Optimization in Business – Linear Programming, Marginal Analysis & Profit Maximisation

1. What is Optimisation?

Optimisation is the mathematical process of finding the best possible value (maximum or minimum) of an objective function subject to a set of constraints. In business the objective is usually profit, cost, revenue or utility, while constraints represent limited resources such as capital, labour, raw material, or time.

Key Definitions

2. Linear Programming (LP)

Linear programming solves optimisation problems where both the objective function and all constraints are linear.

2.1 Standard Form

$$\begin{array}{rl}\text{Maximise }& Z=c_1{x}_1+c_2{x}_2+\cdots +c_n{x}_n\\ \text{subject to }& \{\begin{array}{l}{a}_{11}{x}_1+a_{12}{x}_2+\cdots +a_{1n}{x}_n\le b_1\\ a_{21}{x}_1+a_{22}{x}_2+\cdots +a_{2n}{x}_n\le b_2\\ ⋮\\ a_{m1}{x}_1+a_{m2}{x}_2+\cdots +a_{mn}{x}_n\le b_m\\ x_j\ge 0(j=1,\dots ,n)\end{array}\end{array}$$

2.2 Graphical Solution (Two‑Variable Case)

1. Plot each constraint as a straight line.
2. Shade the region that satisfies all “≤” (or “≥”) inequalities – this is the feasible region.
3. Identify corner (extreme) points of the feasible region.
4. Substitute each corner point into the objective function; the highest (or lowest) value is the optimum.

Worked Example – Daraz Order‑Fulfilment

A small seller on Daraz sells T‑shirts (product $x$) and Mugs (product $y$).

• Profit per T‑shirt = Rs 250, per Mug = Rs 150.
• Daily labour: 8 hours. Each T‑shirt needs 0.2 h, each Mug needs 0.1 h.
• Daily packaging material: 30 units. Each T‑shirt uses 1 unit, each Mug uses 0.5 unit.

Formulate LP

$$\begin{array}{rl}\text{Maximise }Z& =250x+150y\\ \text{s.t. }& \{\begin{array}{l}0.2x+0.1y\le 8\text{(labour)}\\ x+0.5y\le 30\text{(material)}\\ x,y\ge 0\end{array}\end{array}$$

Convert to standard form for graphing

$$\begin{array}{rl}x+0.5y& \le 30\\ 2x+y& \le 80(\text{multiply labour constraint by 10})\end{array}$$

Plotting

Corner points (found by solving line intersections):

1. Intersection of material line with axes → (30,0) and (0,60).
2. Intersection of labour line with axes → (40,0) and (0,80).
3. Intersection of the two lines: solve

$$\{\begin{array}{l}x+0.5y=30\\ 2x+y=80\end{array}\Rightarrow x=20,y=20.$$

Evaluate Z

Optimal solution: Produce 0 T‑shirts and 80 Mugs per day → maximum profit Rs 12,000.

2.3 Simplex Method (More than Two Variables)

When $n>2$ the graphical method is impractical. The Simplex algorithm moves from one basic feasible solution to another, improving the objective value at each step until no further improvement is possible.

Key Steps

1. Convert all constraints to equalities by adding slack, surplus and artificial variables.
2. Set up the initial simplex tableau.
3. Identify entering variable (most positive coefficient in maximisation).
4. Identify leaving variable (minimum ratio test).
5. Pivot to obtain a new tableau.
6. Repeat until all objective‑function coefficients are ≤ 0 (maximisation).

Tip: In exams you are rarely required to complete a full simplex for >3 variables; focus on interpreting the final tableau.

3. Marginal Analysis

Marginal analysis examines the effect of a small change in output on revenue, cost or profit. It is the calculus‑based counterpart of LP for continuous decisions.

3.1 Marginal Revenue (MR) and Marginal Cost (MC)

$$\text{MR}(q)=\frac{dR(q)}{dq},\text{MC}(q)=\frac{dC(q)}{dq}$$

The profit‑maximising output $q^{\ast }$ satisfies

$$\text{MR}(q^{\ast })=\text{MC}(q^{\ast }).$$

3.2 Worked Example – Bank Loan Interest

A bank offers a loan product where total revenue from interest is

$$R(q)=1200q-5q^2,$$

and total cost of processing the loan is

$$C(q)=200q+2q^2,$$

where $q$ (in thousands of rupees) is the loan amount.

Step 1 – Compute MR and MC

$$\begin{array}{rl}\text{MR}(q)& =\frac{d}{dq}(1200q-5q^2)=1200-10q,\\ \text{MC}(q)& =\frac{d}{dq}(200q+2q^2)=200+4q.\end{array}$$

Step 2 – Set MR = MC

$$1200-10q=200+4q\Rightarrow 1000=14q\Rightarrow q^{\ast }=\mathrm{71.43.}$$

Thus the bank maximises profit when the loan size is ≈ Rs 71,430.

Step 3 – Verify with second‑derivative test

$$\frac{d^2\Pi }{d{q}^2}=\frac{d}{dq}(\text{MR}-\text{MC})=-10-4=-14<0,$$

confirming a maximum.

