We consider the discrete optimal problem on networks with integral-time cost criterion by a trajectory when the starting and final states of the system are fixed. A polynomial-time algorithm for solving this problem is proposed

A. Iusiumbeli

D. Lozovanu

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Computer Science Journal of Moldova, vol.10, no.3(30), 2002Algorithms for solving discrete controlproblems on networks∗A. Iusiumbeli, D. LozovanuAbstractWe consider the discrete optimal problem on networks withintegral-time cost criterion by a trajectory when the starting andfinal states of the system are fixed. A polynomial-time algorithmfor solving this problem is proposed.1 IntroductionIn this paper we consider discrete control problems on networks from[1,2]. The dynamics of the system in such problems is described bya directed graph of passages [2]. The vertices of the graph in suchproblems correspond to the states of system and its edges signify thepossibility of the system passage from one state to another. Moreover,on the edges of the graph the cost functions are defined, which dependon time and express the expenditure or the income when the dynamicsystem passes from one state to another.We study the discrete control problem on networks in the casewhen the cost functions on edges of the graph are positive and non-decreasing. A new polynomial-time algorithm for solving the problemis proposed. The algorithm can be used for checking the optimizationprinciple for dynamic networks[3].c©2002 by A. Iusiumbeli, D. Lozovanu∗Supported by CRDF BGP MM2-3018333A. Iusiumbeli, D. Lozovanu2 Problem formulation and dynamical pro-gramming methodWe shall consider the optimal control problem on networks from [3,4].Let L be a dynamical system with finite set of states X, |X|=N and atevery discrete moment of time t = 0, 1, 2, . . . the state of the system Lis x(t) ∈ X. Note that here we associate x (t) with an abstract element.Two states xs and xf are chosen in X, where xs is a starting state of thesystem L, xs = x(0) and xf is the final state of the system, i.e. xf is thestate into which the system should be brought. The dynamics of thesystem is described by a directed graph of passages G=(X,E ), |E| = m,an edge e=(x,y) which signifies the possibility of passage of the systemL from the state x=x(t) to the state y=x(t+1) at any moment of timet = 0, 1, 2, . . .. That means that the edges e=(x,y) ∈ E can be regardedas the possible values of the control parameter u(t) when the state ofthe system is x=x(t), t = 0, 1, 2, . . .. The next state y=x(t+1) of thesystem L is determined uniquely by x=x(t) at the moment of time tand an edge e=(x,y) ∈ E(x), whereE(x) = {(x,y) ∈ E | y ∈ X}So E(x)=E(x(t)) correspond to the admissible set Ut(x(t)) for the con-trol parameter u(t) at every moment of time t. To each edge e=(x,y)a function ce(t) is assigned, which reflects the cost of system’s passagefrom the state x(t) = x ∈ X to the state x(t + 1) = y ∈ X at anymoment of time t = 0, 1, 2, . . ..We consider the discrete optimal problem on networks [1,2,3] forwhich the sequence of system’s passages(x(0), x(1)), (x(1), x(2)), . . . , (x(T − 1), x(T )) ∈ Ewhich transfers the system L from the state xs = x(0) to the state xf =x(T ) with minimal integral-time cost of the passages by a trajectoryxs = x(0), x(1), . . . , x(T ) = xf must be found.Here may be two variants of the problem:• the number of the stages (time T ) is fixed334Algorithms for solving discrete control problems on networks• T is unknown and it must be foundLet us consider that T is fixed. Denote byFxsxf (T ) = minxs=x(0),x(1),...,x(T )=xfT−1∑t=0cx(t),x(t+1)(t)the minimal integral-time cost of system’s passages from xs to xf . Ifthe state xf couldn’t be reached by using T stages, i.e. in G there existno path P ∗(xs, xf ) from xs to xf which contains exactly T edges, thenwe put Fxsxf (T ) = ∞.It is easy to observe that using the dynamical programming