Characterization of Minimum Time-Fuel Optimal Control for LTI Systems
11 Characterization of Minimum Time-FuelOptimal Control for LTI Systems
Rajasree Sarkar, Deepak U. Patil and Indra Narayan Kar
Abstract
A problem of computing time-fuel (cid:82) t f ( k + | u ( t ) | ) dt optimal control for state transfer of a singleinput linear time invariant (LTI) system to the origin is considered. The input is assumed to be bounded( | u ( t ) | ≤ ). Since, the optimal control is bang-off-bang in nature, it is characterized by sequences of +1 , and − and the corresponding switching time instants. All (candidate) sequences satisfying thePontryagin’s maximum principle (PMP) necessary conditions are characterized. The number of candidatesequences is obtained as a function of the order of system and a method to list all candidate sequencesis derived. Corresponding to each candidate sequence, switching time instants are computed by solvinga static optimization problem. Since the candidate control input is a piece-wise constant function, thetime-fuel cost functional is converted to a linear function in switching time instants. By using a simplesubstitution of variables, reachability constraints are converted to polynomial equations and inequalities.Such a static optimization problem can be solved separately for each candidate sequence. Finally, theoptimal control input is obtained from candidate sequences which gives the least cost. For each sequence,optimization problem can be solved by converting it to a generalized moment problem (GMP) and thensolving a hierachical sequence of semidefinite relaxations to approximate the minima and minimizer[1]. Lastly, a numerical example is presented for demonstration of method. Index Terms
Time-fuel optimal control, Sparsity, Optimization, Semidefinite program.
I. I
NTRODUCTION
Optimal utilization of resources for performing any control task necessitates maximizing theoff-duration of control. As a result, problem of computing sparse control for state transfer has
Authors are with the Department of Electrical Engineering, Indian Institute of Technology Delhi, New Delhi, 110016, India.e-mail: { Rajasree.Sarkar, deepakpatil, ink } @ee.iitd.ac.in. February 23, 2021 DRAFT a r X i v : . [ m a t h . O C ] F e b gathered a lot of attention over the past few years. Sparse control is of interest in variety ofdomains, namely, networked, multi-agent control system [2], [3], transportation systems [4],etc. One way to obtain sparse control is by computing a control input that achieves the requiredstate transfer with the least L -norm (non-zero control duration) [5], [6]. Such an optimal controlproblem is difficult to solve because of non-convex and discontinuous nature of the cost function.However, in [5], under certain normality conditions, an equivalence is established between thesolution of L and L norm optimal control problems. In [7], the authors pose a combination of L and L cost for getting continuous control inputs along with computational benefits. Further,in [8], a time- L -norm optimal control problem has also been shown to be equivalent to time- L -norm (i.e., (cid:82) t f ( k + | u ( t ) | ) dt ) optimal control problem. The benefit of such equivalence is that time- L -norm optimal control problem is tractable compared to the time- L -norm counterpart.The time- L -norm optimal control problem has received a lot of attention in past [9]–[11] andis well-known in literature as time-fuel optimal control problem [12], [13].Both time optimal and time-fuel optimal control problem has undergone extensive researchin last decade [14]–[17]. However, in comparison to time optimal control problem, computationof open loop as well feedback time-fuel optimal control is a challenging task. Unlike, the timeoptimal control [14], [16], analytical solutions for time-fuel optimal control cannot be obtaineddirectly from classical control methods like Pontryagin’s minimum principle (PMP) even forlinear systems except double integrator [17] and a class of second order systems [10], [11]. Thisis primarily because of two reasons, namely, (i) time optimal control is of bang-bang natureexhibiting transitions only between two constant levels i.e. +1 and − . On the other hand, time-fuel optimal control shows bang-off-bang nature that changes among three levels i.e. +1 , and − , and (ii) for real eigenvalues the maximum switching for time optimal control is limited to atmost n − switches [15] whereas the maximum switches for fuel optimal control is atmost n [18] . These drawbacks are seen in various versions of the time-fuel optimal control problem oflinear systems that considers free final time [9], bounded time constraints [11], and [12], whereauthors treat only the fuel optimal control problem for a fixed final time. The work presented in[19] achieves closed loop time-fuel optimal control for a second order system. The work in [19]uses all possible sequences of +1 , and − , that the optimal control for a second order systemfollows to constructs state-dependent switching rules. An extension to triple integrator appearsin [13] in which the following quote appears highlighting difficulty of obtaining a switching rulefor a general n -th order systems. February 23, 2021 DRAFT “the possibility of obtaining similar results for other third- and higher order systemsseems remote and, in this sense, the approach lacks generality.”As noted even