
Section 08: Financial Mathematics
Test your understanding of Annuities
Calculate the future value of saving 250/month for 15 years at 4% annual interest.
Find the monthly payment for a 30,000 loan at 5.4% for 4 years.
After 12 payments on a 15,000 loan at 6% annual for 3 years, what is the outstanding balance?
What is the present value of receiving 1,500/month for 10 years at 6% annual?
Exam-Critical Topic
These pricing concepts appear frequently on the Feststellungsprufung!
Every business has two types of costs:
Cost Components
Fixed Costs (\(K_f\) or \(FC\)): Costs that don’t change with production quantity
Variable Costs (\(K_v(x)\) or \(VC(x)\)): Costs that change with quantity produced
\[K(x) = K_f + K_v(x)\]
Common cost function forms:
Linear: \(K(x) = 500 + 3x\)
Quadratic: \(K(x) = 400 + 10x + 0.02x^2\)
Cubic: \(K(x) = 1000 + 50x - 0.5x^2 + 0.01x^3\)

How much does each unit cost on average?
Average Cost (Durchschnittskosten)
\[\bar{K}(x) = \frac{K(x)}{x} = \frac{K_f + K_v(x)}{x} = \frac{K_f}{x} + \frac{K_v(x)}{x}\]
This can be written as: \[\bar{K}(x) = \underbrace{\frac{K_f}{x}}_{\text{Avg. Fixed Cost}} + \underbrace{\frac{K_v(x)}{x}}_{\text{Avg. Variable Cost}}\]
Variable Cost Per Unit (Stuckkosten variabel)
\[k_v(x) = \frac{K_v(x)}{x} = \frac{K(x) - K_f}{x}\]
This is the average variable cost - what each unit costs in variable expenses alone.
Example: \(K(x) = 500 + 10x + 0.01x^2\)
\[k_v(x) = \frac{(500 + 10x + 0.01x^2) - 500}{x} = \frac{10x + 0.01x^2}{x} = 10 + 0.01x\]
Critical insight for pricing decisions:
The variable cost per unit is the absolute minimum you should charge!
Short-Term Lower Limit Price (Kurzfristige Preisuntergrenze)
The minimum price at which a company should continue production in the short term.
\[p_{min,short} = \min\{k_v(x)\} = \text{minimum variable cost per unit}\]
At this price, the company covers its variable costs but not fixed costs.
When to use:
Method: Use calculus to find the minimum of \(k_v(x)\)
Step 1: Calculate \(k_v(x) = \frac{K(x) - K_f}{x}\)
Step 2: Find \(k_v'(x)\) and set equal to zero
Step 3: Solve for \(x\)
Step 4: Calculate \(k_v(x)\) at this value
Alternatively, use: \(k_v'(x) = \frac{K'(x) \cdot x - (K(x) - K_f)}{x^2}\)
Setting this to zero: \(K'(x) \cdot x = K(x) - K_f\)
Given: \(K(x) = 800 + 20x - 0.2x^2 + 0.002x^3\)
Step 1: Variable cost function \[K_v(x) = 20x - 0.2x^2 + 0.002x^3\]
Step 2: Variable cost per unit \[k_v(x) = \frac{20x - 0.2x^2 + 0.002x^3}{x} = 20 - 0.2x + 0.002x^2\]
Step 3: Find minimum \[k_v'(x) = -0.2 + 0.004x = 0\] \[x = 50\]
Step 4: Calculate minimum variable cost per unit \[k_v(50) = 20 - 0.2(50) + 0.002(50)^2 = 20 - 10 + 5 = 15\]
Short-term lower limit price = 15 Euro per unit
The company should not sell below 15 Euro, even in the short term!

Long-Term Lower Limit Price (Langfristige Preisuntergrenze)
The minimum price for sustainable long-term production.
\[p_{min,long} = \min\{\bar{K}(x)\} = \text{minimum average total cost}\]
At this price, the company covers ALL costs (both fixed and variable).
When to use:
Same method, but for average TOTAL cost:
Step 1: Calculate \(\bar{K}(x) = \frac{K(x)}{x}\)
Step 2: Find \(\bar{K}'(x)\) and set equal to zero
Step 3: Solve for \(x\)
Step 4: Calculate \(\bar{K}(x)\) at this value
Same cost function: \(K(x) = 800 + 20x - 0.2x^2 + 0.002x^3\)
Step 1: Average total cost \[\bar{K}(x) = \frac{800 + 20x - 0.2x^2 + 0.002x^3}{x} = \frac{800}{x} + 20 - 0.2x + 0.002x^2\]
Step 2: Find minimum \[\bar{K}'(x) = -\frac{800}{x^2} - 0.2 + 0.004x = 0\]
Multiply by \(x^2\): \[-800 - 0.2x^2 + 0.004x^3 = 0\] \[0.004x^3 - 0.2x^2 - 800 = 0\]
Solving \(0.004x^3 - 0.2x^2 - 800 = 0\) (or \(x^3 - 50x^2 - 200000 = 0\)):
Using numerical methods or calculator: \(x \approx 76.4\)
Step 4: Calculate minimum average cost \[\bar{K}(76.4) = \frac{800}{76.4} + 20 - 0.2(76.4) + 0.002(76.4)^2 \approx 17.44\]
Long-term lower limit price = 17.44 Euro per unit
For sustainable production, the company must charge at least 17.44 Euro!

Three Pricing Zones
| Price Level | Decision |
|---|---|
| \(p < k_v^{min}\) | STOP production - losing on every unit |
| \(k_v^{min} \leq p < \bar{K}^{min}\) | Continue short-term - covers variable costs, contributes to fixed |
| \(p \geq \bar{K}^{min}\) | Sustainable - covers all costs |
Short-term lower limit price:
Long-term lower limit price:
A company has excess capacity. A customer offers to buy 100 units at 16 Euro each.
Using our cost function with:
Analysis:
Decision: Accept the order!
While 16 < 17.44 means we’re not covering all costs, each unit sold at 16 contributes 1 Euro toward fixed costs. Better than nothing!
Contribution Margin (Deckungsbeitrag)
\[\text{Contribution per unit} = p - k_v(x)\]
This is how much each unit “contributes” toward covering fixed costs.
At optimal production level (x = 50):
For 100 units: 100 Euro contribution toward fixed costs!
Work in pairs
Problem 1: Given \(K(x) = 1200 + 30x + 0.1x^2\)
Problem 2: A company has cost function \(K(x) = 500 + 15x\)
FSP exam problems often combine these elements:
Make sure you can distinguish between these questions:
Essential Formulas for Cost Analysis
| Concept | Formula |
|---|---|
| Total Cost | \(K(x) = K_f + K_v(x)\) |
| Variable Cost per Unit | \(k_v(x) = \frac{K(x) - K_f}{x}\) |
| Average Total Cost | \(\bar{K}(x) = \frac{K(x)}{x}\) |
| Short-term Limit | \(\min\{k_v(x)\}\) |
| Long-term Limit | \(\min\{\bar{K}(x)\}\) |
| Contribution Margin | \(p - k_v(x)\) |
Section Complete!
You’ve completed Section 08: Financial Mathematics. Review all formulas before the exam!
These pricing decisions are frequently tested! Practice until you can solve them confidently.
Session 08-03 - Cost Analysis & Pricing Decisions | Dr. Nikolai Heinrichs & Dr. Tobias Vlćek | Home