,εvn) related to v as

    εvj=(vj˘vj)(ˆvjvj),jJ. (3.13)

    Note that ˘vjvˆv(1,,1). It could be evidently found that 0εvj1 for all jJ. Hence, we have εvV.

    Next, we provide some properties on the above-obtained vector εv.

    Proposition 5. Let εvV be the vector defined by (3.13), related to the given solution v. Then there is v+εvS(B,c,d), i.e., v+εv satisfies system (1.6).

    Proof. Since 0εvj1 and

    εvj=(vj˘vj)(ˆvjvj)ˆvjvj,jJ.

    we have

    vjvj+εvjˆvj,jJ.

    That is vv+εvˆv. Since v,ˆvS(B,c,d), following Proposition 2 we have v+εvS(B,c,d).

    Proposition 6. Let εvV be the vector defined by (3.13), related to the given solution v. Then there is vεvS(B,c,d).

    Proof. Since 0εvj1 and

    εvj=(vj˘vj)(ˆvjvj)vj˘vj,jJ,

    we have

    ˘vjvjεvjvj,jJ.

    That is ˘vvεvv. According to Theorem 3 and the given condition, both v and ˘v are solutions to system (1.6). It follows from Proposition 2 that vεvS(B,c,d).

    According to the above Propositions 5 and 6, one easily discovers the Corollary 1 below.

    Corollary 1. Let εvV be the vector defined by (3.13), related to the given solution v. Then there is [vεv,v+εv]S(B,c,d), i.e., [vεv,v+εv] is a symmetrical interval solution regarding v.

    It is shown in Corollary 1 that [vεv,v+εv] is a symmetrical interval solution regarding v. Next, it will be further verified to be the widest one.

    Proposition 7. Let uS(B,c,d) be a solution of (1.6) with uv. Then it holds w[u,v]w[˘v,v].

    Proof. Select arbitrarily jJ. Now let us examine the inequality w[u,v]vj˘vj.

    Case 1. If Ivj=, then ˘vj=0uj by (3.8). Hence

    w[u,v]=kJ(vkuk)vjujvj˘vj. (3.14)

    Case 2. If Ivj, then ˘vj=iIvjci by (3.8). Accordingly, there is iIvj, satisfying

    ˘vj=ci. (3.15)

    At the same time, by (3.7) we have

    jvi=j, (3.16)

    since iIvj. Note that uS(B,c,d). u satisfies the ith inequality from (1.6), i.e.,

    ci(bi1u1)(bi2u2)(binun)di.

    As a result, there is jJ satisfying

    bijujci. (3.17)

    Considering uv, we have bijvjbijujci. According to (3.5), it holds that jJvi. Moreover, Inequality (3.17) implies that

    ujbijujci. (3.18)

    Observing (3.6) and (3.16), we have j=jvi=argmax{vl|lJvi}. Thus,

    vjvl,lJvi.

    Thereby, jJvi indicates

    vjvj. (3.19)

    Combining (3.15), (3.18), and (3.19), we have

    vj˘vj=vjcivjujvjujkJ(vkuk)=w[u,v]. (3.20)

    According to Inequalities (3.14) and (3.20), for any j in J, there is vj˘vjw[u,v]. So we have

    w[˘v,v]=jJ(vj˘vj)w[u,v].

    The proof is complete.

    Theorem 4. Let εvV be the vector defined by (3.13), related to the given solution v. Then [vεv,v+εv] is the widest symmetrical interval solution regarding v.

    Proof. It has been verified in Corollary 1 that [vεv,v+εv] is an SIS regarding v.

    Let [vε,v+ε] be an arbitrary SIS regarding the solution v, where εV=[0,1]n. According to the definition of interval solution, it holds

    [vε,v+ε]S(B,c,d),

    That is to say, both v+ε and vε are solutions. Since ˆv is maximum in S(B,c,d), it holds that v+εˆv. Thus, for any jJ,

    vj+εjˆvj,

    or written as

    ˆvjvjεj.

    As a result,

    jJεjjJ(ˆvjvj). (3.21)

    In addition, since vεS(B,c,d) and vεv, it follows from Proposition 7 that

    w[˘v,v]w[vε,v]. (3.22)

    According to (3.1) and (3.2), w[vε,v]=ε1ε2εn=jJεj, and w[˘v,v]=jJ(vj˘vj). The above inequality (3.22) turns out to be

    jJ(vj˘vj)jJεj. (3.23)

    Combining Inequalities (3.21), (3.23), and (3.13), it holds

    jJεvj=jJ[(ˆvjvj)(vj˘vj)]=[jJ(ˆvjvj)][jJ(vj˘vj)]jJεj. (3.24)

    Consequently, it follows from (3.3) and (3.24)

    w[vεv,v+εv]=jJεvjjJεj=w[vε,v+ε].

    Following Definition 5, [vεv,v+εv] is the widest symmetrical interval solution regarding v.