for second and third order example systems the approach to time-fuel optimalfeedback control synthesis is varied. For obtaining a time-fuel optimal feedback control, it isneccessary to find a commonnality that is dependent only on the order of system. Hence, aresult that characterizes all possible time optimal control functions that satisfy PMP neccessaryconditions can be useful. For a general n -th order system [18] has studied analytical propertiesof the fuel optimal controls and associated reachability sets with fixed final time.In this article, assuming free final time, we utilize the bang-off-bang property and PMP tocharacterize sequences of +1 , and − , which are identified with candidate time-fuel optimalcontrols for an n th order LTI system. The sequences that meet the criterion laid by the PMPneccesary conditions are counted and listed. Such a characterization of optimal control candidatesis advantageous mainly because, only the knowledge of the order of system is sufficient forlisting all possible candidate sequences. It is important, mainly because a list of all possibleoptimal control candidates can be used as a prior knowledge in formulating any control policyfor control problems with time-fuel considerations. To the best of our knowledge, a list andcount of all possible time-fuel optimal control candidates for a general n -th order LTI system isnot available. Further, for each candidate sequence a static non-linear program (NLP) to obtainthe time instants at which the optimal control input switch between +1 , , − is also formulated.The desired control input is obtained by solving several NLPs corresponding to each candidatesequence. The candidate sequence that leads to least time-fuel cost gives the optimal controlinput. By simple substitution of variables, cost functional can be transformed in to a rationalfunction in decision variables and constraints can be represented by polynomial equations andinequalities. Such a NLP with rational cost function and semi-algebraic constraints has beenshown to be equivalent to a generalized moment problem (GMP) in [1], [20]. Further GMPcan be solved by constructing a hierarchy of semidefinite programs [1], [20] using a solverGloptipoly [21]. However, this method is not scalable to large problems, and other standardnon-linear programming solvers like fmincon , SNOPT, IPOPT, etc can also be used. But, theseother solvers cannot guarantee global optima.It is important to note that direct methods like collocation, and discretization also obtain anNLP, but by approximating the original optimal control problem [22], [23]. Whereas, indirectmethods like shooting method work by numerically solving an ordinary differential equation February 23, 2021 DRAFT obtained from the PMP. These indirect methods are dependent heavily on the choice of initialco-state guess by the user [23] and there are no convergence guarantees available for time-fueloptimal control of a general n -th order linear time invariant (LTI) systems. In our approach, we do not use approximation at any stage of NLP formulation. Thus, in a way, our approach toobtain time-fuel optimal control is exact , in contrast to both direct and indirect methods. Theonly stage where approximation of solutions takes place is when using solver for NLP. But, notethat in solver Gloptipoly [21], the GMP problem obtained from the NLP with rational cost andpolynomial cost function is solved by relaxing it to a semidefinite program (SDP) of a finite sizecalled relaxation order. Increasing relaxation order improves the approximation and successivesolutions to the increasing sequence of SDP relaxations converges asymptotically to globallyoptimal solutions [20]. A preliminary version of this article is published in [24] where similarresults were obtained for second order LTI systems.II. M ATHEMATICAL N OTATION AND P RELIMINARIES
The L norm of a continuous-time measurable function u ( t ) over the time interval [0 , T ] ,is defined as (cid:107) u ( t ) (cid:107) = (cid:82) T | u ( t ) | dt. For any set S , we define the number of elements in S as its cardinality denoted as N ( S ) . We are interested in a set of finite-length sequences over Z := {− , , } . For example a set S = { (1 , , − , (1 , , , ( − , , − } is a set of sequencesof length 3. Consider A and B as two finite length sequence set. Then A is said to be a sub-sequence set of B if for every element a ∈ A there exists an element b ∈ B such that a is asub-sequence of b . Equivalently, B is termed as the super-sequence set of A . For any sequence,its conjugate sequence is obtained by reversing the sign of each its elements. For a sequence set S , a set of conjugates of all sequences in S is called the conjugate set of S and is representedas ¯ S . For example, consider a sequence a = ( − , , ∈ S = { (1 , , − , (1 , , , ( − , , − } .The conjugate sequence of a is ¯ a = (1 , , − . Similarly, the conjugate set of S is ¯ S = { ( − , , , ( − , − , , (1 , − , } .III. P ROBLEM F ORMULATION
