Practical Need for Equation-Type Interval Equations and for Their Extended-Zero Solutions

One of the main problems in interval computations is solving systems of equations under interval uncertainty. Usually, interval computation packages consider united, tolerance, and control solutions. In this paper, we explain the practical need for equation-type solutions, when we look for solutions for which both sides are equal. In situations when such a solution is not possible, we provide a justification for extended-zero solutions, in which we ignore intervals of the type [−a, a]. 1 Practical Need for Solving Interval Systems of Equations: What Is Known Need for data processing. In many practical situations, we are interested in the values of quantities y1, . . . , ym which are difficult – or even impossible – to measure directly. For example, we can be interested in a distance to a faraway star or in tomorrow’s temperature at a certain location. Since we cannot measure these quantities directly, to estimate these quantities we must: – find easier-to-measure quantities x1, . . . , xn which are related to yi by known formulas yi = fi(x1, . . . , xn), – measure these quantities xj , and – use the results x̃j of measuring the quantities xj to compute the estimates for yi: ỹi = f(x̃1, . . . , x̃n). Computation of these estimates is called indirect measurement or data processing. Need for data processing under uncertainty.Measurements are never 100% accurate. Hence, the measurement result x̃j is, in general, different from the 2 L. Dymova, P. Sevastjanov, A. Pownuk, and V. Kreinovich actual (unknown) value xj of the corresponding quantity; in other words, the measurement errors ∆xj def = x̃j − xj are, in general, different from 0. Because of the non-zero measurement errors, the estimates ỹi are, in general, different from the desired values yi. It is therefore desirable to know how accurate are the resulting estimates. Need for interval uncertainty and interval computations. The manufacturer of the measuring instrument usually provides us with an upper bound ∆j on the measurement error: |∆xj | ≤ ∆j ; see, e.g., [7]. If no such upper bound is known, i.e., if the reading of the instrument can be as far away from the actual value as possible, then this is not a measuring instrument, this is a wild-guessgenerator. Sometimes, we also know the probabilities of different values ∆xj within this interval; see, e.g., [7, 13]. However, in many practical situations, the upper bound is the only information that we have [7]. In this case, after we know the result x̃j of measuring xj , the only information that we have about the actual (unknown) value xj is that this value belongs to the interval [xj , xj ], where xj def = x̃j −∆j and xj def = x̃j +∆j . In this case, the only thing that we can say about each value yi = fi(x1, . . . , xn) is that this value belongs to the range {fi(x1, . . . , xn) : x1 ∈ [x1, x1], . . . , xn ∈ [xn, xn]}. Computation of this range is one of the main problems of interval computations; see, e.g., [3, 6]. Sometimes, we do not know the exact dependence. The above text described an ideal case, when we know the exact dependence yi = fi(x1, . . . , xn) between the desired quantities yi and the easier-to-measure quantities xj . In practice, often, we do not know the exact dependence. Instead, we know that the dependence belongs to a finite-parametric family of dependencies, i.e., that yi = fi(x1, . . . , xn, a1, . . . , ak) for some parameters a1, . . . , ak. For example, we may know that yi is a linear function of the quantities xj , i.e., that yi = ci + n ∑ j=1 cij · xj for some coefficients ci and cij . The presence of these parameters complicates the corresponding data processing problem. Depending on what we know about the parameters, we have different situations. Simplest situation, when we know the exact values of all the parameters. The simplest situation is when we know the exact values of these parameters. In this case, the dependence of yi on xj is known, and we have the same problem of computing the range as before. Specific case: control solution. Sometimes, not only we know the values al of these parameters, but we can also control these values, by setting them to any Practical Need for Equation-Type Interval Equations 3 values within certain intervals [al, al]. By setting the appropriate values of the parameters, we can change the values yi. This possibility naturally leads to the following problem: – we would like the values yi to be within some given ranges [yi, yi]; for example, we would like the temperature to be within a comfort zone; – in this case, it is desirable to find the range of possible values of xj for which, by applying appropriate controls ai ∈ [al, al], we can place the values yi within these intervals. In the degenerate case, when all the intervals for yi and al are just points, this means solving the system of equations y = f(x, a), where we denoted y def = (y1, . . . , ym), x def = (x1, . . . , xn), and a def = (a1, . . . , ak). From this viewpoint, the above problem can be viewed as an interval generalization of the problem of solving a system of equations, or, informally, as a problem of solving the corresponding interval system of equations. The setX of all appropriate values x = (x1, . . . , xn) can be formally described as X = {x : for some al ∈ [al, al], fi(x1, . . . , xn, a1, . . . , ak) ∈ [yi, yi] for all i}. This set is known as the control solution to the corresponding interval system of equations [3, 12]. Situation when we need to find the parameters from the data. Sometimes, we know that the values ai are the same for all the cases, but we do not know these values. These values must then be determined based on measurements: we measure xj and yi several times, and we find the values of the parameters al that match all the measurement results. Let us number all membership cycles by values c = 1, . . . , C. After each cycle of measurements, we conclude that: – the actual (unknown) value of x (c) j is in the interval [x (c) j , x (c) j ] and – the actual value of y (c) i is in the interval [y (c) i , y (c) i ]. We want to find the set A of all the values a for which y = f(x, a) for some x and y: A = {a : ∀c∃x j ∈ [x (c) j , x (c) j ] ∃y (c) i ∈ [y (c) i , y (c) i ] (f(x , a) = y)}. This set A is known as the united solution to the interval system of equations [3, 12]. What can we do once we have found the range of possible values of a. Once we have found the set A of possible values of a, we can first find the range of possible values of yi based on the measurement results, i.e., find the range {fi(x1, . . . , xn, a) : xj ∈ [xj , xj ] and a ∈ A}. 4 L. Dymova, P. Sevastjanov, A. Pownuk, and V. Kreinovich This is a particular case of the main problem of interval computations. If we want to make sure that each value yi lies within the given bounds [y i , yi], then we must find the set X of possible values of x for which fi(x, a) is within these bounds for all possible values a ∈ A, i.e., the set X = {x : ∀a ∈ A∀i (fi(x, a) ∈ [yi, yi])}. This set is known as the tolerance solution to the interval system of equations [3, 12]. Sometimes, we know that the values a may change. In the previous text, we consider the situations when the values al are either fixed forever, or can be changed by us. In practice, these values may change in an unpredictable way – e.g., if these parameters represent some physical processes that influence yi’s. We therefore do not know the exact values of these parameters, but what we do know is some a priori bounds on these values. We may know bounds [al, al] on each parameter, in which case the set A of all possible combinations a = (a1, . . . , ak) is simply a box: A = [a1, a1]× . . .× [ak, ak]. We may also have more general sets A – e.g., ellipsoids. In this case, we can still solve the same two problems whose solutions we described above; namely: – we can solve the main problem of interval computations – the problem of computing the range – and find the set Y of possible values of y; – we can also solve the corresponding tolerance problem and find the set of values x that guarantee that each yi is within the desired interval. Is this all there is? There are also more complex problems (see, e.g., [12]), but, in a nutshell, most practical problems are either range estimation, or finding control, united, or tolerance solution. These are the problems solved by most interval computation packages [3, 6]. Is there anything else? In this paper, we show that there is an important class of practical problems that does not fit into one of the above categories. To solve these practical problems, we need to use a different notion of a solution to interval systems of equations. 2 Remaining Problem of How to Find the Set A Naturally Leads to Equality-Type Solutions to Interval System of Equations Finding the set A: formulation of the problem. In the previous text, we assumed that when the values of the parameter a can change, we know the set A of possible values of the corresponding parameter vector. But how do we find this set? Practical Need for Equation-Type Interval Equations 5 What information we can use to find the set A. All the information about the real world comes from measurements – either directly from measurements, or by processing measurement results. The only relation between the parameters a and measurable quantities is the formula y = f(x, a). Thus, to find the set A of possible values of a, we need to use measurements of x and y. We can measure both x and y many times. As a result, we get: – the set X of possible values of the vector x and – the set Y of possible values of the vector y. Both sets can be boxes, or they can be more general sets. Based on these two sets X and Y , we need to find the set A. In this problem, it is reasonable to assume that x and a are independent in some reasonable sense. Let us formulate this requirement in precise terms. Independence: towards a formal definition. The notion of independence is well known in the probabilistic case, where it means that probability of getting a value x ∈ X does not depend on the value a ∈ A: P (x | a) = P (x | a′) for all a, a′ ∈ A. An interesting corollary of this definition is that, in spite of being formulated in a way that is asymmetric with respect to x and a, this definition is actually symmetric: one can prove that a is independent of x if and only if x is independent of a. In the interval case, we do not know the probabilities, we only know which pairs (x, a) are possible and which are not. In other words, we have a set S ⊆ X × A of possible pairs (x, a). It is natural to say that the values x and a are independent if the set of possible values of x does not depend on