    This subsection provides an algorithm to find the widest symmetrical interval solution regarding a given solution. Moreover, some numerical examples are given for verifying our proposed algorithm.

    Algorithm Ⅰ: solving the widest symmetrical interval solution regarding v

    Step 1. Following (3.5), calculate the indicator set Jvi for any iI.

    Step 2. Following (3.6), select the indicator jvi from the indicator set Jvi for any iI.

    Step 3. Following (3.7), construct the indicator set Ivj for any jJ.

    Step 4. Following (3.8), generate the vector ˘v=(˘v1,˘v2,,˘vn), where the ˘vj is defined by (3.8).

    Step 5. Following (3.13), calculate the vector εv=(εv1,εv2,,εvn), where the εvj is defined by (3.13).

    Step 6. Construct the widest symmetrical interval solution as [vεv,v+εv], according to (3.2).

    Computational complexity of Algorithm Ⅰ: In Algorithm Ⅰ, Step 1 requires 2mn operations for calculating the indicator sets. Moreover, Steps 2–4 have an identical calculation amount as mn. Besides, Steps 5 and 6 require 3n and 2n operations, respectively. As a result, applying Algorithm Ⅰ to calculate the widest symmetrical interval solution regarding v, it requires totally

    2mn+mn+mn+mn+3n+2n=5mn+5n

    operations. The computational complexity of Algorithm Ⅰ is O(mn).

    Here, we have to mention that the reference [36] also proposes an algorithm to calculate the widest symmetrical interval solution regarding a given solution. However, the algorithm presented in [36] requires totally

    12m2n+172mn+6n1

    operations. The computational complexity is O(m2n). Obviously, our proposed algorithm Ⅰ is superior to the algorithm presented in [36], considering the computational complexity.

    Example 1. Consider the max-min FRIs system as follows:

    {0.4(0.1y1)(0.7y2)(0.5y3)(0.3y4)(0.3y5)0.7,0.2(0.3y1)(0.5y2)(0.8y3)(0.2y4)(0.1y5)0.6,0.2(0.6y1)(0.1y2)(0.3y3)(0.5y4)(0.5y5)0.8,0.5(0.3y1)(0.2y2)(0.6y3)(0.8y4)(0.2y5)0.7. (3.25)

    It is easy to check that both v1=(0.8,0.6,0.6,0.6,0.3), v2=(0.6,0.45,0.3,0.65,0.8) and v3=(0.4,0.6,0.52,0.2,0.4) are solutions of system (3.25). Next, try to obtain the widest symmetrical interval solution regarding vt(t=1,2,3) and calculate the (biggest) symmetrical width regarding vt(t=1,2,3).

    Solution: According to (2.3) and (2.4), we find

    ˆv=(1,1,0.6,0.7,1).

    as the maximum solution for (3.25) after calculation.

    (ⅰ) Find the WSIS regarding v1=(0.8,0.6,0.6,0.6,0.3).

    Steps 1 and 2. Following (3.5) and (3.6), we have

    Jv11={jJ|b1jv1jc1}={jJ|b1jv1j0.4}={2,3},Jv12={jJ|b2jv1jc2}={jJ|b2jv1j0.2}={1,2,3,4},Jv13={jJ|b3jv1jc3}={jJ|b3jv1j0.2}={1,3,4,5},Jv14={jJ|b4jv1jc4}={jJ|b4jv1j0.5}={3,4},

    and

    jv11=argmax{v1j|jJv11}=argmax{v12,v13}=2 or 3,jv12=argmax{v1j|jJv12}=argmax{v11,v12,v13,v14}=1,jv13=argmax{v1j|jJv13}=argmax{v11,v13,v14,v15}=1,jv14=argmax{v1j|jJv14}=argmax{v13,v14}=3 or 4.

    Step 3. If we take jv11=2, jv12=1, jv13=1, jv14=4 and combine (3.7), we can obtain that

    Iv11={iI|jv1i=1}={2,3},Iv12={iI|jv1i=2}={1},Iv13={iI|jv1i=3}=,Iv14={iI|jv1i=4}={4},Iv15={iI|jv1i=5}=.

    Step 4. Following (3.8), we find

    ˘v11=iIv11ci=c2c3=0.2,˘v12=iIv12ci=c1=0.4,˘v14=iIv14ci=c4=0.5,˘v13=˘v15=0.

    Step 5. As a result, we get the vector ˘v1=(0.2,0.4,0,0.5,0). Moreover,

    εv11=(ˆv1v11)(v11˘v11)=(10.8)(0.80.2)=0.20.6=0.2,εv12=(ˆv2v12)(v12˘v12)=(10.6)(0.60.4)=0.40.2=0.2,εv13=(ˆv3v13)(v13˘v13)=(0.60.6)(0.60)=00.6=0,εv14=(ˆv4v14)(v14˘v14)=(0.70.6)(0.60.5)=0.10.1=0.1,εv15=(ˆv5v15)(v15˘v15)=(10.3)(0.30)=0.70.3=0.3.

