Operations 6 min read

How to Optimize Monthly Oil Purchases and Production for Maximum Profit

Given spot and futures prices for vegetable and non‑vegetable oils, refining capacity limits, storage constraints, and a required product hardness range, this model determines the optimal monthly buying, refining, and inventory decisions that maximize profit for a food‑oil manufacturer over a six‑month horizon.

Model Perspective
Model Perspective
Model Perspective
How to Optimize Monthly Oil Purchases and Production for Maximum Profit

Problem Description

Manufacturer needs to refine several raw oils and blend them to produce a sellable food product. Two categories of oils are required: vegetable oils (VEG 1, VEG 2) and non‑vegetable oils (OIL 1, OIL 2, OIL 3).

Spot and futures prices (USD/ton) for each oil by month (January–June) are given:

January: VEG1 110, VEG2 120, OIL1 130, OIL2 110, OIL3 115

February: VEG1 130, VEG2 130, OIL1 110, OIL2 90, OIL3 115

March: VEG1 110, VEG2 140, OIL1 130, OIL2 100, OIL3 95

April: VEG1 120, VEG2 110, OIL1 120, OIL2 120, OIL3 125

May: VEG1 100, VEG2 120, OIL1 150, OIL2 110, OIL3 105

June: VEG1 90, VEG2 100, OIL1 140, OIL2 80, OIL3 135

Additional considerations:

Final food product sells for $150 per ton.

Vegetable and non‑vegetable oils are refined on separate production lines.

Refining capacity is limited to 200 t of vegetable oil and 250 t of non‑vegetable oil per month.

No waste in refining; total refined oil equals total refined product.

Refining cost is negligible.

Storage limit per oil type is 1000 t; storage cost is $ ? per ton per month (value omitted).

Food product hardness must be between 3 and 6. Hardness is a linear blend of oil hardnesses: VEG1 = 8.8 VEG2 = 6.1 OIL1 = 2.0 OIL2 = 4.2 OIL3 = 5.0

Initial inventory: 500 t of each oil at the beginning of January; also 500 t of each oil must be on hand at the end of June.

Mathematical Model

Sets

T: set of months.

V: set of vegetable oils.

N: set of non‑vegetable oils.

O: set of all oil types (V ∪ N).

Parameters

p = 150 USD/t (selling price of final product).

s₀ᵢ: initial storage of oil i (500 t).

sᴛᵢ: target storage of oil i at end of horizon (500 t).

cᵢₜ: purchase price of oil i in month t (from the price tables).

hᵢ: hardness of oil i (values above).

γᵥ = 200 t (vegetable refining capacity per month).

γₙ = 250 t (non‑vegetable refining capacity per month).

H_min = 3, H_max = 6 (hardness limits).

k: storage cost per ton per month (not specified).

Decision Variables

xᵢₜ: amount of oil i purchased in month t.

yᵢₜ: amount of oil i refined (consumed) in month t.

zᵢₜ: inventory of oil i at end of month t.

qₜ: amount of final product produced in month t.

Objective

Maximize total profit over the planning horizon:

<code>∑ₜ (p·qₜ – ∑ᵢ cᵢₜ·xᵢₜ – k·zᵢₜ)</code>

Constraints

Inventory balance: zᵢ₍ₜ₋₁₎ + xᵢₜ = yᵢₜ + zᵢₜ for all i, t.

Refining capacity: ∑_{i∈V} yᵢₜ ≤ γᵥ and ∑_{i∈N} yᵢₜ ≤ γₙ for all t.

Hardness: H_min ≤ (∑ᵢ hᵢ·yᵢₜ) / qₜ ≤ H_max for all t.

Mass balance: qₜ = ∑ᵢ yᵢₜ for all t.

Initial and final inventory: zᵢ₀ = s₀ᵢ, zᵢ_{|T|} = sᴛᵢ.

The model determines the optimal monthly purchase and refining plan that maximizes profit while respecting capacity, storage, and product quality constraints.

operations researchLinear Programmingsupply chain optimizationoil blendingproduction planning
Model Perspective
Written by

Model Perspective

Insights, knowledge, and enjoyment from a mathematical modeling researcher and educator. Hosted by Haihua Wang, a modeling instructor and author of "Clever Use of Chat for Mathematical Modeling", "Modeling: The Mathematics of Thinking", "Mathematical Modeling Practice: A Hands‑On Guide to Competitions", and co‑author of "Mathematical Modeling: Teaching Design and Cases".

0 followers
Reader feedback

How this landed with the community

login Sign in to like

Rate this article

Was this worth your time?

Sign in to rate
Discussion

0 Comments

Thoughtful readers leave field notes, pushback, and hard-won operational detail here.