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25 November 2005 Foundations of Logic and Constraint Programming 1 Constraint (Logic) Programming An overview Boolean Constraints Complete Solvers for SAT Boolean Unification Complete Solvers for Linear Constraints Applications [Dech04] Rina Dechter, Constraint Processing,

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25 November 2005 Foundations of Logic and Constraint Programming 2 Boolean Constraints Boolean domains (often referred to as 0/1 variables) has many applications, namely In Digital Circuits oExemple: Half-adder In problems modelling binary choices oExemple: N-Queens In problems dealing with sets B G1G1 G2G2 A C S

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25 November 2005 Foundations of Logic and Constraint Programming 3 Approaches to Boolean Constraint Solving The satisfaction of Boolean constraints may be addresses in various and alternative ways Simbolically Boolean Unification Complete Solver SAT All constraints are converted into Clauses (CNF) Solution construction (backtracking) or repairing (local search) Finite Domains Booleans are just an instance of a Finite domain (0/1) Techniques common to other Finite Domains

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25 November 2005 Foundations of Logic and Constraint Programming 4 Boolean Constraints To handle Boolean satisfaction simbolically, it is convenient to convert all Boolean constraints using exclusively equality constraint on Boolean Formulas. Boolean Formulas are composed of Two Boolean operators # (exclusive or) and · (conjunction), Boolean constants 0 e 1 ( and possibly other constants, domain dependent) variábles denoted by upper case letters This conversion is always possible, since the set {0, 1, #, ·} is complete. Given arbitrary terms a and b, all the usual operators and constants may be expressed using exclusively the “#” and “ · ” operators. a= 1 # a ; a b = a · b ; a b= a # b # a · b a b = 1 # a # a · b ; a b = 1 # a # b

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25 November 2005 Foundations of Logic and Constraint Programming 5 Boolean Rings More formally, the tuple, where A any domain that includes two elements 0 and 1, is a boolean ring, if the following properties are satisfied Associativity a#(b#c) = (a#b)#ca·(b·c) = (a·b)·c Comutativity a # b = b # aa · b = b · a Distribution a #(b·c) = (a#b)·(a#c)a·(b#c) = a·b#a·c Neutral Element a#0 = aa·1 = a Exclusivity and Idempotence a#a = 0 a·a = a Absorbing Element a·0 = 0

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25 November 2005 Foundations of Logic and Constraint Programming 6 Boolean Constraints on Sets Although Boolean constraints are usually considered on domains with only two elements, another interesting domain of application is Sets. More precisely, considering a domain P of all all sets composed of elements a 1, a 2, a 3,..., a n Constant 1 corresponds to the Universal set {a1, a2, a3,..., an} Constant 1 corresponds to the Empty set { } Operator “#” corresponds to the Set Exclusive Union Operator “·” corresponds to Set Intersection Other constraints on setsmay be converted into equality. For example, given the equivalence a b a b = a a b = b the includes constraint “ “can be rewritten as a # b # a·b = b a # b # b # a·b = b # b a # a·b = 0 a·b = a

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25 November 2005 Foundations of Logic and Constraint Programming 7 Constraint Solving through Boolean Unification The implementation of a symbolical and complete constraint solver fo Booleans is the core of CLP(B), the instance of the CLP(X) scheme to the Boolean Domain. In this solver, Boolean constraints are maintained in a solved form, obtained through Boolean unification. A Boolean constraint t 1 = t 2 (where terms t 1 e t 2 are exclusively written has the form with operators “#” and “ · ”. Such constraint can be satisfied iff there is a Boolean Unifier for terms t 1 and t 2. A Boolean unifier is a set of pairs x / t where x and t are respectively Boolean variables and terms (much in the same way of unifiers in logic programming), and where none of the variables may appear in the terms. As in logic programming, Boolean unifiers will be denoted by greek letters.

