We’ll explore how Linear Programming (LP) helps make critical decisions in different areas of a business.
Marketing: How to design a cost-effective survey.
Finance: How to invest millions for maximum return.
Operations: How to schedule a workforce to meet production goals.
Marketing Application: MSI Research
The Scenario: Market Survey Inc. (MSI) needs to conduct 1000 door-to-door interviews for a new product. They want to do it as cheaply as possible, but they have a lot of rules to follow.
The Puzzle: How do you plan the interviews to meet all the requirements at the minimum cost? This is a classic resource allocation problem perfect for LP.
MSI: Breaking Down the Problem
First, let’s identify the core components of our LP model.
1. Decisions (Variables) We need to decide how many interviews to conduct in each category.
\(DC\): Daytime, with Children
\(EC\): Evening, with Children
\(DNC\): Daytime, No Children
\(ENC\): Evening, No Children
2. Objective (The Goal) Minimize the total survey cost.
Interview Type
Cost
Daytime, with Children
$20
Evening, with Children
$25
Daytime, No Children
$18
Evening, No Children
$20
Minimize Cost:20DC + 25EC + 18DNC + 20ENC
MSI: The Rules (Constraints)
These are the rules MSI must follow, translated into math.
Rule in Plain English
Mathematical Constraint
1. Conduct exactly 1000 interviews in total.
\(DC+EC+DNC+ENC = 1000\)
2. At least 400 households with children.
\(DC+EC \geq 400\)
3. At least 400 households without children.
\(DNC+ENC \geq 400\)
4. Evening interviews must be \(\geq\) daytime.
\(EC+ENC \geq DC+DNC\)
5. \(\geq 40\%\) of “with children” interviews in evening.
\(EC \geq 0.4(DC+EC)\)
6. \(\geq 60\%\) of “no children” interviews in evening.
\(ENC \geq 0.6(DNC+ENC)\)
MSI: The Optimal Solution
After running this model through a solver, we get the cheapest plan.
Optimal Interview Plan:
Variable
Description
Number of Interviews
DC
Daytime, with Children
240
EC
Evening, with Children
160
DNC
Daytime, No Children
240
ENC
Evening, No Children
360
Minimum Total Cost:$20,320
You can verify the cost with this simple calculator:
Financial Application: Winslow Portfolio
The Scenario: Winslow Savings has $20 million to invest over the next 4 months. They want to maximize their interest earnings.
The Puzzle: How should they allocate their money each month between different investment options to get the highest return, while making sure they have enough cash for a big project later? This is a multi-period planning problem.
Winslow: Variables and Objective
This model is dynamic, so our variables need to track investments in each month i (where i=1, 2, 3, 4).
\(G_i\): New investment in Government bonds in month i. (2% return over 2 months)
\(C_i\): New investment in Construction loans in month i. (6% return over 3 months)
\(L_i\): Liquid funds invested locally in month i. (0.75% return per month)
Total Gov’t bonds held at any time must not exceed 8M.
\(C_1+C_2+C_3 \le 8,000,000\), etc.
Total Const. loans held at any time must not exceed 8M.
The Optimal Solution: The solver provides a detailed month-by-month investment plan that results in a maximum interest of $1,429,214.
Operations Application: NWC Workforce
The Scenario: National Wing Company (NWC) has a contract to produce a specific number of airplane wings each month for three months. They need to manage their workforce to meet these targets.
The Puzzle: How can NWC hire, train, and assign workers each month to meet the production schedule at the minimum possible payroll cost?
NWC: The Workforce Flow
A key part of this problem is understanding how the workforce changes over time. A new recruit isn’t immediately productive.
graph TD
subgraph Month 1 - April
R1(Recruits)
T1(Trainers)
end
subgraph Month 2 - May
A2(Apprentices)
R2(Recruits)
T2(Trainers)
end
subgraph Month 3 - June
A3(Apprentices)
QW3(Qualified Workers)
end
R1 -- 1 month training --> A2
A2 -- 1 month experience --> QW3
T1 -- Train --> R1
T2 -- Train --> R2
A Recruit needs one month of training to become an Apprentice.
An Apprentice needs one more month to become a fully Qualified Worker.
NWC: Variables and Objective
Let’s define our decision variables and goal for April (i=1), May (i=2), and June (i=3).
Production Targets: The number of wings produced each month must meet the contract. \(0.6P_1 + 0.3T_1 + 0.05R_1 \ge 20\) (April’s production)
Workforce Balance: The number of qualified workers flows from one month to the next. \(P_2 + T_2 = P_1 + T_1\) (Qualified workers don’t disappear) \(A_2 = R_1\) (April’s recruits become May’s apprentices)
Training Capacity: Each trainer can handle two recruits. \(2T_1 \ge R_1\)
Boundary Conditions: Start with 100 workers, end with at least 140. \(P_1 + T_1 = 100\)\(P_3 + A_3 \ge 140\)
NWC: The Optimal Workforce Plan
The LP solver gives NWC a detailed, cost-effective staffing plan for the entire quarter.
Optimal Plan Summary:
Employee Type
April
May
June
Producers
100
80
100
Trainers
0
20
0
Apprentices
0
0
40
Recruits
0
40
0
Minimum Total Wage Cost:$1,098,000
Insight: The model shows it’s cheapest to start with everyone producing. Then, in May, some producers switch to being trainers to hire and train 40 new recruits, who become apprentices in June, helping to meet the final production targets and the goal of having 140 workers by July.