CS计算机代考程序代写 matlab algorithm THIS PAPER MUST NOT BE

THIS PAPER MUST NOT BE
Class Test (optimisation), S1 2019
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THE UNIVERSITY OF QUEENSLAND
School of Information Technology & Electrical Engineering
Class Test (Optimisation), S1 2019
ENGG7302
Advanced Computational Techniques in Engineering
(MEngSc) CLOSED BOOK
TIME: NINETY minutes for working
FIVE minutes for perusal before examination begins
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ENGG7302 Advanced Computational Techniques in Engineering
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Page 1 of 5

Class Test (optimisation), S1 2019 ENGG7302 Advanced Computational Techniques in Engineering
Question 1. (60 marks)
For each question, please select the correct option (only one option is correct among the
four options)
1.1 Consider the function f (x) = x x3 , the function may have the following properties: [1] the point x = 0 is a critical point; [2] the point x = 0 is a stationary point; [3] the point x = 0 is
a saddle point; [4] the function is unimodal in the domain x £ 1; [5] it is a coercive
function. Then it is in general correct to say that
(a) ONLY [1], [2], [3] and [4] are correct
(b) [1] ~ [5] are all correct
(c) ONLY [1], [2], [3] are correct
(d) ONLY [4], [5] are correct
1.2 Gradient Decent Method (Steepest Decent Method) may have the following property [1] it is unnecessary to have an initial guess; [2] it requires performing line search, which may be solved by one-dimensional Newton’s method; [3] it requires to calculate the Hessian matrix; [4] it involves the function fitting in each iteration; [5] one may consider to use a fixed step size. Then it is in general correct to say that
(a) [1] ~ [5] are all correct
(b) ONLY [1], [4] and [5] are incorrect
(c) ONLY [1] and [4] are incorrect
(d) ONLY [1], [3] and [4] are incorrect
1.3 Among the following optimization methods, [1] Newton’s method; [2] quadratic interpolation method; [3] Steepest descent method; [4] Conjugate gradient descent; [5] Linear programming method; [6] compressed sensing optimisation method, which ones rely on the derivative of the objective function
(a) ONLY [1], [3] and [4]
(b) ONLY [1], [2] and [3]
(c) ONLY [1], [3], [4] and [6]
(d) [1] ~ [6]
1.4 In simulated annealing, during the simulation of cooling, at a fixed temperature T, if we perturb (randomly) the current state to a new state, then we need to evaluate a system parameter DE , the difference in energy between current and new state. It may have the
following properties, [1] The new state must be lower ( DE < 0 ) than current state; [2] If DE < 0 ( new state is lower), accept new state as current stage; [3] If DE 3 0 ( new state is higher), accept new state with probability; [4] If DE 3 0 ( new state is higher), accept new state as current stage; [5] If temperature T=0, it will not accept new state. Then it is in general correct to say that (a) ONLY [1] is incorrect (b) ONLY [2], [3] are correct (c) ONLY [2] [3] [5] are correct Page 2 of 5 Class Test (optimisation), S1 2019 ENGG7302 Advanced Computational Techniques in Engineering (d) [1]~[5] are all incorrect 1.5 The compressed sensing (CS) algorithm may have some properties: [1] it may require uniformly, subsample the signal/data; [2] wavelet transform is the only option for sparsifying transforms; [3] it requires nonlinear optimization, which may be handled by conjugate gradient method; [4] one may use Lagrange multiplier to solve the involved optimisation problem. It is correct to say that (a) (b) (c) (d) [1] ~[4] are all correct ONLY [2], [4] are correct ONLY [3], [4] are a correct ONLY [1], [3], [4] are correct 1.6 Using Newton’s method to minimise (𝑥) = (𝑥& + 5𝑦&)/2 , to calculate the Newton step, we may find the linear system to calculate the Newton step. If we start at (𝑥 = 5; 𝑦 = 1), then the corresponding linear system for finding the Newton step is (a) .1 00 𝑠 = −[5]; 055 (b) .1 00𝑠=[5]; 055 (c) .5 00 𝑠 = −[5]; 015 (d) .5 00𝑠=[5]; 015 1.7 In an experiment, we need to measure the resistances of two resistors (resistor A and resistor B). Individually, resistors A and B is measured as R1=10 Ω and R2=20 Ω , respectively. But in another measurement, the difference between these two resistors is R1-R2=5 Ω . To find the least-square (LS) approximation of the resistances R1, R2, the over-determined system can be set up, and then the LS solution can be written as (a) (b) (c) !1 0$!10$ !R1$# &# & #R2&=#0 1&#20& " % # 1 1 &# 25 & " %" % ( !1 0$+−1( (!1 0$+−1(!10$+ ! R1 $ *# &- *# &- #R2&=*#0 1 &- *#20&- " % *# 1 −1 &- *# 5 &- )" %,)"%, !10$+ 1 $# &- ! R1 $ *! 