methodwe could tabulate the values Fxsx(t)(t), t = 0, 1, 2, . . . , T (Fxsx(0)(0) =0) because for Fxsx(t)(t) the following recurrent formula can be writtenFxsx(t)(t) = minx(t−1)∈X−G (x(t)){Fxsx(t)(t− 1) + cx(t−1),x(t)(t− 1)},whereX−G (y) = {x ∈ X | e = (x, y) ∈ E}So, if T is fixed, then problem can be solved in time O(N2T ) (herewe do not take into consideration the number of operations for calcu-lations the value of functions ce(t) for given t).In the case when T is unknown we shall consider T ∈ [T1, T2] whereT1 and T2 are given. The problem in this case can be reduced toT2 − T1 + 1 problems with T = T1, T = T1 + 1, . . . , T = T2 respec-tively. Obviously, for positive and non-decreasing functions on edgesthe problem can be solved in time O(N3).Further we shall study the problem when T is not fixed. We denoteby T ∗ the optimal time for the following discrete control problem.3 The main results and algorithmWe have formulated the problem for positive and non-decreasing func-tions ce(t) on edges e ∈ E which coincides with the discrete optimalcontrol problem on G with starting state xs and final state xf . In this335A. Iusiumbeli, D. Lozovanucase the optimal trajectory xs = x(0), x(1), . . . , x(T ) = xf correspondsin G to the directed path P ∗(xs, xf ) from xs to xf . We call this paththe optimal path for the dynamic network. For the path P ∗(xs, xf )contains no more than N-1 edges, the problem can be solved in finitetime by using dynamical programming techniques.Here we propose a more simple algorithm for solving this problem.3.1 An algorithm for solving the problem with givenoptimal value of stages T ∗Algorithm 1Preliminary step(Step 0)Set X0 = xf , E0 = ∅. Assign to every vertex x ∈ X two labels t(x)and R(x) as follows:R(xf ) = 0, t(xf ) = T,R(x) = ∞, t(x) = ∞ ∀x ∈ X\{xf}General step(Step k)Find the setXk = {x ∈ X\Xk−1 | (x, y) ∈ E, y ∈ Xk−1},E′= {(x∗, y∗) ∈ E(Xk−1) | c(x∗,y∗)(t(y∗)− 1) + R(y∗)} =miny∈Xk−1minx∈V +(y)\Xk−1{c(x,y)(t(y)− 1) + R(y)}whereE(Xk−1) = {(x, y) ∈ E | y ∈ Xk−1, x ∈ X\Xk−1},V +(y) = {x ∈ X | (x, y) ∈ E}Find the set of verticesX′= {x ∈ Xk | ∃y , (x, y) ∈ E′}336Algorithms for solving discrete control problems on networksFor every x′ ∈ X ′ select one edge (x′ , y′) ∈ E′ and build the union E′kof such edges. After that change the labels t(x′) and R(x′) for everyvertex x′ ∈ X ′ are as followsR(x′) = R(y′) + c(x′ , y′ )(t(y′)− 1),t(x′) = t(y′)− 1, ∀(x′ , y′) ∈ E′kReplace the set Xk by Xk−1⋃X′and Ek by Ek−1⋃E′k. Fix setsY = Xk, E∗ = Ek and the tree H = (Y, E∗). If (xs ∈ Xk or k = T )then STOP, otherwise go to the next step k + 1.Note that the tree H = (Y, E∗) contains paths from every x ∈ Yto xf . The labels R(x), x ∈ Y , indicate the cost of optimal path fromx ∈ Y to xf of the optimal path P ∗(xs, xf ) which contains x and t(x)represents the time moment at which this path passes through vertex x.In [3,4] have been shown that if the optimization principle fordynamical network is satisfied then a more effective algorithm forsolving the optimal control problem on network can be elaborated.Note that the optimization principle on dynamic network is satisfiedif the following condition holds: Let us consider that the cost func-tions ce(t), e ∈ E, in the dynamic network have the property thatif P ∗(xs, x) is an arbitrary optimal path from xs to x which can berepresented as P ∗(xs, x) = P ∗1 (xs, y)⋃P ∗2 (y, x), where P∗1 (xs, y) andP ∗2 (y, x) have no common edges, than a leading part P∗1 (xs, y) of thepath P ∗(xs, x) is also an optimal path of the problem in G with givenstarting state xs and final state y.Theorem 1 Let (G, c(t), xs, xf ) be a dynamic network, wherethe vector-function c(t) = (ce1(t), ce2(t), . . . , cem(t)) has positive andbounded components for t ∈ [0, N − 1]. If T = T ∗ then k = T ∗ andt(xs) = 0. Let H∗ = (X, E∗) be the tree obtained by Algorithm 1,then an arbitrary path P ∗(x, xf ) in the H∗ represents the optimal pathfrom x to xf for optimal control problem