Why does my plot not display the 50 values of t_operation?

I want to plot t_operation for 100<x<150. I tried this code, but it does not work. I do not understand why.

clear all; close all; clc;
hold on
v_tank = 40;
v_trailertank = 80;
v_trailer = 100;
t_loading = 1/12;
t_unloading = 1/12;
L = 400;
for x = 100 : 150
    t_tank1(x)=(t_loading+(x/v_trailertank)+t_unloading)+((L-x)/v_tank);

    t_tank2(x)=(t_loading+(x/v_trailertank)+t_unloading)+x/v_trailer+(t_loading+L/(v_trailertank)+t_unloading);

    t_operation(x)=max(t_tank1(x),t_tank2(x));
end
plot(x,t_operation)

Please help.

Answer 1

[the cyclist]

The reason is that when you reach that plotting statement, x is simply the last value of your loop. Try this:

xrange = 100:150;
plot(xrange,t_operation(xrange))

FYI, you can write that code in vectorized fashion, without using the for loop at all:

xrange = 100 : 150;

t_tank1(x)=(t_loading+(x/v_trailertank)+t_unloading)+((L-x)/v_tank);
t_tank2(x)=(t_loading+(x/v_trailertank)+t_unloading)+x/v_trailer+(t_loading+L/(v_trailertank)+t_unloading);
t_operation(x)=max(t_tank1(x),t_tank2(x));

plot(xrange,t_operation(xrange))

[Steven Lord]

That code makes t_tank1, t_tank2, and t_operation start with 99 0's that you don't care about. It also won't work if x has a granularity finer than 1 unit (contains non-integer values.) I would instead omit the indexing with x.

% Define constants
v_tank = 40;
v_trailertank = 80;
v_trailer = 100;
t_loading = 1/12;
t_unloading = 1/12;
L = 400;

% Define the range of x 
x = 100 : 150;
t_tank1=(t_loading+(x/v_trailertank)+t_unloading)+((L-x)/v_tank);
t_tank2=(t_loading+(x/v_trailertank)+t_unloading)+x/v_trailer+...
        (t_loading+L/(v_trailertank)+t_unloading);
t_operation=max(t_tank1,t_tank2);

% Plot all three lines
plot(x, t_tank1, 'r-', 'DisplayName', 'tank 1');
hold on
plot(x, t_tank2, 'c-', 'DisplayName', 'tank 2');
plot(x, t_operation, 'k:', 'LineWidth', 4, 'DisplayName', 'maximum')
legend show

I chose to plot all three lines in different colors and with the operation variable with a thicker line for illustration purposes.

If I wanted to rerun that model with a finer-grained x, the only line that needs to change is the line defining x.

x = 100:0.5:150;

Rerun the rest of the lines from "t_tank1 = ..." as before.

[Mike Mierlo van]

Both, thanks a lot!

[Mike Mierlo van]

Now I want to know the value of ‘x’ when t_operation is at a miniumum. I changed the code for more accuracy. I know the minium of t_operation is 8.4405, but I do not know to what ‘x’ this corresponds.

The code now:

clear all; close all; clc;

hold on

v_tank = 40000;

v_trailertank = 80000;

v_trailer = 100000;

t_loading = 1/12;

t_unloading = 1/12;

L = 400000;

% Opdracht 1a: plot t_operation(x) voor 100 =< x =< 150

x = 100000:150000;

t_tank1(x)=(t_loading+(x/v_trailertank)+t_unloading)+((L-x)/v_tank);

t_tank2(x)=(t_loading+(x/v_trailertank)+t_unloading)+x/v_trailer+(t_loading+L/(v_trailertank)+t_unloading);

t_operation(x)=max(t_tank1(x),t_tank2(x));

plot(x,t_operation(x))

%geef minimum

t_operationmin = min(t_operation(x))

%now: to what ‘x’ corresponds t_operationmin?

[Mike Mierlo van]

Never mind: I already found it out. It works with: X_min=find(t_operation(x)==t_operationmin)

[the cyclist]

You can do the same thing more efficiently with

[t_operationmin, x_min_index] = min(t_operation(x))

But be careful. The find command will tell you which element (e.g. the "5th element in the vector"), not the actual value of x that that corresponds to. I think you'll need x(x_min_index) for that. For example, in your case, the 5th element of x would be the value x=100004.

As Steven pointed out, you are using x in an odd way, as both an index and a variable. So you need to be careful.