    Thus,

    εv1=(0.2,0.2,0,0.1,0.3).

    Step 6. Based on the above-obtained εv1, we further obtain the widest symmetrical interval solution regarding v1 as

    [v1εv1,v1+εv1]=([0.6,1],[0.4,0.8],0.6,[0.5,0.7],[0,0.6]).

    The (biggest) symmetrical width regarding v1 is

    w[v1εv1,v1+εv1]=2(εv11εv12εv13εv14εv15)=2(0.20.200.10.3)=0.

    In fact, since \(v_{3}^{1} = \hat{v}_{3}^{1} = 0.6\), it can be concluded from equations (3.3) and (3.13) that there must be \(w[ v^1-\varepsilon^{v^1}, v^1+\varepsilon^{v^1} ] = 0\).

    (ⅱ) Find the WSIS regarding v2=(0.6,0.45,0.3,0.65,0.8).

    Steps 1 and 2. Following (3.5) and (3.6), we find

    Jv21={jJ|b1jv2jc1}={jJ|b1jv2j0.4}={2},Jv22={jJ|b2jv2jc2}={jJ|b2jv2j0.2}={1,2,3,4},Jv23={jJ|b3jv2jc3}={jJ|b3jv2j0.2}={1,3,4,5},Jv24={jJ|b4jv2jc4}={jJ|b4jv2j0.5}={4},

    and

    jv21=argmax{v2j|jJv21}=argmax{v22}=2,jv22=argmax{v2j|jJv22}=argmax{v21,v22,v23,v24}=4,jv23=argmax{v2j|jJv23}=argmax{v21,v23,v24,v25}=5,jv24=argmax{v2j|jJv24}=argmax{v24}=4.

    Step 3. Following (3.7), we can obtain that

    Iv21={iI|jv2i=1}=,Iv22={iI|jv2i=2}={1},Iv23={iI|jv2i=3}=,Iv24={iI|jv2i=4}={2,4},Iv25={iI|jv2i=5}={3}.

    Step 4. According to (3.8), we further obtain

    ˘v21=˘v23=0,˘v22=iIv22ci=c1=0.4,˘v24=iIv24ci=c2c4=0.5,˘v25=iIv25ci=c3=0.2.

    Step 5. As a result, we obtain the vector ˘v2=(0,0.4,0,0.5,0.2). Moreover,

    εv21=(ˆv1v21)(v21˘v21)=(10.6)(0.60)=0.40.6=0.4,εv22=(ˆv2v22)(v22˘v22)=(10.45)(0.450.4)=0.550.05=0.05,εv23=(ˆv3v23)(v23˘v23)=(0.60.3)(0.30)=0.30.3=0.3,εv24=(ˆv4v24)(v24˘v24)=(0.70.65)(0.650.5)=0.050.15=0.05,εv25=(ˆv5v25)(v25˘v25)=(10.8)(0.80.2)=0.20.6=0.2.

    Thus,

    εv2=(0.4,0.05,0.3,0.05,0.2).

    Step 6. Based on the above-obtained εv2, we further obtain the widest symmetrical interval solution regarding v2 as

    [v2εv2,v2+εv2]=([0.2,1],[0.4,0.5],[0,0.6],[0.6,0.7],[0.6,1]).

    The (biggest) symmetrical width regarding v2 is

    w[v2εv2,v2+εv2]=2(εv21εv22εv23εv24εv25)=2(0.40.050.30.050.2)=0.1.

    (ⅲ) Find the WSIS regarding v3=(0.4,0.6,0.52,0.2,0.4).

    Steps 1 and 2. Following (3.5) and (3.6), we find

    Jv31={jJ|b1jv3jc1}={jJ|b1jv3j0.4}={2,3},Jv32={jJ|b2jv3jc2}={jJ|b2jv3j0.2}={1,2,3,4},Jv33={jJ|b3jv3jc3}={jJ|b3jv3j0.2}={1,3,4,5},Jv34={jJ|b4jv3jc4}={jJ|b4jv3j0.5}={3},

    and

    jv31=argmax{v3j|jJv31}=argmax{v32,v33}=2,jv32=argmax{v3j|jJv32}=argmax{v31,v32,v33,v34}=2,jv33=argmax{v3j|jJv33}=argmax{v31,v33,v34,v35}=3,jv34=argmax{v3j|jJv34}=argmax{v33}=3.

    Step 3. Following (3.7), we can obtain that

    Iv31={iI|jv3i=1}=,Iv32={iI|jv3i=2}={1,2},Iv33={iI|jv3i=3}={3,4},Iv34={iI|jv3i=4}=,Iv35={iI|jv3i=5}=,

    Step 4. According to (3.8), we further obtain

    ˘v31=˘v34=˘v35=0,˘v32=iIv32ci=c1c2=0.4,˘v33=iIv33ci=c3c4=0.5.

