Optimal Control of process

Transcription

1 VYSOKÁ ŠKOLA BÁŇSKÁ TECHNICKÁ UNIVERZITA OSTRAVA FAKULTA METALURGIE A MATERIÁLOVÉHO INŽENÝRSTVÍ Optimal Control o process Study Support Milan Heger Ostrava 8

2 Title: Optimal Control o process Code: Author: Milan Heger Edition: irst, 8 Number o pages: 9 Academic materials or the N39 Economics and Management o Industrial Systems study programme at the Faculty o Metallurgy and Materials Engineering. Prooreading has not been perormed. Eecution: VŠB - Technical University o Ostrava

3 STUDY INSTRUCTIONS Optimal Control o process You have received a study package containing an integrated university mimeographed or the combined study o the course o Optimal Control Theory or the nd semester o the ield o study o 39T4 - Automation and Computer Technology in Industrial Technologies, which also contains the study instructions. Prerequisites Completing the courses o Automatic Control I and II o bachelor's study is recommended to study this subject, but it is not a prerequisite. Course objective and learning outputs The course objective is to become acquainted with the issue o optimal control and optimization methods in general. Ater studying this course, students should be able to: Knowledge outputs: Students will know the basic terms and relations o optimal control theory. Students will know the analytical and numerical methods o one-dimensional and multidimensional static optimization and the classical etreme regulation theory. Student will know the principles o linear programming and dynamic optimization. Skills outputs: Students will be able to classiy and apply the individual methods o optimal control theory in practice. Students will be able to design procedures to optimize the control o the individual technological units. Students will be able to calculate the etremes o unctions and unctionals during the solution o optimization tasks dealing with the control o technological units. Who is the course intended or The course is in the master s study o the ield o 39T4 - Automation and Computer Technology in Industrial Technologies o the program o study o N39 - Economics and Management o Industrial Systems, but it can be studied also by applicants rom any other ield o study i they meet the required prerequisites. 3

4 The study support is divided into parts, chapters, which logically divide the studied matter but are not equally comprehensive. The estimated study time o a chapter may vary considerably, which is why large chapters are urther divided into numbered sub-chapters and they correspond to the structure described below. We recommend the ollowing procedure to be used to study each charter: Careully read the theoretical part o the chapter. Use a computer to immediately try all, even i only partial eamples. Create all the programs that are in the assignment o the tasks to solve and try to modiy them in a creative way. Method o communication with the educators: More detailed instructions, as well as tasks, programs and projects will be assigned by the teacher at the beginning o direct contact teaching. The results will be checked according to teacher s instructions. Consultations with the teacher can be arranged with the teacher directly during lectures or by , which can be ound in VŠB-TU Ostrava contacts. The ollowing icons can serve or your orientation in the tet: Study time: hours At the beginning o the chapter, there is time necessary to study the subject matter. The time is approimate and can serve as a rough guideline or the study arrangement o the whole course or chapter. The time spent on each chapter highly dependent on the number o eamples you will solve independently on a computer and on the depth o their elaboration. Objective: Ater studying this paragraph, you will be able to describe... deine... solve... First, you will become amiliar with the objectives you should achieve ater reading the chapter. They are speciic skills, knowledge and practical eperience that you acquire by studying the chapter. 4

5 Eplication It is ollowed by the Eplication o the studied matter, introduction o new terms, their eplanations, where all is accompanied by igures, tables, eamples and links to tutorials with animations. Summary o terms Important passages and terms you are supposed to learn are briely repeated at the end o the chapter. Questions There are several theoretical as well as practical questions to check that you have mastered the subject matter well and thoroughly. Task to solve Because most theoretical terms o this course are o immediate practical importance and usage, there are practical tasks to solve presented to you at the end. The main purpose o the course, the ability to apply the newly acquired knowledge in solving real-world situations, can be ound in them. Connection with the educator The lecturers and teachers o eercises are ready to consult the studied issues with the students o combined study by , which is easy to ind in the VŠB-TU Ostrava contacts section. 5

6 The author o this tutorial wishes you a successul and pleasant study o this tetbook M. Heger Milan Heger Content. ONE-DIMENSIONAL STATIC OPTIMIZATION Introduction Analytical methods o static optimization Numerical methods o static optimization.... MULTIDIMENSIONAL STATIC OPTIMIZATION Introduction Multi-dimensional static optimization task without restriction Multidimensional static optimization tasks with a restriction in the orm o equality Multidimensional static optimization tasks with a restriction in the orm o inequality LINEAR PROGRAMMING The theory o linear programming Production programming Nutritional problem Distribution problem Cutting plan DYNAMIC OPTIMIZATION Dynamic optimization theory Calculus o variations Dynamic programming Minimum (maimum) principle

7 . One-dimensional static optimization.. Introduction Study time:.5 hour Objective Ater studying this paragraph, you will be able to deine the term o one-dimensional static optimization describe the issue o the creation o objective unctions and limitations make a graphic solution o the issue o static systems optimization Eplication In technical practice, we oten come across the request or process optimization, which can be described mathematically by means o a generally nonlinear unction () o one variable. Optimization in this case means inding the etreme (E) o unction () in the interval o easible solutions (closed interval), a ; b, where "a" is the let interval boundary (the lowest value), and "b" is the right interval boundary (the highest value). An etreme must be understood as the maimum or minimum o unction (), which is known as an objective unction. According to Weierstrass theorem, every unction in a closed interval must always attain at least one minimum and one maimum. There are always local constrained etremes appearing on the borders o interval and i there is an etreme inside the interval, we talk about a local ree etreme. Global maimum is the highest o the local maima and, on the contrary, global minimum is to the lowest o the local minima. Constrained etremes are easy to ind, they are always the terminal points o the easible solutions interval. The task o optimization, however, is to ind the ree etremes, which can be located at any point o the easible solutions interval. 7

8 Locating ree local etremes The ollowing igures show three dierent unctions having three qualitatively dierent minima. The derivations o the unctions are shown in the igures as well.,8,6,4, - -, -,4 -,6 -,8 - () '() () '() () '() In the irst case, the irst derivation value o the independent variable *, i.e. the minimum point, is zero. In two ollowing cases, derivation at point * doesn t eists. You could thereore say that the minimum is ound at the point where the irst derivation o the objective unction equals zero or does not eist. The ollowing igures, however, also include points, where the irst derivation is zero or does not eist, but it is visible here that the minimum is not present at these points. This means that only the necessary condition (NC) o the eistence o an etreme has been met. I there is an etreme at the given point, the necessary condition has been met, but i the NC has been met at the given point, an etreme does not have to necessarily be present at this point. 8

