optimization-ai999x: offline, verified optimization solver with step-by-step answers

Desktop app & command line · Compute in your browser · Install (Python) · Guides · Evaluation · AI999X on Hugging Face · Website · Support: Buy Me a Coffee

License: Apache-2.0 · Author: Ibnul Jaif Farabi (creator and engineer) · Organization: AI999X · Version 1.5.0

Field: Mathematics & Statistics · Tasks: Question Answering

optimization-ai999x is an offline, deterministic optimization solver written in Rust: a desktop app, a command-line program, a browser module and a Python package. It answers optimization questions with verified results from 72 certified capabilities, gives 240 documented answers identically on Windows and Linux, and refuses rather than guesses when an answer cannot be verified.

optimization-ai999x reads a question in plain English (or a typed call), picks a certified optimization capability, computes the answer exactly, checks it, and returns the result with its steps, units, LaTeX and assumptions, or a refusal that says what is missing. It runs fully offline on any CPU, needs no GPU, no API key and no internet, and is built to give zero wrong answers: when it is not sure, it refuses instead of guessing.

Run it your way. optimization-ai999x comes as the AI999X desktop app and as the command-line program optimization for Windows and Linux, runs in your browser as WebAssembly, and installs as a Python package. The desktop app, the command line and the browser run the same compiled Rust program on your own device: offline, no Python needed.

Get it

Way to run it	For	Get it
AI999X Optimization desktop app 1.5.0	Windows, Linux. This model in its own app: search, ask in words, forms, live figures, explanations, offline	Windows (x86-64) · Linux (x86-64)
AI999X desktop app 1.5.0	Windows, Linux. Every AI999X model in one app: search, ask in words, forms, live figures, explanations, offline	Windows (x86-64) · Linux (x86-64)
Command line (optimization)	Windows, Linux. One program per model, offline, no Python	Windows (x86-64) · Linux (x86-64)
In your browser	any device. Nothing to install	Compute on the model page
Python	Python 3.10+	pip install optimization-ai999x

Checksums: SHA256SUMS · every version: https://downloads.ai999x.com/desktop/latest/ · install guide: desktop & command line

Quick start

Desktop app

Download AI999X Optimization for your system (above), unpack it and start it; type a question in words or pick a capability, fill in the form and press Compute. Figures and an explanation of the answer appear beside it. The AI999X desktop app has the same, with Optimization among all AI999X models.

Linux, from a terminal:

curl -LO https://downloads.ai999x.com/desktop/1.5.0/ai999x-optimization-1.5.0-linux-x86_64.tar.gz
curl -LO https://downloads.ai999x.com/desktop/1.5.0/SHA256SUMS
sha256sum -c --ignore-missing SHA256SUMS
tar xzf ai999x-optimization-1.5.0-linux-x86_64.tar.gz
./ai999x-optimization-1.5.0-linux-x86_64/ai999x-optimization

Command line

Linux:

curl -LO https://downloads.ai999x.com/desktop/1.5.0/ai999x-cli-1.5.0-linux-x86_64.tar.gz
curl -LO https://downloads.ai999x.com/desktop/1.5.0/SHA256SUMS
sha256sum -c --ignore-missing SHA256SUMS
mkdir -p ~/.local/share && tar xzf ai999x-cli-1.5.0-linux-x86_64.tar.gz -C ~/.local/share
echo 'export PATH="$HOME/.local/share/ai999x-cli-1.5.0-linux-x86_64:$PATH"' >> ~/.bashrc && source ~/.bashrc
optimization --version

Windows (PowerShell):

$F = "ai999x-cli-1.5.0-windows-x86_64.zip"
Invoke-WebRequest "https://downloads.ai999x.com/desktop/1.5.0/$F" -OutFile $F
Invoke-WebRequest "https://downloads.ai999x.com/desktop/1.5.0/SHA256SUMS" -OutFile SHA256SUMS
(Get-FileHash $F -Algorithm SHA256).Hash.ToLower() -eq ((Select-String -SimpleMatch $F SHA256SUMS).Line -split ' ')[0]
Expand-Archive $F -DestinationPath "$env:LOCALAPPDATA\AI999X" -Force
$D = "$env:LOCALAPPDATA\AI999X\ai999x-cli-1.5.0-windows-x86_64"; $env:Path += ";$D"
[Environment]::SetEnvironmentVariable("Path", [Environment]::GetEnvironmentVariable("Path", "User") + ";$D", "User")
optimization --version

Every command below was run as shown; the text under it is what the program printed.

Ask in words: Write the question as you would say it: the program reads the quantities and their units, picks the capability and answers.

$ optimization ask "Solve the transportation problem with supply [20, 30], demand [10, 25, 15] and cost matrix [[2, 3, 1], [5, 4, 8]]."
Read as: Transportation problem (minimum shipping cost)
min_cost = 150
shipments = [[5.0,0.0,15.0],[5.0,25.0,0.0]]
plan_unique = yes
dual = [-3.0,0.0,5.0,4.0,4.0]
  by: Transportation problem (minimum shipping cost)  (the transportation LP solved exactly (rational simplex), certified by its dual (u_i + v_j <= c_ij))
  assumes: total supply >= total demand (surplus stays at the sources at no cost); at most 60 routes

What a capability needs: The equation or method, every input and output with its SI unit. Read this before solve: it tells you the input names.