    Step 5. As a result, we obtain the vector ˘v3=(0,0.4,0.5,0,0). Moreover,

    εv31=(ˆv1v31)(v31˘v31)=(10.4)(0.40)=0.60.4=0.4,εv32=(ˆv2v32)(v32˘v32)=(10.6)(0.60.4)=0.40.2=0.2,εv33=(ˆv3v33)(v33˘v33)=(0.60.52)(0.520.5)=0.080.02=0.02,εv34=(ˆv4v34)(v34˘v34)=(0.70.2)(0.20)=0.50.2=0.2,εv35=(ˆv5v35)(v35˘v35)=(10.4)(0.40)=0.60.4=0.4.

    Thus,

    εv3=(0.4,0.2,0.02,0.2,0.4).

    Step 6. Based on the above-obtained εv3, we further obtain the widest symmetrical interval solution regarding v3 as

    [v3εv3,v3+εv3]=([0,0.8],[0.4,0.8],[0.5,0.54],[0,0.4],[0,0.8]).

    The (biggest) symmetrical width regarding v3 is

    w[v3εv3,v3+εv3]=2(εv31εv32εv33εv34εv35)=2(0.40.20.020.20.4)=0.04.

    Then, the problem has already been solved.

    Remark 2. In fact, the widest symmetrical interval solution regarding an identical solution might be not unique. For example, considering v1=(0.8,0.6,0.6,0.6,0.3) in system (3.25), we find jv11=2 or 3, jv12=1, jv13=1, and jv14=3 or 4 in Example 1. Taking jv11=2, jv12=1, jv13=1, and jv14=4, we find the widest symmetrical interval solution regarding v1 as

    [v1εv1,v1+εv1]=([0.6,1],[0.4,0.8],0.6,[0.5,0.7],[0,0.6]),

    with the (biggest) symmetrical width w[v1εv1,v1+εv1]=0. However, when taking jv11=3, jv12=1, jv13=1, jv14=4, the corresponding widest symmetrical interval solution is

    ([0.6,1],[0.2,1],0.6,[0.5,0.7],[0,0.6]).

    The corresponding (biggest) symmetrical width is also w([0.6,1],[0.2,1],0.6,[0.5,0.7],[0,0.6])=0. That is to say, we find two different widest symmetrical interval solutions regarding the solution v1=(0.8,0.6,0.6,0.6,0.3). But they have the same (biggest) symmetrical width.

    Example 2. Consider the max-min FRIs system as follows:

    {0.37(0.35y1)(0.48y2)(0.87y3)(0.63y4)(0.26y5)(0.74y6)0.82,0.34(0.28y1)(0.92y2)(0.76y3)(0.81y4)(0.53y5)(0.19y6)0.87,0.25(0.89y1)(0.76y2)(0.42y3)(0.35y4)(0.56y5)(0.75y6)0.79,0.36(0.33y1)(0.85y2)(0.27y3)(0.49y4)(0.93y5)(0.35y6)0.85,0.41(0.59y1)(0.28y2)(0.18y3)(0.66y4)(0.83y5)(0.91y6)0.89,0.32(0.29y1)(0.23y2)(0.31y3)(0.73y4)(0.56y5)(0.82y6)0.86. (3.26)

    It is easy to check that system (3.26) is consistent and v=(0.65,0.48,0.51,0.43,0.39,0.45) is one of its solutions. Apply Algorithm Ⅰ to find the widest symmetrical interval solution regarding v and calculate the (biggest) symmetrical width regarding v.

    Solution: According to (2.3) and (2.4), we find

    ˆv=(0.79,0.87,0.82,1,0.85,0.89).

    as the maximum solution for (3.26) after calculation.

    Step 1. Following (3.5), we have

    Jv1={2,3,4,6},Jv2={2,3,4,5},Jv3={1,2,3,4,5,6},Jv4={2,4,5},Jv5={1,4,6},Jv6={4,5,6}.

    Step 2. Following (3.6), we have

    jv1=argmax{vj|jJv1}=3,jv2=argmax{vj|jJv2}=3,jv3=argmax{vj|jJv3}=1,jv4=argmax{vj|jJv4}=2,jv5=argmax{vj|jJv5}=1,jv6=argmax{vj|jJv6}=6.

    Step 3. Following (3.7), we have

    Iv1={iI|jvi=1}={3,5},Iv2={iI|jvi=2}={4},Iv3={iI|jvi=3}={1,2},Iv4={iI|jvi=4}=,Iv5={iI|jvi=5}=,Iv6={iI|jvi=6}={6}.

    Step 4. Following (3.8), we have

    ˘v1=iIv1ci=c3c5=0.41,˘v2=iIv2ci=c4=0.36,˘v3=iIv3ci=c1c2=0.37,˘v4=˘v5=0,˘v6=iIv6ci=c6=0.32.

    Hence, we obtain the vector ˘v=(0.41,0.36,0.37,0,0,0.32).

    Step 5. Following (3.13), we have

    εv1=(v1˘v1)(ˆv1v1)=(0.650.41)(0.790.65)=0.240.14=0.14.

