---
title: "Static mathematical model with example"
subject: "Simulation and Modeling"
chapter: 2
key: "semester_8/simulation_and_modeling:static-mathematical-model"
source: https://www.bctengineer.com/q/semester_8/simulation_and_modeling/static-mathematical-model
site: BCT Engineering
---

# Static mathematical model with example

Simulation and Modeling · Chapter 2: Physical and Mathematical models

- Asked in: 2072 Ashwin, 6 marks
- Video: [Implementation of Static Mathematical Model | Simple Market Model || Linear Market Model](https://www.youtube.com/watch?v=dldy7qKn-8M) — CSEian's Diary
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/simulation_and_modeling/static-mathematical-model
- Whole chapter: https://www.bctengineer.com/md/semester_8/simulation_and_modeling/chapter-2

## Question

Explain the static mathematical model with suitable example.

## Short answer

A static mathematical model uses algebraic equations without time, having no memory. Example: Ohm’s law V = IR (V = 10 V for I = 0.5 A, R = 20 Ω) or break-even quantity x = F/(s – v) = 500 units when fixed cost Rs 10,000, price Rs 50, variable cost Rs 30.

## Answer

A **static mathematical model** is a set of algebraic relationships between variables that describes the system at a given instant or at equilibrium. Time does not appear in the equations, so the output depends only on the present value of the inputs. It has no memory and needs no initial conditions.

#### General form
`y = f(x1, x2, ..., xn)`, where y is the output and the x are inputs and parameters.

#### Characteristics
- Algebraic equations (linear or nonlinear).
- Gives steady-state or one-time behaviour.
- Easy to solve and can be evaluated directly or by solving simultaneous equations.
- Cannot predict transient behaviour or how the system evolves.

#### Examples
1. **Ohm's law:** `V = I R`. If R = 20 ohm and I = 0.5 A, V = 10 V. No time term appears.
2. **Break-even model:** profit `P = (s - v) x - F`. With selling price s = Rs 50, variable cost v = Rs 30, fixed cost F = Rs 10,000, break-even quantity `x = F/(s - v) = 10,000/20 = 500` units.
3. **Economic order quantity:** `Q = sqrt(2 D S / H)`. For D = 1,000 units/year, S = Rs 100 per order, H = Rs 5 per unit per year, `Q = sqrt(40,000) = 200` units.
4. **Static queue result (steady state):** M/M/1 utilization `rho = lambda/mu`; with lambda = 4 and mu = 6 per hour, rho = 0.667.

#### Uses and limits
Static models are used for design, sizing and what-if analysis when the time pattern is unimportant. When the time behaviour matters (response after a disturbance, growth, oscillation) a dynamic model is needed. A static model can be seen as a dynamic model in its steady state, where all derivatives are zero.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which is true of a static mathematical model?
   A. It has no memory, needs no initial conditions and time does not appear
   B. It predicts transient behaviour
   C. It needs initial conditions
   D. It uses time derivatives
   Answer: A. Static models are algebraic; output depends only on present inputs. (Key point: Static model characteristics)

2. With selling price Rs 50, variable cost Rs 30 and fixed cost Rs 10,000, the break-even quantity is:
   A. 200 units
   B. 500 units
   C. 333 units
   D. 1,000 units
   Answer: B. x = F/(s - v) = 10,000/20 = 500 units. (Key point: Break-even calculation)

3. For D = 1,000 units/year, S = Rs 100 per order and H = Rs 5 per unit per year, Q = sqrt(2DS/H) equals:
   A. 100 units
   B. 400 units
   C. 200 units
   D. 40,000 units
   Answer: C. Q = sqrt(2 x 1000 x 100 / 5) = sqrt(40,000) = 200 units. (Key point: EOQ calculation)

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