9 () '() () '() () '() Locating points where the necessary condition has been met is the irst step in the search or objective unction etremes. That is how we acquire the so-called suspicious points. The second step is the veriication whether the identiied suspicious point is the minimum, maimum, or is not an etreme at all. That is where the suiciency conditions (SC) come to their use. The rule here is that i the suiciency condition has been met, the given point * contains an etreme (SC E). The solution o algebraic epressions can take advantage o analytical methods. In such case, we obtain an absolutely accurate solution. Where the application o an analytical solution is not possible, we use a numerical solution. The accuracy o the solution here is limited by the parameters speciied by us (e.g. to one-hundredth). Some special cases also use graphicalnumerical methods or eperimental methods. Summary o terms o the chapter (sub-chapter) objective unction, easible solutions interval, etreme, maimum, minimum, constrained etreme, ree etreme, local etreme, global maimum, global minimum, necessary conditions (NC), suiciency conditions (SC), suspicious point. Questions to the presented subject matter. What does one-dimensional static optimization mean?. What does an objective unction with a ree etreme look like? 3. How do we solve an optimization task o one-dimensional static optimization? 9

10 .. Analytical methods o static optimization Study time: hour Objective Ater studying this paragraph, you will be able to deine the term analytical one-dimensional static optimization describe the issue o the identiication o suspicious points ind an analytic solution o the tasks o one-dimensional static optimization Eplication I a derivation o an objective unction is an algebraic epression, we can use analytic calculation methods to locate the objective unction etreme. Calculation procedure: calculate the irst derivation o the objective unction, identiy the suspicious points, where the irst derivation o the objective unction is zero or does not eist (NC), check in which o the suspicious points there is an etreme (SC). The veriication o suicient conditions can be based on three approaches (methods):. Comparison o unction values on the let ([*] -) and right ([*] +) rom the suspicious point. I [*] -> (*) and simultaneously [*] +> (*), then there is a local minimum present at the given point. I the inequalities are reversed, then there is a local maimum in the given point.. Comparison o polarity o unction derivations on the let ([*] -) and right ([*] +) rom the suspicious point. I [*] - < and simultaneously [*] +>, then there is a local minimum present at the given point. I the inequalities are reversed, then there is a local maimum, or strong local maimum, in the given point. 3. A method o higher derivations, when the objective unction is gradually derived and the value o the suspicious point n ( * ) is introduced into these derivations. The

11 procedure is repeated until the moment when the n-th derivation in the suspicious point is non-zero n(*). I "n" is an odd number, then it is not an etreme. I it is an even one, then it depends on the polarity o the n-th derivation. I the n-th derivation is positive, then there is a local minimum present in the suspicious point, in opposite case, there is a local maimum. Tasks to solve Eample.. Calculate the maimum o objective unction () = 3-5 on the easible solutions interval (uncertainty interval): ; Solution We make the irst derivation o objective unction () = 3-5: () = 3-5 make it equal zero: 3( * ) - 5 = ( * ) = 5/3 the suspicious point: * 5, and the objective unction value at this point: ( * ) -4, According to Weierstrass theorem, other suspicious points are the terminal points o the uncertainty interval, i.e.: * = ( * ) = 3 * = (3 * ) = - where local minima or maima will certainly be present. Now, we will veriy whether the suspicious point *, is a local minimum:. We compare the unctional values on the let and right o this point. Because the closest suspicious points are the end points o the uncertainty interval, we can use them to compare the unctional values. Both are higher than the unction value at point. This means that it is a local and global minimum at the same time. *

12 () (). I we calculate the irst derivation at the terminal points -5 7, the derivation gradually changes rom negative to positive value. It means that it is the local and global minimum at the same time. 3. To use the method o higher derivation, we irst calculate the second derivation: 6 and or suspicious point je >. Because the irst non-zero derivation at the suspicious point is even and positive, there is a strong local minimum at the given point, which is also the global minimum at the same time. () * ( * ) Summary o terms o the chapter (sub-chapter) At the end o the chapter, there is a list o the terms that have been discussed in the tet and that you must know. The method o comparison o unction values, the method o comparison o the irst derivation sign, the method o higher derivations. Questions to the presented subject matter 4. What is the necessary condition o the eistence o an etreme and how is it calculated? 5. Which methods o suiciency conditions veriication do you know? 6. How do we obtain an analytical solution o one-dimensional static optimization?.3. Numerical methods o static optimization Study time: hours Objective Ater studying this paragraph, you will be able to deine the term numerical one-dimensional static optimization describe the issue o numerical identiication o suspicious points numerically solve the tasks o one-dimensional static optimization

13 Eplication Unless the derivation o the objective unction is an algebraic epression, we can use the numerical calculation methods to locate the objective unction etreme. Numerical methods can be divided into dierential and direct. Dierential ones require the knowledge o at least the irst derivation, while direct ones, on the contrary use only the knowledge o the unction values. The dierential methods include: Bolzano's method, Newton's method, secant method. Direct methods are urther divided into interpolation and comparative. Quadratic interpolation method is used as the interpolation one. Comparative methods can be divided into passive and adaptive. The representative o the passive methods is the uniorm comparative method, while the well-known adaptive methods include the golden section method and the Fibonacci method. The principle o numerical methods is based on the gradual iteration cycle o approimation to the real solution o the task. The iteration cycle is terminated when the desired solution accuracy has been achieved. Accuracy is mostly measured in a deined dierence between the last (n-th) solution and the actual solution, which is, o course, unknown: n * * ;, where ε is an arbitrarily small number. That is why the condition to achieve the given accuracy is determined by the relation n - n- < ε. Some methods are based on a gradual reduction o the uncertainty interval a ; n b n, which must always contain the point *. In this case, the calculation is terminated by the condition o bn - an < ε. All procedures will be eplained on the task dealing with the determination o a minimum. The maimizing task will be converted to minimization by means o a simple adjustment: * arg ma arg min Bolzano's method (dichotomy method, interval bisection method) 3

14 This method belongs to dierential methods. The solution is based on the knowledge o the irst derivation o the objective unction, which is used to decide on the net solution procedure in the net step. The method is based on the gradual shortening o the uncertainty interval Ii = a i ; b i so that it still contains the point where the searched etreme is still present. The ratio o reduction o the length o two consecutive intervals li:li+ is :. It can be concluded that: lim I n n lim n n * l The procedure o inding a minimum o objective unction by means o Bolzano s method We deine the desired calculation accuracy by using the value o ε. In the irst step, we select the uncertainty interval in compliance with the task assignment a = a b = b and we calculate its mid-point: = (a + b)/ a = a * ( ) ( ) a ; b b = b I the irst interval length is: l = b - a a = a b = l = l/ I we calculate the irst derivation at point : () I the value o derivation () >, then the minimum must be located to the let o this point, and we thereore select the ollowing uncertainty interval rom point a to point, thus obtaining the deault parameters o the interval or the second step. I the value o derivation () <=, then the minimum must be located to the right o this point, and we thereore select the ollowing uncertainty interval rom point to point b. We proceed in a similar manner in the ollowing steps. The calculation is terminated when the ollowing condition is met: bn - an <= ε 4