    In a similar way, we further have εv2=0.12, εv3=0.14, εv4=0.43, εv5=0.39, and εv6=0.13. Then we find the vector εv=(0.14,0.12,0.14,0.43,0.39,0.13).

    Step 6. According to (3.2), we can obtain the widest symmetrical interval solution regarding v as

    [vεv,v+εv]=([0.51,0.79],[0.36,0.6],[0.37,0.65],[0,0.86],[0,0.78],[0.32,0.58]).

    Moreover, the (biggest) symmetrical width regarding v is

    2(εv1εv6)=0.140.120.140.430.390.13=0.24.

    In the previous section, we have studied the symmetrical interval solution having the biggest width, which is named WSIS. The width of a WSIS reflects its stability, i.e., the ability to bear the tolerable variation of a provided feasible scheme. For characterizing the most stable feasible scheme, we further define the concept of a centralized solution regarding the max-min FRI system (1.6). We will proposed an effective method for solving the centralized solution.

    Definition 6. (Centralized solution regarding system (1.6)) Let vCS(B,c,d) be a solution of (1.6). The WSIS regarding vC is [vCεvC,vC+εvC]. Then vC is said to be a centralized solution regarding system (1.6), if

    w[vCεvC,vC+εvC]w[vεv,v+εv]

    holds for any solution vS(B,c,d), where [vεv,v+εv] is the WSIS regarding v.

    Define the index set

    Jˆvi={jJ|bijˆvjci}, (4.1)

    where iI, in accordance with the maximum solution ˆv. Moreover, the following m indices are induced by the above index sets:

    jˆvi=argmax{ˆvj|jJˆvi}, (4.2)

    where iI. In addition, based on the indices jˆv1,jˆv2,,jˆvm, we could define

    Iˆvj={iI|jˆvi=j}, (4.3)

    where jJ. Accordingly, we construct the vector ˇv=(ˇv1,ˇv2,,ˇvn) by

    ˇvj={iIˆvjci,ifIˆvj,0,ifIˆvj=,jJ. (4.4)

    Proposition 8. Let ˇv be defined by (4.4). Then it holds ˇvS(B,c,d), serving as a solution of (1.6).

    Proof. The given conditions indicate ci[0,1], iI. According to (4.4), it is evident that ˇvj[0,1], jJ.

    For arbitrary iI, denote j=jˆvi. By (4.2), it is clear j=jˆviJˆvi. According to (4.1), it holds

    bijbijˆvjci. (4.5)

    On the other hand, j=jˆvi indicates iIˆvj, by (4.3). Hence, Iˆvj and by (4.4), we have ˇvj=iIˆvjci. It follows from iIˆvj that

    ˇvj=iIˆvjcici. (4.6)

    Combining (4.5) and (4.6), we have

    jJ(bijˇvj)bijˇvjcici=ci. (4.7)

    Note that Inequality (4.7) holds for all iI. Therefore, ˇv is a solution in the inequalities (1.6).

    Proposition 9. Let uS(B,c,d) be a solution of (1.6). Then it holds that w[ˇv,ˆv]w[u,ˆv].

    Proof. Let j be an arbitrary index in J. Next, we first examine that w[u,ˆv]ˆvjˇvj in two cases.

    Case 1. If Iˆvj=, then by (4.4), ˇvj=0. The given condition uS(B,c,d) indicates uj[0,1]. Hence

    ˆvjˇvj=ˆvjujˆvj0kJ(ˆvkuk)=w[u,ˆv]. (4.8)

    Case 2. If Iˆvj, then by (4.4), ˇvj=iIˆvjci. There exists iIˆvj such that

    ci=iIˆvjci=ˇvj. (4.9)

    Meanwhile, iIˆvj indicates jˆvi=j by (4.3). The given condition uS(B,c,d) also indicates

    (bi1u1)(bi2u2)(binun)ci. (4.10)

    There is jJ such that

    bijujci. (4.11)

    Inequality (4.11) implies that

    ujbijujci. (4.12)

    Since ˆv is maximum and uS(B,c,d), we have ˆvjuj. Hence by (4.11) we further have

    bijˆvjbijujci. (4.13)

    According to (4.1), it holds that jJˆvi. Considering jˆvi=j and jJˆvi, it follows from (4.2) that

    ˆvjˆvj. (4.14)

    Using the inequalities (4.9), (4.12), and (4.14), we have

    ˆvjˇvj=ˆvjciˆvjujˆvjujkJ(ˆvkuk)=w[u,ˆv]. (4.15)

    Considering the arbitrariness of the index j in J, by (4.8) and (4.15) we have ˆvjˇvjw[u,ˆv], jJ. Hence it holds w[ˇv,ˆv]=jJ(ˆvjˇvj)w[u,ˆv].

    In accordance with the solutions ˆv and ˇv, we denote

    vC=ˆv+ˇv2,εvC=ˆvˇv2. (4.16)

    Some properties are investigated for the vectors vC and εvC as follows.

