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# Matlab Tutor

## Peterson N

Master of Architecture, Mechanical Engineering, University of Melbourne, Australia

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I have more than 10 years of experience as a Matlab tutor. Before I joined Matlab Assignment Experts as an academic assistant, I was a lecturer in a well-known university here in Adelaide where I used to teach Simulink and a variety of data analysis classes. Since 2018, I have been offering expert Matlab assignment help online to students not only on topics related to Simulink but also those issued from data analysis units such as quantitative methods, econometrics, financial forecasting, mathematical statistics, and more. My goal is to help as many students as I can by sharing my knowledge so that they can achieve academic excellence

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Mathematical Computing
clc, clear all, close all %% Part 1 %% Question 1 syms q p C(q) = 610 - 0.2*q + 0.15*q^2; % Cost function D(p) = 260 - 20*p; % Demand function % Revenue is equal to the amount of units multiplied by the demand (price) % function R(p) = p*D(p) % The profit is equal to the Revenue minus the Cost P(p) = R(p) - subs(C(q), q, D(p)) % The marginal cost is the first derivative of the cost function MC(q) = diff(C, q) % The average cost is the Cost function divided by the total units AC(q) = C(q)/q % The marginal revenue is the first derivative of the Revenue function MR(p) = diff(R,p); %% Question 2 figure ps = 0:0.1:13; plot(ps, D(ps), 'linewidth', 2); xlabel('Price ($)'); ylabel('Demand (q)'); title('Demand vs. Price'); grid on %% Question 3 figure subplot(1,2,1) qs = 1:1:D(0); plot(qs, C(qs), 'linewidth', 2), grid on xlabel('Quantity (q)'); ylabel('Cost ($)'); title('Cost, Marginal Cost, Marginal Revenue and Average Cost vs. Quantity'); hold on plot(qs, MC(qs), 'linewidth', 2); plot(qs, AC(qs), 'linewidth', 2); plot(qs, MR(qs), 'linewidth', 2); legend('Cost', 'Marginal Cost', 'Average Cost', 'Marginal Revenue'); subplot(1,2,2) plot(qs, MC(qs), 'linewidth', 2), hold on; plot(qs, AC(qs), 'linewidth', 2); plot(qs, MR(qs), 'linewidth', 2); legend('Marginal Cost', 'Average Cost', 'Marginal Revenue'); xlabel('Quantity (q)'); ylabel('Cost ($)'); title('Marginal Cost, Average Cost and Marginal Revenue vs. Quantity'); grid on %% Question 4 % To calculate the point of elasticity of demand, we select 2 points of the % curve % For example, we selec the points (8, 100) and (9, 80) % Change in quantity q_change_perc = (9-8)/(9+8)/2 *100; p_change_perc = (80-100)/(80+100)/2 *100; p_elast_D = abs(q_change_perc/p_change_perc) % Because the value is < 1, the demand is inelastic % Now, we will calculate the intervals from P = 1 to 13, where demand is % elastic and inelastic pvec = 0:0.01:13; N = length(pvec); elast_vec = []; for i = 1:N-1 P1 = [pvec(i) D(pvec(i))]; P2 = [pvec(i+1), D(pvec(i+1))]; q_change_perc = (P2(1)-P1(1))/(P2(1)+P1(1))/2 *100; p_change_perc = (P2(2)-P1(2))/(P2(2)+P1(2))/2 *100; p_elast_D = abs(q_change_perc/p_change_perc); elast_vec = [elast_vec, p_elast_D]; end temp = pvec(find(elast_vec > 1)); elastic_interval = [temp(1), temp(end)] temp = pvec(find(elast_vec < 1)); inelastic_interval = [temp(1), temp(end)] temp = pvec(find(elast_vec == 1)); if length(temp)>0 unitary_interval = [temp(1), temp(end)] else unitary_interval = [] end %% Question 6 % The revenue is a negative parabola, so we find its vertex a = -20; b = 260; c = 0; pmax = -b/(2*a); qmax = subs(D,p,pmax); %% Question 7 % The Average Cost is a function that is inversely