15 and the result is: * = n ± ε Other suitable numerical methods o one-dimensional optimization can be ound on the study portal o VŠB-TU Ostrava [ Tasks to solve Eample.. Use the accuracy o ε =.3 to calculate the minimum o objective unction ()= 3-5 on the easible solution interval (uncertainty interval): ; Solution We create the irst derivation or the objective unction () = 3-5: () = 3 5 step : a = a = b = b = and we calculate its mid-point (): = (a + b)/ = (+)/ = the irst interval length is: l = b - a = - = Since the condition to terminate the calculation has not been met ( bn - an <= ε ), we will continue with the net step. We calculate the irst derivation at point : () = 3 5 () = 3() 5 = - < The value o the irst derivation at point is a negative number, which is why the second interval I is chosen on the right rom point. I ; b step : 5

16 a = = b = b = and we calculate its mid-point (): = (a + b)/ = (+)/ =,5 the irst interval length is: l = b a = - = Since the condition to terminate the calculation has not been met ( bn - an <= ε ), we will continue with the net step. We calculate the irst derivation at point : () = 3 5 () = 3(,5) 5 =,75 > The value o the irst derivation at point is a positive number, which is why the third interval I3 is chosen on the let rom point. I3 a ; step 3: a3 = a = b3 = =,5 and we calculate its mid-point (3): 3 = (a3 + b3)/ = (+,5)/ =,5 the irst interval length is: l3 = b3 a3 =,5- =,5 Since the condition to terminate the calculation has been met ( bn - an <= ε ), we terminate the calculation: * = n ± ε * =,5 ±,3 6

17 Summary o terms o the chapter (sub-chapter) At the end o the chapter, there is a list o the terms that have been discussed in the tet and that you must know. Dierential and direct methods, interpolation and comparative methods, passive and adaptive methods, Bolzano's method, accuracy ε. Questions to the presented subject matter 7. What does numerical method calculation accuracy mean? 8. What numerical methods do you know? 9. How do we acquire a numerical solution o one-dimensional static optimization? 7

18 . Multidimensional static optimization.. Introduction Study time:,5 hours Objective Ater studying this paragraph, you will be able to deine the term multidimensional static optimization describe the issue o division o multidimensional o static optimization tasks Eplication Multidimensional static optimization is a task deined in n-dimensional Euclidean space R n. Independent variable is a vector (indicated in bold) consisting o n components: Objective unction is an n-dimensional unction (), where the set o easible solutions can be represented by the entire n-dimensional Euclidean space X = R n, then we talk about tasks without restrictions, and i the set o the easible solutions is a subset o the Euclidean space X R n, deined by restrictive unctions, we talk about tasks with restrictions that may be divided into two types. According to the shape o the restrictive unctions, the tasks are divided into two types: task with restrictions in the orm o equality restrictive unctions with the orm o gj() = bj, where j m 8

19 task with restrictions in the orm o inequality restrictive unctions with the orm o, e.g. gj() bj, where j m The task o optimization can be epressed similarly to one-dimensional optimization, using the ollowing entry: * arg min X Summary o terms o the chapter (sub-chapter) At the end o the chapter, there is a list o the terms that have been discussed in the tet and that you must know. Euclidean space, tasks without restriction, tasks with restriction in the orm o equality, tasks with restriction in the orm o inequality. Questions to the presented subject matter. What does linear Euclidean space mean?. What are tasks without restrictions, tasks with restrictions in the orm o equity and restriction in the orm o inequality?.. Multi-dimensional static optimization task without restriction Study time: 5 hours Objective Ater studying this paragraph, you will be able to deine the term optimization multi-dimensional tasks without restrictions describe the issue o solving multi-dimensional tasks without restrictions illustrate the task o optimization multi-dimensional tasks without restrictions in Ecel 9

20 Eplication The procedure used to ind the etremes o multidimensional static optimization involves the ollowing steps. First, we calculate the irst partial derivation o objective unction and create a unction gradient: T n,...,, The stationary point (suspicious point) is a point in which the gradient value is zero. A necessary condition or the eistence o an etreme is that all partial derivations are equal to zero. NC o the st order: i = To ind out whether the suspicious points contain some type o etreme or not, we will use Hessian matri o second partial derivations, which has the ollowing orm: n n n n H Ater calculating the individual second partial derivations o Hessian matri and substitution o the suspicious point value, we receive the ollowing matri:

21 n n n n H Subsequently, we will eamine the deiniteness o Hessian matri that is crucial or determining whether there is a minimum, a maimum, or a saddle point without an etreme at the suspicious point. The deiniteness o the matri H() (at the stationary point ) is determined by means o Sylvester s criterion using the ollowing procedure: We mark the main minors (subdeterminants) o the matri using symbols H, H,, Hn, then Hessian matri is positively deinite H()>, when all its main corner subdeterminants are positive. Hessian matri is positively semideinite H(), when at least one main corner subdeterminant is zero Hi =, while all its main subdeterminants are non-negative. I the matri H() is positively (semi-) deinite, then matri - H() is negatively (semi-) deinite and vice versa. I the matri is neither positively (semi-) deinite nor negatively (semi-) deinite, it is known as indeinite. I matri H() >, there is strong local minimum at the suspicious point. I matri H(), there is non-strong local minimum at the suspicious point. I matri H() <, there is strong local maimum at the suspicious point.

22 I matri H(), there is non-strong local maimum at the suspicious point. I matri H() is indeinite, there is no etreme at the suspicious point. Tasks to solve Eample.. We calculate the etreme o a multidimensional objective unction or the ollowing assignment: R Solution We calculate the irst partial derivations o objective unction: and The stationary points: and then the stationary point: = [;] We continue by the calculation o the Hessian matri: H in this case, the stationary point has the same Hessian matri: H We determine the deiniteness o the matri according to Sylvester s criterion:

23 H = > and H = 4 > Hessian matri is positively deinite H() > and, at point * = [;] there is strong local minimum.,5,5-,5,,8 -,5,5- -,5 - -,8 -,6 -,4 -,,,4,6,8 - -,4 Eample.. We calculate the etreme o a multidimensional objective unction or the ollowing assignment: ( ) R Solution We calculate the irst partial derivations o objective unction: and The stationary points: Then the stationary point: = [;] and We continue by the calculation o the Hessian matri: 3