    Theorem 5. Let vC and εvC be defined by (4.16). Then there is vCS(B,c,d). Moreover, [vCεvC,vC+εvC] is the widest symmetrical interval solution regarding vC.

    Proof. According to Proposition 8, it holds ˇvS(B,c,d). ˆv is the maximum solution. It is clear that ˆvˇv. Then we have

    vC=ˆv+ˇv2[ˇv,ˆv]. (4.17)

    Hence, vCS(B,c,d) is a solution. Note that

    {ˇv=ˆv+ˇv2ˆvˇv2=vCεvCS(B,c,d),ˆv=ˆvˇv2+ˆv+ˇv2=vC+εvCS(B,c,d). (4.18)

    It follows from Definitions 3 and 4 that [vCεvC,vC+εvC] is an SIS regarding the solution vC.

    Let [vCε,vC+ε] be an arbitrary SIS regarding vC, where εV. Following Definitions 3 and 4, we obtain

    vCε,vC+εS(B,c,d). (4.19)

    Thus, vCεvC+εˆv. This indicates

    w[vCε,ˆv]w[vCε,vC+ε]. (4.20)

    On the other hand, since vCεS(B,c,d), we have

    w[vCεvC,vC+εvC]=w[ˇv,ˆv]w[vCε,ˆv], (4.21)

    by Proposition 9. Inequalities (4.20) and (4.21) imply that w[vCεvC,vC+εvC]w[vCε,vC+ε].

    As a result, by Definition 5 we know that [vCεvC,vC+εvC] is a widest symmetrical interval solution regarding vC.

    Proposition 10. Let uS(B,c,d) be an arbitrary solution in (1.6). Then it holds w[u,ˆv]w[ˇv,ˆv].

    Proof. Let j be an arbitrary index in J. Now we examine the inequality ˆvjˇvjw[u,ˆv].

    Case 1. If Iˆvj=, then ˇvj=0 by (4.4). Hence

    w[u,ˆv]=kJ(ˆvkuk)ˆvjujˆvj0=ˆvjˇvj. (4.22)

    Case 2. If Iˆvj, then ˇvj=iIˆvjci by (4.4). Obviously, there exists iIˆvj, such that

    ˇvj=ci. (4.23)

    At the same time, by (4.3) and iIˆvj we have

    jˆvi=j. (4.24)

    Note that uS(B,c,d). Obviously, u satisfies the ith inequality in (1.6), i.e.,

    ci(bi1u1)(bi2u2)(binun)di.

    As a result, there is jJ satisfying

    bijujci. (4.25)

    Considering uˆv since ˆv is the maximum solution, we obtain bijˆvjbijujci. According to (4.1), it holds jJˆvi. Moreover, Inequality (4.25) implies that

    ujbijujci. (4.26)

    Observing (4.2) and (4.24), we have j=jˆvi=argmax{ˆvl|lJˆvi}. Thus,

    ˆvj=ˆvjˆviˆvl,lJˆvi.

    Since jJvi, it holds

    ˆvjˆvj. (4.27)

    Combining (4.23), (4.26), and (4.27), we have

    ˆvjˇvj=ˆvjciˆvjujˆvjujkJ(ˆvkuk)=w[u,ˆv]. (4.28)

    Inequalities (4.22) and (4.28) contribute to ˆvjˇvjw[u,ˆv], jJ. So we have

    w[ˇv,ˆv]=jJ(ˆvjˇvj)w[u,ˆv].

    Theorem 6. vC and εvC are defined by (4.16). Then it holds that vC is a centralized solution regarding system (1.6).

    Proof. Let vS(B,c,d) be an arbitrary solution in (1.6). [vεv,v+εv] is the WSIS regarding v. Since [vεv,v+εv] is an interval solution of (1.6); by Definition 3, it holds

    [vεv,v+εv]S(B,c,d). (4.29)

    That is vεv,v+εvS(B,c,d). According to Proposition 10,

    w[ˇv,ˆv]w[vεv,ˆv]. (4.30)

    Since ˆv is the maximum solution, it is clear ˆvv+εv. Thus

    w[vεv,ˆv]w[vεv,v+εv]. (4.31)

    Inequalities (4.30) and (4.31) imply that

    w[ˇv,ˆv]w[vεv,v+εv]. (4.32)

    By (4.16), it is clear ˇv=vCεvC and ˆv=vC+εvC. Inequality (4.32) could be further written as

    w[vεv,v+εv]w[vCεvC,vC+εvC]. (4.33)

    According to Definition 6, vC is a centralized solution regarding system (1.6). Moreover, the WSIS regarding vC is [vCεvC,vC+εvC]=[ˇv,ˆv].

    This subsection provides an algorithm to find the centralized solution regarding a given max-min system. Moreover, some numerical examples are given for verifying our proposed algorithm.

