Root-finding and optimisation
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1 Root-finding and optimisation Reference tets: Chapra, S., R. Canale. Numerical Methods for Engineers, 3rd ed., McGraw-Hill, New York, Parts 2 and 4. Root-finding Simple mathematical equations have eplicit solutions, meaning that the variable we need to calculate can be isolated on the left hand side of the equation. Many problems we face as engineers involve solution of implicit equations. For eample, in the following polynomial, if is known then y can be solved eplicitly, whereas if y is known, solving for is an implicit problem. y = There are various ways of solving an implicit equation. The most common is to consider it as a root-finding problem. The roots of a function are where that function equals zero. For eample, using the polynomial above, solve for y = 5. This is equivalent to solving f() = 0, where f() is defined as, f ( ) = If you plot f() against you will see there are 2 roots at approimately = -0.3 and = 4.7. There are a range of root-finding methods available. Here we introduce two methods using the same eample f()
2 Newton-Raphson This requires an initial estimate of the root (for our eample, we ll choose 1 = 0.75), and estimates the gradient of the function f() at this value of. The gradient is then projected to the intersection with the -ais, which gives the second estimate of the root, 2. A third estimate is achieved in the same way, etc, until the estimate converges to the true root, in our case, f() The advantage of this method is that it usually converges to an accurate estimate very quickly. However, there are a number of eamples where it takes a long time to converge or finds the wrong root (see Chapra, pages ). For eample, what would have happened if we had used 1 = 1? Also, in general, our problems are more comple than this polynomial and we do not want to have to plot out f() every time. Therefore we generally do not know whether more than one root eists, and we might not get the answer we are looking for. Also, some functions we use are discontinuous i.e. there are some values of where gradients do not eist, in which case Newton-Raphson should not be used. Bisection method First, upper and lower bounds of are chosen (u 1 and l 1 ), so that we know the root lies within these bounds (in practical problems this is usually easy, because there are usually physical constraints on the value of ). f() is calculated at these two values of and also at their mid-point m 1. In our eample below, we have chosen l 1 = -0.85, u 1 = 0 and m 1 = The three values of f() are used to identify the sub-interval within which f() changes from positive to negative. This sub-interval is used to define the
3 second estimates of l and u. In our eample, l 2 = and u 2 = 0, so m 2 becomes This process continues until the interval converges on the root. f() -1 l 1 m 1 u 1 =0 m 2 3rd interval 2nd interval initial interval The advantage of this method is that it avoids the need to calculate gradients, and will always find a root which is between the initial estimates l 1 and u 1 (whereas the Newton-Raphson method may find a root which is far away from 1 ). However, in most problems this bisection method is less efficient that Newton-Raphson. Chapra s book describes improved versions of both these methods of root-finding, and gives instructions for developing computer code. Root-finding using Ecel: Manning s equation eample For steady flow in a rectangular open channel, Manning's equation can be used to determine the flow velocity as a function of channel characteristics during uniform, steady flow: 1 2 / 3 1/ 2 U = R S (1) n where U = mean velocity [m/s]; n = Manning roughness coefficient ; R = hydraulic radius [m] = A c /P; A c = cross-sectional area [m 2 ]; B = width [m]; H = mean depth [m]; P =
4 wetted perimeter [m]; and S = slope. Note that A c and P are related to the more fundamental parameters B and H by A c = BH (2) P = B + 2H (3) P A c B S H Q, U The flow is equal U multiplied by A c, Q = UA c (4) where Q = flow rate [m 3 /s]. Given the channel characteristics (H, B, S and n), these two equations can be solved directly for flow and velocity. A more interesting problem involves determine the depth and velocity if Q, B, S and n are given. This can be illustrated by eliminating U by substituting Eq. (1) into Eq. (4) Q = S n 12 / 53 / ( BH) ( B+ 2H) 23 / (5) Although we have one equation with one unknown, it is impossible to solve eplicitly for H. There is no way to algebraically manipulate Eq. (5) so that H is isolated on one side of the equation. However, the depth can be determined by root-finding. This is done by subtracting Q from each side to yield, f ( H) = 12 / 53 / S n ( BH) ( B+ 2H) 23 / Q = 0 (6) The value of H that makes this equation equal to zero is the root. For eample, if Q = 5 m 3 /s, B = 20 m, n = 0.03 and S = , the equation is
