OPTIMAL INTERVAL ENCLOSURESFOR FRACTIONALLY-LINEAR FUNCTIONS,AND THEIR APLICATION TO INTELLIGENT CONTROLRobert N. Lea1, Vladik Kreinovich2, Raul Trejo31NASA Johnson Space Center, Houston TX 77058, USAemail blea@gothamcity.jsc.nasa.gov2Computer Science Department, University of Texas at El PasoEl Paso TX 79968, USA, email vladik@cs.ep.utexas.edu3Sistemas de Informacion, Division de Ingeneria y Ciencias,ITESM (Instituto Technologico de Monterrey),Campus Estado de Mexico, Apdo. Postal 2,Modulo de Servicio Postal, Atizapan, Mexico 52926email rtrejo@campus.cem.itesm.mxAbstract. One of the main problems of interval computations is, given a functionf(x1; :::; xn) and n intervals x1; :::;xn, to compute the range y = f(x1; :::;xn). Thisproblem is feasible for linear functions f , but for generic polynomials, it is known to becomputationally intractable. Because of that, traditional interval techniques usually com-pute the enclosure of y, i.e., an interval that contains y. The closer this enclosure to y,the better. It is desirable to describe cases in which we can compute the optimal enclosure,i.e., the range itself.In this paper, we describe a feasible algorithm for computing the optimal enclosurefor fractionally linear functions f . Applications of this result to intelligent control aredescribed.1. INTRODUCTIONOne of the main objectives of interval computations is: given a function y = f(x1; :::; xn)of n variables and n intervals xi = [x�i ; x+i ], 1 � i � n, to estimate the intervaly = f(x1; :::;xn) of the possible values of y. A typical application is when f describeshow a physical quantity y depends on the physical quantities xi, intervals xi describe pos-sible values of xi, and the desired interval y consists of possible values of y. In general, theproblem of computing y is computationally intractable (NP-hard) [G85]. Crudely speak-ing, this means that if we can solve this problem in feasible time (i.e., in time that does notexceed a polynomial of the length of the input), then we would be able to solve practicallyall discrete problems in polynomial time, which is usually believed to be impossible.In view of this negative result, traditional algorithms of interval computations (e.g.,naive interval computations or more sophisticated methods like a centered form [H69],[N75], [M79], [N80], [AH83], [RR84], [N85]), provide an enclosure of the desired interval,i.e., an interval Y with the property y � Y. For several simple classes of functions, exact(\optimal") enclosure Y = y can be computed. E.g., for a linear function f(x1; :::; xn) =a0 + a1x1 + ::: + anxn, the desired interval is y = a0 + a1 � x1 + ::: + an � xn, wherea � [x�; x+] = [min(a � x�; a � x+);max(a � x�; a � x+)].1

In many practical applications, the function f is fractionally linear, i.e., it is equal tothe ratio of two linear functions:y = f(x1; :::; xn) = a0 + a1x1 + :::+ anxnb0 + b1x1 + :::+ bnxn(see, e.g., [KKM86], [KKM87], [GK88], [GK88a], [MK88], [MK88a], [TG95]). For suchfunctions, naive interval computations and central form do not lead to the exact enclosure.In this paper, an optimal enclosure for such f will be described.We also describe a \continuous" version of this enclosure, and its application to intel-ligent control.2. DEFINITIONS AND THE MAIN RESULTDe�nition 1.� By a fractionally linear function, we mean a tuple (n;~a;~b), where:� n is a positive integer;� ~a = (a0; a1; :::; an) is a tuple of real numbers;� ~b = (b0; b1; :::; bn) is a tuple of real numbers;The value of this function for ~x = (x1; :::; xn) is de�ned asy = f(x1; :::; xn) = a0 + a1x1 + :::+ anxnb0 + b1x1 + :::+ bnxn :� By a basic interval computation problem for a fractionally linear function f (or simplya problem, for short), we mean a tuple (f;x1; :::;xn), where f is a fractionally linearfunction, and xi = [x�i ; x+i ] are intervals. We say that an algorithm computes the opti-mal enclosure for the problem (f;x1; :::;xn) if it computes the range y = f(x1; :::;xn).� We say that a problem (f;x1; :::;xn) with f = (n;~a;~b) is non-degenerate if0 62 b0 + b1 � x1 + ::+ bn � xn:Comment. If a problem is not non-degenerate, then the interval of possible values of thedenominator contains 0, and therefore, the set of all possible values of the fractions contains1. So, the range is an interval only if the problem is non-degenerate.THEOREM 1. There exists an algorithm that computes the optimal enclosure for anarbitrary non-degenerate fractionally linear problem in quadratic time (i.e., in computationtime � Cn2).Comment. So, the basic problem of interval computations is feasible for fractionally linearfunctions.Let us describe the algorithm (the proof that it always computes the optimal enclosureis given in Section 5). 2

ALGORITHM.� Step 1: making a denominator positive. If b0+P bix�i < 0, change the signs ofall the coefficients, i.e., set ai := �ai and bi := �bi.� Step 2: making the coe�cients in the denominator non-negative. For alli = 1; :::; n, if bi < 0, replace xi with the new variable yi = �xi, changethe signs of the coefficients ai and bi, and change xi = [x�i ; x+i ] to yi =[�x+i ;�x�i ].� Step 3: eliminating degenerate variables. If ai=bi = aj=bj for some i 6= j, andjbij � jbj j, replace variables xi and xj with a single variable yi, forwhich bi and ai stay the same as before, but for which yi = xi+(bj=bi)xj.� Step 4: ordering the variables. Order the variables xi in the increasing or-der of the corresponding ratios ai=bi, so that:a1b1 < a1b2 < ::: < anbn :� Final step: computing y�. Compute y+ = max(y+0 ; y+1 ; :::; y+n ), wherey+k = a0 + a1x�1 + :::+ akx�k + ak+1x+k+1 + :::+ anx+nb0 + b1x�1 + :::+ bkx�k + bk+1x+k+1 + :::+ bnx+n :Compute y� = min(y�0 ; y�1 ; :::; y�n ), wherey+k = a0 + a1x+1 + :::+ akx+k + ak+1x�k+1 + :::+ anx�nb0 + b1x+1 + :::+ bkx+k + bk+1x�k+1 + :::+ bnx�n :Example. Let n = 2, x1 = x2 = [1; 2],f(x) = 1 + x1 + x21 + x1 � 4x2 :In this case, the interval 1 + [1; 2]� 4[1; 2] = [�6;�1] does not contain 0, so, the problemis non-degenerate. Let us apply the above algorithm:� Step 1. Since b0 +P bix�i = 1 + 1 � 4 = �2 < 0, we change the signs of all thecoe�cients. As a result, we arrive at the following problem: x1 = x2 = [1; 2],f(x) = �1� x1 � x2�1� x1 + 4x2 :� Step 2. The coe�cient bi is negative for i = 1, so for this i, we introduce the newvariable, and correspondingly change the coe�cients a1; b1 and the interval x1. As aresult, we get the following problem: x1 = [�2;�1], x2 = [1; 2],f(x) = �1 + x1 � x2�1 + x1 + 4x2 :3