Consider a n th order single input LTI system defined as: ˙ x ( t ) = Ax ( t ) + B u ( t ) , u ( t ) ∈ R (1)where x ( t ) ∈ R n represents the state variable of the system and A ∈ R n × n and B ∈ R n × arethe system and input matrices respectively. Assume that the pair ( A , AB ) is controllable and February 23, 2021 DRAFT the eigenvalues of A are non-zero, real. Further, for purpose of solving a particular optimizationproblem in section V, we will need assumption that eigenvalues are distinct and rational i.e. ( λ ( A ) ∈ Q −{ } ) . Hence the eigenvalue λ i with numerator n i and denominator d i for i = 1 , ..., n are expressed as λ i = n i /d i = c i /l where c i = n i l/d i , l = LCM ( d , ..., d n and n i , d i ∈ Z − { } .Note that such an assumption is not restrictive since rational numbers are dense in the set ofreal numbers and therefore, any real number can be approximated by a rational number uptoarbitrary precision. Later, in Section V, rational eigenvalue assumption helps in converting theoptimal control problem into a tractable optimization problem with rational cost function andconstraints described by polynomial inequalities. Further, without loss of generality, we assumethat A is in diagonal form as A = diag ( λ ( A )) and B = [ b , ..., b n ] T with b , ..., b n (cid:54) = 0 . Let theinput u ( t ) be constrained as | u ( t ) | ≤ . Thus, the set of the admissible controls is U = { u : [0 , ∞ ) → [ − , | u is measurable and | u ( t ) | ≤ almost everywhere t ∈ [0 , ∞ ) } Our objective is to choose a control u ( t ) ∈ U that steers the system (1) from initial condition x (0) = x to the origin i.e. x ( t f ) = 0 with least possible (cid:107) u ( t ) (cid:107) in finite time t f . To meetthis objective, it is necessary that the initial condition x lies in the Reachable Set defined next.The reachable set R is the set of all initial conditions x ∈ R n transferable to the origin byusing an admissible control, R = (cid:110) x | x = − (cid:82) t e − A τ B u ( τ ) dτ, u ( t ) ∈ U (cid:111) . The minimizationof (cid:107) u ( t ) (cid:107) along with the assurance of finite t f ≥ is formally stated as follows: Problem 1 (Time Fuel Optimal Control) . Find a control u ( t ) ∈ U that steers system (1) from x ∈ R to the origin while minimizing J = (cid:82) t f ( k + | u ( t ) | ) dt, t f : free where k > is aweighing parameter to be appropriately chosen. A larger k places more weight on time compared to (cid:107) u ( t ) (cid:107) and thus the control input obtainedby solving Problem 1 will be such that the system trajectory reaches the origin faster with anincreased (cid:107) u ( t ) (cid:107) . On the other hand, smaller k is expected to give control input such thatstate trajectory reaches slowly to the origin, but, also a reduced (cid:107) u ( t ) (cid:107) . The choice of k in thecost function determines the trade-off between the fuel consumption and the speed of systemresponse. Also, k > ensures that the solution to Problem 1 drives the state trajectory to theorigin in finite time. February 23, 2021 DRAFT
A. Solution To Problem 1
The optimal solution to Problem 1 necessarily satisfies the conditions of Pontryagin MinimumPrinciple (PMP). These conditions will be utilized next for characterizing the candidate functionsfor the optimal control. We state the Pontryagin Minimum Principle (PMP) as follows [14]:
Theorem 1 (Pontryagin Minimum Principle (PMP)) . Let u ∗ ( t ) be the optimal control functionthat transfers the initial condition x to the origin with minimum cost J . Let x ∗ ( t ) be thetrajectory followed by the system (1) on application of u ∗ ( t ) with x ∗ (0) = x and x ∗ ( t f ) = .Then x ∗ ( t ) and u ∗ ( t ) satisfy the following conditions:(a) u ∗ ( t ) ∈ U is such that for each t ∈ [0 , t f ] it minimizes the Hamiltonian H , defined as H ( x ( t ) , p ( t ) , u ( t ) , k ) := k + | u ( t ) | + p T ( t )( Ax ( t ) + B u ( t )) , where p ( t ) = [ p ( t ) , ..., p n ( t )] T is thecostate,(b) corresponding to u ∗ ( t ) and x ∗ ( t ) , there exists an associated optimal costate trajectory p ∗ ( t ) which solves the cannonical system: ˙ x ∗ ( t ) = ∂H∂ p ( x ∗ ( t ) , p ∗ ( t ) , u ∗ ( t ) , k ) (2) ˙ p ∗ ( t ) = − ∂H∂ x ( x ∗ ( t ) , p ∗ ( t ) , u ∗ ( t ) , k ) (3) with boundary conditions: x ∗ (0) = x and x ∗ ( t f ) = and(c) terminal condition: H ( x ∗ ( t ) , p ∗ ( t ) , u ∗ ( t ) , k ) | t = t f = 0 . For ( A , AB ) being controllable, the optimal control u ∗ ( t ) that satisfies all the conditions ofTheorem 1 is given by: u ∗ ( t ) = − sgn ( ψ ( t )) if | ψ ( t ) | > , if | ψ ( t ) | < , (4)where ψ ( t ) = (cid:104) B , p ∗ ( t ) (cid:105) = b p ( t ) + ... + b n p n ( t ) . With A = diag ( c i /l ) and solution of (3), ψ ( t ) is further expressed as ψ ( t ) = b π e c l t + ... + b n π n e cnl t where ( π , ..., π n ) is the initial conditionof p ∗ ( t ) . The initial costate values are unconstrained and as a result, we are unable to determinethe optimal control function u ∗ ( t ) . However, we note that psi ( t ) is a linear combination ofseveral exponential terms. Thus, the following lemma from [14] can be utilized to characterizethe the optimal control function candidates by exploiting the number of roots of functions ψ ( t ) , ψ ( t ) − and ψ ( t ) + 1 . February 23, 2021 DRAFT
Lemma 1.
Let η , η , ..., η m be distinct real numbers and let f ( t ) , f ( t ) , ..., f m ( t ) be polynomials(with real coefficients) of degree d , d , ..., d m respectively. Then the function f ( t ) e η t + f ( t ) e η t + ... + f m ( t ) e η m t has at most d + d + ... + d m + m − real roots. We define the set of real roots of functions ψ ( t ) + 1 , ψ ( t ) and ψ ( t ) − as ψ − ( j ) := { t ≥ | ψ ( t ) = j } for j = − , , +1 respectively. Lemma 1 combined with (4) helps in concludingthat u ∗ ( t ) is necessarily a piecewise constant function with finitely many switches between +1 , and − . Theorem 2.