5 53 / ( 20H) f ( H) = = 0 (7) 23 / ( H) If you didn t know how to find roots using numerical methods, you could obtain a solution by trial-and-error. That is, you could guess values of H until you determined a value that drove the function to zero. Although this is an inefficient approach, eventually you would determine H = m. The result can be checked by substitution into Eq. (7) to give, ( ) f ( H ) = ( ) 5 / 3 2 / 3 5 = which is very close to zero. Our other unknown, the velocity can be determined by substitution back into Eq. (4), Q U = BH = 5 20(0.702) = 0.36 m/s An easier way to find the root is to use Ecel s Goal-Seek (from the Tools menu). Goalseek is based on the Newton-Raphson method. Try this yourself. Set cell refer to the cell in which you have calculated the value of f(h). To value enter the target value of f(h) = 0. By changing cell refer to the value of H (which will also be refered to within your f(h) formula). Eercise: Set up the polynomial equation in Ecel, and find both roots using Goal-seek.
6 Optimisation Optimisation is one of the most important challenges in hydrology and environmental engineering. The term optimisation algorithm covers a wide range of numerical methods of finding an optimal (i.e. maimised or minimised) solution. A optimisation algorithm consists of at least 2 components: 1. a series of formulae that produce an output given inputs such as data and parameters; 2. a measure of how good the output is (called one of the following: objective function; criteria; goodness of fit; cost function). The aim is usually to identify the input values which produce the optimum (maimum or minimum) objective function value. There are many different methods of optimisation that range from simple automated trial and error to very comple algorithms. In this workshop we introduce 2 simple types and use the optimisation tool provided in Ecel. Before we start, some of you will realise that you have already done some optimisation - in the 2 nd Ecel workbook. There, you used regression to optimise the parameters of a pipe orifice equation. In that case, the objective function which you minimised was the errors between the observed and calculated orifice flow. Regression, however, is limited to fitting linear models to data we need to look at more general methods. Newton s method This is almost the same as the Newton-Raphson method which we used for root-finding. When a continuous function reaches a maimum or minimum, its first derivative becomes zero. For a univariate function (i.e. a function with only one variable input,, and objective function y), y = f () dy = f '( ) = 0 d So all we have to do in order to optimise y is locate the root of f (). To maimise y from the previous polynomial eample, y = dy d 3 2 = f '( ) =
7 f'() The major problem in this eample is that there are three roots and two maima! This is a general problem with the Newton method, and so it should only be used if you know your starting point 1 is close to the optimum. Another disadvantage is the need to calculate derivatives, so the function needs to be continuous in the vicinity of the root. Ecel includes an optimisation tool called Solver 1. Solver uses a comple version of Newton s method, which combines a quasi-newton algorithm with a conjugate gradient search (see Chapra, Chapters 14 and 15). Solver does not have problems with distinguishing between maima and minima, and allows multivariate optimisation (i.e. of functions with more than one variable input). However, it still has problems distinguishing between two maima, as in our polynomial eample. Make a copy of your spreadsheet which includes the polynomial equation. Use Solver to maimise the value of y. Go to Tools, Solver: 1 You might have to install Solver. Go to Tools on the Ecel menubar, and choose Add-ins, and tick Solver add-in, then OK.