� Step 3. The values a1=b1 = 1 and a2=b2 = �1=4 are di�erent, so, we do nothing atthis step.� Step 4. Since a1=b1 > a2=b2, we change the order of the variables. As a result, we getthe following problem: x1 = [x�1 ; x+1 ] = [1; 2], x2 = [x�2 ; x+2 ] = [�2;�1],f(x) = �1� x1 + x2�1 + 4x1 + x2 :� Final step. We compute y+ as max(y+0 ; y+1 ; y+2 ), where:y+0 = �1� 2 + (�1)�1 + 8 + (�1) = �46 = �23 ;y+1 = �1� 1 + (�1)�1 + 4 + (�1) = �32 = �32 ;y+2 = �1� 1 + (�2)�1 + 4 + (�2) = �41 = �4:Hence, y+ = �(2=3). We also compute y� as min(y�0 ; y�1 ; y�2 ), where:y�0 = �1� 1 + (�2)�1 + 4 + (�2) = �41 = �4;y�1 = �1� 2 + (�2)�1 + 8 + (�2) = �55 = �1;y�2 = �1� 2 + (�1)�1 + 8 + (�1) = �46 = �23 :So, y� = �4, and y = [�4;�(2=3)].3. CONTINUOUS VERSION OF THIS ALGORITHMLet us describe a version of this algorithm that can be used for a continuous version ofthe above problem, in which, instead of n intervals xi, we have an interval-valued function.The motivation for this case will be given in Section 4.De�nition 2.� We say that a function f : R! R is piecewise-continuous if it is continuous everywhereexcept maybe �nitely many points x, in which both one-sided limits f(x � 0) =limh!0;h>0 f(x� h) and f(x+ 0) = limh!0;h>0 f(x+ h) exist (but may be di�erentfrom each other), and f(x) coincides with one of these limits.� Let real numbers a < b be �xed. By F , we will denote a class of all piecewise-continuous functions f : [a; b] ! [0; 1] for which f(a) = f(b) = 0 and f(x) 6= 0 forsome x.� Let f 2 F . By B�(f), we will denote the set of all functions g 2 F for whichg(x) 2 [f(x)� �; f(x) + �] for all x. 4

� For a function f 2 F , we denoteD(f) = R xf(x) dxR f(x) dx :By D(B�(f)), we denote the range of D, i.e., fD(g) j g 2 B�(f)g.THEOREM 2. For given (a; b), f(x) and �:� if f(x) � � for all x, then D(B�(f)) = (a; b);� else, D(B�(f)) = [u�; u+], where u� is the solution of the equation F�(u) = 0, u+ isthe solution of the equation F+(u) = 0,F�(u) = R ua (u� x)min(1; f(y) + �) dx� R bu (x� u)max(0; f(x)� �) dx;and F+(u) = R ua (u� x)max(0; f(x)� �) dx� R bu (x� u)min(1; f(x) + �) dx:(All the proofs are given in Section 5).Comment. Both functions F�(u) and F+(u) are increasing, so the values u� and u+ canbe computed by, e.g., a binary search algorithm. This means that to determine u+ and u�with precision ", we need no more than Cj log2(")j computational steps. In other words,to get k digits of u+ and u�, we need � Ck computational steps; this is a quite feasiblealgorithm.Remarks.1) Why cannot we just use interval analysis? For every x from (a; b), possiblevalues of g(x) form an interval (f(x) � �; f(x) + �): Therefore, we can apply thegeneral methodology of interval analysis (initiated in [M66]; see, e.g., [M79], [AH83],[RR84]) to get the interval estimates for R xg(x) dx and R g(x) dx, and then apply theinterval division rule to get an interval that contains [u�; u+].The main reason why we did not apply this approach is that the resulting intervalis larger than [u�; u+], and we will see in Section 4, in the intended applications,overestimating (\overshoot") can diminish the reliability and quality of the resultingcontrol.Let us give a simple example where simple interval computations really \overshoots":take f(x) = x for 0 � x � 1, f(x) = 2 � x for 1 � x � 2, and f(x) = 0 else.Such functions are among the most commonly used in fuzzy control (they are calledtriangular because their graphs form a triangle). For this function, the above-describedinterval estimate leads to an interval [1� 4� + o(�); 1 + 4� + o(�)] (computations aregiven in Section 3), while the actual bounds (computed by using the above Theorem)are [1��+o(�); 1+�+o(�)]. So, for a reasonable membership function, and for small�, we have a 4 times \overshoot".2) The above simple interval estimates can be reasonably improved if we usea continuous version of what is called a centered from in interval mathematics (see,e.g., [H69], [N75], [M79], [N80], [AH83], [RR84], [N85]), namely, if we �rst represent5

D(g) in the form D(f)+A(g�f)=B(g�f), and then apply the interval computationstechnique. As a result (for details see Section 5), we get an interval[D(f)� �(D(f)(a+ b)� 1=2(a2 + b2))=(R f(x) dx) + o(�);D(f) + �(D(f)(a+ b)� 1=2(a2 + b2))=(R f(x) dx) + o(�)].This estimate gives the right asymptotics for small �, but still gives an overshoot. Andsince we explained why overestimate is dangerous, it is much better to use a preciseestimate from Theorem 1.4. APPLICATIONS TO INTELLIGENT CONTROL4.1. What is intelligent control. Successes of intelligent control methodologyTraditional control theory is not always applicable. In case we do not have theprecise knowledge of a controlled system, we are unable to apply traditional control theory.Such situations occur, e.g., when we are devising a control for a future Martian rover, ora control for any other space mission into the unknown. In such cases, we often havea skillful operator who is good at making control decisions in uncertain environments.This operator can communicate his skills only in terms of natural-language rules that usewords like \small", \medium", etc. So, it is desirable to transform these rules into aprecise control. Such a methodology was �rst outlined by L. Zadeh [Z71, CZ72, Z73] andexperimentally tested by E. Mamdani [M74] in the framework of fuzzy set theory [Z65],therefore the whole area of research is now called fuzzy control (it is also called intelligent,or rule-based control). For the current state of fuzzy control, the reader is referred to thesurveys [S85], [L90] and [B91].Let us explain its main ideas on the following simple example. The goal of a thermostatis to keep a temperature T equal to some �xed value T0, or, in other words, to keep thedi�erence x = T � T0 equal to 0. To achieve this goal, one can switch the heater (orthe cooler) on and o� and control the degree of cooling or heating. We actually controlthe rate with which the temperature changes, i.e., in mathematical terms, a derivative_T of temperature with respect to time. So, if we apply the control u, the behavior ofthe thermostat will be determined by the equation _T = u. In order to automate thiscontrol, we must come up with a function u(x) that describes what control to apply if thetemperature di�erence x is known.Why can't we extract u(x) from an expert? We are talking about a situation wheretraditional control theory does not help, so we must use the experience of an expert todetermine the control function u(x). Why can't we just ask the expert questions like\suppose that x is 5 degrees; what do you do?", write down the answers, and thus plotu(x)? It sounds reasonable at �rst glance, until you try applying the same idea to a skillin which practically all American adults consider themselves experts: driving a car. If youask a driver a question like: \you are driving at 55 mph when the car in 30 ft. in front ofyou slows down to 47 mph., for how many seconds do you hit the brakes?", nobody willgive a precise number. You might install measuring devices into a car or a simulator, andsimulate this situation, but what will happen is that the amount of time to brake will bedi�erent for di�erent simulations. The problem is not that the expert has some precise6