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25 November 2005 Foundations of Logic and Constraint Programming 8 Boolean Unification: Examples Example 1: Terms t 1 =1 # A and t 2 = A·B admit the following Boolean unifier = {A/1, B/0} In fact, denoting by t (or t ) the application of substitution to term t, t 1 = (1#A) {A/1, B/0}= 1 # 1 = 0 t 2 = (A·B) {A/1, B/0}= 1 · 0 = 0 which guarantees that constraint t 1 = t 2 is satisfiable. Example 2: Terms t 1 = 1 # A·B and t 2 = C # D admit unifiers = {C/1#A·B#D} and = {C/1#D, A/0} since t 1 = (1#A·B) {C/1#A·B#D} = 1#A·B t 2 = (C#D ) {C/1#A·B#D} = 1#A·B#D#D = 1#A·B and t 1 = (1#A·B) {C/1#D, A/0} = 1#0·B= 1 t 2 = (C#D) {C/1#D, A/0} = (1#D#D)= 1

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25 November 2005 Foundations of Logic and Constraint Programming 9 Multiple Boolean Unifiers Like in LP it is convenient to consider a hierarchy of unifiers, related by the “more general” relationship. Unifier is more general then there is a substitution such that Example: As seen above, terms t 1 = 1 # A·B and t 2 = C # D, admit unifiers = {C/1#A·B#D} and = {C/1#D, A/0} In this case, is more general than , since for {A/0}, it is . = {C/1#A·B#D } o {A / 0} = {C/1#D, A/0} =

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25 November 2005 Foundations of Logic and Constraint Programming 10 Most General Boolean Unifiers In general, given two Boolean terms, t 1 e t 2, there might be not only more than one unifiers, but also more than one most general unifier. Example: Boolean terms t 1 = 1 # A·B and t 2 = C # D may be unified by the most general unifiers 1 = { C / 1 # A·B # D} 2 = { D / 1 # A·B # C} 3 = { A / 1 # C # D, B / 1} 4 = { A / 1, B / 1 # C # D} None of the unifiers may be obtained by the composition of any of the others and a substitution.

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25 November 2005 Foundations of Logic and Constraint Programming 11 Boolean Unification Algorithm Taking into account that constraint t 1 = t 2 is equivalent to t 1 #t 2 = 0, the unification of terms t 1 and t 2 is equivalent to constrain term t 1 #t 2 to 0. The conditions necessary to zero a term t, may be analysed follows: 1.Zeroing a constant term is trivially checkable a.If t = 0 the term is already zero; b.If t = 1 term cannot be zeroed 2.Given distribution, a non constant term t, can always be written in the form a·U#b (where U is one of the variables occurring in t) factoring t in order to U. 3.A term t = a·U # b may onlty be zeroed if either a = 1 or b = 0, or both. Otherwise, i.e. a = 0 and b = 1, it would be t = 1 0, whatever the value chosen for U. 4.Such condition ( a = 1 or b = 0 ) is guaranteed iff the term (1 # a)·b is zeroed.

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25 November 2005 Foundations of Logic and Constraint Programming 12 Boolean Unification Algorithm 5.Once condition a = 1 or b = 0 is guaranteed, I.if a = 0 (and b = 0 ), then variable U may take an arbitraru value, since 0 = 0·U # 0 ) II.if a = 1, then variable U must take valur b (in this case, 0 = 1·U # b ) 6.The assignment to U of these terms can be implemented by means of a fresh variable, not occurring in t. Denoting by _U this new variable, the previous condition is equivalent to U = (1#a)·_U # b 7.In fact, a.If a = 0 (and b = 0 ) then U = _U, i.e. U may take any arbitrary value, since variable _U, being a fresh variable, is not constrained anywhere. b.If a = 1 then U = (1#1)·_U # b, i.e. U = b, as intendend.

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25 November 2005 Foundations of Logic and Constraint Programming 13 Boolean Unification Procedure The reasoning above can be used in the implementation of predicate unif_bool, shown below. predicate unif_bool(in: t 1, t 2 ; out: ); t t 1 # t 2 ; unif_bool zero(t, ); end predicate. The predicate inputs terms t 1 and t 2, that are to be unified, and succeeds if predicate zero succeedes, returning unifier obtained from this predicate.