The optimal control function u ∗ ( t ) that steers states of system (1) from x ∈ R tothe origin with minimum J satisfies the following conditions (c.f. [18]):(i) u ∗ ( t ) is a piecewise constant function on an interval t ∈ [0 , t f ] switching between +1 , and − . Moreover, switching always takes place between +1 to , to +1 , − to and to − . Direct switching between +1 to − and vice-a-versa is not possible.(ii) u ∗ ( t f ) (cid:54) = 0 and is always equal to +1 or − ,(iii) The function ψ ( t ) that defines u ∗ ( t ) is such that(a) N ( ψ − (0)) ≤ n − (b) N ( ψ − (+1)) ≤ n ,(c) N ( ψ − ( − ≤ n ,(iv) u ∗ ( t ) has at most n discontinuities.Proof. The proof of each statement in the theorem is provided in separate ordered arguments:(i) From (4), we see that u ∗ ( t ) is a piecewise constant function switching between +1 , and − values determined by ψ ( t ) . Also, let [ τ , τ ] be an interval such that ψ ( τ ) = − and ψ ( τ ) = +1 . By continuity of ψ ( t ) , there exists a time τ ∈ [ τ , τ ] such that ψ ( τ ) =0 . Thus, there exists a sub interval in [ τ , τ ] in which the | ψ ( t ) | ≤ and as a resultcorresponding input u ∗ ( t ) = 0 on that sub-interval. Thus, the corresponding u ∗ ( t ) transitsbetween the − to +1 values through zero. Similarly it can be shown that u ∗ ( t ) transitsbetween the +1 to − values also through zero.(ii) If u ∗ ( t f ) is zero, then H ( x ∗ ( t ) , p ∗ ( t ) , u ∗ ( t ) , k ) | t = t f = k , thus, violating condition (c) inTheorem 1. Therefore, u ∗ ( t f ) (cid:54) = 0 . From (i), it follows that u ∗ ( t f ) is equal to ± . February 23, 2021 DRAFT (iii) From Lemma 1, we note that ψ ( t ) = 0 has n − real roots. Thus, N ( ψ − (0)) ≤ n − .Next, let us define ψ +1 ( t ) := ψ ( t ) − b π e c l t + ... + b n π n e cnl t − e cn +1 l t ψ − ( t ) := ψ ( t ) + 1 = b π e c l t + ... + b n π n e cnl t + e cn +1 l t where c n +1 = 0 . Recall that λ ( A ) was assumed to be non-zero distinct rational. Therefore, c , ..., c n , c n +1 are distinct real numbers. Again from Lemma 1, we conclude that ψ +1 ( t ) and ψ − ( t ) has n real roots respectively. Thus, N ( ψ − (+1)) ≤ n and N ( ψ − ( − ≤ n .(iv) Note from (4) that for the function u ∗ ( t ) switch happens only when ψ ( t ) = +1 or ψ ( t ) = − . Hence, by using (iii), the number of discontinuities in u ∗ ( t ) is N ( ψ − (+1))+ N ( ψ − ( − ≤ n + n = 2 n .Theorem 2, gives necessary conditions that optimal control candidates must satisfy. FromTheorem 2, the resulting form of u ∗ ( t ) can be expressed as follows: u ( t ) = ... ... t ∈ ( t m , t m +1 ) − t ∈ ( t m +1 , t m +2 )0 t ∈ ( t m +2 , t m +3 )+1 t ∈ ( t m +3 , t m +4 ) ... ... ± t ∈ ( t f − , t f ) (5) where, switching time instances satisfy t < ... < t m < t m +1 < ... < t f . Let
U ⊂ U be the set ofall possible inputs of the form given by (5) i.e., U = { u : [0 , t f ] → Z | u is piecewise constant } .Further, let the set of all inputs of the form (5) satisfying all the conditions of Theorem 2 bedenoted as U ∗ ⊂ U . In other words, U ∗ is the set of all control inputs that satisfy the PMPnecessary conditions. B. Correspondence between input set and sequence set
Note that if we ignore the values of switching instants in (5) and consider constant valuesarranged in their temporal order, then each element of U can be represented by a finite lengthsequence over Z = {− , , } . For example input of the form (5) can be compactly represented February 23, 2021 DRAFT by the following sequence ( ..., , +1 , , − , ..., ± This allows us to define an equivalencerelation among elements of U defined as follows: Definition 1.
Two inputs u ( t ) , u ( t ) ∈ U are equivalent if their corresponding sequences arethe same. This equivalence relation divides U into disjoint equivalence classes. Moreover, a sequenceover Z represents all equivalent inputs belonging to that respective equivalence class. Forexample all inputs, u ( t ) = − , t ∈ [0 , t )0 , t ∈ [ t , t )1 , t ∈ [ t , t ) with switching instants satisfying < t < t < t < ∞ are equivalent and are represented by asequence ( − , , . Consequently, sequence ( − , , forms an equivalence class of all inputswith different values of the time instants in above mentioned form. Thus, a bijective map canbe set up between the set of equivalence classes of U and the set of sequences over Z .Since, all conditions of Theorem 2 put restrictions only on the number of switching events andtemporal order in which +1 , − and appear in optimal control candidates, it is easier to dealwith sequences rather than piecewise continuous functions. Hence, to characterize all possibleoptimal control candidates that satisfy conditions of Theorem 2, we list all the equivalence classesto which optimal control candidates belong to. We call the sequence for which the correspondingequivalent inputs satisfy the conditions of Theorem 2 as a candidate sequence. TABLE IN
UMBER OF TIMES THE THREE LEVELS OF ψ ( t ) ARE CROSSED FOR CERTAIN SUB - SEQUENCE
Sub-sequence N ( ψ − (+1)) min N ( ψ − (0)) N ( ψ − ( − , (1 , (0 , − ( − , (1 , , − ( − , , +1) February 23, 2021 DRAFT0
C. Candidate Sequence Properties
Candidate sequences for optimal control u ∗ ( t ) are sequences obtained by arranging +1 ’s, ’sand − ’s in various combinations that satisfy the conditions of Theorem 2. This section describesthe structure of such candidate sequences.