number in mind (like 1.453 sec) that he cannot express in words; the problem is that onetime it will be 1.3, another time it may be 1.5, etc.An expert usually expresses his knowledge in words. An expert cannot expresshis knowledge in precise numeric terms (such as \hit the brakes for 1.43 sec"), but whathe can say is \hit the brakes for a while". So the rules that can be extracted from himare \if the velocity is a little bit smaller than maximum, hit the breaks for a while". Let'sillustrate the rules on the thermostat example. If the temperature T is close to T0, i.e., ifthe di�erence x = T � T0 is negligible, then no control is needed, i.e., u is also negligible.If the room is slightly overheated, i.e., if x is positive and small, we must cool it a littlebit (i.e., u = _x must be negative and small). If the temperature is a little lower, then weneed to heat the room a little bit. In other terms, if x is small negative, then u must besmall positive, etc. So we have the following rules:1) if x is negligible, then u must be negligible;2) if x is small positive, then u must be small negative;3) if x is small negative, then u must be small positive;etc.Brief description of fuzzy control methodology. First, we combine all the rules intoone statement relating x and the control u. If we know x, what control u should we apply?u is a reasonable control if either:� the �rst rule is applicable (i.e., x is negligible) and u is negligible; or� the second rule is applicable (i.e., x is small positive), and u is small negative; or� the third rule is applicable (i.e., x is small negative), and u is small positive; or� one of the other rules is applicable.Summarizing, we can say that u is an appropriate choice for a control if and only ifeither x is negligible and u is negligible, or x is small positive and u is small negative, etc. Ifwe use the denotations C(u) for \u is an appropriate control",N(x) for \x is negligible", SPfor \small positive, SN for \small negative" and use the standard mathematical notations& for \and", _ for \or" and � for \if and only if", we come to the following informal\formula":C(x) � (N(x)&N(u)) _ (SP (x)&SN(u)) _ (SN(x)&SP (u)) _ ::: (1)How do we formalize this combined statement: four stages of fuzzy controlmethodology. In order to formalize statements like the one we just wrote down, we�rst need to somehow interpret what notions like \negligible", \small positive", \smallnegative", etc., mean. The main di�erence between these notions and mathematicallyprecise (\crisp") ones like \positive" is that any value is either positive or not, while forsome values it is di�cult to decide whether they are negligible or not. Some values are sosmall that practically everyone would agree that they are negligible, but the bigger is thevalue, the fewer experts that will say that it is negligible, and the less con�dent they willbe in that statement. For example, if someone is performing a complicated experimentthat needs �xed temperature, then for him 0.1 degree is negligible, but 1 degree is not.For another expert �5 degrees is negligible. 7

First stage: describing the degree of con�dence. This degree of con�dence (alsocalled degree of belief, degree of certainty, truth value, certainty value) can take all possiblevalues from \false" to \true". Inside the computer, \false" is usually described by 0,\true" by 1. Therefore it is reasonable to use intermediate values from the interval (0,1)to describe arbitrary degrees of certainty. This idea appeared in fuzzy logic [Z65], andthat's why the resulting control is called fuzzy control. So the �rst stage of a fuzzy controlmethodology is to somehow assign values from the interval [0,1] to di�erent statementslike \0.3 is negligible" or \0.6 is small positive". There are several ways to do that [DP80].For example ([BW73], [B74], [DP80, IV.1.d], [KF88]) we can take several (N) experts,and ask each of them whether he believes that a statement is true (for example, that 0.3is negligible). If M of them answer \yes", we take M=N as a desired certainty value.Another possibility is to ask one expert and express his degree of con�dence in terms ofthe so-called subjective probabilities [S54].Second stage: forming membership functions. The procedure described above al-lows us to get the truth values of, for example, N(x) for di�erent values of x. But evenif we spend a lot of time questioning experts, we can only ask a �nite amount of ques-tions. Therefore, we will only get the values N(x) for �nitely many di�erent values of x:x1; x2; :::; xn, so we must somehow extrapolate the known truth values of N(xi) to comeout with a function that, for every possible x, gives a value from the interval [0,1] thatexpresses our degree of con�dence that this property is true for x. Such a function iscalled a membership function and is usually denoted by �(x). A membership function ofthe property N is denoted by �N (x), a membership function of the property SP (x) by�SP (x), etc.Third stage: & and _ operations. After the second stage we are able to assign truthvalues to the statements N(x), SP (x), etc. Our goal is to describe the possible values ofcontrol. In formula (1), control is represented by the statement C(u), meaning \u is anappropriate value of control". To get the truth value of this statement for di�erent valuesof C, we must somehow interpret the operations \and" and \or" that relate them to thevalues that we already know.Suppose that we have already chosen some rules to process & and _. Namely, we havechosen a procedure that allows us, given the truth values a and b of some statements Aand B, to compute the truth values of A&B and A _ B. Let's denote the resulting truthvalue of A&B by f&(a; b), and the truth value of A _B by f_(a; b). Now we can computethe truth value of C(x) for every x, i.e., a membership function of the property C.In particular, for our thermostat example the resulting membership function is�C(u) = f_(f&(�N (x); �N(u)); f&(�SP (x); �SN(u)); f&(�SN (x); �SP (u)); :::):Typical choices for f_(a; b) are max(a; b) or a+ b, for f&(a; b): min(a; b) or ab, etc.Fourth stage: defuzzi�cation. After the �rst three stages we have the \fuzzy" in-formation about the possible controls: something like \with degree of certainty 0.9 thevalue u = 0:3 is reasonable, with degree of certainty 0.8 the value u = 0:35 is reason-able, etc". We want to build an automatic system, so we must choose one value �u. So8

we must somehow transform a membership function �(u) into a single value. Such aprocedure is called defuzzi�cation. The most commonly used defuzzi�cation is a centroidD(�) = (R x�(x) dx)=(R �(x) dx).Successes of intelligent control. Fuzzy control is applied to control trains, appliances(dishwashers, laundry machines, camcorders, etc), manufacturing processes, etc (see, e.g.,[S85], [L90], [B91]). Computer simulations show that intelligent control is a very e�cientway to control space missions [L88, LTTJ89].4.2. The problems of intelligent control, and what is necessary for their solutionThe number is generated; with what precision shall we implement it? Supposethat this methodology generates a control value �0:310745. Does this mean that we reallyneed to generate such a precise control? Of course not, because the whole methodology isbased on the experts' estimates of membership functions, and they are never that precise.But what precision should we keep? Is �0:3 su�cient, or we need to achieve �0:31?This is a problem of e�ciency. This is not a purely academic question, because if itis su�cient to use �0:3, then we can install a simple controlling devices that can generatecontrols with a precision 0.1, but if we really need a 1% precision, then we need a muchmore complicated and more expensive device. For example, it is relatively easy to get a10o precision in the orientation of a space station, but to get 1o or less, one needs to usecomplicated and expensive super-precise engines, and in addition, be very careful aboutall possible experiments that would interfere with these engines.So, if we implement the control in too crude a manner, we will get only a crudeapproximation to the desired control, and thus loose some of this control's quality. If wetry to implement it with too much of a precision, we will waste e�orts and money onunnecessary precision. Therefore, we must know with what precision the control shouldbe implemented.Another e�ciency-related issue: do we need control at all? Another precision-related e�ciency problem appears in the situations, when the intelligent control method-ology leads to the values that are very close to 0 (here, 0 value of control means that nocontrol is necessary). In many cases, it would be more e�cient not to control at all thanto apply a very small control. For example, if we are controlling the spaceship, then somefuel and energy is spent just for keeping the control engines working. In these cases, it isbetter to switch the control o� whenever possible. So, we would like to apply 0 controlif the methodology prescribes to use a small control (that may be di�erent from 0 justbecause of the uncertainty in the initial data).Here again we have a dilemma: if we set this threshold too low, we do many unneces-sary control actions, and waste e�orts and energy (in case of a spaceship, fuel). If we setthis threshold too high, we do not control in the situations where we really need to, andthus we loose stability and/or other desired properties of control. So, ideally we should setthis threshold \just right". 9