24 - -,8 -,6 -,4 -,,,4,6,8 H We determine the deiniteness o the matri according to Sylvester s criterion: H = - < and H = 4 > Matri is not positively deinite, and that is why we create a negatively ormed matri by multiplying all the members o the original matri minus one, and the stationary point in this case has the ollowing Hessian matri: H We determine the deiniteness o the matri H according to Sylvester s criterion: H = > and H = 4 > Hessian matri is positively deinite H >, which is why the original matri negatively deinite and there is strong local maimum at point * = = [;]. H is -,5 - -,4,,8 -, ,5 -,5-- -,5 ---,5-4

25 5 Eample.. Calculate the etreme o a multidimensional objective unction or the ollowing assignment: R Solution We calculate the irst partial derivations o objective unction: and The stationary points: and Then the stationary point: = [;] We continue by the calculation o the Hessian matri: H We determine the deiniteness o the matri according to Sylvester s criterion: H = - < and H = -4 < The matri is not positively deinite, and that is why we create negatively ormed matri by multiplying all the members o the original matri minus one, and the stationary point in this case has the ollowing Hessian matri: H We determine the deiniteness o the matri H according to Sylvester s criterion:

26 H = > a H = -4 < is not positively deinite, which is why the original matri cannot be negatively deinite and there is no etreme at point * = = [;], but there is a saddle point. Hessian matri H H,5 -,5 - -,8 -,6 -,4 -,,,4,6,8 - -,4,,8,5- -,5 -,5- ---,5 - Eample.. Calculate the etreme o a multidimensional objective unction or the ollowing assignment: 4 4 R Solution We calculate the irst partial derivations o objective unction: 3 4 a 3 4 The stationary points: and Then the stationary point: = [;] We continue by calculation o the Hessian matri: 6

27 - -,8 -,6 -,4 -,,,4,6,8 H In this case, there is the ollowing Hessian matri or the stationary point: H We determine the deiniteness o the matri according to Sylvester s criterion: H = and H = Hessian matri at point * = [;] is semi-deinite, which is why we have to monitor the matri deiniteness in the surrounding o the suspicious point = * ± ε ( = ± ε as well as = ± ε) The power epression i suggests that in the surroundings o point = * ± ε, there will always be positive values, which is why the Hessian matri is positively semi-deinite H() and at point * = [;], there is non-strong local minimum.,5,5-,5,,8 -,5,5- -,5 -,4 - The illustration o the chart took advantage o the surace type, deined on the ollowing table ragment: 7

28 B C D E F G H - -,9 -,8 -,7 -,6 3 - =($B3^4+C$^4) =($B3^4+D$^4) =($B3^4+E$^4) =($B3^4+F$^4) =($B3^4+G$^4) 4 -,9 =($B4^4+C$^4) =($B4^4+D$^4) =($B4^4+E$^4) =($B4^4+F$^4) =($B4^4+G$^4) 5 -,8 =($B5^4+C$^4) =($B5^4+D$^4) =($B5^4+E$^4) =($B5^4+F$^4) =($B5^4+G$^4) 6 -,7 =($B6^4+C$^4) =($B6^4+D$^4) =($B6^4+E$^4) =($B6^4+F$^4) =($B6^4+G$^4) 7 -,6 =($B7^4+C$^4) =($B7^4+D$^4) =($B7^4+E$^4) =($B7^4+F$^4) =($B7^4+G$^4) 8 -,5 =($B8^4+C$^4) =($B8^4+D$^4) =($B8^4+E$^4) =($B8^4+F$^4) =($B8^4+G$^4) 9 -,4 =($B9^4+C$^4) =($B9^4+D$^4) =($B9^4+E$^4) =($B9^4+F$^4) =($B9^4+G$^4) Summary o terms o the chapter (sub-chapter) At the end o the chapter, there is a list o the terms that have been discussed in the tet and that you must know. Gradient, Hessian matri, matri deiniteness, resolver. Questions to the presented subject matter. What does unction gradient mean? 3. What is the Hessian matri? 4. How do we solve multidimensional tasks without restrictions?.3. Multidimensional static optimization tasks with a restriction in the orm o equality Study time: 5 hours Objective Ater studying this paragraph, you will be able to deine the term optimization multidimensional tasks with a restriction in the orm o equality describe the issue o the solution o multidimensional tasks with a restriction in the orm o equality solve multidimensional tasks with a restriction in the orm o equality in Ecel. 8

29 Eplication Optimization tasks with a restriction in the orm o equality (these are conventional tasks involving a constrained etreme) can be epressed as ollows: where * arg min X X is a subset o R n deined by the m-dimensional set o restricting unctions in the orm o general nonlinear equations: gj() = bj, while j m The solution must comply with the restrictive conditions while optimizing the objective unction. The relationship between the dimension o the task "n" and the number o restrictive unctions "m" indicates the degree o task reedom, and it is deined as a = n - m. An objective unction is optimized only i s>, which means, that the number o restrictive unctions "m" is smaller than the dimension o task "n". In case that s =, the number o restrictive unctions "m" is the same as the dimension o task "n". The solution we come to is a solution o a system o restrictive unctions not dependent on the orm o the objective unction. In the event that a <, the number o restrictive unctions "m" is higher than the dimension o task "n". This eample generally doesn t have any solution. I s =, it is possible to substitute the epressions arising by adjusting the restrictions instead o n- unknown quantities in the objective unction, thus converting this task into a onedimensional one. Generally, however, the solution o an optimization task with constraints in the orm o equality usually requires the creation o the so-called Lagrangian unction, which converts the given task to "n + m" dimensional task without restrictions, which we are already capable o solving. 9

30 The Lagrangian unction has the ollowing orm: m L(,, n,p,p, pm) = (,, n,p,p, pm) + g,,..., p j j n j The necessary conditions o the irst order can be acquired rom the conditions o stationarity o the Lagrangian unction: * * * * * * L,,..., n, p, p,..., pm = this means that all partial derivations o the Lagrangian unction according to the individual i and pj will be equal to zero: L i L p * * * g,,..., b j j n j where pj variable is known as the Lagrangian multiplier. This way, analogous to the optimization o the tasks without restrictions, we obtain the stationary points (suspicious points) and, subsequently, according to the deiniteness o the Hessian matri, calculated or the original objective unction (), we determine the type o etreme. The determination o the type o etremes is then the same as in case o tasks without restrictions, which we already know rom the previous chapter. b j Eample.. Tasks to solve We calculate an etreme o multidimensional task with restriction in the orm o equality or the ollowing assignment: with a restriction unction in the orm o: 3