8 Your target cell is the one you want to be maimised (i.e. the one with the polynomial formula in it), and the cell you want to change will have the value of in it. There are no constraints in this problem. Note that if you use a starting point of less than about 1.85, then Solver will not find the true maimum. (Note - Solver can also be used to find roots by specifying a value ( Value of: ) rather than a maimum or minimum). Optimisation using random sampling The first step in this method is to define upper and lower limits of all the input variables, between which you know the optimum values lie. The second step is to generate a large number of random samples from within these limits and run the formulue using each sample. Finally, locate the optimum value of the objective function and corresponding input values (e.g. by sorting). The sampled optimum will be an approimation of the true optimum. A smaller number of samples will tend to produce a poorer approimation, and so will a wide difference between the upper and lower limits. Therefore, after applying the method initially, it is common to then go back and refine the estimates of the upper and lower limits and repeat. This method is very robust, because it does not rely on gradients and has no problems when there are multiple maima and minima it will (approimately) find them all. In the polynomial eample, lets assume upper and lower limits, u = 5 and l = 0, and take 8 random samples of : f() l u I have been lucky and one of my samples gave a good approimation of the maimum y (note - in practical problems we would not usually know the true maimum, so we would not know how good our approimation was). However, it can be seen that many more samples would be required to be sure of a good approimation of the maimum y. In
9 comple practical problems, hundreds of thousands of samples and function evaluations might be needed, in which case this method may be too computationally inefficient. Constrained optimisation: eample using Ecel s Solver In this eample, you will optimise a simplified water abstraction and wastewater treatment system to minimise the associated costs. The system is illustrated below, Drinking water abstraction = A Sewage treatment plant: pollution load = W flow = Q flow = Q-A Distance downstream of STP = The input variables which you will optimise are: A percentage reduction in the abstraction rate A A percentage reduction in the pollution load W The associated costs are: The cost of other sources of drinking water to replace the reduction in A The cost of installing a new STP to reduce W The cost due to loss of amenity because of pollution in the river Cost functions Assume that the cost of replacing the abstraction is, 2 XA = 2000R where R is the percentage reduction in A. Assume that the cost of the new STP is, XW = S 0.75
10 where S is the percentage reduction in W. Assume that the cost associated with river pollution is, XC = 1000D where D is the length of river in which the dissolved oygen concentration is less than 5g/m 3. Mathematical model We need a mathematical model in order to estimate D. The model to use is called the Streeter-Phelps model of river dissolved oygen, = c k L k d 0 c s ka d e kd v e ka v where c (g/m 3 ) is the concentration of dissolved oygen at distance downstream of the STP, c s (g/m 3 ) is the natural concentration of dissolved oygen, k d is the pollution degradation rate (s -1 ), k a is the natural rearation rate (s -1 ), L 0 is the concentration of pollution in the river at the outfall of the STP (i.e. at = 0), and v (m/s) is the river velocity. Using a simple mass balance equation where the STP outfall meets the river, L 0 can be calculated as, W ' L 0 = Q' where W is the pollution load after the reduction, and Q is the flow after the abstraction, W ' = W (1 S /100) Q' = Q A ( 1 R /100) Assume v is the following function of flow, v = 0.2 Q [ '] Use the following inputs and parameter values,
11 c s = 10 g/m 3 kd = 2 day -1 [= s -1 ] ka = 5 day -1 [= s -1 ] W = kg/day Q = 5 m 3 /s A = 2 m 3 /s Ecel spreadsheet First you should set up a spreadsheet which contains entries for all the parameters and inputs. Also include initial estimates for S and R, and the equations for XA, XW, W, Q, L 0 and v. For eample: You then need to calculate D. To do so, calculate c for a column of values, with ranging from 0 up to, say, 50000m at intervals of 500m. You should find that c(=0) = 10g/m 3 and then c decreases to a minimum then increases again back up to 10g/m 3. Using Ecel s IF function, calculate D and hence XC. Hence calculate the total cost = XC + XA + XW. You can now try to use Solver to minimise this cost by changing S and R. You should use the following constraints in Solver: 0 R S Min ( c ) 0 [My answer for the optimum cost was just under 5m, with S = 52% and R = 0]
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