on G.Proof. We prove the theorem by using the induction principle onthe number of steps k of the algorithm. In the case when k = 0 thetheorem is evident.337A. Iusiumbeli, D. LozovanuLet us consider the theorem holds for any k ≤ r and let us showit is true for k = r + 1. If Hr = (Xr, Er) is the tree obtained after rsteps and Hr+1 = (Xr+1, Er+1) is the tree obtained after r + 1 stepsof the algorithm, then X0 = Xr+1\Xr and E0 = Er+1\Er representsthe vertex set and the edge set obtained by the algorithm at the stepr +1. Let us show that if y′is an arbitrary vertex of X0, then in Hr+1the unique directed path P ∗(y′ , xf ) from y′to xf is optimal. Indeed,if this is not the case, then there exists an optimal path Q(y′, xf ) fromy′to xf , which does not contain the edge e = (y′, z′) ∈ E0. The pathQ(y′, xf ) can be represented as Q(y′, xf ) = Q1(y′, z′) ∪ {(z′ , x′)} ∪Q2(x′, xf ), where x′is the first vertex of the path Q(y′, xf ) belongingto Xr when we pass from y′to xf . Let us show thatcost(Q(y′, xf )) > cost(P ∗(y′, xf ))wherecost(Q(y′, xf )) =T−1∑t = T−mQ−1cet(t),eT−mQ−1, eT−mQ , . . . , eT−1 are the corresponding edges of the directedpath Q(y′, xf ) when we pass from y′to xf .cost(P ∗(y′, xf )) =T−1∑t = T−mP−1ce′t(t),e′T−mP−1, e′T−mP , . . . , e′T−1 are the corresponding edges of the directedpath P ∗(y′ , xf ) when we pass from y′to xf .Note by R(x′) = cost(Q2(x′, xf ))According to the algorithm, we can stateR(x′) + c(y′ , x′ )(t(x′)− 1) > R(z′′) + c(y′ , z′′ )(t(z′′)− 1) = R(y′)where e′= (y′, z′′) is the first edge of the path P ∗(y′ , xf ). Thencost(Q2(x′, xf ) ∪ {(z′ , x′)}) > cost(P ∗(y′ , xf ))338Algorithms for solving discrete control problems on networksR(x′) + c(y′ , x′ )(t(x′)− 1) = cost(Q2(x′ , xf ) ∪ {(y′ , x′)})R(x′) = cost(P ∗(x′, xf ))The cost functions ce(t),∀e ∈ E, are positive, thereforecost(Q(y′, xf )) = cost(Q1(y′, z′) ∪ {(z′ , x′)} ∪Q2(x′ , xf )) ≥cost(Q2(x′, xf ) ∪ {(z′ , x′)}) > cost(P ∗(y′ , xf ))i.e. Q(y′, xf ) is not an optimal path from y′to xf . That means thatthe tree Hr+1 = (Xr+1, Er+1) contains an optimal path from everyy′ ∈ Xr+1 to xf .If k = T ∗ then the fact that t(xs) = 0 is evident (conform algorithm).3.2 An algorithm for solving the problem when the num-ber of stages is unknownAlgorithm 2We study the problem when T is not fixed. In this case when T isunknown we shall assume T ∈ [1, N − 1]. The problem in this casecan be reduced to N − 1 problems with T = 1, T = 2, . . . , T = N − 1respectively. For every T i fixed from [1,N-1] we calculate ti(xs) andRi(xs), i ∈ [1, N − 1].FindR∗(xs) = mini ∈ [1,N−1]{Ri(xs) | ti(xs) = 0},which represented the minimal integral-time cost of system’s passagesfrom xs to xf andT ∗ = {T i ∈ [1, N − 1] | ti(xs) = 0 and Ri(xs) = R∗(xs)}References[1] R. Bellman,R. Kalaba, Dynamic programming and modern controltheory. Academie press, New-York and London,1974.339A. Iusiumbeli, D. Lozovanu[2] D. Lozovanu, The network models of discrete optimal control anddynamic games with p players. Discrete mathematics and applica-tions. 13, N04, 2001 pp.126–143.[3] D. Lozovanu, Dynamic games with p players on networks. Bul. ofAcademy of Sciences of Moldova. Ser. Mathematics, 2000 1(32)41–54.[4] D. Lozovanu, S. Pickl, Discrete optimal control problems and dy-namic games with p players. Discrete Aplied Mathematic, 2002(will be published)A. Iusiumbeli, D. Lozovanu, Received July 5, 2002A. Iusiumbeli,The Department of Mathematics and Computer ScienceThe State University of Moldova60, Mateevici str.,277709, Chishinau, MoldovaE–mail : iutol@rambler.ruD. Lozovanu,Institute of Mathematics and Computer ScienceAcademy of Sciences of Moldova5, Academy str., ChisinauMD2028, MoldovaPhone: 73–35–83E–mail : lozovanu@math.md340

Algorithms for solving discrete control problems on networks

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Algorithms for solving discrete control problems on networks

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