31 Solution We create a Lagrangian unction in the orm o: L(,,p) = p calculate all partial derivations o the Lagrangian unction according to the individual i and p and make them equal zero: * * * p L * * * p L * L p * * * * g,,..., b n By solving the presented set o three equations with three unknowns we come to the solution: * = [ ; ] Now, we use the Hessian matri to determine the deiniteness o the original objective unction (). H, which is positively deinite at all points, and that is why there is a strong local minimum at the suspicious point * = [; ]. However, compared to the minimum rom the tasks without restriction, this one is shited in such a way to lie on the curve horizontal projection deined by relation, which is based on the shape o the restriction unction, while being located at the lowest position o the objective unction (). Now, the task with the same assignment will be calculated in Ecel: For this purpose, we use two cells or the values o and (C4: D4), we name them or the sake o clarity in cells (C3: D3) and cell B4 will be indicated by the tet "setting o the values o i ". In cell C5, we create a ormula corresponding to objective unction (), and cell B5 will be indicated by the tet "the value o objective unction (). 3

32 In cell C6, we create a ormula corresponding to the let side o the restriction unction g(), and cell B6 will be indicated by the tet "let side o restriction unction g()j. In cell C7, we type in a constant corresponding to the right side o the restriction unction g() and cell B7 will be indicated by the tet 'let side o restriction unction g()j. B C D 3 4 setting the values o i 5 value o objective unction () =C4*C4+D4*D4 6 let side o restriction unction g() j =D4-C4^ 7 right side o restriction unction b j We use a solver to solve the given task. Its setting can be seen in the dialog window copy: The solution outcome is clear rom the ollowing table: setting the values o i value o objective unction () let side o restriction unction g() j right side o restriction unction b j It clearly shows that the solution matches the solution by means o Lagrangian unction * = [; ]. The minimum value o the objective unction () =. The last two lines o the table show that the restrictive condition has been met as well, since the value on the let and right sides o the restriction unctions are identical. 3

33 Summary o terms o the chapter (sub-chapter) At the end o the chapter, there is a list o the terms that have been discussed in the tet and that you must know. Lagrange multiplier, Lagrangian unction, type o restriction, solver. Questions to the presented subject matter 5. What does Lagrange multiplier mean? 6. What is the Lagrangian unction? 7. How do we solve a task o multidimensional optimization with a restriction in the orm o equality?.4. Multidimensional static optimization tasks with a restriction in the orm o inequality Study time: 5 hours Objective Ater studying this paragraph, you will be able to deine the term optimization multidimensional tasks with a restriction in the orm o inequality describe the issue o the solution o multidimensional tasks with a restriction in the orm o inequality solve multidimensional tasks with a restriction in the orm o inequality in Ecel. Eplication Optimization tasks with a restriction in the orm o inequality (these are unconventional tasks involving a constrained etreme) can be epressed as ollows: where * arg min X 33

34 X - is a subset o R n deined by the m-dimensional set o restricting unctions in the orm o general nonlinear equations: gj() bj, while j m we can easily change the reverse inequality into a suitable type by multiplying it by -: -gj() -bj, The solution must comply with the restrictive conditions while optimizing the objective unction. The relationship between the dimension o the task "n" and the number o restrictive unctions "m" in case o tasks with a restriction in the orm o inequality is not restricted in any way, i.e. the number o restrictive unctions m in this case can be even higher than the dimension o the task "n". Generally, however, to solve an optimization task with restrictions in the orm o inequality, we create the so-called Lagrangian unction, which also converts the task into n + m - dimensional task without restrictions. The Lagrangian unction has the ollowing orm: m L(,, n,p,p, pm) = (,, n,p,p, pm) + g,,..., p j j n j To solve a task with a restriction in the orm o inequality, we use the so-called Kuhn - Tucker conditions: L i L p j * * * * * *,,..., n, p, p,..., pm * * * * * *,...,, p, p p,...,, n m * * * * * * * i L,,..., n, p, p,..., pm = i p * * * * * * * L,,..., n, p, p,..., pm = i p * i j b j 34

35 * p i Calculation procedure We calculate all solutions o the equations o Kuhn - Tucker conditions (gradually considering zero solutions o i and pj as well). We veriy or which solutions the inequalities o Kuhn - Tucker conditions and the conditions o non-negativity are met. Now, we choose the solutions or which the objective unction has a minimum. Eample.. Tasks to solve We calculate an etreme o multidimensional task with restriction in the orm o inequality or the ollowing assignment: with a restriction unction in the orm o: Solution We create a Lagrangian unction in the orm o: L(,,p) = p calculate all partial derivations o the Lagrangian unction according to the individual i and p and make the relations or Kuhn-Tucker conditions: * * * p L * * * p L * 35

36 L p * * * * g,,..., b L n * * * * p * L * * * p * p * * * L p * * * * g,,..., b p * n By solving the presented set o three equations with three unknowns, we come to the solution: * = [ ; ] Summary o terms o the chapter (sub-chapter) At the end o the chapter, there is a list o the terms that have been discussed in the tet and that you must know. Constrained etreme, Kuhn Tucker conditions, solver. Questions to the presented subject matter 8. What does the term Kuhn Tucker conditions mean? 9. How do we solve a task o multidimensional optimization with a restriction in the orm o inequality? 3. Linear programming 3.. The theory o linear programming 36

37 Study time: 5 hours Objective Ater studying this paragraph, you will be able to deine the term linear programming describe the issue o the creation o objective unction and restrictions graphically solve the issue o optimization o static linear systems Eplication Linear programming typically belongs to multidimensional optimization tasks, generally with a restriction in the orm o inequality, where both the objective unction and the restrictions consist o linear unctions. The objective unction is called linear and has the orm o: where z = c.+c.+. cn.n, ci are invariables, i are independent variables, n Euclidean space dimension in which the task is being solved (number o variables). According to the physical nature o the task, we look or the linear orm minimum or maimum so as to meet the restrictive conditions. The restrictions are in the orm o linear inequalities, where the number o restrictions can be even higher than the dimension o the task and they can, or eample, have the orm o: a.+a.+. an.n > l a.+a.+. an.n l am.+am.+. amn.n < lm It is usually typical or the solvability o the tasks that, in case o minimizing, the restrictions are in the orm o and, in case o maimizing, they are in the orm o. The conditions o 37

38 non-negativity o variables are typical as well, because in practice, they involve, or eample, the numbers o products, the numbers o technological operations or the weight quantities o materials. All ollow the rule saying that: i. Two-dimensional linear programming tasks are easy to eplain using a graphical solution. This means that the task contains only two inputs and, and the number o restrictions can be arbitrary. The conditions o non-negativity are deined by the irst quadrant as a set o easible solutions. I the restrictions use equalities instead o inequalities, we acquire a set o lines that are easy to draw. Inequalities, however, mean that the easible solutions always lie in one hal-plane determined by each o the lines drawn in this way. Which o the hal-planes it is can be easily determined i we enter the beginning o coordinates into the inequalities, i.e. point [,]. I the inequality is met, we can choose the hal-plane in which the beginning o the coordinates is located. The set o easible solutions is the intersection o all such graphically created sets. I the linear orm z =, we obtain a line again, this time passing through the beginning o the coordinates. I it is a maimization task, we look or a parallel to this line, which lies in the easible solutions area and is the urthest rom the original line at the same time (see the ollowing igure). In our case, the result would be only, which is dierent rom zero. * p z // z z = p Figure: Graphical solution o linear programming Tasks to solve Eample.. 38