1 0 1 $# &- *! 1 0 #R2&=*#0 1 −1&#0 1 &- *#0 1 −1&#20&- " % *" %# 1 −1 &- *" %# 5 &- ) "%,) "%, Page 3 of 5 Class Test (optimisation), S1 2019 ENGG7302 Advanced Computational Techniques in Engineering ( !1 0$+−1( !10$+ ! R1 $ *! 1 0 −1 $# &- *! 1 0 −1 $# &- (d)#R2&=*#0 1 1 &#0 1 &- *#0 1 1 &#20&- " % *" %# 1 −1 &- *" %# 5 &- ) "%,) "%, 1.8 Use Gauss-Newton method to fit nonlinear model function 𝑓(𝑡, 𝑥) = 𝑥8exp (𝑥&𝑡) to data If we take the initial guess (x1, x2) = (1,0), then the initial residual function between the data and model result will be (a) r = [1 -0.3 -0.7 -0.9]; (b) r = [1 0.09 0.49 0.81]; (c) r = [-1 -0.09 -0.49 -0.81]; (d) r = [-1 -0.3 -0.7 -0.9]; 1.9 Given the following two individuals in a Genetic Algorithms: X1=[1 0 1 0 1 0 1 0 1 0]; X2=[0 0 0 0 0 1 1 1 1 1] Perform a one-point crossover with crossover point between genes 5 and 6, and also mutation on gene 2 of X1. Then the new child individuals will be: (a) X1=[1010101010];X2=[0000011111] (b) X1=[1110111111];X2=[0100001010] (c) X1=[1010111111];X2=[0000001010] (d) X1=[1110111111];X2=[0000001010] 1.10 Determine the stationary values of function z = -x4 - y4 + 4xy. At the critical point (x = 0, y = 0), the stationary value of z is (a) a true minimum; (b) a true maximum; (c) a saddle point; (d) it is inconclusive for the evaluation of extrema. Question 2. (20 marks) Company A is going to build a new swimming pool for UQ. Can you please help design the swimming pool as shown in Figure 1? 2.1 Suppose three sides of the pool satisfy the following condition: L + W + H =100 metre. You are asked to maximise the volume of the pool. Use the Lagrange multipliers to solve the above constrained optimisation problem. Please write down the final Lagrange equations (please show detailed solution procedure but DON’T solve the final Lagrange equations). In questions 2.2(a) and 2.2(b), please pick up ONLY one to answer (DON’T work on both), show detailed solution procedure, but DON’T solve it. Page 4 of 5 Class Test (optimisation), S1 2019 ENGG7302 Advanced Computational Techniques in Engineering 2.2(a) Suppose three sides of the pool satisfy the following condition: L + W + H =100 metre, you are asked to use the linear programming (LP) approach to minimise the longest the edge length of the pool. 2.2(b) Suppose the height (H) of the pool is fixed to be 1 metre, and two sides of the pool satisfy the following condition: L ≥ 2W metre, you are asked to use the linear programming (LP) approach to maximise the volume of the pool. Figure 1. The rectangular swimming pool to be built at UQ. L W, H are the edge lengths of the pool, that is, L: length; W: width; H: height. Question 3. (20 marks) 3.1 Please set up an appropriate linear programming (LP) model and write down the simplex tableau (please do not solve it). 3.2 Solve the above LP problem using the graphical approach. 3.3 Consider using the Matlab function linprog( ) to solve the above LP problem. Please write down the objective function vector 𝑓, matrices A, A , and vectors b, b , LB,UB (in eq eq Paul has $20,000 to invest in three funds F1, F2 and F3. Fund F1 offers a return of 2% and has a low risk. Fund F2 offers a return of 4% and has a medium risk. Fund F3 offers a return of 5% but has a high risk. To be on the safe side, Paul invests no more than $3000 in F3 and at least twice as much as in F1 than in F2. Assuming that the rates hold till the end of the year, what amounts should he invest in each fund in order to maximise the year end return? Matlab format). linprog is a matlab function for solving the linear programming problem: minfT*x subjectto: A*x<=b x LB<=x <=UB x = linprog( f, A, b, Aeq, beq, LB, UB ) x- the design variables f- linear objective function vector A- matrix for linear inequality constraints b- vector for linear inequality constraints Aeq- matrix for linear equality constraints, it is an empty matrix for this case beq- vector for linear equality constraints, it is empty for this case LB- vector of lower bounds UB- vector of upper bounds END OF EXAMINATION Page 5 of 5