proportional to the % quantity % To minimize this function, we will use the MATLAB's function fmincon % We set the bounds for the quantity to: 0 < q < Inf type ObjectiveAverageCost [qmin, fval, exiflag] = fmincon('ObjectiveAverageCost', 10, [], [], [], [], 0, Inf); qmin_int = round(qmin) pmin = double(solve(D==qmin)) pmin_int = double(solve(D==qmin_int)) %% Question 8 % The profit function is a negative parabola, so the price that maximizes % it is its vertex a = -80; b = 1816; p_profit_max = -b/(2*a) q_profit_max = D(p_profit_max) %% Part 3 %% Question 10 % This time, the supply function is a price function, where the independent % variable is the quantity. % we can express the quantity in function of price as: S(p) = p-8; % The equilibrium point is where the demand and supply are the same p_equil = solve(D == S) q_equil = D(p_equil) figure ps = 0:0.1:20; plot(ps, D(ps), 'linewidth',2), hold on plot(ps, S(ps), 'linewidth', 2) plot(p_equil, D(p_equil), 'r*', 'linewidth', 2); xlabel('Price ($)'); ylabel('Quantity (q)'); title('Supply and Demand vs. Price'); legend('Demand', 'Supply'); grid on %% Question 11 % The consumer surplus is the integral between the equilibrium point and % the Demand function CS = int((D(p)-q_equil), p, 0, p_equil) %% Question 12 % The producer surplus is the integral between the Supply function and the % equilibrium point PS = int((q_equil - S(p)), p, 0, p_equil) function f = ObjectiveAverageCost(x) f = 3*x/20 + 610/x - 1/5; end 
Numerical Computing in Matlab
% The code you used to get the answer: n = 10; digitsWhole = digits(n+(numel(num2str(floor(pi))))); MyPi = vpa(pi); digits_after =char(rem(MyPi,1)); % 10 digits as required ref = strfind(digits_after,'.'); tenth_digit = str2num(digits_after(ref+n)); % The 10th digit of pi after the decimal is: fprintf('The 10th digit of pi after the decimal is: %g \n' ,tenth_digit) % The value of y didn't appear in the command window because: % *It was suppressed with a semicolon (;)* % Your code: horizontal_vec = [8 2 6 10] % Your code: vertical_vec = [5; 9; -9; 0] %OR: vertical_vec = [5 9 -9 0]' % Your code: k=1; for i=98:-1:42 if mod(i,2)~=0 && i~=1 odd(k)=i; k=k+1; end end odd % expected horizontal vector num2str(odd) % Your code: k=1; for i=5:1:101 if mod(i,2)==0 && i~=1 even(k)=i; k=k+1; end end even % expected horizontal vector num2str(even) % The code you used to get the answer: X = 10:1.5:175; Number_of_element_in_X = numel(X); % The answer is: fprintf('The answer is: %g \n', Number_of_element_in_X) % Your code: n=36; a= 23; b = 27; list = linspace(a,b,n)'; num2str(list) % The code you used to create a 5 x 4 matrix of random numbers between -2 % and 5: a = -2; b = 5; m= 5; n=4; Mat_X = a + (b-a).*rand(m,n) % The code you used to determine the number in the 2nd row and 4th column: Mat_X(2,4) % The code you used to create a 3 x 7 matrix of random integers between -8 % and -2: a = -8; b = 2; m= 3; n=7; Mat_Y = randi([a,b], m, n) % The code you used to slice out the 2nd and 3rd rows: j = 1; % removes 2nd and 3rd rows Mat_Y_new = Mat_Y(j,:) % The code you used to create a 3 x 4 matrix of 0's: Mat_A = zeros(3,4); % The code you used to create a 3 x 4 matrix of 6's: Mat_B = 6* ones(3,4); % The code you used to combine them horizontally: Mat_C = horzcat(Mat_A, Mat_B) % The code you used to combine them vertically: Mat_D = vertcat(Mat_A, Mat_B) % Your code: m = 3; n = 700; for i = 1:m for j = 1:n Mat_E(i,j) = j; end end Mat_E % Alternatively: % Mat_E = [1:700].*ones(3,1) % Your code: m = 850; n = 850; for i = 1:m for j = 1:n Mat_F(j,j) = j; end end Mat_F; % Alternatively: % Mat_F = diag([1:850])