39 Draw a graphical solution o the ollowing assignment: p: 5 + p:,5 +5 z = + minimum Solution 5 p * p z // z 4 z = solution o equations: results in: 5 + =,5 +5 = =,5 =,5 Summary o terms o the chapter (sub-chapter) At the end o the chapter, there is a list o the terms that have been discussed in the tet and that you must know. Linear programming, restricting conditions, linear orm, polyedr. Questions to the presented subject matter. What does linear programming mean?. What is linear orm?. How do we acquire a graphical solution o simple linear programming tasks? 3.. Production programming Study time: 5 hours Objective Ater studying this paragraph, you will be able to deine the term production programming describe the issue o creating restrictions and linear orms or production programming solve the production programming tasks in Ecel 39

40 Eplication The individual typical linear programming tasks will be presented here using elementary eamples, which can include other applications with the same mathematical basis. An enterprise manuactures two products successively on two machines; each product V and V consumes certain time o machines S and S. Both machines have a limited number o hours they are able to provide during production throughout a year. The products are sold at prices c and c. The ollowing table shows the concrete values. V V Limit S a a l S a a l price c c z You can use the table to create a mathematical model that includes the conditions o nonnegativity, restrictions and objective unction. The objective unction is called a linear orm and it is: z = c.+c. we are looking or the maimum o the linear orm so as to meet the restrictive conditions: The restrictions are in the orm o linear inequalities, where the number o restrictions can be even higher than the dimension o the task and may, or eample, have the orm o: a.+a. l a.+a. l The non-negativity conditions mean that all : and It is suitable to use a "Simple algorithm" or the solution, but or the sake o simplicity the eamples will be solved in Ecel. 4

41 The result will be in the orm o * = [ *, * ]. The interpretation o the result is as ollows - we will produce * o V products and * o V products. Eample.. Tasks to solve Calculate the solution o production programming or the ollowing assignment: V V Limit S 4 5 S 6 3 Price 5 ZISK Solution We create a mathematical restriction model: We create a mathematical linear orm model: z =.+5. And the conditions o non-negativity: a We create an Ecel spreadsheet and then we use the Solver supplement to make the actual optimization: B C D E F 3 V V Limit 4 S 4 5 4

42 5 S price 5 PROFIT V V Limit Reality S =C$*C4 =D$*D4 =SUM(C:D) 3 S =C$*C5 =D$*D5 =SUM(C3:D3) 4 price =C$*C6 =D$*D6 =SUM(C4:D4) Solver setting options: In setting options, we set Linear model and Non-negative numbers. Solver setting: Set a cell this identiies the cell in which there is the ormula or the calculation o turnover (proit) Set Ma or the maimization task (proit) Adjusted cells we identiy the cells in which the independent i variable values are. Restricting conditions - creates relations between the limit and reality cells (machine time). 4

43 The given task will be solved ater pressing the Solve button. The outcome is a table: A B C D E F 9 4 V V Limit Reality S 3 S 4 Price The result has the ollowing interpretation: We will produce only 4 pcs o product and the turnover will come to Kč. Summary o terms o the chapter (sub-chapter) At the end o the chapter, there is a list o the terms that have been discussed in the tet and that you must know. Production programming, type o restriction, type o linear orm, solver. Questions to the presented subject matter 43

44 3. What does production programming mean? 4. What is a solver? 5. How do we solve the production programming tasks? 3.3. Nutritional problem Study time: 5 hours Objective Ater studying this paragraph, you will be able to deine the term nutritional problem describe the issue o creating a restriction and a linear orm or the nutritional problem solve the nutritional problem tasks in Ecel Eplication Nutritional (blending) problem represents a dual task o production programming. Dual means that the mathematical model is dual, meaning that the inequalities are reversed and the linear orm is minimized. We will also use a practical eample here to eplain the principles and procedure. An enterprise purchases three raw materials S, S and S3. The enterprise obtains three elements P, P and P3 rom them. Production needs to ensure at least l units o P element and, gradually, at least l3 units o element P3. The raw materials are bought at prices c to c3. The ollowing table shows the common values. S S S3 Limit P a a a 3 l P a a a 3 l P3 a 3 a 3 a 33 l 3 Price c c c 3 z 44

45 We can create a mathematical model, including the non-negativity conditions, restrictions and objective unction, according to the table. The objective unction is called a linear orm and it is: z = c.+c. +c3.3 we are looking or the maimum o the linear orm so as to meet the restricting conditions: The restrictions are in the orm o linear inequalities, where the number o restrictions can be even higher than the dimension o the task and may, or eample, have the orm o: a.+a.+a3.3 l a.+a.+a3.3 l a3.+a3.+a33.3 l3 The non-negativity conditions mean that all : a and 3 It is suitable to use "Dual Simple algorithm" or the solution, but or the sake o simplicity the eamples will also be solved in Ecel. The result will be in the orm o * = [ *, *, 3 * ]. The interpretation o the result is as ollows - we will purchase * o S raw material and * o S raw material and 3 * o S3 raw material. Eample.. Tasks to solve Calculate the solution o a blending problem or the ollowing assignment: S S S3 Limit 45

46 P P 3 5 P3 6 3 price 5 5 PROFIT Solution We create a mathematical restriction model: We create a mathematical linear orm model: z = And the non-negativity conditions: a a 3 We create an Ecel spreadsheet and then we use Solver supplement to make the actual optimization: B C D E F G 3 S S S3 Limit 4 P P P price 5 5 PROFIT =C3 =D3 =E3 Limit Reality 3 =B4 =C$*C4 =D$*D4 =E$*E4 =F4 =SUM(C3:E3) 4 =B5 =C$*C5 =D$*D5 =E$*E5 =F5 =SUM(C4:E4) 5 =B6 =C$*C6 =D$*D6 =E$*E6 =F6 =SUM(C5:E5) 46

47 6 price =C$*C7 =D$*D7 =E$*E7 =SUM(C6:E6) Solver setting options: In setting options, we set Linear model and Non-negative numbers. Solver setting: Set a cell this identiies the cell in which there is the ormula or the calculation o costs Set Min or the cost minimization task Adjusted cells we identiy the cells in which the independent i variable values are. Restricting conditions - creates relations between limit and reality cells. 47