1) Candidate Sequence Structure:
Consider the following sub-sequences: (a) (+1 , , (b) ( − , , (c) (0 , +1 , , (d) (0 , − , , (e) (0 , +1) , (f) (0 , − , (g) (+1) and (h) ( − . A finiteconcatenation of these sub-sequences yields another sequence. To be consistent with conditions(i), (ii), (iii-b) and (iii-c) of Theorem 2, the concatenation of these elements should satisfy thefollowing requirements.1) Terminal sub-sequence should end with a non-zero value,2) any two consecutive sub-sequences should be such that, the first sub-sequence ends withzero value and the next sub-sequence starts with zero.Further, to satisfy condition (iii) of Theorem 2, we would subsequently consider the number oftimes ψ ( t ) crosses +1 , and − corresponding to each sub-sequence (shown in Table I). Now, inthe resulting sequence obtained from such a concatenation, we combine two consecutive zeroesinto a single zero. For example, a concatenation of the sub-sequences that satisfy the aboveconditions, is done as follows: (0 , +1 , , (0 (cid:124) (cid:123)(cid:122) (cid:125) , +1 , , (0 (cid:124) (cid:123)(cid:122) (cid:125) , − , , (0 (cid:124) (cid:123)(cid:122) (cid:125) , − The sequence extracted will be (0 , +1 , , +1 , , − , , − . Let S be the set of all sequencesobtained by finite concatenation of sub-sequences (a)-(h). Note that any sequence in S alreadysatisfies condition (i) of Theorem 2.To characterize a candidate sequence we must further identify sequences from S that satisfyremaining conditions of Theorem 2. For that, we decompose a candidate sequence into thefollowing three segments1) Beginning segment:
A candidate sequence begins with any one of the sub-sequences (+1 , , ( − , , (0 , +1 , and (0 , − , . We denote N ( ψ − (+1)) in this segment as β + and N ( ψ − ( − as β − . The various values of ( β + , β − ) for these four beginning sub-sequencesare (1 , , (0 , , (2 , and (0 , respectively.2) Middle segment:
This segment is a finite concatenation of sub-sequences that both start andend with zero values, namely, (0 , +1 , and (0 , − , . Let this segment be a concatena- February 23, 2021 DRAFT1 tion of γ + and γ − numbers of (0 , +1 , and (0 , − , sub-sequences respectively. Then N ( ψ − ( − and N ( ψ − (+1)) in this segment are γ + and γ − respectively.3) End segment:
This segment is formed out of sub-sequences (0 , +1) and (0 , − to satisfycondition (ii) of Theorem 2. Let N ( ψ − (+1)) = (cid:15) + and N ( ψ − ( − (cid:15) − . The values of ( (cid:15) + , (cid:15) − ) for these possible transitions are (1 , and (0 , respectively.Let us denote the set of sequences in S for which N ( ψ − (+1)) = p and N ( ψ − ( − q by S p,q ⊂ S . Also, let S + p,q be the set of sequences in S p,q that begin with sub-sequences (+1 , or (0 , +1 , and similarly, S − p,q be the set of sequences that begin with ( − , or (0 , − , in S p,q . Then, S p,q = S + p,q ∪ S − p,q and S + p,q ∩ S − p,q = ∅ . Moreover, for any sequence in S + p,q , we have β + ∈ { , } and β − = 0 . Similarly for any sequence in S − p,q , we have β − ∈ { , } and β + = 0 .Therefore, for any sequence in S p,q with p + q > , the quantities ( β + , (cid:15) + , γ + , β − , (cid:15) − , γ − ) mustsatisfy the following set of equations and inequalities. β + + 2 γ + + (cid:15) + = q, γ + ∈ Z ≥ (6a) β − + 2 γ − + (cid:15) − = p, γ − ∈ Z ≥ (6b) β + β − = 0 , β + , β − ∈ { , , } (6c) β + + β − (cid:54) = 0 (6d) (cid:15) + (cid:15) − = 0 , (cid:15) + , (cid:15) − ∈ { , } (6e) (cid:15) + + (cid:15) − = 1 (6f)The solution set for equation (6) is listed in Table II for all cases of p and q . For the remainingsequences that arise when p + q ≤ , we directly use Table I to obtain the following sequencesets: (i) S +0 , = { (+1) } , (ii) S − , = { ( − } , (iii) S +0 , = { (0 , +1) } , (iv) S − , = ∅ , (v) S +1 , = ∅ ,(vi) S − , = { (0 , − } .Next, we note the symmetry in the set S p,q with p + q > . Flipping the sign of each non-zero element that appear in a sequence in set S + p,q gives us a sequence in S − q,p . In other words,conjugate set (recall from section II) S + p,q ⊂ S − q,p . Similarly, each sequence in set S − q,p is also asequence in S + p,q , giving us S − q,p ⊂ S + p,q . Proposition 1. S − q,p = S + p,q , p + q > Note from Proposition 1, that we only need to list sequences in S + p,q . Thus, solving thefollowing system of equations to realize the sequence structure in S + p,q for p + q > suffices to February 23, 2021 DRAFT2
TABLE IIS
OLUTION SET OF SYSTEM OF EQUATIONS (6)Set p q β + (cid:15) + γ + β − (cid:15) − γ − S + p,q even, ≥ odd, ≥ q − p odd, ≥ even, ≥ q − p − odd, ≥ odd, ≥ q − p − even, ≥ even, ≥ q − p S − p,q even, ≥ odd, ≥ q − p − odd, ≥ even, ≥ q p − odd, ≥ odd, ≥ q − p − even, ≥ even, ≥ q p − construct the set S p,q . β + + 2 γ + + (cid:15) + = q, γ + ∈ Z ≥ , β + ∈ { , } (7a) γ − + (cid:15) − = p, γ − ∈ Z ≥ (7b) (cid:15) + (cid:15) − = 0 , (cid:15) + , (cid:15) − ∈ { , } (7c) (cid:15) + + (cid:15) − = 1 (7d)The feasible solution for equations (7) are shown in Table II. Recall that each solution of (7)corresponds to a particular sequence in set S + p,q . D. Computation of Number of Candidate Sequences
In this subsection, we aim to compute the number of candidate sequences. At first, we proposethe following lemma that gives a count on the number of sequences in S p,q . Lemma 2.
