Please Help Me Simple Question

Hello, i am newbie

I need a simple program that can calculate cut (digging) price,

because the difficulty of cut is increased per m depth, the price can't be equal.

For example:

for the first 1m depth of cut

the price for cut is $10/m^3

for 1m-2m depth of cut

the price for cut is $14/m^3

for 2m-3m depth of cut

the price for cut is $18/m^3

and so on (constant addition).

But, the value of the depth is random and can go really high, let's say 30 m, so using "if" is not edible,

Can anyone can help me to calculate this with a couple lines of codes only?

Thank you very much for your help.

[Image Analyst]

If the depth is 1.5 m, is the price $14 for the whole depth, or $10 for the first one meter, and $14 for the next half meter? It makes a difference.

What is the price after 3 m? Does it stay at $18, OR does it increase by $4 for every meter of depth?

The price is per cubic meter, so to know the total price we need not only the depth, but also the area of the cut, and an assumption that the cut has perfectly vertical sides. So, what is the area of the cut?

[Antoni Ho]

the first one is $10 and next one $14 meter next one is $18,

always increase $4 per meter (to infinity actually)

the area of cut is 1 unit (constant) (m2)

it is like this

if the depth is 4.5m

total price = area x height x unit price

total price = 1 x 1 x 10 + 1 x 1 x 14 + 1 x 1 x18 + 1 x 1 x 22 + 1 x 0.5 x 26

Answer 1

[TADA]

This is a simple linear equation:

price = 10 + 4*floor(depth);

[Antoni Ho]

it is not like that

the first one meter is $10, next $1.4, next $1.8m

it is like this

if the depth is 4.5m

total price = area x height x unit price

total price = 1 x 1 x 10 + 1 x 1 x 14 + 1 x 1 x18 + 1 x 1 x 22 + 1 x 0.5 x 26

yes, the area is constant

[TADA]

Something like this should do it:

s = [ones(1,floor(depth)), mod(depth, 1)]; % cut portion of cubic meter depth
x = 1:numel(s);
ppcm = cumsum(10+x*4); % price per cubic meter
price = sum(ppcm.*s);

[TADA]

I made a mistake, cumsum is not needed:

depth = 4.5;
s = [ones(1,floor(depth)), mod(depth, 1)]; % cut portion of cubic meter depth
x = 0:(numel(s)-1);
price = sum(s.*(10+x*4))

    price =

      77

Answer 2

[Star Strider]

Try this:

digprice = @(depth) 6 + ceil(fix((depth+1)))*4;                 % Digging Price As A Function Of Depth

dep = linspace(0, 30, 150);                                     % Depth Vector

figure
plot(dep, digprice(dep))
grid
xlabel('Depth (m)')
ylabel('Digging Price ($)')

Experiment to get the result you want.

[Antoni Ho]

i am sorry, but i don't understand that.

but, i think it also wrong (sorry to say that)

[Star Strider]

I am not certain what you want.

Run my code, and view the plot to see what it does.

A sample of the output:

    Depth     Price
      0.0     $ 10
      0.5     $ 10
      1.0     $ 14
      1.5     $ 14
      2.0     $ 18
      2.5     $ 18
      3.1     $ 22
      3.6     $ 22
      4.1     $ 26
      4.6     $ 26
      5.1     $ 30
      5.6     $ 30
      6.1     $ 34

The code that produced it:

Result = [dep(:) digprice(dep(:))];
fprintf('    Depth     Price\n')
fprintf('     %4.1f     $%3d\n', Result')