Reliability. Intelligent control is not based on our precise knowledge of the controlledsystem, it is just a translation of the expert operator's control into mathematical terms.Since even the best human operator can err, this control that simulates his behavior mayalso err. This raises the problem of reliability.This problem is extremely important for intelligent control, because, e.g., in spaceapplications the doubts in reliability are the main reason why, in spite of the very promisingresults of computer simulations ([L88, LJB90], etc), fuzzy control techniques are not yetwidely applied to space missions.How to solve this problem? One way to do it is to take into the consideration thefact that fuzzy control is just a simulation of an expert's control. If we are not sure thatan expert is giving us the right advice, what do we do? Ask another expert or experts; iftheir opinions coincide, we believe in them much more than we believed in a single expert'sopinion. If they di�er, but there is a strong majority in favor of one of them, we go witha majority.The similar approach can be applied to fuzzy control: namely, instead of implementingthe rules of a single expert controller, let us implement several fuzzy control systems thatrepresent opinions of several experts. These systems can have di�erent sets of rules, usedi�erent fuzzy notions. For every situation, each of these systems will generate somecontrol value. If these values coincide, then this is the right value, and this value is muchmore reliable than in the case when we have only one fuzzy control system. If they di�er,but the majority of these systems prompt the same control, we control according to themajority.This sounds like a reasonable idea, but how to interpret the word \coincides"? Fuzzycontrol systems produce real values, and it is hardly unprobable that systems with di�erentrules, that use di�erent notions, will produce precisely the same real numbers. So, if the�rst systems generates the value 0.310987, and the second one the value 0.345901, do thesecontrol recommendations coincide or not? It depends on what is the precision of fuzzycontrol recommendations. If it is 0.1, then these controls are in good accordance, and wemust apply it. If the precision is 0.01, then there is a disagreement between them, and webetter consult the human operator before actually controlling the spaceship.In this case, it is also important not to underestimate and not to overestimate thisprecision. If we underestimate it (e.g., estimate it at 0.01, while in reality it is 0.1), thenin many cases, when the control decisions are really in good precision, and we can rely onthe automated control system, we would call the human operator. If we overestimate (e.g.,estimate it at 0.1, while in reality precision is 0.01), and use this \overshooting" estimate todetermine consistency, then we will occasionally erroneously apply the automated controleven when the results of several fuzzy systems disagree with each other (e.g., if they are0.3 and 0.35). In the �rst case, we do not use the entire potential and ability of theautomated control system; in the second case, we loose in reliability. Therefore, it isextremely important to get the precise estimates.10

4.3. Interval estimates help to solve the problems of intelligent controlInformal summary of the problem. The above examples show that it is very importantthat a fuzzy control system would not only generate a number u of the recommendedcontrol, but would also generate an interval [u�; u+] of possible reasonable controls.Motivation of the following mathematical formalization. How to estimate theseintervals? As we have already mentioned, the main reason why the resulting estimatesfor a control are not precise is that the entire methodology of intelligent control is basedon the experts' estimates of membership functions, and these estimates are not precise.In other words, if we apply the same procedure to the same expert (or the same groupof experts) twice, for the same statement we can get di�erent degrees of belief p 6= p0.This uncertainty can be easily determined: namely, we apply the same procedure two,three, or more times to the same expert or group of experts, and compute the di�erencesp � p0 between the values that were obtained on di�erent repetitions of this procedure.The biggest value jp�p0j is the desired estimate of the experts' uncertainty. Let us denotethis estimate by �.This number means that when we produce a membership function f(x), for the sameexperts it could as well be any function g(x) such that jf(x) � g(x)j � �. In the above-described fuzzy control methodology, we �rst process membership functions, and then usea centroid defuzzi�cation procedure D(f) = (R xf(x) dx)=(R f(x) dx) to produce a singlevalue of control. So, the problem is: knowing that g(x) can be any function such thatjf(x)� g(x)j � �, what are the possible results D(g) of defuzzifying g(x)?Additional remark: membership functions must have compact support. Wehave already formulated the mathematical problem, but we want to add one more remarkbefore we turn to a mathematical formulation. The parameters that we can control areusually bounded by technical reasons: we cannot get arbitrary acceleration, since ourabilities are limited by the existing boost engines; we cannot get unlimited rotation speed,etc. In all cases, there exists an interval (a; b) of control values that are technically and/orphysically feasible. Values of control outside this interval are simply impossible, therefore,for x 62 (a; b) we must have f(x) = 0, and g(x) = 0 for all possible membership functionsg. In mathematical terms, this restriction means that we limit ourselves to functions g(x)with compact support supp(f) � [a; b].The membership functions are usually assumed to be piecewise continuous and notnecessarily everywhere continuous is that we want to include the case, when the knowledgeis not fuzzy, and the membership function f(x) is equal to 1 or 0 for all x.As a result, we can apply Theorem 2 to compute the set of possible values of thecontrol.4.4. Other defuzzi�cation rulesCentroid of largest area. Centroid is just one of the possible defuzzi�cation rules.Sometimes, it does not work �ne, therefore, we need to apply more complicated rules11

(see, e.g., [YPL92, KQLFLKBR92]). In [KQLFLKBR92] we proved that the reasonabledemands select either this centroid rule, or an alternative rule called centroid of largestarea, where we �rst restrict a membership function �C(x) for control to some interval I,and only then apply a defuzzi�cation. So, the resulting control is equal to �u = D((�C)jI).In this case, the only step where we go from membership functions to an actual controlis also a centroid, therefore, we can apply Theorem 2 to compute the interval of possiblecontrols for this case as well.Center-of-Maximum. Another frequently used defuzzi�cation rule is as follows:De�nition 3. By a center-of-maximum defuzzi�cation we mean a mapping that trans-forms a function �(x) from F into a number DCOM (�) = 1=2(m� + m+), wherem� = inffx : �(x) = maxy(�(y))g, and m+ = supfx : �(x) = maxy(�(y))g.A natural property of the defuzzi�cation procedure d is that if we know the member-ship function with better and better accuracy (i.e., if � ! 0, then the resulting interval ofpossible values d(B�(f)) must converge to the point d(f)). In other words, the mapping dmust be continuous w.r.t. the metric �(f; g) = sup jf(x)� g(x)j.THEOREM 3.� Centroid defuzzi�cation D is continuous for all f .� Center-of mass defuzzi�cation is not everywhere continuous.4.5. ConclusionsWhen applying fuzzy control, we suggest not only to compute the recommended value, butalso to compute the interval [u�; u+] of possible values. To compute this interval, we mustdetermine the uncertainty � with which we can estimate the experts' degrees of belief (seeSection 1.3), and then apply the algorithm from Theorem 2. These interval estimates willenable us to do the following:� Increase e�ciency: Explain what control to use: any value from u� to u+ is �ne. Thissolves the problem of how precisely we must follow the recommendation u, and thusenables to choose the least costly way of following these control recommendations.� Solves the problem of whether to apply a control or not: if u� � 0 � u+, we do notneed to apply any control; else we must. This will enable us to avoid unnecessarycontrol actions.� Increase reliability: if we have several fuzzy control systems, and for each of them wehave an interval of recommended controls, then:� if the intersection of all these intervals is non-empty (i.e., all these intervals havea common point), then we take any control from this intersection as a reliablecontrol value;� if there is no point common to all these intervals, but the majority of them havea non-empty intersection, then we choose a control from this intersection.5. PROOFSProof of Theorem 1. Let us denote the numerator of the function f by A, and itsdenominator by B. Let us �rst prove that Steps 1{4 do not change the problem:12