48 The given task will be solved ater pressing the Solve button. The outcome is a table: 3 3,89 38,89 S S S3 Limit Reality P 55,56 94,4 5 5 P 4,67 94,4 36 P3 83,33 6,7 price 77,8 97, 5 The interpretation o the result is as ollows: We will purchase only 3.89 units o raw material S and units o raw material S, the costs will be 5 Kč. Summary o terms o the chapter (sub-chapter) At the end o the chapter, there is a list o the terms that have been discussed in the tet and that you must know. Nutritional problem, model, type o restriction, type o linear orm, solver. 48

49 Questions to the presented subject matter 6. What does nutritional problem mean? 7. What does a model o nutritional problem look like? 8. How do we solve a nutritional problem task? 3.4. Distribution problem Study time: 5 hours Objective Ater studying this paragraph, you will be able to deine the term distribution problem describe the issue o creating a restriction and a linear orm or the distribution problem solve the distribution problem task in Ecel Eplication A distribution (transportation) problem is slightly more comple, especially in terms o the size o the task. The inequalities are to be represented in both variants here and the linear orm will be minimized. We will also use a practical eample to eplain the principles and procedures. In three warehouses o three suppliers D, D and D3, there are certain amounts o available material ld, ld, ld3, which is intended or distribution. On the other side, there are three customers O, O and O3 who need the material in the amounts o lo, lo, lo3. The transport o a unit o material rom the i-th suppliers to the j-th customer represents a cost at the price o cij (c to c33). The make the task solvable, we must keep the condition o ΣlDi = ΣlOj. The ollowing table shows the common values. 49

50 O O O3 Limit D i D c c c 3 l D D c c c 3 l D D3 c 3 c 3 c 33 l D3 limit O j l O l O l O3 PROFIT We can create a mathematical model, including the non-negativity conditions, restrictions and objective unction, according to the table. The objective unction is called a linear orm and it is: z = c.+c. + + cij.ij+ + c33.33 we are looking or the minimum o the linear orm so as to meet the restricting conditions: The restrictions are in the orm o linear inequalities, where the number o restrictions can be even higher than the dimension o the task and may, or eample, have the orm o: in case o the supplier: ++3 ld ++3 ld ld3 in case o the customer: ++3 lo ++3 lo lo3 The non-negativity conditions mean that all : ij It is suitable to use "Vogel s method" or the solution, but or the sake o simplicity the eamples will also be solved in Ecel. 5

51 The result will be in the orm o * = [ *, *,, 33 * ]. The interpretation o the result is as ollows - we will transport ij * units o material rom the i-th supplier to the j-th customer. Eample.. Tasks to solve Calculate a distribution problem solution or the ollowing assignment: O O O3 Limit D i D D D3 6 3 limit O j 5 PROFIT Solution We create a mathematical restriction model: in case o the supplier: in case o the customer: We create a mathematical linear orm model: z = And the conditions o non-negativity: a a 3 5

52 We create an Ecel spreadsheet and then we use Solver supplement to make the actual optimization: A B C D E F G ij cij.ij 3 6 =C4*C =D4*D =E4*E 7 =C5*C =D5*D =E5*E 8 3 =C6*C3 =D6*D3 =E6*E3 9 =C3 =D3 =E3 Limit Reality =B4 =C =D =E =F4 =SUM(C:E) =B5 =C =D =E =F5 =SUM(C:E) 3 =B6 =C3 =D3 =E3 =F6 =SUM(C3:E3) 4 Limit =C7 =D7 =E7 =SUM(C6:E8) 5 Reality =SUM(C:C3) =SUM(D:D3) =SUM(E:E3) Solver setting options: In setting options, we set Linear model and Non-negative numbers. 5

53 Solver setting: Set a cell this identiies the cell in which there is the ormula or the calculation o costs Set Min or the cost minimization task Adjusted cells we identiy the cells in which the independent i variable values are. Restricting conditions - creates relations between limit and reality cells. The given task will be solved ater pressing the Solve button. The outcome is a table: A B C D E F G ij cij.ij O O O3 Limit Reality D D 3 D3 4 Limit Reality 5 53

54 The interpretation o the results is: The ollowing transactions will be eecuted: D O: D O: D O: D3 O: 5 units units units units The costs o all transactions will amount to 5 Kč. Everybody has been satisied according to the requirements. Summary o terms o the chapter (sub-chapter) At the end o the chapter, there is a list o the terms that have been discussed in the tet and that you must know. Transportation problems, model, type o restriction, type o linear orm, solver. Questions to the presented subject matter 9. What does transportation problem mean? 3. What does a transportation problem model look like? 3. How do we solve a transportation problem task? 3.5. Cutting plan Study time: 3 hours Objective Ater studying this paragraph, you will be able to deine the term cutting plan describe the issue o creating a restriction and a linear orm or the cutting plan solve the cutting plan tasks in Ecel 54

55 Eplication Cutting plan is the most comple area o linear programming. The compleity does not lie in the compleity o the actual optimization, but in the necessity to create a generator o cutting plan set. Its programming is not an easy task and, in Ecel, this would mean using programming in MS Visual Basic. We will use a practical eample to eplain the principles and procedures as well. The company has a steel pipe with the length o m. The production, however, requires pipes T, T and T3, with the lengths o.6 m,.4 m and.3 m, which must be cut by the company. The production needs to ensure at least lunits o T element and up to at least l3 units o element T3. The goal o optimization is to minimize the amount o waste cuttings Z to Zn in meters. The irst task will be to generate all the possible options or cutting the initial two-meter pipes by creating the so-called cutting plans P to Pn: The ollowing table shows the common values. P P P3 P4 P5 P6 P7 P8 P9 P P Limit T a, a, a,3 a,4 a,5 a,6 a,7 a,8 a,9 a, a,n l T a, a, a,3 a,4 a,5 a,6 a,7 a,8 a,9 a, a,n l T3 a3, a3, a3,3 a3,4 a3,5 a3,6 a3,7 a3,8 a3,9 a3, a3,n l3 rest Z Z Z3 Z4 Z5 Z6 Z7 Z8 Z9 Z Zn waste We can create a mathematical model, including the non-negativity conditions, restrictions and objective unction, according to the table. The objective unction is called a linear orm and it is: z = z.+z. +z zn.n we are looking or the maimum o the linear orm so as to meet the restricting conditions: The restrictions are in the orm o linear inequalities, where the number o restrictions can be even higher than the dimension o the task and may, or eample, have the orm o: 55

