Linear Programming using the simplex method

[arl2267]

The Chemistry department at a local college decides to stock at least 900 small test tubes and 600 large test tubes. It wants to buy at least 2700 test tubes to take advantage of a special price. Since the small test tubes are broken twice as often as the large, the department will order at least twice as many small test tubes as large.a) If the small test tubes cost 18 cents each and the large ones cost 15 cents each, how many of each size should be ordered to minimize cost?b) Suppose the minimum number of test tubes is increased to 3000. Use shadow costs to calculate the total cost in this case.

I understand how to set up the tableau, I just have a hard time forming the minimization problem and the constraints. Here is my attempt so far:

Minimum: w= 18x₁+ 15x₂
Subject to: 2x₁+ x<sub>2 is greater than or equal to 900
x greater than or equal to 600

I'm pretty sure that's wrong, but if someone can help me with the equations, I can do the rest. Thanks.</sub>

 

[CaptainBlack]

The Chemistry department at a local college decides to stock at least 900 small test tubes and 600 large test tubes. It wants to buy at least 2700 test tubes to take advantage of a special price. Since the small test tubes are broken twice as often as the large, the department will order at least twice as many small test tubes as large.a) If the small test tubes cost 18 cents each and the large ones cost 15 cents each, how many of each size should be ordered to minimize cost?b) Suppose the minimum number of test tubes is increased to 3000. Use shadow costs to calculate the total cost in this case.

I understand how to set up the tableau, I just have a hard time forming the minimization problem and the constraints. Here is my attempt so far:

Minimum: w= 18x₁+ 15x₂
Subject to: 2x₁+ x<sub>2 is greater than or equal to 900
x greater than or equal to 600

I'm pretty sure that's wrong, but if someone can help me with the equations, I can do the rest. Thanks.</sub>

Minimise: w= 18x₁+ 15x₂

Subject to:
x1 >= 900
x2 >= 600
x1+x2>=2700
2x1>=x2

CB

 

[Sudharaka]

Minimise: w= 18x₁+ 15x₂

Subject to:
x1 >= 900
x2 >= 600
x1+x2>=2700
2x1>=x2

CB

Hi arl2267, :)

I think CaptainBlack's answer can be improved improved a little. :) Adding the third and fourth constraints will give the first constraint. Hence the first constraint can be dropped to get,

\(\mbox{Minimize : }w= 18x_{1}+ 15x_{2}\)

\(\mbox{Subject to:}\)

\(x_2\geq 600\)
\(x_1+x_2\geq 2700\)
\(2x_1\geq x_2\)
\(x_1,\,x_2\geq 0\)

Kind Regards,
Sudharaka.

 

[CaptainBlack]

Hi arl2267, :)

I think CaptainBlack's answer can be improved improved a little. :) Adding the third and fourth constraints will give the first constraint. Hence the first constraint can be dropped to get,

\(\mbox{Minimize : }w= 18x_{1}+ 15x_{2}\)

\(\mbox{Subject to:}\)

\(x_2\geq 600\)
\(x_1+x_2\geq 2700\)
\(2x_1\geq x_2\)
\(x_1,\,x_2\geq 0\)

Kind Regards,
Sudharaka.

The non-negativity constraints are redundant as we already have lower bounds on x1 and x2.

Also, it is better to retain the redundant constraint on the minimum for x1 so that the model does not need to be redone when numbers are changed

CB

 

[arl2267]

Thank you both for your help! I really appreciate it.

 

[arl2267]

Wait, how am I supposed to enter 2x₁>=x₂ into a tableau? I haven't seen that before on any other of my problems.

 

[Sudharaka]

Wait, how am I supposed to enter 2x₁>=x₂ into a tableau? I haven't seen that before on any other of my problems.

Well, you can rearrange that to, \(2x_1-x_2\geq 0\).