� Step 1. If we change the signs of all the coe�cients ai and bi, then both numeratorand denominator will change signs, and the ratio will remain unchanged.� Step 2. If we rename the variable xi = �yi, then the values ai and bi, and the intervalof possible values of yi must be changed accordingly.� Step 3. If ai=bi = aj=bj , then aj=ai = bj=bi. Therefore,aixi + ajxj = ai(xi + (aj=ai)xj) = ai(1 + (bj=bi)xj);and bixi + bjxj = bi(xi + (bj=bi)xj). Therefore, we can replace the terms aixi + ajxjand bixi + bjxj that depend on xi and xj with aiyi and biyi, where the new variableyi is equal to yi = xi + (bj=bi)xj . If xi 2 xi and xj 2 xj , then the set yi of possiblevalues of the new variable is equal to yi = xi + (bj=bi)xi.� Step 4. Renaming the variables does not change the problem.In view of Steps 1{4, we can assume that bi � 0, and that the ratio ai=bi is increasing asi increases.After Step 1, we can be sure that the value of the denominator B is positive at leastfor one combination of xi 2 xi; since the problem is non-degenerate, the denominatorcannot attain 0, and hence, it is always positive.The function f is a continuous function de�ned on a compact x1� :::�xn. Therefore,its maximum y+ is attained at some point (x1; :::; xn).The function f is smooth; therefore, if the maximum is attained for xi inside theinterval [x�i ; x+i ] (i.e., for xi 2 (x�i ; x+i )), we will have @f=@xi = 0. Applying the formulafor the derivative of the fraction, we conclude that ai�B�bi�A = 0, hence, ai=bi = A=B = f .If we replace xi with x�i , then, the new value B0 of the denominator B will be equal toB+bi(x�i i�xi). The new value A0 of A will be A+ai(x�i �xi). Since A = bf and ai = bif ,we conclude that A0 = Bf + bif(x�i � xi) = f(B + bi(x�i � xi)) = fA0. Hence, the newvalue of the ratio f 0 = A0=B0 is equal to the old value f = A=B. Therefore, we can changethis xi to x�i without changing the optimal value of f . So, without losing generality, wecan assume that for every i, xi is either equal to x�i , or to x+i .If xi = x�i , then, since we have achieved the maximum of f for xi, an increase inxi can only decrease f . So, the partial derivative @f=@xi must be non-negative at thepoint (x1; :::; xn). This partial derivative is equal to (aiB� biA)=B2, so, the fact that it isnon-negative, means that aiB � biA � 0, which is equivalent to aiB � biA. Since B > 0and bi � 0, we can divide both sides of this inequality by biB, resulting in ai=bi � A=B.Similarly, if xi = x+i , a decrease in xi can only decrease f . So, we will get @f=@xi � 0,and ai=bi � A=B.So, for every i, either xi = x�i and ai=bi � A=B, or xi = x+i and ai=bi � A=B. Hence,if ai=bi < A=B, we have xi = x�i , and if ai=bi > A=B, we have xi = x+i . If ai=bi = A=B,then, as above, we can switch xi to x�i without changing the value of f .Since after Step 4, the variables are ordered in the order of the ratio ai=bi, this meansthat for all variables x1; :::; xk up to some k�th one, we have xi = x�i , and for the other13

variables xk+1; :::; xn, we have xi = x+i . In other words, we conclude that y+ = y+k forsome k. Hence, y+ � max(y+0 ; :::; y+n ):On the other hand, each value y+k is a possible value of the function f and is therefore,not exceeding the maximum y+ of the function f . So, y+k � y+ for all k; hence,max(y+0 ; :::; y+n ) � y+:From these two inequalities, we conclude that y+ = max(y+0 ; :::; y+n ):In this proof, we used transformations that, strictly speaking, make sense only ifbi 6= 0. However, the case when bi = 0 for some i, can be handled in the same manner, ifwe allow expressions like ai=0 (meaning +1 or �1 depending on the sign of ai).The proof for y� is similar.To complete the proof, we must now show that the algorithm described in Section 2requires quadratic time. Indeed, initial steps 1 and 2 require the number of operations thatis linear in n. Sorting (Step 4) can be done in time n log2(n) � n2 (see, e.g., [CLR90]),and the Final step requires us to compute 2(n+ 1) expressions y�k each of which requires4n+1 arithmetic operations: 2n multiplications, 2n additions, and 1 division. Totally, weneed � 2(n+ 1)(4n+ 1) = O(n2) arithmetic operations. Q.E.D.Proof of Theorem 2. Let us denote the desired set D(B�(f)) by C.1. Let us �rst notice that D(f) � b. Indeed, since x � b for x 2 [a; b], we havexf(x) � bf(x), hence R ba xf(x) dx � R ba bf(x) dx = b R ba f(x) dx. Dividing both sides ofthis inequality by R ba f(x) dx, we conclude that D(f) � b.Likewise, once can prove that D(f) � a.2. Let us now prove that D(f) < b.Since we have already proved that D(f) � b, it is su�cient to prove that D(f) 6= b.Let us prove it by reduction to a contradiction. Suppose that D(f) = b. Then, multi-plying both sides by R ba f(x) dx, we conclude that R ba xf(x) dx = b R ba f(x) dx. Therefore,R ba xf(x) dx = R ba bf(x) dx. Moving all the terms to the right-hand side, we conclude thatR ba (b� x)f(x) dx = 0. We assumed that f(x) is a piecewise-continuous non-negative func-tion; therefore, the function (b � x)f(x) is also non-negative and piecewise-continuous.This means that we can divide the interval (a; b) into �nitely many intervals, on each ofwhich (b� x)f(x) is non-negative and continuous. The integral of (b� x)f(x) over (a; b)is equal to the sum of the integrals of this functions over these integrals. Since we areintegrating a non- negative function, all these integrals are non-negative. Hence, their sumcan be equal to 0 only in one case: if all the terms in the sum (i.e., all the integrals) areequal to 0. But an integral of a continuous non-negative function is equal to 0 only if thisfunction is identically 0. So, (b � x)f(x) = 0 for all x, hence, f(x) = 0 for all x 2 (a; b).14

