# an Unbounded Feasible Region app — the minimum cost

SOLVING MINIMIZATION PROBLEMS

Many linear programming problems involve minimizing an objective such as cost instead of maximizing a profit function. A restaurant, for example, may wish to develop a work schedule to meet staffing needs while minimizing the total number of employees. Also, a manufacturer may seek to distribute its products from several factories to its many regional warehouses in such a way as to minimize total shipping costs.

Minimization problems can be solved graphically by first setting up the feasible solution region and then using either the corner-point method or an iso-cost line approach (which is analogous to the iso-profit approach in maximization problems) to find the values of X1 and X2 that yield the minimum cost.

Iso-cost
An approach to solving a linear programming minimization problem graphically.

📌 Example 1: A minimization problem with two variables
Example 1 shows how to solve a minimization problem.

Cohen Chemicals, Inc., produces two types of photo-developing fluids. The first, a black-and-white picture chemical, costs Cohen \$2,500 per ton to produce. The second, a color photo chemical, costs \$3,000 per ton.

Based on an analysis of current inventory levels and outstanding orders, Cohen’s production manager has specified that at least 30 tons of the black-and-white chemical and at least 20 tons of the color chemical must be produced during the next month. In addition, the manager notes that an existing inventory of a highly perishable raw material needed in both chemicals must be used within 30 days. To avoid wasting the expensive raw material, Cohen must produce a total of at least 60 tons of the photo chemicals in the next month.

We may formulate this information as a minimization LP problem. Let

X1= number of tons of black-and-white picture chemical produced
X2= number of tons of color picture chemical produced

Subject to:
X1≥30 tons of black-and-white chemical
X2≥20 tons of color chemical
X1+X2≥60 tons total
X1,X2≥ \$0 nonnegativity requirements

To solve the Cohen Chemicals problem graphically, we construct the problem’s feasible region, shown in Figure 1.

Figure 1 ■ Cohen Chemicals’ Feasible Region The area is not bounded to the right in a minimization problem as it is in a maximization problem.

Minimization problems are often unbounded outward (that is, on the right side and on the top), but this characteristic causes no problem in solving them. As long as they are bounded inward (on the left side and the bottom), we can establish corner points. The optimal solution will lie at one of the corners.

In this case, there are only two corner points, a and b, in Figure 1. It is easy to determine that at point a, X1=40 and X2=20, and that at point b, X1=30 and X2=30. The optimal solution is found at the point yielding the lowest total cost.

Thus

Total cost at a=2,500X1+3,000X2
2,500(40)+3,000(20)
=\$160,000
Total cost at b=2,500X1+3,000X2
= 2,500(30)+3,000(30)
=\$165,000

The lowest cost to Cohen Chemicals is at point a. Hence the operations manager should produce 40 tons of the black-and-white chemical and 20 tons of the color chemical.

📌 Question 1. There are two types of fertilizers F1 and F2. F1 consists of 10% nitrogen and 6°/o phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 cost \$6/kg and F2 costs \$5/kg, determine how much of each type of fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

Let the farmer buy X kg of fertilizer F1 and y kg of fertilizer F2. Therefore, x≥0 and y≥0.
The given information can be complied in a table as follows.

 Nitrogen (%) Phosphoric Acid (%) Cost (\$/kg) F1 (x) 10 6 6 F2 (y) 5 10 5 Requirements (kg) 14 14

F1 consists of 10% nitrogen and F2 consists of 5% nitrogen. However, the farmer requires at least 14 kg of nitrogen.
∴10% of x+5% of y≥14

x/10+y/20≥14→2x+y≥280

F1 consists of 6% phosphoric acid and F2 consists of 10% phosphoric acid. However, the farmer requires at least 14 kg of phosphoric acid.
Total cost of fertilizers, Z=6x+5y.
The mathematical formulation of the given problem is minimizing Z=6x+5y … (1)
subject to the constraints,
2x+y≥280…(2)
3x+5y≥700…(3)
x,y≥0…(4)

The feasible region determined by the system of constraints is as follows. It can be seen that the feasible region is unbounded.
The corner points are A(700/3, 0), B(100, 80), and C(0, 280).
The values of Z at these points are as follows.

 Corner point Z=6x+5y A(700/3, 0) 1400 B(100, 80) 1000 → Minimum C(0, 280) 1400

As the feasible region is unbounded, therefore, 1000 may or may not be the minimum value of Z.
For this, we draw a graph of the inequality, 6x+5y<1000, and check whether the resulting half plane has points in common with the feasible region or not. It can be seen that the feasible region has no common point with 6x+5y<1000. Therefore, 100 kg of fertiliser F1 and 80 kg of fertilizer F2 should be used to minimize the cost. The minimum cost is \$1000.
💎 How can we Minimize an Objective Function Using the Iso-Cost Line Approach? 👈

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