Automation in Complex Systems MIE090

Transcription

1 Automation in Complex Systems MIE090 Exam Tuesday May 31, 2016 You may bring the course book and the reprints (defined in the course requirements), but not the solution to problems or your own solutions to simulation tasks and home works or other personal notes. A calculator is also permitted (memory cleared). You may answer in Swedish or in English. Grading: There are 30 points all together. The following grades will apply: Grade 3: at least 15 points Grade 4: at least 20 points Grade 5: at least 25 points Note that all answers should be complete and well motivated. Your line of thought should be easy to follow and hand calculations should be provided for all mathematical problems. Corrections will be completed not later than Tuesday June 20, Good Luck! Problem 1 (2 points) Consider a large process with dynamics acting on a wide range of time scales with many different measurements and controllers. The fact that the process dynamics act in many different time scales is sometimes advantageous but in other cases a difficulty. Discuss the main problems associated with a wide range of time constants and how they can be solved with regard to modelling and simulation of the process. : Many different time scales means a stiff system. Models of stiff system are often slow to simulate. This can be overcome by using stiff solvers. If this is not feasible (e.g. the system uses noise, highly dynamic inputs, event-based controllers, etc., which have a negative impact of stiff solvers), rewriting the system into a DAE will improve the simulation speed significantly (although this model representation is an approximation of the original system) and allows the use of an explicit solver. Problem 2 (4 points) A machine has insufficient production rate in relation to the demand. Incoming jobs can be described as a Poisson process with an arrival rate of one job per 10 minutes (λ). The existing machine needs 12 minutes to complete a job (exponentially distributed). No input buffer size limitation exists. The service discipline is FIFO. There are two alternatives for improvement: i) to replace the existing machine with one new machine with double capacity; ii) to install a second machine (identical to the existing one) in parallel with the existing machine. The two machines would have one common queue. The main goal is to improve (reduce) the waiting time in the system. Which alternative is the best? Demonstrate by calculating:

2 a) the average queue lengths (1 p) b) the total waiting times in the systems. Motivate in words why there is a difference. (1 p) c) Assume we make two separate queues (of unlimited size) for alternative (ii), each with the arrival rate of λ/2. Will this result in a shorter or a longer waiting time compared to the case with one common queue? (1 p) d) How big is the probability to have more than 3 jobs in the queue in the alternative with one fast machine (alternative (i))? (1 p) : a) For the original system the arrival rate is 6 per hour and the production rate is 5 per hour (i.e. for the existing system the buffer size grows towards infinity). System (i) means increasing the production rate to 10 per hour in an M/M/1 system. For such a system L q = ρ 2 1 ρ = 0.62 = 0.9 items For system (ii) we have an M/M/2 system where L q = mm ρ m +1 m! ( 1 ρ) p 2 0 and p 0 = m m ρ m +1 m m! ( 1 ρ) + ( mρ) n 1. The utilization rate is 0.6 and p 0 is n! n= 0 therefore equal to This gives L q = items. So system (ii) has on an average fewer items in the queue but on the other hand it has two items in the machines whereas (i) only has one item in the machine. 1 b) The total waiting time for system (i) is W = µ λ = 1 = 0.25 hours and for system 10 6 (ii) W = W q + 1 µ = L q λ + 1 = /6 +1/5 = hours. The waiting time in the queue is µ actually shorter for (ii) compared to (i), hours versus 0.15 hours. But when the job goes into the machine in (ii) it does not help that much that there are two machines in parallel, the individual job must go through just one machine, and that machine is only working at half the speed of the machine in (i). The time in one machine in (ii) is 1/5 = 0.2 hours and in (i) 1/10 = 0.1 hours. Therefore the total waiting time in the system will be larger in (ii) than in (i). c) System (ii) is now modified into two independent M/M/1 systems working in parallel. W for each of the subsystems is then 1/(5-3) = 0.5 hours, i.e. the total waiting time in the system is increased by 60%. d) The probability to have more than 3 jobs in the queue is equal to the probability of having more than 4 jobs in the system (1 item is in the machine) = p 5 +p 6 +p 7 +p 8 = 1 p 0 p 1 p 2 p 3 p 4 = = where p 0 = 1 ρ and p n = ρ n (1 ρ). Problem 3 (2 points) Consider a system where three independent machine works in parallel, each with the average capacity 20 jobs per hour. The machine outputs are exponentially distributed. Likewise, jobs arrive to the machines stochastically (one common queue) with 48 jobs per hour at an average. Also the job arrival time is exponentially distributed. The service discipline is FIFO. Assume that a maximum of 3 jobs is allowed in the queue. Calculate the average waiting time in the queue (1 p) Calculate the rejection rate, in other words, the number of jobs per hour that are