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25 November 2005 Foundations of Logic and Constraint Programming 14 Boolean Unification Procedure Predicate zero can now be implemented as shpown below predicate zero (in: t; out: ); case t = 0: zero = True, = {}; case t = 1: zero = False; otherwise: A·u # B t; s (1#A)·B; if zero(s,σ) then zero True; {U/(1#A)·_U#B} o σ else zero = False end if end case end predicate

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25 November 2005 Foundations of Logic and Constraint Programming 15 Boolean Unification Procedure: Example {U/(1#A)·_U#B} o σ Notice that substitution returned by predicate zero is obtained through the composition of substitutions {U/(1#A)·U#B} and σ obtained from the recursive call of the predicate. Example: Satisfy constraint X # X·Z = Y·Z # 1 Unify X#X·Z and Y·Z#1 zero X#X·Z#Y·Z#1 = (1#Z)·X#Y·Z#1 % A x ·X#B x zero (1#(1#Z))·(Y·Z#1) % (1#A x )·B x = Z·(Y·Z#1) = Z·Y#Z % A y ·Y#B y zero (1#Z)·Z %(1#A y )·B y = 0

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25 November 2005 Foundations of Logic and Constraint Programming 16 Boolean Unification Algorithm: Example Now let us analyse the unifier that is returned by the sequence of calls Unify X#X·Z and Y·Z#1 zero X#X·Z#Y·Z#1 = (1#Z)·X#Y·Z#1 zero (1#(1#Z))·(Y·Z#1)=Z·Y#Z %(1#A x )·B x zero (1#Z)·Z = 0 %(1#A y )·B y σ z = {} σ y = {Y/(1#A y )·_Y#B y } o σ z = {Y/(1#Z) ·_Y#Z} o {} = {Y/(1#Z) ·_Y#Z} σ x = {X/(1#A x )·_Y#B x } o σ y = {X/(1#1#Z)·_X # Y·Z#1} o {Y/(1#Z)·_Y#Z} = {X/ Z·_X #Y·Z#1} o {Y/(1#Z)·_Y#Z} = {X/ Z·_X #((1#Z)·_Y#Z)·Z#1, Y/(1#Z)·_Y#Z} = {X/ Z·_X #Z#1, Y/(1#Z)·_Y#Z} = {X/ Z·_X #Z#1, Y/(1#Z)·_Y#Z}

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25 November 2005 Foundations of Logic and Constraint Programming 17 Interpreting Unifiers Hence, constraint X#X·Z = Y·Z#1 is satisfiable, since the unification of terms X#X·Z and Y·Z#1 succeeds, returning the most general unifier = {X/Z·_X #Z#1, Y/(1#Z)·_Y#Z} This result can be confirmed: (X#X·Z) = (Z·_X#Z#1)#(Z·_X#Z#1)·Z = (Z·_X#Z#1)#Z·_X = Z#1 (Y·Z#1) = ((1#Z)·_Y#Z)·Z#1) = Z#1 Analysing the unifier we note that the two terms unify whenever Z=0 and X = 1 and Y is arbitrary ( Y =_Y ) Z=1 and Y = 1 and X is arbitrary ( X =_X ) The unifier has therefore the ground instances {X/1, Y/0, Z/0}, {X/1, Y/1, Z/0}, {X/0, Y/1, Z/1}, {X/1, Y/1, Z/1}.

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25 November 2005 Foundations of Logic and Constraint Programming 18 Digital Circuits Application A simple example: 1.Model circuit below with Boolean constraints 2.How does the output relates (symbolically) with the inputs. R 1 : Unifier 1 is obtained by solving R 1 C = 1 # A·B 1 ={C / 1 # A·B} R 2 : Then 1 is applied to R 2 R2’: R2 1 :D = (1#A·C) {C/1#A·B} D = (1#A·(1#A·B)) D = 1 # A # A·B R 1 : C = 1 # A·B R 2 : D = 1 # A·C R 3 : E = 1 # B·C R 4 : F = 1 # D·E C R1R1 F R4R4 D R2R2 E R3R3 B A

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25 November 2005 Foundations of Logic and Constraint Programming 19 Digital Circuits Application Solving R2’: D = 1 # A # A·B unifier ’ 2 is obtained ’ 2 = {D / 1 # A # A·B} 2 is obtained by composition of 2 ’ with 1 2 = 1 o 2 ’ = {C/1#A·B} o {D/1#A#A·B} = {C/1#A·B, D/1#A#A·B} R 3 : Applying 2 to R 3, then R3’: R3 2 : E = (1#B·C) {D/1#A#A·B, C/1#A·B} : E = 1#B·(1#A·B) Solving R 3 ’, unifier 3 ’ is obtained 3 ’ = {E/1#B#A·B} Now, 3 ’ is composed with 2 3 = 2 o 3 ’ = {E/1#B#A·B} o {D/1#A#A·B} o {C/1#A·B} = {E/1#B#A·B, D/1#A#A·B, C/1#A·B}