Let p ≤ n and q ≤ n be fixed. Then, the number of sequences in S p,q for p + q > ,is N ( S p,q ) = N ( S + p,q ) + N ( S − p,q ) with N ( S + p,q ) = ( γ + p,q + γ − p,q ) C γ + p,q and N ( S − p,q ) = ( γ + q,p + γ − q,p ) C γ + q,p .Proof. In a solution to (7) for a fixed value of p ≤ n and q ≤ n , the values of β + , β − and (cid:15) + , (cid:15) − are also fixed as per Table II. This means that the starting and end segments are samefor all sequences in S + p,q for that p and q . As a consequence, the only distinguishing factorto separate one sequence from another in S + p,q is the arrangement of sub-sequences in their February 23, 2021 DRAFT3 middle segments. Hence, the number of sequences in S + p,q is determined by computing thenumber of possible concatenations of γ + p,q number of (0 , +1 , sub-sequences and γ − p,q numberof (0 , − , sub-sequences which is N ( S + p,q ) = ( γ + p,q + γ − p,q ) C γ + p,q . Similarly, for the set S − p,q = S + q,p (from Proposition 1), the number of possible concatenations of γ + q,p (0 , − , sub-sequences,and γ − q,p (0 , +1 , sub-sequences is N ( S − p,q ) = ( γ + q,p + γ − q,p ) C γ + q,p . Since S + p,q ∩ S − p,q = ∅ , we get N ( S p,q ) = N ( S + p,q ) + N ( S − p,q ) .Note that sequences in S + p,q for any p, q ≤ n satisfy condition (i), (ii), (iii-b) and (iii-c) in Theo-rem 2. However, it is to be ascertained whether all sequences also satisfy condition (iii-a) in The-orem 2. Note that, since condition (iii-a) restricts number of roots of ψ ( t ) and does not affect theswitching transitions for optimal input, it is impossible to verify condition (iii-a) directly for all S + p,q with p + q > , except the one sequence ˜ u + = (+1 , , − , , ( − , . . . , ( − n − , , ( − n ) in S + n,n , which is a concatenation of n sequences (+1 , , − with alternating signs. Clearly ˜ u + will lead to N ( ψ − (0)) = n and thus violating condition (iii-a) of Theorem 2. Similarly, for S − n,n , the sequence ˜ u − can be obtained which violates the condition (iii-a) of Theorem 2. Forrest of the sequences, the minimum of N ( ψ − (0)) < n can be ensured and hence condition(iii-a) of Theorem 2 is satisfied.Let S + = (cid:83) np =0 (cid:83) nq =0 S + p,q for p + q > . The set S + can be divided into four disjoint sets,namely, S +1 , S +2 , S +3 , S +4 defined as: S +1 = (cid:8) s ∈ S + | β + = 2 , (cid:15) + = 1 , (cid:15) − = 0 (cid:9) S +2 = (cid:8) s ∈ S + | β + = 2 , (cid:15) + = 0 , (cid:15) − = 1 (cid:9) S +3 = (cid:8) s ∈ S + | β + = 1 , (cid:15) + = 0 , (cid:15) − = 1 (cid:9) S +4 = (cid:8) s ∈ S + | β + = 1 , (cid:15) + = 1 , (cid:15) − = 0 (cid:9) Thus, we now get following theorem to count all the candidate sequences.
Theorem 3.
The number of candidate sequences in S + is n C n − + 2 n C n + n +1 C n − for n even n C n − + n +1 C n +12 − for n oddProof. Using Table II and Lemma 2, the number of sequences in S +1 for odd n is sum of p + q − C q − over p = 0 , , ..., n − nested with q = 3 , , ..., n . We simplify this sum by substituting February 23, 2021 DRAFT4 i = p/ and j = ( q + 1) / as follows: n − (cid:88) i =0 n +12 (cid:88) j =2 i + j − C i (8)Since, (cid:80) pk =0 r + k C k = r + p +1 C p , we get n +12 (cid:88) j =2 j − n − C n − = n +12 (cid:88) j =1 j − n − C n − − Therefore, (cid:80) n − j =0 j + n − C j − n C n − − . Following similar mathematical operations, the totalnumber of sequences in S + i , i = 2 , , are n C n − − , n +1 C n +12 − and n C n − − respectively.Identical manipulations can be made for n even and number of sequences in S + i , i = 1 , , , are n C n − − , n C n − , n C n − and n +1 C n − − respectively. By eliminating ˜ u + and consideringthe sequences in S + p,q with p + q ≤ , the theorem statement follows.IV. C HARACTERIZATION OF T IME - FUEL O PTIMAL C ONTROL
Recall the definition of super-sequence set from Section II. The super-sequence set for S + i , i =1 , .., sets are summarized in Table III and IV for n even and n odd respectively. TABLE IIIS
UPER - SEQUENTIAL S ET OF E QUIVALENCE C LASSES OF S + FOR n = EVEN
Equiv. Super Starting Intermittent Seg. EndClass Set Segment γ + γ − Segment S +1 S + n,n − (0 , +1 , n − n (0 , +1) S +2 S + n − ,n (0 , +1 , n − n − (0 , − S +3 S + n − ,n − (+1 , n − n − (0 , − S +4 S + n,n (+1 , n − n (0 , +1) Thus, the set of all candidate sequences in S + is S + n,n ∪ S + n − ,n ∪ S + n,n − ∪ S + n − ,n − −{ ˜ u + } . Notethat for n > , the set S + n − ,n ∪ S + n,n − { ˜ u + } forms the super-sequence set of S + n,n − , S + n − ,n − and therefore also includes all candidate sequences. But, for n = 2 , S +2 , − { ˜ u + } = ∅ , so, S , is considered to be the super-sequence of S +4 for n = 2 . Therefore, we represent the set of allcandidate sequences in S + as: S + n − ,n ∪ S +0 ,n for n = 2 S + n − ,n ∪ S + n,n − { ˜ u + } for n > (9) February 23, 2021 DRAFT5
TABLE IVS
UPER - SEQUENTIAL S ET OF E QUIVALENCE C LASSES OF S + FOR n = ODD
Equiv. Super Starting Intermittent Seg. EndClass Set Segment γ + γ − Segment S +1 S + n − ,n (0 , +1 , n − n − (0 , +1) S +2 S + n,n − (0 , +1 , n − n − (0 , − S +3 S + n,n (+1 , n − n − (0 , − S +4 S + n − ,n − (+1 , n − n − (0 , +1) For n = 2 , we get only two sequences (0 , +1 , , − ∈ S +1 , and (+1 , , +1) ∈ S +0 , . For n > , all elements of the set S + n,n − { ˜ u + } are the subsequences of the following sequence, (+1 , , s , , ..., s n − , , ( − n ) (10)with (cid:80) n − m =1 s m = − (for n even) and (cid:80) n − m =1 s m = 0 (for n odd) with s m − for m =1 , ..., n − . Similarly, all elements of the set S + n − ,n are the subsequences of the followingsequence, (0 , +1 , , s , , ..., s n − , , − ( − n ) (11)where s m − for m = 1 , ..., n − , (cid:80) n − m =1 s m = 0 (for n even) and (cid:80) n − m =1 s m = − (for n odd). We combine (10) and (11) to construct the following function u + c ( t ) from which all othertime-fuel optimal control candidates (with +1 appearing as the beginning non-zero input) canbe obtained by putting equality constraints on the consecutive time instants. u + c ( t ) = t ∈ [0 , t ]+1 t ∈ [ t , t ]0 t ∈ [ t , t ] s t ∈ [ t , t ] ... ... t ∈ [ t n − , t n − ] s n − t ∈ [ t n − , t n ]0 t ∈ [ t n , t n +1 ]( − n t ∈ [ t n +1 , t n +2 ] (12) February 23, 2021 DRAFT6 where , s i = 1 ∀ i = 1 , . . . , n − (cid:80) n − i =1 s i = for n odd − for n even , ≤ t ≤ t ≤ ... ≤ t n +1 ≤ t n +2 (13)and at least two pairs of switching time instances t j for j = 1 , ..., n + 2 are equal. Let usdenote the set of all inputs of the form (12) satisfying constraints (13) by U + c . Similarly, weobtain u − c ( t ) and the corresponding set U − c . Finally, note that U ∗ ⊂ U + c ∪ U − c .V. O PTIMAL C ONTROL P ROBLEM TO O PTIMIZATION P ROBLEM