3 never admitted to the system (1 p) : a) This is a traditional M/M/3/6 system. The utilization rate is 48/60 = 0.8. The probability vector is calculated as ( ) p 0 = mm ρ m+1 1 ρ K m m! ( 1 ρ) + ( p n = mρ ) n p 0 (n=1,2,3) n! p n = mm ρ n p 0 (n=4,5,6) m! m n=0 ( mρ) n n! 1 which gives (after some calculations) p = [ ]. In this case we only need p 0 and p 6. L q = mm ρ m +1 ( )( K m)ρ K m m! 1 ρ [ ] p 0 which gives L q = jobs. From ( ) 2 1 ρ K m 1 ρ here we calculate W q as W q = L q λ where λ = λ ( 1 p 6)= h -1 and W q = hours. The rejection rate is λ λ = jobs per hour. Problem 4 (3 points) Consider a system described as a birth-death process with an arrival rate of 10 jobs per hour and a production rate of 20 jobs per hour. What is the minimum number of states we need to represent the system if we can allow a maximum rejection probability of 10% on an average? Also provide the complete probability vector in stationarity for your selected number of states. : This is a classical birth-death process with x states. There can be x-1 jobs in the system, each having the probability p0, p1, p2, etc. respectively. The departure rate is all the time µ = 20 per hour. The arrival rate is all the time λ = 10 per hour. The following condition must always hold: p0 + p1 + p2 + p3 + p(x-1) = 1. The general steady state condition for this type of system is: k p = λ k + 1 pk which gives us µ k + 1 p1 = 10/20*p0; p2 = 10/20*p1 = (10/20) 2 *p0; p3 = (10/20) 3 *p0; etc. So the full probability vector as a function of p0 is: p = p0 * [ 1 10/20 (10/20) 2 ] = p0 * [ ]. If we normalize this vector (since p0 + p1 + p2 + p3 + = 1) we know the values within [ ] will become smaller than the values given above (since the sum of all the values is larger than 1) and we can conclude that certainly no more than 5 states will be required ( < 0.1). If we write the normalised p vector based on five states we get [ ]. But then we see that also p3 < 10% and it must therefore be enough to use less than 5 states to represent the system. Normalising the vector based on the first four states instead gives [

4 ]. p2 is now larger than 10% and will only become larger if we normalise the vector based on the first 3 states. Consequently we need four states to represent this birth-death process to meet the defined criteria in terms of rejection probability and the statioary probability vector is p = [ ]. Problem 5 (4 points) Numerical solvers are an important issue when simulating dynamic systems. a) We distinguish between explicit and implicit methods for numerical integration. Explain the advantages and disadvantages for each one of the principle methods. (1 p) b) One of the most popular explicit solver is the 4 th order Runge-Kutta method. Explain in words and in graphs in detail how this method calculates its next value x n+1 for a fixed h. (1 p) c) A new solver method is proposed as: x n +1 = x n + h f ( x n +1,t n +1) 18 f ( x n,t n ) + 9 f ( x n 1,t n 1 ) +15 f ( x n 2,t n 2 ) + 7 f ( x n 3,t n 3 ) ( ) What order is this solver? What type of solver is it? Does it work (motivate)? (1 p) d) A simple differential equation is given by dx(t)/dt = 5x(t) + 2. Simulate by hand calculations this equation forward in time using both Euler backward and forward approximations (5 time steps is enough). Use h = 1 and x(0) = 0. (1 p) : a) Explicit solver, e.g. Runge-Kutta, only take into account previous function values when determining in which direction to move for the next integration step. They are generally robust but sensitive to integration step length (too long steps may lead to instability). For stiff systems they may run into problems since they must adjust the integration step length to the fastest dynamics of the simulated system. On the other hand they have no major difficulties handling noise, highly dynamic inputs and hybrid systems (e.g. mix of time-driven and event-driven systems), which is often a problem for implicit solvers. Implicit solvers (e.g. Gear) take into account also future values x(n+1) when determining in which direction to move. Therefore at least one algebraic equation must be solved for each integration step, and this may require some kind of built in iterative solution method for non-linear systems (e.g. Newton-Raphson). On the other hand the implicit solver can take long integration steps to compensate for this extra computational burden. In particular all stiff solvers are based on implicit solvers. Implicit solvers sometimes use a built-in explicit solver to calculate a good initial value for the next iteration, so called predictor-corrector methods. The more sophisticated explicit AND implicit solvers used today are variable step-size solvers that adjust the integration step according to the selected tolerances and the system dynamics on-line. b) The equations for the 4 th order RK is given on page 302. It is a multi-step method although it only uses values between t and t+h. First the derivative f 1 (direction) is calculated in the starting point (x n,t n ), i.e. the result of the previous integration step. Secondly, RK calculates the derivative at the time point t n +h/2 for the x-value x n +f 1 h/2, i.e. it updates x in the direction previously calculated, and the result is a new direction f 2. RK repeats the same principle again but this time updates the value of x in the f 2 direction. The result is a new value of the derivate, f 3. In the final iteration RK calculates the derivative at time t n +h for the x-value x n +f 3 h, i.e. a value of the derivative for the final destination point of this integration step. Finally, the method weighs the four calculated derivatives (directions) together in a normalized manner (1/3+1/6+1/6+1/3=1), multiplies with the integration step h and adds the result to the

