As we have explained in last article ( Introduction to Optimization), Constrained is similar to unconstrained optimization in some points like below:

  • Requires a prescriptive model.
  • Uses an objective function.
  • Solution is an extreme value.

And different mainly in:

  • Multiple decision variables.
  • Constraints on resources.

We will study business situation through simple example and explain to to optimize the solution step by step. Our motivation problem is called Banner Chemicals


  • Banner Chemicals manufactures specialty chemicals. One of their products comes in two grades, high and supreme. The capacity at the plant is 110 barrels per week.
  • The high and supreme grade products use the same basic raw materials but require different ratios of additives. The high grade requires 3 gallons of additive A and 1 gallon of additive B per barrel while the supreme grade requires 2 gallons of additive A and 3 gallons of additive B per barrel.
  • The supply of both of these additives is quite limited. Each week, this product line is allocated only 300 gallons of additive A per week and 280 gallons of additive B.
  • A barrel of the high grade has a profit margin of $80 per barrel while the supreme grade has a profit margin of $200 per barrel.

Question: How many barrels of High and Supreme grade should Banner Chemicals produce each week?

Problem Formulation Process

  1. Determine the decision variables
  • What are you trying to decide?
  • What are their upper or lower bounds?
  1. Formulate the objective function
  • What are we trying to minimize or maximize?
  • Must include the decision variables and the form of the function determines approach (linear for LP)
  1. Formulate each constraint
  • What is my feasible region?  What are my limits?
  • Must include the decision variables and will almost always be linear functions
Step 1.  Determine Decision Variables     
  • XH = Number of High grade barrels to produce per week
  • XS = Number of Supreme grade barrels to produce per week
  • Bounds XH ≥ 0     XS ≥ 0
Step 2. Formulate Objective Function
  • Profit = 80XH + 200XS
  • Maximize z(XH, XS) = 80XH + 200XS
Step 3. Formulate Constraints  

Plant Capacity is 110 barrels    : XH + XS ≤ 110


  • 3XH + 2XS ≤ 300
  • XH + 3XS ≤ 280




Additive A

3 gal

2 gal

300 gal

Additive B

1 gal

3 gal

280 gal


Graphical Representation

Sensitivity Analysis

Formal Notation


A number of anomalies can occur in LPs:

  • Alternate Optimal Solutions
  • Redundant Constraints
  • Infeasibility

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