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Expert: - 1/31/2012


Question
QUESTION: Solve the following L P problem through SIMPLEX method
Maximize Z = 5x1 + 6x2
Subject to
  2x1 + 3x2    ≤  3000
  5x1 + 7x2    ≤  1000
  x1 +  x2        ≤    500
  x1 and  x2    ≥       0

ANSWER: The variable to add first on a straight forward SIMPLEX problem like this is the variable with the largest coefficient in the Z row.  In this case, since 6 is bigger that 5, the variable to add would be x2.

In the 1st equation, the value of x2 would be 3000/3 = 1000.
In the 2nd equation, the value of x2 would be 1000/7 = 142 6/7.
Since the 2nd value is smaller, add that one first.

Subtract 6/7 of the 2nd equation from the Z-row.
Subtract 3/7 of the 2nd equation from the 1st equation.

Since the Z equation still has a 5/7 as the x1 coefficient, add that one next.
Since the 2nd equation is the only equation with a positive coefficient for x1,
zero out the x1 coefficient in the Z-row by adding the appropriate multiple times the 2nd equation.

In the end, this says to stick to x1 with a value of 200.
Note that the 1st equation has 5x1, which is 1000, and that's the value on the right.
Note that the 2nd equation has only x1, which is 200, so the value of the slack variable in this equation is 300.

Even though x2 is the first variable to add, the fact that the 2nd equation says 7x2 ≤ 500 puts to much of a constraint on how high x2 can get for it to make the Z equation have that large a value.


---------- FOLLOW-UP ----------

QUESTION: Dear Mr.Wislon,

Thank you for taking your time to answer my question. Would it be possible for you to put the solution in numbers than words?

Answer
LP Problem
LP Problem  
I'll attach it in a file.

The 1st column is the value of the variables for each equation.
The 2nd column is the multipiers to multiply the main row by.

As can be seen, the only variable that ends up included is the 1st variable.
The value of it is 200.

This makes the 1st slack variaable have a value of 2600
and the 3rd slack variable have a value of 300.

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