    Algorithm Ⅱ: solving the centralized solution regarding system (1.6)

    Step 1. Following (4.1), calculate the indicator set Jˆvi for any iI.

    Step 2. Following (4.2), select the indicator jˆvi from the indicator set Jˆvi for any iI.

    Step 3. Following (4.3), construct the indicator set Iˆvj for any jJ.

    Step 4. Following (4.4), generate the vector ˇv=(ˇv1,ˇv2,,ˇvn), where the ˇvj is defined by (4.4).

    Step 5. Following (4.16), calculate the vectors vC and εvC. Then we find the centralized solution regarding system (1.6) as vC and the corresponding (biggest) symmetrical width as 2(εvC1εvC2εvCn).

    Computational complexity of Algorithm Ⅱ: In Algorithm Ⅱ, Step 1 requires 2mn operations for calculating the indicator sets. Moreover, Steps 2–4 have an identical calculation amount as mn. Besides, Step 5 requires 4n+2 operations. As a result, applying Algorithm Ⅱ to calculate the widest symmetrical interval solution regarding v, it requires totally

    2mn+mn+mn+mn+4n=5mn+4n+2

    operations. The computational complexity of Algorithm Ⅱ is also O(mn).

    Example 3. Consider the max-min FRIs system as follows:

    {0.3(0.2y1)(0.4y2)(0.5y3)(0.1y4)(0.6y5)(0.7y6)0.8,0.4(0.3y1)(0.6y2)(0.2y3)(0.8y4)(0.5y5)(0.1y6)0.9,0.1(0.7y1)(0.3y2)(0.4y3)(0.2y4)(0.5y5)(0.6y6)0.6,0.5(0.8y1)(0.1y2)(0.3y3)(0.4y4)(0.7y5)(0.2y6)0.9,0.2(0.5y1)(0.2y2)(0.1y3)(0.9y4)(0.6y5)(0.4y6)0.8. (4.34)

    Try to find the centralized solution regarding the system (4.34).

    Solution:

    Notice that

    ˆv=(0.6,1,1,0.8,1,1).

    is the maximum solution of system (4.34). Combining (4.1) and (4.2), we have

    Jˆv1={jJ|b1jˆvjc1}={jJ|b1jˆvj0.3}={2,3,5,6},Jˆv2={jJ|b2jˆvjc2}={jJ|b2jˆvj0.4}={2,4,5},Jˆv3={jJ|b3jˆvjc3}={jJ|b3jˆvj0.1}={1,2,3,4,5,6},Jˆv4={jJ|b4jˆvjc4}={jJ|b4jˆvj0.5}={1,5},Jˆv5={jJ|b5jˆvjc5}={jJ|b5jˆvj0.2}={1,2,4,5,6},

    and

    jˆv1=argmax{ˆvj|jJˆv1}=argmax{ˆv2,ˆv3,ˆv5,ˆv6}=2 or 3 or 5 or 6,jˆv2=argmax{ˆvj|jJˆv2}=argmax{ˆv2,ˆv4,ˆv5}=2 or 5,jˆv3=argmax{ˆvj|jJˆv3}=argmax{ˆv1,ˆv2,ˆv3,ˆv4,ˆv5,ˆv6}=2 or 3 or 5 or 6,jˆv4=argmax{ˆvj|jJˆv4}=argmax{ˆv1,ˆv5}=5,jˆv5=argmax{ˆvj|jJˆv5}=argmax{ˆv1,ˆv2,ˆv4,ˆv5,ˆv6}=2 or 5 or 6.

    Take jˆv1=2, jˆv2=5, jˆv3=3, jˆv4=5, jˆv5=6. Combining (4.3) and (4.4), we can obtain that

    Iˆv1={iI|jˆvi=1}=,Iˆv2={iI|jˆvi=2}={1},Iˆv3={iI|jˆvi=3}={3},Iˆv4={iI|jˆvi=4}=,Iˆv5={iI|jˆvi=5}={2,4},Iˆv6={iI|jˆvi=6}={5},

    and

    ˇv1=0,ˇv2=iIˆv2ci=c1=0.3,ˇv3=iIˆv3ci=c3=0.1,
    ˇv4=0,ˇv5=iIˆv5ci=c2c4=0.5,ˇv6=iIˆv6ci=c5=0.2.

    Then we find the vector ˇv=(0,0.3,0.1,0,0.5,0.2). In accordance with (4.16), we further have

    vC=ˆv+ˇv2=(0.3,0.65,0.55,0.4,0.75,0.6),εvC=ˆvˇv2=(0.3,0.35,0.45,0.4,0.25,0.4).

    Then it holds that vC=(0.3,0.65,0.55,0.4,0.75,0.6) is a centralized solution regarding system (4.34). The corresponding widest symmetrical interval solution regarding vC is

    [vCεvC,vC+εvC]=([0,0.6],[0.3,1],[0.1,1],[0,0.8],[0.5,1],[0.2,1]), (4.35)

    with the (biggest) symmetrical width w[vCεvC,vC+εvC]=0.5.

