clear

clc

close all

%Fixed Cost

M = {'Concrete',12, 30, 85000, 900, 4, 4;'Wood',18, 50, 100000, 800, 12, 10; 'Adobe',15,42, 55000, 600, 4, 5};

choice = menu('choose a type of material',M(1:3,1));

choice = {M{choice,1:7}};

SurfaceArea = 3000; %ft^2

Thickness = choice{1,2}/12; %converts to ft

MaterialCost = choice{1,3}; %$/ft^3

MicellaneousMaterial = choice{1,4}; %$

LaborCost = choice{1,5}; %$ /(person week)

Laborers = choice{1,6}; %# of laborers(people)

Weeks = choice{1,7}; %weeks of construction

MCost = Thickness*SurfaceArea*MaterialCost; %Total cost of matieral,$

LCost = LaborCost*Laborers*Weeks; %Total cost of labor,$

FixedCost = MCost + LCost + MicellaneousMaterial; %Fixed cost, $

%Variable Costs and Revenue

%Cost Variables

Energycost = input('Energy Cost:'); %$

Laborcost = input('Labor Cost:'); %$

Landfillcost = input('Land Fill Cost:'); %$

Maintenancecost = input('Maintenance Cost:'); %$

%Revenue Variables

Priceadmission = input('Price admission per person:'); %$

People = input('# of people per week:'); %$

Donations = input('Expected donations per week:'); %$

%Other Variables

OWeeks = input('Operating weeks:'); %Operating weeks per year

Years = input('# of years:'); %Years to operate construction

N = (0:Years); %Years

%Process

TFixedCost = FixedCost; %Fixed cost already calculated

VariableCost = Energycost + Laborcost + Landfillcost + Maintenancecost; %Total cost to operate enclosure per week

TotalCost = TFixedCost + VariableCost * N * OWeeks ; %Cost to operate for # of years

Revenue = ((Priceadmission * People * N) + (Donations * N)) * OWeeks; %Revenue from customers for # of years

Profit = (Revenue - TotalCost) .* N; %Total profit per year

BreakEvenY = TFixedCost ./ (TotalCost - VariableCost); %Breakeven point in years

BreakEvenM = BreakEvenY * 12; %Breakeven in months

MProfit = Profit / BreakEvenM(1,5)

WM = 4.2; %Weeks per month

DonationsBreak = Donations * WM * 4; %$ from donations needed to breakeven in 4 months

%Summary of Results

fprintf('Material: %s \t Operating %0.0f weeks per year will generate per year: \t\t Revenue: $%0.0f \t\t Cost: $%0.0f \t The breakeven time is %0.2f months \t The total profit after %0.0f years is $%1.3e2.',choice{1:1},OWeeks,Revenue(1,2),TotalCost(1,2),BreakEvenM(1,Years),Years,Profit(1,Years))

fprintf(' It will take a one-time donation of $%5.2f to breakeven in four months.',DonationsBreak)

%Figures

figure('color','w');

plot(N,TotalCost,'r--','linewidth',3) ;

hold on

plot(N,Revenue,'b-','linewidth',3) ;

hold on

plot(BreakEvenM(1,5),'o k','Markerfacecolor','k');

grid on

title('Revenue and Total Cost')

xlabel('Number of Years (Y), [Y]')

ylabel('Revenue and Cost (R),(C) [$]')

legend('Total Cost','Revenue','location','northwest')

figure('color','w')

plot(N, Profit,'g-.','linewidth',2.5);

hold on

plot(BreakEvenM(1,5),Profit(1,2),'o k','Markerfacecolor','k')

grid on

title('Profit')

xlabel('Number of Years (Y), [Y]')

ylabel('Profit of the construction choice (P), [$]')

legend('Profit','location','northwest')

Need help with the Option to evaluate, the above code is what we have so far .Please help urgent

Optlons to Evaluate The zoo has asked that you consider running your analysis for at least 5 years. For each construction cost, the user should be able to choose the material from a menu then manually enter all other inputs First, the concrete has several advantages and some disadvantages. Concrete is the least expensive material with a moderate operating cost. It saves on energy costs due to its ability to insulate. However, it is also the least aesthetically appealing material and does not keep with the theme of the zoo. Because of this, fewer people will visit the enclosure, the price per ticket must be decreased, and the amount of donations will decrease. The particular details are as follows: Name: Concrete Energy Cost: $5,00o/week Maintenance Cost: s3,oo0/week Number of weeks per year: 50 .Operational Labor Cost: $4,ooo/week .Landfill Cost: $3,ooo/week Number of years for analysis: 5 Admission Price: $2/persorn Number of People: 10,000/week Donations per week: s400/week Second, the wood has several advantages and some disadvantages as well. Wood keeps well with the theme of the zoo and is very attractive. Therefore, more people will want to see the enclosure at an even higher price and will be more prone to donate even more to help with the research. However, it is also the most expensive material and has a higher maintenance cost and energy cost. The particular details are as follows: Optlons to Evaluate The zoo has asked that you consider running your analysis for at least 5 years. For each construction cost, the user should be able to choose the material from a menu then manually enter all other inputs First, the concrete has several advantages and some disadvantages. Concrete is the least expensive material with a moderate operating cost. It saves on energy costs due to its ability to insulate. However, it is also the least aesthetically appealing material and does not keep with the theme of the zoo. Because of this, fewer people will visit the enclosure, the price per ticket must be decreased, and the amount of donations will decrease. The particular details are as follows: Name: Concrete Energy Cost: $5,00o/week Maintenance Cost: s3,oo0/week Number of weeks per year: 50 .Operational Labor Cost: $4,ooo/week .Landfill Cost: $3,ooo/week Number of years for analysis: 5 Admission Price: $2/persorn Number of People: 10,000/week Donations per week: s400/week Second, the wood has several advantages and some disadvantages as well. Wood keeps well with the theme of the zoo and is very attractive. Therefore, more people will want to see the enclosure at an even higher price and will be more prone to donate even more to help with the research. However, it is also the most expensive material and has a higher maintenance cost and energy cost. The particular details are as follows