56 a. +a. +a a n. n l a. +a. +a a n. n l a 3. +a 3. +a a 3n. n l 3. a m. +a m. +a m a mn. n l m The non-negativity conditions mean that all : and to n It is suitable to use "Dual Simple algorithm" or the solution, but or the sake o simplicity the eamples will also be solved in Ecel. The result will be in the orm o * = [ *, *,, n * ]. The interpretation o the result is as ollows - we will eecute * times the cutting plan P and * times P up to n * times Pn. Eample.. Tasks to solve Calculate a cutting plan solution or the ollowing assignment: P P P3 P4 P5 P6 P7 P8 P9 P P P Limit T 3 5 T T rest,,,,,,,, WASTE Solution We create a mathematical restriction model: We create a mathematical linear orm model: z =,. +. +, ,. 56

57 And the non-negativity conditions: a to We create an Ecel spreadsheet and then we use the Solver supplement to make the actual optimization: B C D E F G H I J K L M 3 P P P3 P4 P5 P6 P7 P8 P9 P P 4 T 3 5 T T rest =- C4*$R4- C5*$R5- C6*$R6 =- D4*$R4- D5*$R5- D6*$R6 =- E4*$R4- E5*$R5- E6*$R6 =- F4*$R4- F5*$R5- F6*$R6 =-G4*$R4- G5*$R5- G6*$R6 =- H4*$R4- H5*$R5- H6*$R6 =- I4*$R4- I5*$R5- I6*$R6 =- J4*$R4- J5*$R5- J6*$R6 =- K4*$R4- K5*$R5- K6*$R6 =- L4*$R4- L5*$R5- L6*$R6 =-M M6* =C3 =D3 =E3 3 =B4 =C$*C4 =D$*D4 =E$*E4 =F$*F4 =G$*G4 =H$*H4 =I$*I4 =J$*J4 =K$*K4 =L$*L4 =M$ 4 =B5 =C$*C5 =D$*D5 =E$*E5 =F$*F5 =G$*G5 =H$*H5 =I$*I5 =J$*J5 =K$*K5 =L$*L5 =M$ 5 =B6 =C$*C6 =D$*D6 =E$*E6 =F$*F6 =G$*G6 =H$*H6 =I$*I6 =J$*J6 =K$*K6 =L$*L6 =M$ 6 REST =C$*C7 =D$*D7 =E$*E7 =F$*F7 =G$*G7 =H$*H7 =I$*I7 =J$*J7 =K$*K7 =L$*L7 =M$ Solver setting options: In setting options, we set Linear model and Non-negative numbers. 57

58 Solver setting: Set a cell this identiies the cell in which there is the ormula or the calculation o costs Set Min or the cost minimization task Adjusted cells we identiy the cells in which the independent i variable values are. Restricting conditions - creates relations between limit and reality cells. The given task will be solved ater pressing the Solve button. The outcome is a table: P P P3 P4 P5 P6 P7 P8 P9 P P P Limit Reality T T 5 35 T3 rest The interpretation o the results is: We will carry out only two cutting plans: P will be repeated 5 times and P will be repeated 5 times. This way, there will be no waste, but the unilateral restrictions lead to the act that we will have a lot more T pipes than we need or production. By introducing restrictions or the limit o T = and using integral programming, we obtain the ollowing solutions: 58

59 P P P3 P4 P5 P6 P7 P8 P9 P P P Limit Reality T T T3 rest 6,8 6,8 We will repeat P 84 times and P will be repeated 5 times. This will, however, produce waste in the amount o 84 pipes with the length o. m, but only two 6 cm pipes T. Summary o terms o the chapter (sub-chapter) At the end o the chapter, there is a list o the terms that have been discussed in the tet and that you must know. Cutting plan, model, cutting plan generator, type o restriction, type o linear orm, solver. Questions to the presented subject matter 3. What does a cutting plan mean? 33. How are the individual cutting plans generated? 34. What does a cutting plan model look like? 35. How do we solve a cutting plan task? 59

60 4. Dynamic optimization 4.. Dynamic optimization theory Study time: 5 hours Objective Ater studying this paragraph, you will be able to deine the term dynamic optimization describe the issue o creating objective unctionals solve the issue o dynamic system optimization Eplication A controlled system is described by a dynamic state equation in a vector orm: t t, u t, t and some o the ollowing parameters are entered: (t) initial state t initial time (t) inal state t inal time Any dierential equation o the system can be rewritten using a state-space representation or this type o dierential equations system o the irst order. It is necessary to deine the criterion o optimality i the control is to be optimal. This criterion will generally take the orm o an objective unctional: J t I t, ut, tdt Ft, t t o It clearly shows that the objective unctional can be generally dependent on the state o the system, the control, time and unction o the inal parameters F. Deinition o optimal control task It is a minimization, which can be written in the orm o: 6

61 min t t u t I U, u, tdt F t, t the system is described by equations in the orm o: t t, u t, t when t t (t) (t ) * (t) (t ) t t Fig. 4. Etended state space The situation can be illustrated in (n + ) dimensional etended state space (see Fig. 4.). Ω - set o easible trajectories. The task o optimal control is to select such a trajectory rom the easible state trajectories that minimizes the objective unctional "J". Basic tasks There will generally be three mutually equivalent tasks rom the point o view o the objective unctional orm: ) Bolza problem: J t u t It, u t, tdt Ft t ) Lagrange problem: J t u t It, u t t 3) Mayer problem: u t F J t,t, tdt, t Lagrange problem is the most requently used one in practice. 6

62 The best possible control circuit properties can be chosen rom a larger number o objective unctionals, however, those shown here are the most requent. Types o objective unctionals ) Sub-integral unction I, then: u t t J dt (Lagrange problem) t whose solution will lead to: J u t t t (Mayer problem) The control deviation has to be removed in the shortest possible time - we talk about time-optimal task. This is the most common task in practice, and there is a signiicant actuating variable limiting actor the actuator energy potential: t u U ut present during the optimization resulting rom ) Sub-integral unction actual state: I w t t I depends on the dierence between the desired state and the Here, we want state to optimally ollow the desired state w, which is epressed by the square o their dierences, which should be minimal. 3) Sub-integral unction I u t I depends on the control square: Here, we want the generalized energy, necessary to control the transition o the system rom the initial state to the inal one, to be minimal. 4) Sub-integral unction depends on the control: I u t 6

63 Here, we want to minimize the amount o media (e.g. gas) entering the system. Other criteria include, e.g. reliability, eiciency etc. Solution methods The most common methods used to solve these optimization tasks can be divided into three groups: ) Calculus o variations ) Dynamic programming 3) The principle o minimum (maimum) Tasks to solve Eample.. Draw a state trajectory o a pendulum. Solution (t) (t) Summary o terms o the chapter (sub-chapter) At the end o the chapter, there is a list o the terms that have been discussed in the tet and that you must know. State trajectory, control trajectory, state description, objective unctional. Questions to the presented subject matter 36. What does a state trajectory mean? 37. What is an objective unctional? 38. How do we get a state-space representation? 63