In this section, we use U ± c obtained in Section IV and convert Problem 1 to an equivalentoptimization problem. A. Constraints
The set of initial sets that can be steered to the origin using inputs from U + c is X + c = { x = − (cid:82) t e − At Bu ( t ) dt, u ( t ) ∈ U + c } . Note that since the input is in piecewise constant form, the term − (cid:82) t e − At Bu ( t ) dt is a sum of exponentials of parameters t , ..., t n +2 . Recall, A is in diagonalform with λ ( A ) along the diagonal. Therefore, X + c can be alternately represented in terms of e λ t , ..., e λ n t . Using λ i = c i /l , i = 1 , ..., n and performing a substitution as follows: a j = e tjl for j = 1 , ..., n + 2 , (14)the parametric representation of X + c is achieved in terms of a j ’s. This representation is polynomialif all the eigenvalues i.e. λ i ’s have same sign. However, λ i ’s with both positive and negativesigns result in a rational parametric representation of X + c . Therefore, the expression for eachcomponent of x , in general, is written as x i, = N i ( a , ..., a n +2 ) /D i ( a , ..., a n +2 ) = N i, /D i where N i ( a , ..., a n +2 ) is the numerator and D i ( a , ..., a n +2 ) is the denominator polynomialof the rational function x i, for i = 1 , ..., n . Rearranging, a polynomial equality constraint isobtained as x i, D i − N i = 0 . Note that substitution (14) translate the inequality constraint ≤ t ≤ ... ≤ t n +2 to ≤ a ≤ ... ≤ a n +2 . B. Cost function
Using the piecewise constant nature of u ( t ) ∈ U + c , the cost function J in Problem 1 isexpressed as a weighted combination of final time and time duration for which u ( t ) is non-zero. February 23, 2021 DRAFT7
The cost function with u ( t ) as (10), denoted by J , is J = kt n +1 + t − t + t .... − t n + t n +1 − .Subsequently, the cost function with u ( t ) as (11), denoted as J , is J = kt n − t + t − t − .... − t n − + t n With substitution (14), and by monotonically increasing nature of logarithms,the cost function is expressed as J = a a . . . a k +12 n +1 a a . . . a n , and J = a a . . . a k +12 n a a . . . a n − C. Time-fuel optimization problem
With the constraint and cost function defined above, we are required to solve two sets ofoptimization problem for n > as:Minimize J = a a . . . a k +12 n +1 a a . . . a n (OP1)Subject to x i, = − b i lc i (cid:2) − a − c i − s a − c i + s a − c i −· · · − s n − a − c i n − + s n − a − c i n − − ( − n a − c i n + ( − n a − c i n +1 ] , i = 1 , ..., n,a j − a j +1 ≤ , ∀ j = 1 , . . . , n, n − (cid:88) m =1 s m = − for n even for n odd ,a ≥ , s m ∈ { +1 , − } Minimize J = a a . . . a k +12 n a a . . . a n − (OP2)Subject to x i, = − b i lc i (cid:2) − a − c i + a − c i − s a − c i + s a − c i · · · − s n − a − c i n − + s n − a − c i n − +( − n a − c i n − − ( − n a − c i n ] , i = 1 , ..., na j − a j +1 ≤ , ∀ j = 1 , . . . , n − , n − (cid:88) m =1 s m = for n even − for n odd ,a ≥ , s m ∈ { +1 , − } February 23, 2021 DRAFT8
D. Discussion
Note that problems (OP1) and (OP2) are mixed-integer nonlinear programming problems(MINLP), which are in general computationally difficult to solve even with the available solvers.Therefore, we treat these MINLP’s as a collection of multiple optimization problems by puttingthe values of the integer variable s m , m = 1 , , ... , in the constraints. Based on the values of s m and eliminating one corresponding to ˜ u + , the number of optimization problems that are requiredto be solved in the form (OP1) is n − C n − for n even (and n − C n − − for n odd). Similarly, thenumber of optimization problems that are required to be solved in the form (OP2) is n − C n − for n even (and n − C n − for n odd). Note that these two sets of optimization problems areobtained for u ( t ) ∈ U + c . Two more sets of such problems can be formulated for u ( t ) ∈ U − c insimilar manner. Therefore, we are required to solve, in total, the following number of non-linearprograms for n ≥ with rational cost function and semi-algebraic constraints, (cid:16) n − C n − n − C n − (cid:17) for n even (cid:16) n − C n − − n − C n − (cid:17) for n odd . The time-fuel optimal control is obtained by solving these optimization problems. Each problemcan be solved by converting it into a generalized moment problem and approximating it by ahierarchy of semidefinite programs (See [20] for more details). After solving all the optimizationproblems of the form (OP1) and (OP2), there is a possibility of multiple solutions yielding thesame minimum cost. All solutions which yields minimum cost are selected and depending onrequirements in terms of number of switchings, time of state-transfer and the L norm of input,a suitable optimal solution can be chosen. Also, since we are required to solve each optimizationproblem separately, the computation can done in a distributed manner. We also note finally thatthe existence of solution for at least one problem is guaranteed if and only if x ∈ R .