5 previous value of x n, which produces the new value of the state variable x n+1. The whole procedure is then repeated for every new integration step. The solution will benefit from a figure, plotting x as a function of t and demonstrating the different directions the RK method will move as it progresses. c) Fifth order implicit solver. Should NOT work since it is not properly normalized: =25 whereas the solver is normalised based on the value 61. Probably the solver should use a value of +18 rather than 18, which would yield the correct normalisation. As written now the solver will include a gain =25/61 in each integration step. d) Euler backwards > dx/dt (x(t) x(t-h))/h gives after rewriting: x(t) = (x(t-h) + 2h)/(1-5h). Using the values in the text: x(0)=0; x(1)= 0.5; x(2)=-0.375; x(3)= ; x(4)= ; x(5)= Euler forwards > dx/dt (x(t+h) x(t))/h gives after rewriting: x(t) = (1+5h) x(t-h) +2h. Using the values in the text: x(0)=0; x(1)=2; x(2)=14; x(3)=86; x(4)=518; x(5)=3110. The may appear Euler backwards is finding the correct solution whereas the Euler forward goes towards infinity as a result of using a too large integration step. But this is actually the correct solution. We can see the potential problem for the Euler backwards: if h=1/5 the denominator is zero and if h is smaller than 1/5 the expression can never produce negative numbers if we start in x(0)=0. The analytical solution to the equation is x(t) = 0.4 (e 5t 1), i.e. goes towards infinity. Problem 6 (1 points) Explain why many industrial systems are satisfactorily described by only a few independent variables (e.g. principal components) although many more measurement variables are generally available. Give at least two reasons! Redundant measurements for safety reasons, many variables are coupled (e.g. pressure in a pipe at different locations), the measured variables are often indirect measures of a few underlying main mechanisms driving the process and methods like PCA reshapes the system description to describe those more fundamental mechanisms. Problem 7 (2 points) Consider a 2nd order model of the rotation of a stiff cylinder, Newton s 2nd law. Two identical simulation modules are now connected by simply connecting the two stiff cylinders with a stiff axis. a) What is the order of the resulting dynamic system? (1 p) b) How would you handle this in Matlab/Simulink? (1 p) a) The order of the total system is 2 (both the angles and the angular velocities are the same. b) Matlab has no automatic elimination of the equations. This has to be done manually.