4. Profit‑Maximisation Graphs

Profit $\Pi (q)=R(q)-C(q)$. The maximum profit occurs where the vertical distance between the revenue and cost curves is greatest.

4.1 Steps to Sketch

1. Plot $R(q)$ and $C(q)$.
2. Identify intersection points (break‑even).
3. Locate the point where the slope of $R$ equals the slope of $C$ (MR = MC).
4. Draw a vertical line at this $q^{\ast }$; the height difference gives ${\Pi }_{\max }$.

4.2 Worked Example – Ncell Data Plan

A telecom company sells a data plan.

• Revenue per GB: $R(q)=500q-2q^2$.
• Cost per GB: $C(q)=150q+q^2$.

Find profit‑maximising GB $q^{\ast }$.

$$\begin{array}{rl}\text{MR}& =500-4q,\\ \text{MC}& =150+2q,\\ 500-4q& =150+2q\Rightarrow 350=6q\Rightarrow q^{\ast }=58.33\text{ GB}.\end{array}$$

Maximum profit

$${\Pi }_{\max }=R(58.33)-C(58.33)=(500\cdot 58.33-2\cdot 58.33^2)-(150\cdot 58.33+58.33^2)\approx Rs13,889.$$

5. Sensitivity (Shadow Price) Analysis

After solving an LP, the shadow price tells how much the objective function would improve per unit increase in a right‑hand‑side resource.

5.1 Example from Daraz LP

In the Daraz example, the optimal solution used the full material constraint (30 units) but only 8 hours of labour (since 0 Mugs use 0 labour). The shadow price for material is the increase in profit if one extra unit of material becomes available.

From the final tableau (omitted for brevity) the shadow price = Rs 150 per extra material unit. Hence, buying an additional packaging unit yields an extra Rs 150 profit, justifying a small purchase cost.

6. Comparison of Optimisation Techniques

| Technique | When to Use | Strengths | Weaknesses |
|-----------|-------------|-----------|------------|
| Graphical LP (2‑var) | Small‑scale problems, teaching | Intuitive visual insight | Limited to ≤2 decision variables |
| Simplex LP | Medium‑large linear models | Guarantees optimal solution, handles many variables | Requires tableau manipulation, may be computationally heavy |
| Marginal Analysis (Calculus) | Continuous decisions, smooth cost/revenue functions | Direct use of derivatives, easy for single‑product cases | Not applicable when functions are non‑differentiable or constraints are discrete |
| Integer Programming | Problems where variables must be whole numbers (e.g., number of trucks) | Provides realistic solutions for discrete decisions | NP‑hard; often solved by branch‑and‑bound, not by hand |

7. Applications in Business

• Inventory Management: Determine order quantity that minimises holding + ordering cost (EOQ model – a special case of optimisation).
• Transport & Logistics: Linear programming for vehicle routing (e.g., Pathao delivery fleet).
• Workforce Scheduling: Allocate shifts to meet demand while minimising overtime.
• Pricing Strategy: Use marginal analysis to set price where marginal revenue equals marginal cost.
• Capital Budgeting: Choose project mix that maximises NPV under budget constraints (LP formulation).

Real‑World Visuals

8. In the real world

1. eSewa Payment Gateway – Uses linear programming to allocate server capacity among concurrent transactions while minimising latency. The objective function maximises the number of processed payments per second, subject to CPU, memory and network bandwidth constraints.
2. Khalti’s Transaction Fee Optimisation – Applies marginal analysis: the fee structure is set where the marginal revenue from an additional transaction equals the marginal cost of fraud detection and settlement.
3. NEPSE Portfolio Allocation – Fund managers employ LP to maximise expected return of a stock portfolio under risk‑budget constraints; the shadow price of the risk limit indicates how much extra return could be earned by relaxing the risk cap.

Worked Real Situation:
A small retailer in Kathmandu wants to decide how many units of hand‑woven scarves ($x$) and hand‑made mugs ($y$) to produce daily.

• Profit per scarf = Rs 300, per mug = Rs 180.
• Scarves need 0.3 h labour, mugs need 0.2 h. Daily labour = 6 h.
• Fabric for scarves = 2 m² each, clay for mugs = 0.5 m² each. Daily material = 12 m².

Formulating and solving the LP (as shown earlier) yields an optimal mix of 12 scarves and 6 mugs, giving a daily profit of Rs 4,560. This mirrors the Daraz example but with locally recognisable products.

9. Summary of Steps for Solving Optimisation Problems

Exam tip

• Read the question twice. Identify whether the problem asks for a maximum or minimum and note all given constraints.
• List variables first. A clear table of decision variables, objective function and constraints earns marks even before solving.
• For LP with ≤ 2 variables: always sketch the feasible region; the optimum lies at a corner point. Mark each corner, compute the objective value, and state the best one.
• For calculus‑based marginal problems: write revenue and cost functions, differentiate, set MR = MC, and verify the second‑derivative sign. Show the algebraic steps; a missing derivative sign costs marks.
• Shadow price: if the exam provides a final simplex tableau, locate the RHS column’s “dual” values – they are the shadow prices. Mention what a positive/negative value implies.
• Time management: allocate ~5 minutes for setting up the model, ~10 minutes for solving, and ~2 minutes for checking calculations.

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