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25 November 2005 Foundations of Logic and Constraint Programming 20 Digital Circuits Application R 4 : Finally, before solving R 4, 3 is applied R 4 ’: R 4 3 : F = (1#D·E) {E/1#B#A·B, D/1#A#A·B, C/1#A·B} : F = 1#(1#A#A·B)·(1#B#A·B) : F = 1#1#B#A·B#A#A·B#A·B#A·B#A·B#A·B : F = A # B Now, R 4 ’ is solved and unifier 4 ’ is obtained 4 ’ = {F/A#B} Composing 4 ’ with 3, it is obtained 4, a unifier for all the constraints 4 = 3 o 4 ’ = {E/1#B#A·B, D/1#A#A·B, C/1#A·B}o{F/A#B} = {E/1#B#A·B, D/1#A#A·B, C/1#A·B, F/A#B} By interpreting 4 we may conclude that this circuit with 4 nand gates implements the exclusive-or, since F / A # B.

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25 November 2005 Foundations of Logic and Constraint Programming 21 CLP(B) in SICStus Once loaded SICStus Prolog the module on Boolean Constraints is loaded with directive :- use_module(library(clpb)). Boolean Constraints may be specified with built-in predicate sat(E), where equality is specified as ‘=:=’ and several Boolean operators can be used, namely C = or(A,B) specified asC =:= A+B C = nor(A,B) specified asC =\= A+B C = xor(A,B) specified asC =\= A#B C = and(A,B) specified asC =:= A*B C = nand(A,B) specified asC =\= A*B C = not(A) specified asC =\= A

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25 November 2005 Foundations of Logic and Constraint Programming 22 CLP(B) in SICStus Some Examples: 1.?- sat(A#B=:=F). sat(A =:= B#F)? % Constraint A#B = F is satisfiable with unifier A / B+F 2.?- sat(A*B=:=1#C*D). sat(A=\=C*D#B)? % Constraint A·B=1+C·D is satisfiable with unifier A/1+C·D+B 3.?- sat(A#B=:=1#C*D), sat(C#D=:=B). sat(B=:=C#D), sat(A=\=C*D#C#D) ? % Constraints A+B=1+C·D and B=C+D are satisfiable with unifier {A/1+C·D+C+D, B/C+D}

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25 November 2005 Foundations of Logic and Constraint Programming 23 CLP(B) in SICStus A more complete example, related to the previous circuit | ?- circuit(A,B,[C,D,E],1). C = 1, E = A, sat(B=\=A), sat(D=\=A) ? Note:Compare this answer with unifier 5 = {E/1+B, D/B, C/1, F/1, A/1+B } found before. C G1G1 F G4G4 D G2G2 E G3G3 B A :- use_module(library(clpb)). nand_gate(X,Y,Z):- sat(X*Y =:= 1#Z). circuit(A,B,[C,D,E],F):- nand_gate(A,B,C), nand_gate(A,C,D), nand_gate(B,C,E), nand_gate(D,E,F).

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25 November 2005 Foundations of Logic and Constraint Programming 24 CLP(Q) Application Another domain for which a complete solver exists is that of linear constraints over the rationals or reals. The integration of such a solver with LP forms the CLP(Q) or CLP(R) instance of the CLP(X) scheme. CLP(Q) and CLP(R) differ in their approach handle arithmetics: If the coeficients are rational (division between integers) the solutiona are also rational. Hence CLP(Q) avoids rounding errors by handling properly with arithmetics (“+”,” –”, “*”, “/”) over rationals (with arithmetics without rounding CLP(R) rounds rationals as floating point numbers, and is more “efficient” if less “precise” These constraints are quite common in applications. Even when constraints are not linear, problems are often “linearised” (i.e. modelled with linear constraints such taht the errors are acceptable), given the very efficient way of solving these constraints, based on the SIMPLEX algorithm.

