112k views
2 votes
Simple Minimization Problem Part 2

minimize f=5x1+3x2
subject to 1x1 + 1x2>=4
3x1+ 1x2>= 6
0x1 + 7x2>=14
x1>=0, x2>=0
using the dual simplex method rather than the two phase method

1 Answer

4 votes

Final answer:

The question involves solving a linear programming minimization problem using the dual simplex method. The student must first convert the inequality constraints to a suitable form and create an initial simplex tableau. The dual simplex method then iteratively updates this tableau while maintaining the feasibility of the dual problem to reach the optimal solution.

Step-by-step explanation:

The student is asking about solving a linear programming problem using the dual simplex method. In the given problem, we want to minimize the objective function f = 5x1 + 3x2 subject to a set of constraints. These constraints are:

  • x1 + x2 ≥ 4
  • 3x1 + x2 ≥ 6
  • 7x2 ≥ 14
  • x1 ≥ 0, x2 ≥ 0

However, the dual simplex method requires the constraints to be in the form of ≤ (less than or equal to). Therefore, the constraints need to be multiplied by -1 to convert the inequalities. Next, we would add slack variables and construct the initial simplex tableau. From there, the dual simplex algorithm proceeds by iteratively updating this tableau to find the optimal solution while ensuring that the pivot elements are chosen so that the non-basic variables (those outside the current solution) satisfy their non-negativity restrictions.

The solution process involves finding the most negative element in the right-hand side of the tableau (this element corresponds to the variable that will enter the basis), and then determining the pivot row such that the ratio of the pivot row's right-hand element to its corresponding pivot column element (which must be negative) is the least (in absolute value). We then perform pivot operations until all the right-hand side entries are non-negative, which indicates that an optimal solution has been found. Throughout this process, maintaining a feasible solution to the dual problem ensures optimization of the primal problem.