Since we demanded that f(x) = 0 outside the interval (a; b), we conclude that f(x) isidentically 0, which contradicts to our de�nition of a membership function.This contradiction proves that our assumption was wrong, and so, D(f) < b.3. Likewise, we can prove that D(f) > a. Hence, C � (a; b).4. Now, let us consider the case when f(x) � � for all �. We have just proved thatC � (a; b), so, in order to prove that C = (a; b), we must prove that any number c from theinterval (a; b) is a possible control.Indeed, suppose that c 2 (a; b). In this case, c� a > 0 and b � c > 0. Let us denotethe smallest of these two numbers by � = min(c � a; b � c). Let us take g(x) = � ifjx � cj < �, and g(x) = 0 else. Then g(x) � �, hence g(x)� f(x) � g(x) � �. Likewise,from f(x) � �, we conclude that f(x)�g(x) � �. So, jf(x)�g(x)j � �, and g(x) is possible.Then, R ba xg(x) dx = R c+�c�� x� dx = �x2=2 jc+�c�� = �=2((c + �)2 � (c � �)2) = 2�c�. Thedenominator of D(g) is equal to R ba g(x) dx = R c+�c�� � dx = �((c + �) � (c � �)) = 2��therefore, D(f) = c.5. Let us now consider the case when f(x0) > � for some x0. In this case, if a non-negative function g(x) equals 0 outside (a; b), and satis�es the inequality jf(x)� g(x)j � �for all x, then from jf(x0)� g(x0)j � � and f(x0) > � we can conclude that g(x0) > 0, sog(x) is not identically 0.6. Let us denote by PC the set of all piecewise-continuous non-negative functionson (a; b) that are not identically 0. The expression �(g; h) = supx jg(x) � h(x)j de�nesa metric on PC. The expressions R xg(x) dx and R g(x) dx are both continuous in thismetric. Indeed, if �(g; h) = supx jg(x)� h(x)j � �, thenj R g(x) dx� R h(x) dxj = j R (g(x)� h(x)) dxj � R jg(x)� h(x) dx � R ba � dx = (b� a)�,and (b� a)� ! 0 as � ! 0. Likewise,j R xg(x) dx� R xh(x) dxj � R jxj� dx = �(R ba jxj dx)! 0as � ! 0. Therefore, the ratio D(g) of these two expressions is also continuous in thismetric.7. Let us now denote the set of all possible functions by P. By de�nition, P � PC.It is easy to see that P is a connected set: if g; h 2 P, then for every t 2 [0; 1], theirconvex combination gt(x) = tg(x)+(1�x)h(x) also belongs to P, and this family gt formsa continuous family connecting g and h. Therefore, since D is continuous on P, the setD(P) of all possible values of D(g) is a connected subset of R. Therefore, it is either theentire real line, or a half-line, or an interval (open, close or semi-close).If we denote u� = inffD(g) : g 2 Pg and u+ = supfD(g) : g 2 Pg, then we canconclude that the desired set C coincides with one of the following sets:[u�; u+]; (u�; u+); [u�; u+) or (u�; u+].8. Let us now �nd u+. For every u 2 [a; b], let us de�ne a function f+u (x) as follows:f+u (x) = max(0; f(x)� �) for x < u, f+u (x) = min(1; f(x) + �) for x � u, and f+u (x) = 015

for x 62 (a; b). This is a piecewise-continuous non-negative function, and it is easy to checkthat jf(x)� f+u (x)j � � for all x. So, in view of 5., f+u (x) is a membership function andhence, for all u, it is a possible function.9. Let us prove that for every possible function g, D(g) � D(f+D(g)). This inequalitymeans that to �nd a supremum u+ of D(g) for all possible g, it is su�cient to consideronly possible functions f+u (x) (In other words, u+ = supfD(f+u ) : u 2 [a; b]g).Since g(x) is a possible function, for every x we have 0 � g(x) � 1 and jf(x)�g(x)j � �.Therefore, g(x) � f(x) + � and g(x) � 1, hence g(x) � min(1; f(x) + �). Likewise,g(x) � max(0; f(x)� �).From the de�nition of D(g), we conclude that D(g) = (R ba xg(x) dx)=(R ba g(x) dx),hence, R ba xg(x) dx = D(g) R ba g(x) dx = R ba D(g)g(x) dx,and so, R ba (x � D(g))g(x) dx = 0: If x < D(g), then x � D(g) < 0, and from g(x) �max(0; f(x)� �), we conclude that(x�D(g))g(x) � (x�D(g))max(0; f(x)� �) = (x�D(g))f+u (x).For x � D(g), we have x�D(g) � 0, and from g(x) � min(1; f(x) + �), we conclude that(x�D(g))g(x) � (x�D(g))min(1; f(x) + �) = (x�D(g))f+u (x).So, for all x, we have (x�D(g))g(x) � (x�D(g))f+u (x). Therefore,R ba (x�D(g))g(x) dx � R ba (x�D(g))f+u (x) dx.But we have proved that R ba (x � D(g))g(x) dx = 0, hence, R ba (x � D(g))f+u (x) � 0.Therefore, R ba xf+u (x) dx � D(g) R ba f+u (x) dx. Dividing both sides of this inequality byR ba f+u (x) dx, we conclude that D(f+u ) = (R ba xf+u (x) dx)=(R ba f+u (x) � D(g) dx). Q.E.D.10. Let us now prove that the function u! D(f+u ) is continuous.Indeed, the numerator R ba xf+u (x) dx of the fraction that de�nes D(f+u ) can be repre-sented as the sum of two integrals: I1(u) + I2(u), where I1(u) = R ua xmax(0; f(x)� �) dxand I2(u) = R ba xmin(1; f(x) � �) dx. Both integrals Ii(u) are integrals of a piecewise-continuous function, and the only dependency on u is that u is one of the integrationlimits. It is well known that the value of an integral of a piecewise-continuous functioncontinuously depends on its limits, so both functions I1(u) and I2(u) are continuous in u.Therefore, their sum (i.e., the numerator) is also continuous. Likewise, we can prove thatthe denominator is continuous. Therefore, this ratio D(f+u ) is also a continuous functionof u.11. Since D(f+u ) is a continuous function of u, the supremumu+ = supfD(f+u ) : u 2 [a; b]gis attained for some u: u+ = D(f+u ) for some u. According to 9., for g = f+u , wehave D(g) � D(f+D(g)). Since D(f+u ) already equals the supremum u+, we conclude thatD(f+D(g)) = D(g) = u+, so D(f+u+) = u+. Hence, the desired supremum u+ satis�es theequation D(f+u ) = u. 16