6 Problem 8 (2 points) To control the Swedish electric power grid is definitely a complex problem. Why is it so? You may use the classification and term suggested by Thomas Gillblad but your motivation why it becomes complex is crucial and as many reasons you can come up with should be included. The system itself is "Integrative complexity" according to the classification. Many simple devices connected together in a large system is difficult to handle. However, the uncertainty regarding consumption, and uncontrollable production (solar, wind) gives "Operational complexity" that also makes the overall control a difficult task. Faults, blackouts, partial shutdown and startup are scenarios that also should be handled. Problem 9 (1 points) Climate change is one of the most difficult challenges for mankind for the next decades. From an energy point of view, the global warming has several consequences: How is energy generation influenced by climate change? (i) Higher temperatures: lower energy costs for heating but higher energy costs for air condition and cooling. (ii) Many areas become more arid: lack of cooling water for thermal power plants. Also, higher temperatures mean too warm cooling water. (iii) Hydro plants: evaporation will increase. Lack of water in rivers leading to the hydro dam may also lower the level of the dams. This means less hydropower generated Problem 10 (5 points) a) Describe briefly the need for plant wide control although we have excellent tools from control theory. (1 p) b) In plant wide control there is no such thing as one single optimal control structure. Give at least three examples on different approaches to the optimization problem. (1 p) In designing a control system for a whole production plant, including the local control level, supervisory control level and optimization level, one often discusses the term degree of freedom or DOF. c) What do we mean by degree of freedom in this context? (1 p) d) Why must we distinguish between different types of degrees of freedom? (2 p)

7 a) In a large production system many processes are connected. These processes may operate with different timeconstants, different known and unknown input variations and individual constraints. The interconnection between the state variables in itself very often leads to a combinatorial explosion if everything should be included in one system description. This calls for an approach with a layered control structure like the one used in PWC. b) The optimization could be made for e.g. maximum production, minimal cost per unit or lowest power consumption. c) Number of independently manipulated variables equals the DOF. d) Some of the manipulated variables don t affect the steady state in the process (e.g. the level in a storage tank). Consequently, these may have impact on the dynamics of the process but not on the final production. Constraints have to be put on some DOF. E.g. even if a waste combustion mill has production capacity for a certain electric power capacity it is constrained by the amount of waste coming in. Problem 11 (2 points) A large section of a dairy (mejeri) plant has been redesigned with new machines. You are responsible for the implementation project of the HMI-part of the control room. (Hardware, software platform and application software). You are familiar with the new machines and their way of operating from previous installations. Give a brief plan for the steps you will take in the project until it can be handed over for daily use? A good checklist for the project could be the seven steps described in Hollifield et al although you are not expected to learn them by heart for full credit on this problem. Only the main features from the following list are required. 1. Adopt a High Performance HMI Philosophy and Style Guide Comment: Both the hardware and the software is continuously being improved and the performance is increased. Thus, it is important to stay updated on recent state of the art HMI and the research in the area. Based on this the style and philosophy should be chosen. 2. Assess and benchmark existing graphics against the HMI Philosophy Comment: Here it is of uttermost importance to get a good understanding of not only the current technical function but also the operators view of the current system. If the user can associate the right operation because of previous experience. This must of course not lead into a bad HMI design but in many cases there is no contradiction and only a win-win situation. In this step interviews or questionnaires with operators are valuable tools. 3. Determine specific performance and goal objectives for the control of the process, for all modes of operation Comment: Be sure to understand all operations that have to be done including startup, shutdown and maintenance. They should be included in the design. 4. Perform task analysis to determine the control manipulations needed to achieve the performance and goal objectives Comment: Usually a redesign and new machines are expected to result in higher performance and maybe new types of operation/tools. The HMI design should provide help to the operator to meet these expectations.

8 5. Design and build high performance graphics, using the design principles in the HMI Philosophy and elements from the style guide Comment: Now it is time to implement! 6. Install, commission, and provide training on the new HMI Comment: Training of the operators is often needed and it is important to remember the operator view from step 2 to develop this training program. 7. Control, maintain, and periodically reassess the HMI Comment: This point refers to the daily use and is not included in the current problem. Problem 12 (2 points) OPC servers are common in automation systems since the end of 1990:s. However, there have been major problems to continue the use of them. Describe the background for these problems and briefly describe the current solution to them. The original version of OPC was based on OLE (Object Linking and Embedding) which in its turn is based on COM and DCOM(Distributed Component Object Model). This framework was available in Microsoft Windows until the introduction of the.net and from Windows Vista COM was not included any more. To solve this OPC UA (Unified Architecture) was developed. In this new framework a Service Oriented Architecture (SOA) is used. That makes the definition insensitive to operating system platform.