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25 November 2005 Foundations of Logic and Constraint Programming 25 CLP(Q) : Example Contraints: % Minimum required flow x 6, z 10 % minimum flow % flow capacity of arcs a 5, b 3, c 7, d 2, e 8, f 6 % processing capacity of nodes x 7, y 9, a + b 9, c 8, a + b + d 9, e + f 13 % flow maintenance x = a, y = b + c, a + b + d = e, c = d + f, e + f = z % non-negative constraints x, y,z, a, b, c, d, e, f /a 2/d 8/e 6/f 3/b 7/c x y z The following problem illustrates a typical use of linear constraints: Problem (Network Management): Find the value of traffic from points x,y to z such that the traffic does not exceed the flow capacity of each arc, nor the processing capacity of each node.

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25 November 2005 Foundations of Logic and Constraint Programming 26 CLP(Q) Solver: SIMPLEX The SIMPLEX algorithm aims at converting all linear constraints into a solved form, composed of equality constraints on linear expressions. This conversion is done in two steps. 1.All inequality constraints on decision variables are converted into equality constraints with the addition of “slack” (non-negative) variables a X a n X n ≤ b a X a n X n + S = b a X a n X n ≥ b a X a n X n - S = b 2.In general m constraints on n decision variables are converted into m constraints on k+n non-negative variables, where k m ( k = m if all constraints are inequality constraints). 3.A solved form is obtained if a partition can be found on the k+n variables in m basic variables X i (i 1..n) and p = k+n-m non basic variables Y j (j 1.. p), such that the m constraints can be written as X i = d i + c i1 Y c ip Y p ( for all i 1..n) where all free coefficients d i are non-negative.

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25 November 2005 Foundations of Logic and Constraint Programming 27 SIMPLEX: Geometric Interpretation X2 = 0 X2 X1X1 X3 = 0 X5 = 0 X1 = 0 X4 = 0 X3 = X1 - X2 X4 = -3 + X1 + X2 X5 = 5 + X1 - X2 x1 = 1 - x3/3 + x5/3 x2 = 6 - x3/3 - 2 x5/3 x4 = x3/3 - X5/3 X1 = 5 - X3 - X4 X2 = -2 + X3 + 2 X4 X5 = X3 - 3 X4 X2 = 5 + X1 - X5 X3 = X1 + X5 X4 = X1 - X5 2 X1 + X2 ≤ 8 X1 + X2 ≥ 3 X1 - X2 ≥ -5 X1, X2 ≥ 0 2 X1 + X2 + X3 = 8 X1 + X2 - X4 = 3 X1 - X2 - X5 = -5

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25 November 2005 Foundations of Logic and Constraint Programming 28 SICStus CLP(Q)/CLP(R) Solvers SICStus Prolog supports constraint solving over the Rational/Real numbers. As for Booleans, to handle these constraints, a module for CLP(Q) or CLP(R) must be loaded, with directives: :- use_module(user:library(clpq)). :- use_module(user:library(clpr)). Contraints may now be specified with the usual Prolog syntax but inside { }. For example the previous problem may be directly specified as p1([X1,X2]) :- {2*X1 + X2 = = 3, X1 - X2 >= -5 }, { X1 >= 0, X2 >= 0 }. or specified with explicit slack variables p2([X1,X2,X3,X4,X5]) :- {2*X1+X2+X3 = 8, X1+X2–X4 = 3, X1-X2–X5 = -5 }, { X1 >= 0, X2 >= 0, X3 >= 0, X4 >= 0, X5 >= 0 }.

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25 November 2005 Foundations of Logic and Constraint Programming 29 SICStus CLP(Q)/CLP(R) Solvers Once defined these programs can be run. Answers projected to the called variables, with some extra variables included when necessary p1([X1,X2]) :- {2*X1 + X2 = = 3, X1 - X2 >= -5 }, { X1 >= 0, X2 >= 0 }. | ?- p1([X1,X2]). {X1+1/2*X2=<4}, {X1>=0}, {X2>=0}, {X1+X2>=3}, {X1-X2>= -5} ? X2 >= 0 X2X2 2 X1 + X2 = 8 X1 - X2 = -5 X1 >= 0 X1 + X2 = 3 X1