12. Let us prove that the desired supremum satis�es also the equation F+(u) = 0from the formulation of the Theorem.Indeed, if we substitute the de�nitions of D(g) and f+u into the equation D(f+u ) = u,we conclude that (R ua xmax(0; f(x)� �) dx+ R bu xmin(1; f(x) + �) dx)=(R ua max(0; f(x)� �) dx+ R bu min(1; f(x) + �) dx) = u.Multiplying both sides by the denominator, and moving all terms to the left-hand side, weconclude that F (u) = 0, where we denotedF (u) = R ua xmax(0; f(x)� �) dx+ R bu xmin(1; f(x) + �) dx�u R ua max(0; f(x)� �) dx� u R bu min(1; f(x) + �) dx.Combining integrals from a to u and from u to b, we conclude that F (u) = F1(u)+F2(u),where F1(u) = R ua xmax(0; f(x)� �) dx� u R ua max(0; f(x)� �) dx =R ua xmax(0; f(x)� �) dx� R ua umax(0; f(x)� �) dx = R ua (x� u)max(0; f(x)� �) dxand, likewise, F2(u) = R bu (x�u)min(1; f(x)+�) dx. One can easily see that F (u) = F1(u)+F2(u) = �F+(u), so the equation F (u) = 0 is equivalent to the equation F+(u) = 0 fromthe formulation of the Theorem. Therefore, u+ really satis�es the equation F+(u) = 0.13. Let us now prove that the equation F+(u) = 0 has only one solution, and therefore,this solution coincides with the desired supremum u+.We have F+(u) = �F (u) = �F1(u)�F2(u). In the �rst integral F1(u), the integratedfunction is always non-negative, hence the whole integral is non-negative; likewise, F2(u) �0. So, F+(u) = jF1(u)j � F2(u). We will prove that jF1(u)j is non-decreasing, and thatF2(u) is decreasing, and from that we will conclude that the equation F+(u) = 0 has onlyone solution.14. Let us �rst prove that a function jF1(u)j = R ua (u� x)max(0; f(x)� �) dx is non-decreasing in u. Indeed, suppose that u < v. Then, u � x < v � x, hence for all x � v,(u� x)max(0; f(x)� �) � (v � x)max(0; f(x)� �). Therefore,R va (u� x)max(0; f(x)� �) � R va (v � x)max(0; f(x)� �) = jF1(v)j.But jF1(u)j = R ua (u � x)max(0; f(x)� �) � R va (u � x)max(0; f(x)� �), hence jF1(u)j �jF1(v)j. So, the function jF1(u)j is non-decreasing.15. Let us now prove that a function F2(u) = R bu (x � u)min(1; f(x) + �) dx isstrictly decreasing. Indeed, suppose that u < v. Then, x � u > x � v, and hence(x � u)min(1; f(x) + �) � (x � v)min(1; f(x) + �) for all x � v. From this, we con-clude that R bv (x � u)min(1; f(x) + �) > R bv (x � v)min(1; f(x) + �) = F2(v). Sincemin(1; f(x) + �) � min(1; �) = � > 0, we have R vu (u� x)min(1; f(x) + �) dx > 0, henceF2(u) = R bv (x� u)min(1; f(x) + �) + R vu (u� x)min(1; f(x) + �) dx >R bv (x� u)min(1; f(x) + �) � F2(v), and F2(u) > F2(v).16. For u < v, from jF1(u)j � jF1(v)j and F2(u) > F2(v), we conclude that jF1(u)j �F2(u) < jF1(v)j � F2(v), i.e., that F+(u) < F+(v), and so, F+ is strictly increasing.17

Therefore, there can be only one value u, for which F+(u) = 0, and hence, this value mustcoincide with u+.We have proved the formula for u+. The fact that F+(u) is strictly increasing, explainsthat we can �nd u+ by using a bisection method.17. For u�, the proof is similar, but instead of the functions f+u we must consider thefunctions f�u that are equal to 0 outside (a; b), are equal to min(1; f(x)+ �) for a < x � u,and to max(0; f(x)� �) for u < x < b. Then we prove:� that D(g) � d(f�D(g)), and therefore, while looking for the in�mum u�, it is su�cientto consider only functions of the type f�u ;� that u! D(f�u ) is continuous, and therefore the maximum is attained for some u;� that for this u, D(f�u ) = u.From the last equation, we conclude that F�(u) = 0, and prove that F�(u) is strictlyincreasing and hence, the equation F�(u) = 0 has a unique solution. This solution thuscoincides with u�. Q.E.D.Applying simple interval methods to estimate the uncertaintyLet us apply simple interval computations to the function f that is de�ned as follows:f(x) = x for 0 � x � 1, f(x) = 2� x for 1 � x � 2, and f(x) = 0 else. For this function,D(f) = R 20 xf(x) dx= R 20 f(x) dx = 1=1 = 1. For the numerator R 20 xg(x) dx, the inequalityjf(x)� g(x)j � � leads to a conclusion thatj R xf(x) dx� R xg(x) dxj � R xjf(x)� g(x)j dx � � R 20 x dx = �x2=2 j20 = 2�.Therefore, the interval of possible values of a numerator is contained in [1 � 2�; 1 + 2�].For a denominator, we obtain the similar estimate:j R f(x) dx� R g(x) dxj � R jf(x)� g(x)j dx � � R 20 dx = 2�,so the interval of possible values of the denominator is also contained in [1 � 2�; 1 + 2�].Dividing these two intervals, we conclude that the interval of possible values of D(g) iscontained in [(1� 2�)=(1 + 2�); (1 + 2�)=(1� 2�)]. For small �, these interval bounds areasymptotically equal to [1� 4� + o(�); 1 + 4� + o(�)].Interval estimates for the centered formIn this approach, we represent D(g) as D(f) + A(g)=B(g), where A(g) = R (x �D(f))g(x) dx and B(g) = R g(x) dx, and then apply interval estimates to this form. Then,for A(g), we get an estimatejA(g)j � R ba �jx�D(f)j dx = �(RD(f)a (D(f)� x) dx+ R bD(f)(x�D(f)) dx) =�((x2=2� xD(f)) jD(f)a + (D(f)x� x2=2) jbD(f)) = �(D(f)(a+ b)� 1=2(a2 + b2)),hence the interval for A(g) is[��(D(f)(a+ b)� 1=2(a2 + b2)); �(D(f)(a+ b)� 1=2(a2 + b2))].For B(g), we get the interval [B(f)� �(b� a); B(f) + �(b� a)]. So, the resulting intervalfor D(g) = D(f) +A(g)=B(g) is[D(f)� �(D(f)(a+ b)� 1=2(a2 + b2))=(B(f)� (b� a)�);D(f) + �(D(f)(a+ b)� 1=2(a2 + b2))=(B(f)� (b� a)�)].18

For small �, we get[D(f)� �(D(f)(a+ b)� 1=2(a2 + b2))=B(f) + o(�);D(f) + �(D(f)(a+ b)� 1=2(a2 + b2))=B(f) + o(�)].Proof of Theorem 3.1. Let us �rst prove that D is a continuous mapping. For that, it is su�cient to provethat the mappings � ! R �(x) dx and � ! R x�(x) dx are continuous, then DC will becontinuous as a ratio of two continuous mappings. Indeed, if j�(x) � �0(x)j � � for allx 2 [a; b], then �� � �(x)��0(x) � �, hence ��(b�a) � R (�(x)��0(x)) dx = R �(x) dx�R �0(x) dx � (b � a)�. Therefore, j R �(x) dx � R �0(x) dxj � (b � a)�, and we can easilyprove continuity with � = "=(b � a). Likewise, j R x�(x) dx � R x�0(x) dxj � �(R ba jxj dx);so this expression is also continuous.2. Let us now prove that center-of-maximum is not continuous. Indeed, let us take atrapezoidal function �(x) that is equal to 0 for jxj > 2, to 1 for jxj � 1, to 2 � jxj for1 � jxj � 2. Then m� = �1;m+ = 1, hence DCOM (�) = 0. For every �, we can de�ne anew function ��(x) = �(x)(1� (�=3)(1�jxj)). For this new function, the maximum (equalto 1) is attained in only one point x = 1, so DCOM (��) = 1.Since we are considering only the values from �1 to 2, we have j1 � jxjj � 3, hence(�=3)(1 � jxj) � �, and j�(x) � �0(x)j � �. So, for " = 1=2, no matter how small � > 0we take, we can always �nd a new function �� such that �(�; ��) � �, but DCOM (��) �DCOM (�) = 1 > 1=2. So DCOM is not continuous. Q.E.D.Acknowledgments. This work was supported by a NSF Grant No. CDA-9015006,NASA Research Grants No. 9-482 and No. 9-757, and the Institute for Manufacturingand Materials Management grant. The authors are greatly thankful to Lot� A. Zadeh,Hung T. Nguyen, and Andrew Bernat for valuable discussions, and to the anonymousreferees whose suggestions enabled us to formulate our results in more general intervalterms.REFERENCES[AH83] G. Alefeld, J. Herzberger. Introduction to interval computations, AcademicPress, N. Y., 1983.[B91] H. R. Berenji. Fuzzy logic controllers. In: An Introduction to Fuzzy LogicApplications in Intelligent Systems (R. R. Yager, L. A. Zadeh. eds.), Kluwer AcademicPubl., 1991.[B74] J. M. Blin. Fuzzy relations in group decision theory, J. of Cybernetics, 1974,Vol. 4, pp. 17{22.[BW73] J. M. Blin and A. B.Whinston. Fuzzy sets and social choice. J. of Cybernetics,1973, Vol. 3, pp. 28{36.[CZ72] S. S. L. Chang and L. A. Zadeh. On fuzzy mapping and control, IEEE Trans-actions on Systems, Man and Cybernetics, 1972, Vol. SMC-2, pp. 30{34.19