1) Solver for the optimization problem:
Each optimization problem being defined with arational cost function and semi-algebraic constraints, these optimization problems can be solvedusing a matlab based software package named Gloptipoly 3 (see [21]). Gloptipoly 3 converts theoptimization problem into an equivalent generalized moment problem (GMP) and then computesthe global optimal solution(s) by solving a hierarchy of semidefinite program (SDP) relaxations(see [25], [20], [26] for more details). It is important to note here, that Gloptipoly solverintroduces additional variables as relaxation order increases. If the number of variables in theoriginal optimization problem (e.g., OP1) are η and the relaxation order is ρ , with π number February 23, 2021 DRAFT9 of inequality constraints, the SDP will have π semidefinite constraints with moment matricesof size ω × ω where ω = (cid:0) η + ρρ (cid:1) . Worst case complexity for obtaining an (cid:15) -optimal solution toa SDP with constraints of size ω is O ( √ ω log( (cid:15) )) [27]. Since the growth of ω with relaxationorder ρ is very fast, using present day desktop computers only small examples can be workedout. At this point, note that one can also use any standard NLP solvers such as fmincon , SNOPT,IPOPT, etc., but at a loss of guarantee of globally optimal solutions.
2) Restrictions on number of switchings:
In addition, the computation of time-fuel optimalcontrol with the number of switching or discontinuities restricted as r ≤ n can also be handledin the proposed formulation. In such case, we consider sequences for u ∗ ( t ) from S p,q where p, q are the such that p, q ≤ n and p + q ≤ r . For example, for computing the time-fuel optimalcontrol for a system of order n = 3 with at most r = 4 switchings, we use sequences fromset S , , S , , S , , S , and S , and solve ten optimization problem formulated using u ( t ) as:(1) (+1 , , +1 , , − , (2) (0 , +1 , , − (3) (+1 , , − , , +1) , (4) (+1 , , − , , − and (5) (0 , − , , +1) and their conjugates. E. Example
Let us consider a second order system LTI system with A = diag ( − , − , B = [1 , T and We set the initial and final states as: x (0) = [0 . , . T , x ( t f ) = . The optimal control u ( t ) ∈ U + c = { (0 , +1 , , − , (+1 , , +1) } . Therefore, we achieve two optimizations problemsboth for problem (OP1) and (OP2), one with u ( t ) ∈ U + c and the other with u ( t ) ∈ U − c . Similarly,two other optimization problems can be formulated with u ( t ) ∈ U − c . By solving these fouroptimization problems, the problem (OP1) with u ( t ) ∈ U − c gives minimum cost and is shown inFigure 1. The performance measures for k = 0 , . , , , and minimum time control are shownin Table V. The sparsity in u ∗ ( t ) is computed as the ratio of off-duration of u ∗ ( t ) to t f . Underthe application of u ∗ ( t ) , the state trajectory steers from x to origin and u ∗ ( t ) follows one ofderived candidate sequence − , , +1 as shown in Figure 1.VI. C ONCLUSION
In this article, we computed time-fuel optimal control for LTI systems by characterizing thecontrol in terms of sequences of +1 , , − and switching time instants. A method is devisedto count and derive all candidate sequences (satisfying PMP necessary conditions). Further, allthe candidate sequences are utilized to transform the optimal control problem into multiple February 23, 2021 DRAFT0
TABLE VP
ERFORMANCE MEASURES FOR DIFFERENT VALUES OF k IN E XAMPLE k J ∗ t f Time duration Sparsityfor u ∗ ( t ) (cid:54) = 0 ∞ t (sec) -1.5-1-0.500.511.5 u * ( t ) k=0k=0.5k=1k=2k=3Min. Time Fig. 1. u ∗ ( t ) Trajectory static optimization problems which are tractable. Then, the optimal control input is obtained bysolving each optimization problem and selecting the solution with least cost. The computation canbe distributed as each optimization problem can be solved separately. Such characterization ofcontrol in terms of time instants can be further exploited in aperiodic feedback control techniquessuch as self-triggered feedback control [28].Developing dedicated problem solvers utilizing the structure of cost and constraints is thesubject of current and future research. Recently a way to exploit the sparseness of polynomialconstraints to make this approach scalable for optimal power flow computation appeared in [29].Such ideas utilizing any special sparsity structure can be pursued to alleviate the complexityissues the method currently suffers from. Further, a possible classification of initial conditionslabelled by the valid candidate sequences is also an interesting direction of research. Such aclassification will help in reducing the number of optimization problems that are required to be
February 23, 2021 DRAFT1 t (sec) -0.4-0.3-0.2-0.100.10.20.30.40.50.60.7 x ( t ) k=0k=0.5k=1k=2k=3Min. Time Fig. 2. x ( t ) Trajectory t (sec) -0.4-0.3-0.2-0.100.10.20.30.40.5 x ( t ) k=0k=0.5k=1k=2k=3Min. Time Fig. 3. x ( t ) Trajectory solved. R
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