    Remark 3. In fact, the centralized solution regarding a given system with max-min FRIs might not be unique. For example, it can be easily examined that

    (0.35,0.75,0.75,0.55,0.75,0.75)

    is indeed another centralized solution regarding system (4.34).

    Example 4. Consider the max-min FRIs system as follows:

    {0.37(0.35y1)(0.48y2)(0.87y3)(0.63y4)(0.26y5)(0.74y6)0.82,0.34(0.28y1)(0.92y2)(0.76y3)(0.31y4)(0.53y5)(0.19y6)0.87,0.25(0.89y1)(0.76y2)(0.42y3)(0.18y4)(0.56y5)(0.75y6)0.79,0.36(0.33y1)(0.85y2)(0.27y3)(0.35y4)(0.93y5)(0.35y6)0.85,0.41(0.59y1)(0.28y2)(0.18y3)(0.38y4)(0.83y5)(0.91y6)0.89,0.32(0.48y1)(0.23y2)(0.31y3)(0.26y4)(0.56y5)(0.28y6)0.86. (4.36)

    Try to find the centralized solution regarding system (4.36).

    Solution: After calculation, the maximum solution of system (3.26) can be obtained as ˆv=(0.79,0.87,0.82,1,0.85,0.89). Next, we apply our proposed Algorithm Ⅱ to find the centralized solution regarding (4.36).

    Step 1. Following (4.1), we have

    Jˆv1={2,3,4,6},Jˆv2={2,3,5},Jˆv3={1,2,3,5,6},Jˆv4={2,5},Jˆv5={1,5,6},Jˆv6={1,5}.

    Step 2. Following (4.2), we have

    jˆv1=argmax{vj|jJˆv1}=4,jˆv2=argmax{vj|jJˆv2}=2,jˆv3=argmax{vj|jJˆv3}=6,jˆv4=argmax{vj|jJˆv4}=2,jˆv5=argmax{vj|jJˆv5}=6,jˆv6=argmax{vj|jJˆv6}=5.

    Step 3. Following (4.3), we have

    Iˆv1={iI|jˆvi=1}=,Iˆv2={iI|jˆvi=2}={2,4},Iˆv3={iI|jˆvi=3}=,Iˆv4={iI|jˆvi=4}={1},Iˆv5={iI|jˆvi=5}={6},Iˆv6={iI|jˆvi=6}={3,5}.

    Step 4. Following (4.4), we have

    ˇv1=0,ˇv2=iIˇv2ci=c2c4=0.36,ˇv3=0,ˇv4=iIˇv4ci=c1=0.37,ˇv5=iIˇv5ci=c6=0.32,ˇv6=iIˇv6ci=c3c5=0.41.

    Hence, we obtain the vector ˇv=(0,0.36,0,0.37,0.32,0.41).

    Step 5. Following (4.16), we can calculate the vectors vC and εvC. After calculation, we have

    vC=ˆv+ˇv2=(0.395,0.615,0.41,0.685,0.585,0.65).
    εvC=ˆvˇv2=(0.395,0.255,0.41,0.315,0.265,0.24).

    Hence, the centralized solution regarding system (4.36) is vC=(0.395,0.615,0.41,0.685,0.585,0.65), with the corresponding (biggest) symmetrical width 2(εvC1εvC2εvC6)=0.48.

    Recently, some researchers adopted the max-min FRIs or FREs system to model a P2P educational information sharing system. Any feasible scheme in the P2P network system was found to be a solution to the max-min system, e.g., system (1.6). As a result, we employed the concept of the widest symmetrical interval solution to represent the stability for a feasible scheme. Moreover, reflecting the most stable feasible scheme, the concept of a centralized solution of system (1.6) was further defined. Effective resolution methods were proposed for the widest symmetrical interval solution regarding a provided solution and the centralized solution regarding system (1.6), respectively. The proposed resolution methods have been demonstrated and examined through the numerical examples. In future works, one might further generalize the widest symmetrical interval solution and centralized solution to the addition-min FRIs system, or even to the classical max-t-norm fuzzy relational system.

    Miaoxia Chen: Validation, Investigation, Writing-original draft; Guocheng Zhu: Validation, Methodology; Shayla Islam: Conceptualization, Supervision; Xiaopeng Yang: Conceptualization, Supervision, Funding acquisition, Writing-review & editing. All authors have read and agreed to the published version of the manuscript.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    We would like to express our appreciation to the editor and the anonymous reviewers for their valuable comments, which have been very helpful in improving the paper. This work was supported by the National Natural Science Foundation of China (12271132), the Guangdong Basic and Applied Basic Research Foundation (2024A1515010532, 2024ZDZX1027) and the NSF of Hanshan Normal University (XYN202303, 2023KQNCX041, XSPY202405). This work was supported by the Guangdong Province Quality Project (Data Science Innovation and Entrepreneurship Laboratory).

    The authors declare that they have no conflicts of interest.



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