# Statistical and Optimization Methods for Engineers

Q1: a) Find the maxima and minima, if any, of the function

1. b) Determine the convexity of the following function using the second derivative test.

Q2: Minimize

• Apply the bisection method with initial bounds  and
in a tabular form.
• Apply Newton’s method with an error tolerancePlease present
• your search procedures in a tabular form.

was generated using the model:You are asked to use the dataset
to estimate the two parametersa1 and a2in the model, by treating it as a leastsquares problem.

Q4: Solve the following linear programming problem using the two methodsbelow:

• The graphical solution;
• The Simplex method;

Q5: Find the minimum value of the following function using LagrangeMultipliers:

Solution

“{r setup, include=FALSE}

knitr::opts_chunk$set(echo = TRUE) “$\underline{Question\: 1 : part\: (a)}f(x)=4x^3-18x^2+27x-7 \:\implies f'(x)=12x^2-36x+27$For local minima/maxima ,$f'(x)=0$. So$\: 12x^2-36x+27=0$.$\: \implies 4x^2-12x+9=0 \: \implies(2x-3)^2=0 \: \implies x=2/3$. Now ,$f”(x)=24x-36 \: \implies f”(\frac{2}{3})=0$. So ,$x=\frac{2}{3}$is an inflection point. There is no minima or maxima of this function$\underline{Question \:1 : part \:(b)}$Here$f(x)=x^{2}+x^{4} \:\implies f'(x)=2x + 4x^3 \:\implies f”(x)=2+12x^2 \:\implies f”(x)>0$for all real$x$, since$x^2 \geq 0$. This proves that$\:f(x)$is convex.$\underline{Question\: 2 : part \:(a)}f(x)=0.25x^4+2x^2-x^3-2x$For minimization , we are seeking$x^{*}$such that$f'(x^{*})=0$“{r} derivative<- function(x){-3*x^2-2 + 4*x + x^3} bisection_method<- function(a, b, tol, f ){ if (f(a)*f(b) > 0){ print(“No root found.”) } else while ((b – a)/2.0 >tol){ midpt= (a + b)/2.0 if (f(midpt) == 0){ return(midpt) #The midpoint is the x-intercept/root. } else if (f(a)*f(midpt) < 0){ # Increasing but below 0 case b = midpt print(c(a,b)) } else{ a = midpt print(c(a,b)) } } return((a+b)/2) } bisection_method(0,2.4,tol=0.04,derivative) derivative(1.0125) “ As you observe , The found$\:x^{*}=1.0125$and the tabulation is as follows :- “{r} #tabular form iteration=c(1:5) a=c(0.0,0.6,0.9,0.90,0.975) b=c(1.2,1.2,1.2,1.05,1.050) data.frame(iteration,a,b) “$\underline{Question \:2 : part (b)}$Our recursion is$x_{n+1}=x_{n}-\frac{f'(x_n)}{f”(x_n)}$“{r} derivative<- function(x){-3*x^2-2 + 4*x + x^3} second_derivative<- function(x){3*x^{2}-6*x+4} x1=1.2 ; tol=0.001 for ( n in 1:10){ y=x1 x1=x1-{derivative(x1)}/{second_derivative(x1)} print(x1) if(abs(y-x1)<tol) { break; } } #tabulation of x_n iteration=c(1,2,3) x=c(1.014286,1.000006,1) print(data.frame(iteration,x)) “ Thus the answer turns out to be$x^{*}=1$using Newton’s method.$\underline{Question \: 3}$“{r} library(readr) Data_for_Assessment_2_2_ <- read_csv(“~/Desktop/Advent/Data_for_Assessment 2 (2).csv”) z<-Data_for_Assessment_2_2_ library(nlmrt) myformula<- y~a1*exp(x/a2) parname<-c(“a1″=1,”a2″=1) funss<-model2ssfun(myformula,pvec = parname) fungr<-model2grfun(myformula,pvec = parname) BFGS<-optim(par=parname,fn=funss,gr=fungr,method = “BFGS”,y=z$y,x=z$x) print(BFGS) “ So,$\hat{a_1}=1.398 \:, \: \hat{a_2}=1.999$are the optimal values to minimize the residual sum of squares .$\underline{Question \: 4 \: part \:1}$It’s enough to minimize$-18x_1-15x_2$and add 20 to the answer . “{r} library(intpoint) a <- c(-18,-15) A<- matrix(c(1,1,3,2),nrow = 2,byrow = T) b <-c(5,12) solve2dlp(t=-1,c=a,bm=b,m=A) #t=-1 since it’s a minimization problem. “ The optimal values is attained for$(x_1,x_2)=(2,3)$and min value of$f=-18x_1-15x_2+20$is$-81+20\: =\:-61$.$\underline{Question \: 4 \: part \:2}$We’ll use the simplex method . It’s enough to minimize$-18x_1-15x_2$and add 20 to the answer . “{r} library(boot) a <- c(-18,-15) A<- matrix(c(1,1,3,2),nrow = 2,byrow = T) b <-c(5,12) simplex(a,A,b) “ The optimal values is attained for$(x_1,x_2)=(2,3)$and min value of$f=-18x_1-15x_2+20$is$-81+20\: =\:-61$.$\underline{Question \: 5}$For minimum ,$L=f-\lambda g=2x^2+y^2+3z^2-\lambda(2x-3y-4z-49)\frac{\partial L}{\partial x}=\frac{\partial L}{\partial y} = \frac{\partial L}{\partial z}=\frac{\partial L}{\partial \lambda}=0$. Therefore ,$4x-2\lambda=2y+3\lambda=6z+4\lambda=0$…..(1) So , at the extremum ,$x=\frac{\lambda}{2}, \: y=\frac{-3\lambda}{2} , \: z=\frac{-2\lambda}{3}$…..(2) Putting (*) in$2x-3y-4z=49$, we get$\lambda = 6 \implies \:$optimal solution is,$\:x=3 , \:y=-9 , \:z=-4$. So, minimum value=$18+81+48=147\$