[CLR90] Th. H. Cormen, Ch. L. Leiserson, R. L. Rivest, Introduction to algorithms,MIT Press, Cambridge, MA, 1990.[DP80] D. Dubois and H. Prade. Fuzzy sets and systems: theory and applications.Academic Press, N.Y., London, 1980.[G85] A. A. Gaganov, Computational complexity of the range of the polynomial inseveral variables, Cybernetics, 1985, pp. 418{421.[GK88] A. I. Gerasimov and V. Kreinovich. \Piecewise- fractionally-linear approxi-mation," Leningrad Polytechnical University and National Research Institute for Scienti�cand Technical Information (VINITI), 1988, 15 pp. (in Russian)[GK88a] A. I. Gerasimov and V. Kreinovich. \On the problem of optimal approx-imation choice for metrological characteristics," Leningrad Polytechnical University andNational Research Institute for Scienti�c and Technical Information (VINITI), 1988, 13pp. (in Russian).[H69] E. R, Hansen (ed.). Topics in interval analysis, Oxford University Press, 1969.[KF88] G. J. Klir and T. A. Folger. Fuzzy sets, uncertainty and information. PrenticeHall, Englewood Cli�s, NJ, 1988.[KKM86] I. N. Krotkov, V. Kreinovich and V. D. Mazin, \Methodology of designingmeasuring systems, using fractionally linear transformations,"Measuring Systems. Theoryand Applications. Proceedings of the Novosibirsk Electrical Engineering Institute, 1986, pp.5{14 (in Russian).[KKM87] I. N. Krotkov, V. Kreinovich and V. D. Mazin, \General form of measure-ment transformations which admit the computational methods of metrological analysis ofmeasuring- testing and measuring-computing systems," Izmeritelnaya Tekhnika, 1987, No.10, pp. 8{10 (in Russian); English translation: Measurement Techniques, 1987, Vol. 30,No. 10, pp. 936{939.[KQLFLKBR92] V. Kreinovich, C. Quintana, R. Lea, O. Fuentes, A. Lokshin, S. Ku-mar, I. Boricheva, and L. Reznik. What non- linearity to choose? Mathematical founda-tions of fuzzy control. Proceedings of the 1992 International Fuzzy Systems and IntelligentControl Conference, Louisville, KY, 1992, pp. 349{412.[L88] R. N. Lea. Automated space vehicle control for rendezvous proximity operations.Telemechanics and Informatics, 1988, Vol. 5, pp. 179{185.[LJB90] R. N. Lea, Y. K. Jani and H. Berenji. Fuzzy logic controller with reinforcementlearning for proximity operations and docking. Proceedings of the 5th IEEE InternationalSymposium on Intelligent Control, 1990, Vol. 2, pp. 903{906.[LTTJ89] R. N. Lea, M. Togai, J. Teichrow and Y. Jani. Fuzzy logic approach tocombined translational and rotational control of a spacecraft in proximity of the Space20

Station. Proceedings of the 3rd International Fuzzy Systems Association Congress, 1989,pp. 23{29.[L90] C. C. Lee. Fuzzy logic in control systems: fuzzy logic controller. IEEE Transac-tions on Systems, Man and Cybernetics, 1990, Vol. 20, No. 2, pp. 404{435.[M74] E. H. Mamdani. Application of fuzzy algorithms for control of simple dynamicplant, Proceedings of the IEE, 1974, Vol. 121, No. 12, pp. 1585{1588.[MK88] V. D. Mazin and V. Kreinovich. \An explanation of the universal character ofthe transformation functions y = (a lnx+b)=(c lnx+d)," Leningrad Polytechnical Instituteand National Research Institute for Scienti�c and Technical Information (VINITI), 1988,6 pp. (in Russian).[MK88a] V. D. Mazin and V. Kreinovich. \An important property of fractional-linear transformation functions," Leningrad Polytechnical Institute and National ResearchInstitute for Scienti�c and Technical Information (VINITI), 1988, 13 pp. (in Russian).[M66] R. E. Moore. Interval analysis, Prentice Hall, 1962.[M79] R. E. Moore. Methods and applications of interval analysis, SIAM Studies inApplied Mathematics, 1979.[N75] K. Nickel (ed.) Interval mathematics. Lecture Notes in Computer Science, Vol.29, Springer{Verlag, Berlin{Heidelberg{N.Y., 1975.[N80] K. Nickel (ed.) Interval mathematics 1980. Academic Press, N.Y., 1980.[N85] K. Nickel (ed.) Interval mathematics 1985. Lecture Notes in Computer Science,Vol. 212, Springer{Verlag, Berlin{Heidelberg{N.Y., 1985.[RR84] H. Ratschek and J. Rokne. Computer methods for the range of functions, EllisHorwood and John Wiley, 1984.[S54] L. J. Savage. The foundations of statistics. Wiley, New York, 1954.[S85] M. Sugeno (editor). Industrial applications of fuzzy control, North Holland,Amsterdam, 1985.[TG95] R. Trejo and A. I. Gerasimov, \Choosing interval functions to represent mea-surement inaccuracies: group-theoretic approach", Reliable Computing, 1995, Supplement(Extended Abstracts of APIC'95: International Workshop on Applications of Interval Com-putations, El Paso, TX, Febr. 23{25, 1995), pp. 207{210.[YPL92] J. Yen, N. P uger, and R. Langari. A defuzzi�cation strategy for a fuzzylogic controller employing prohibitive information in command formulation, Proceedingsof IEEE International Conference on Fuzzy Systems, San Diego, CA, March 1992.[Z65] L. Zadeh. Fuzzy sets. Information and control, 1965, Vol. 8, pp. 338{353.[Z71] L. A. Zadeh. Towards a theory of fuzzy systems, In: Aspects of network andsystems theory, R. E. Kalman, N. DecLaris (eds.), Holt. Rinehart, Winston, 1971.[Z73] L. A. Zadeh. Outline of a new approach to the analysis of complex systems anddecision processes, IEEE Transactions on Systems, Man and Cybernetics, 1973, Vol. 3,pp. 28{44. 21