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25 November 2005 Foundations of Logic and Constraint Programming 30 SICStus CLP(Q)/CLP(R) Solvers The solved form is more easily (?) identified with the explicit slack variables. p2([X1,X2,X3,X4,X5]) :- {2*X1+X2+X3 = 8, X1+X2–X4 = 3, X1-X2–X5 = -5 }, { X1 >= 0, X2 >= 0, X3 >= 0, X4 >= 0, X5 >= 0 }. | ?- p2([X1,X2,X3,X4,X5]). {X5=9-3/2*X2-1/2*X3}, {X4=1+1/2*X2-1/2*X3}, {X1=4-1/2*X2-1/2*X3}, {X2+X3=<8}, {X2+1/3*X3=<6}, {X2>=0}, {X3>=0}, {X2-X3>= -2} ? Solved form identifies X1 = 4, X2 = 0 X2 >= 0 X2X2 2 X1 + X2 = 8 X1 - X2 = -5 X1 >= 0 X1 + X2 = 3 X1

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25 November 2005 Foundations of Logic and Constraint Programming 31 SICStus CLP(Q)/CLP(R) Solvers Of course, if the problem has no solutions the answer is no! p3([X1,X2]) :- {2*X1 + X2 >= 8, X1 + X2 =< 3, X1-X2 =< -5 }, { X1 >= 0, X2 >= 0}. | ?- p3([X1,X2]). no ? X2 >= 0 X2X2 2 X1 + X2 = 8 X1 - X2 = -5 X1 >= 0 X1 + X2 = 3 X1

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25 November 2005 Foundations of Logic and Constraint Programming 32 SICStus CLP(Q)/CLP(R): Example Example: A certain amount of money is lent at some interest rate for a number of time periods, to be payed with constant instalments. Model Assuming that I is the number of time periods remaining to pay the debt D i represents the debt at the begining of time period i. R is the interest rate (in percentage points) P is the (constant) instalment to be payed at the end of each time period then we have the following relations: The debt at the begining of a period is the debt at the begining of the previous period, increased by the interest, less the payment made. D i = D i+1 * (1 +R/100) – P for i > 1 No debt should exist at the end of the last time period last time period 0 = D 1 * (1 +R/100) – P

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25 November 2005 Foundations of Logic and Constraint Programming 33 SICStus CLP(Q)/CLP(R): Example These constraints can be easily expressed in CLP(Q) / CLP(R) as shown in the programs below. Notice that such program could not be easily converted to pure Prolog, in statements like V is Exp, since one does not know what values are given and what are know. % 0 = D1 * (1 +R/100) – P debt(Debt, Rate, TimeLeft, Payment):- { TimeLeft = 1, Payment = Debt * (1 + Rate/100) }. % Di = Di+1 * (1 +R/100) – P for i > 1 debt(Debt, Rate, TimeLeft, Payment):- { TimeLeft > 1, NextTimeLeft = TimeLeft - 1, NextDebt = Debt * (1 + Rate/100) – Payment }, debt(NextDebt, Rate, NextTimeLeft, Payment).

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25 November 2005 Foundations of Logic and Constraint Programming 34 SICStus CLP(Q)/CLP(R): Example Some interaction with the program What yearly payment is due when 100 units are lent for 7 years at a 5% interest rate. What is the total payment? | ?- debt(100,5,7,P), {T = 7* P}. P = , T = ? What amount of money can be borrowed by someone that is willing to pay a fixed instalment of 10 units for 7 years, whne the interest rate is 5%? | ?- debt(D,5,7,10). D = ? What is the relationship between the yearly payment and the initial debt, when 7 years are used and the interest rate is 5%? | ?- debt(D,5,7,P). {P= *D} ?

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25 November 2005 Foundations of Logic and Constraint Programming 35 SICStus CLP(Q)/CLP(R): Example It is important to notice that CLP(Q)/CLP(R) is only complete when the constraints are linear. When they are not they are frozen in the constraint store, pending further instantiation What is the relation between the yearly payment and the initial debt, when 7 years are used and the interest rate is R% | ?- debt(D,R,7,10). clpr:{10.0-D-0.01*(R*D)+_A=0.0},% 7 years to go clpr:{ *(_B*R)-_B=0.0},% 6 years to go clpr:{10.0+_C-0.01*(_A*R)-_A=0.0},% 5 years to go clpr:{10.0+_D-0.01*(_C*R)-_C=0.0},% 4 years to go clpr:{10.0+_E-0.01*(_D*R)-_D=0.0},% 3 years to go clpr:{10.0+_F-0.01*(_E*R)-_E=0.0},% 2 years to go clpr:{10.0+_B-0.01*(_F*R)-_F=0.0} ?% 1 years to go

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