a. Thus, we arrive at the following definition. Definition 1. Let S ⊆ X ×A be a set. – We say that a pair (x, a) is possible if (x, a) ∈ S. – Let x ∈ X and a ∈ A. We say that a value x is possible under a if (x, a) ∈ S. The set of possible-under-a values will be denoted by Sa. – We say that the variables x and a are independent if Sa = Sa′ for all a, a ′ from the set A. Proposition 1. Variable x and a are independent if and only if S is a Cartesian product, i.e., S = sx × sa for some sx ⊆ X and sa ⊆ A. Proof. If S = sx × sa, then Sa = sx for each a and thus, Sa = Sa′ for all a, a′ ∈ A. Vice versa, let us assume that x and a are independent. Let us denote the common set Sa = Sa′ by sx. Let us denote by sa, the set of all possible values a ∈ A, i.e., the set of all a ∈ A for which (x, a) ∈ S for some x ∈ X. Let us prove that in this case, S = sx × sa. Indeed, if (x, a) ∈ S, then, by definition of the set sx, have x ∈ Sa = sx, and, by definition of the set sa, we have a ∈ sa. Thus, by the definition of the Cartesian product B × C as the set of all pairs (b, c) of all pairs of elements b ∈ B and c ∈ C, we have (x, a) ∈ sx × sa. 6 L. Dymova, P. Sevastjanov, A. Pownuk, and V. Kreinovich Vice versa, let (x, a) ∈ sx × sa, i.e., let x ∈ sx and a ∈ sa. By definition of the set sx, we have Sa = sx, thus x ∈ Sa. By definition of the set Sa, this means that (x, a) ∈ S. The proposition is proven. As a corollary, we can conclude that the independence relation is symmetric – similarly to the probabilistic case. Corollary. Variables x and a are independent if and only if a and x are independent. Proof: indeed, both case are equivalent to the condition that the set S is a Cartesian product. What can we now conclude about the dependence between A, X, and Y . Since we assumed that a and x are independence, we can conclude that the set of possible values of the pair (x, a) is the Cartesian product X ×A. For each such pair, the value of y is equal to y = f(x, a). Thus, the set Y is equal to the range of f(x, a) when x ∈ X and a ∈ A. The resulting new type of solutions to interval systems of equations. So, we look for sets A for which Y = f(X,A) def = {f(x, a) : x ∈ X and a ∈ A}. We will call the set A satisfying this property an equality-type solution to the interval system of equations. Such solutions for the interval system of equations y = f(x, a), in which we want the interval versions Y and f(X,A) of both sides of the equation to be exactly equal, are known as formal solutions; see, e.g., [1, 2, 5, 10–12]. 3 What If the Interval System of Equations Does Not Have an Equality-Type Solution: A Justification for Enhanced-Zero Solutions But what if an equality-type solution is impossible: analysis of the problem. The description in the previous section seem to make sense, but sometimes, the corresponding problem has no solutions. For example, in the simplest case when m = n = k = 1 and f(x, a) = x + a, if we have Y = [−1, 1] and X = [−2, 2], then clearly the corresponding equation Y = X +A does not have a solution: no matter what set A we take the width of the resulting interval X + A is always larger than or equal to the width w(X) = 4 of the interval X and thus, cannot be equal to w(Y ) = 2. What shall we do in this case? How can we then find the desired set A? Of course, this would not happen if we had the actual ranges X and Y , but in reality, we only have estimates for these ranges. So, the fact that we cannot find A means something is wrong with these estimates. How are ranges X and Y estimated in the first place? To find out what can be wrong, let us recall how the ranges can be obtained from the experiments. Practical Need for Equation-Type Interval Equations 7 For example, in the 1-D case, we perform several measurements of the quantity x1 in different situations. Based on the corresponding measurement results x (c) 1 , we conclude that the interval of possible values must include the set [x1 , x ≈ 1 ], where x1 def = min c x (c) 1 and x ≈ 1 def = max c x (c) 1 . Of course, we can also have some values outside this interval – e.g., for a uniform distribution on an interval [0, 1], the interval formed by the smallest and the largest of the C random numbers is slightly narrower than [0, 1]; the fewer measurement we take, the narrower this interval. So, to estimate the actual range, we inflate the interval [x1 , x ≈ 1 ]. In these terms, the fact that we have a mismatch between X and Y means that one of these intervals was not inflated enough. The values x correspond to easier-to-measure quantities, for which we can make a large number of measurements and thus, even without inflation, get pretty accurate estimates of the actual range X. On the other hand, the values y are difficult to measure; for these values, we do not have as many measurement results and thus, there is a need for inflation. From this viewpoint, we can safely assume that the range for X is reasonably accurate, but the range of Y needs inflation. So how do we find A? In view of the above analysis, if there is no set A for which Y = f(X,A), the proper solution is to inflate each components of the set Y so that the system becomes solvable. To make this idea precise, let us formalize what is an inflation. What is an inflation: analysis of the problem.We want to define a mapping I that transforms each non-degenerate interval x = [x, x] into a wider interval