$ optimization info eoq
Economic order quantity (EOQ)
  minimise D S / Q + H Q / 2 (at the optimum ordering cost equals holding cost)
  in   D        demand per period (units, e.g. per year) [1]
  in   S        fixed cost per order [1]
  in   H        holding cost per unit per period [1]
  out  Q        economic order quantity [1]
  out  total_cost minimum ordering + holding cost per period [1]
  out  orders_per_period number of orders per period [1]
  out  cycle_time time between orders (in periods) [1]

Solve: Inputs as name=value, each with any unit of the right kind. The answer lists every output, the method used, each input converted to SI, and the conditions the answer assumes.

$ optimization solve eoq D=1200 S=50 H=3
Q = 200
total_cost = 600
orders_per_period = 6
cycle_time = 0.1666666667
  by: Economic order quantity (EOQ)  (minimise D S / Q + H Q / 2 (at the optimum ordering cost equals holding cost))
  D = 1200
  S = 50
  H = 3
  assumes: D, S, H > 0; constant demand, instant replenishment, no shortages

Many questions at once: One JSON request per line in, one JSON answer per line out, in order: for scripts, notebooks and agents (the program starts once). In PowerShell: Get-Content cases.jsonl | optimization batch

cases.jsonl:

{"key": "eoq", "inputs": {"D": 1200, "S": 50, "H": 3}}
{"key": "ahp_consistency_ratio", "inputs": {"judgments": "[3, 5, 2]"}}

$ optimization batch < cases.jsonl
{"outcome":"SOLVED","message":"","value":200.0,"unit":"","solved_for":"Q","si":[["D",1200.0],["S",50.0],["H",3.0]],"values":[["Q",200.0,""],["total_cost",600.0,""],["orders_per_period",6.0,""],["cycle_time",0.16666666666666666,""]],"target":"Q","phrase":"economic order quantity","assumptions":["D, S, H > 0","constant demand, instant replenishment, no shortages"],"name":"Economic order quantity (EOQ)","equation":"minimise D S / Q + H Q / 2 (at the optimum ordering cost equals holding cost)"}
{"outcome":"SOLVED","message":"","value":0.0031849983307239735,"unit":"","solved_for":"CR","values":[["CR",0.0031849983307239735,""],["CI",0.0018472990318199045,""],["lambda_max",3.00369459806364,""],["weights",[0.6483290138222366,0.2296507940626371,0.12202019211512621],""],["acceptable","yes",""]],"target":"CR","phrase":"consistency ratio","assumptions":["3 to 10 criteria","RI = 0.58, 0.90, 1.12, 1.24, 1.32, 1.41, 1.45, 1.49 for n = 3..10 (Saaty)"],"name":"AHP consistency ratio","equation":"principal eigenpair by power iteration (certified as an eigenpair), then CI and CR"}

In your browser

Open Optimization on ai999x.com, choose a capability and press Compute: the model runs on your own device as WebAssembly; nothing is sent anywhere. Change any input (any unit of the right kind) and the answer, its explanation and the figures follow.

In your own page: optimization_ai999x.wasm and the loader ax-wasm.js (https://www.ai999x.com/compute/rust/ax-wasm.js).

Python

pip install optimization-ai999x

Worked examples in every form

The same worked example from the command line, from Python and in the browser. Each command-line output is what the program printed.

Absolute extrema of a polynomial on a closed interval

The absolute minimum and maximum of a polynomial on [lo, hi] (closed-interval method: endpoints and critical points), and where they are attained.

Command line:

$ optimization solve polynomial_extrema_interval "coefficients=[1, -3, 0, 2]" lo=-1 hi=3
min_value = -2
max_value = 2
argmin_example = -1
argmax_example = 0
  by: Absolute extrema of a polynomial on a closed interval  (real roots of f' in [lo, hi] (monotone pieces between the roots of f'' bisected to the last float), compared with the endpoints)
  lo = -1
  hi = 3
  assumes: lo < hi; degree 1 to 5

Python:

import optimization_ai999x as m

a = m.solve("polynomial_extrema_interval", coefficients=[1, -3, 0, 2], lo="-1", hi="3")
print(a.values or (a.value, a.unit))

Browser: open this capability on the model page and press Compute in your browser.

What this means: This calculation determines the global extrema of a polynomial on the closed interval from -1 to 3. It evaluates the function at the endpoints and all real critical points where the derivative is zero within the interval. The smallest resulting value is reported as the absolute minimum, -2, attained at -1, while the largest is the absolute maximum, 2, found at 0.

Closed cylinder of least surface area for a given volume

The radius, height and surface area of the closed cylindrical can that holds a volume V with the least material.

Command line:

$ optimization solve can_min_surface V=500
radius = 4.301270069
height = 8.602540138
min_area = 348.7342055
  by: Closed cylinder of least surface area for a given volume  (A(r) = 2 pi r^2 + 2V/r; A'(r) = 0)
  V = 500
  assumes: V > 0; closed top and bottom

Python:

import optimization_ai999x as m

a = m.solve("can_min_surface", V="500")
print(a.values or (a.value, a.unit))

Browser: open this capability on the model page and press Compute in your browser.

What this means: This calculation determines the dimensions for a closed cylindrical container that holds a specific volume while using the least amount of material. Given an input volume of 500, the method finds the optimal radius to be 4.30127 and the corresponding height to be 8.60254. The resulting minimum surface area required is 348.734. This ensures efficient packaging design by balancing the top, bottom, and side surfaces.

Critical point of a two-variable quadratic and its classification

For f(x, y) = a x^2 + b xy + c y^2 + d x + e y + f0: the critical point, its value and whether it is a local minimum, a local maximum or a saddle point (second-derivative test).

Command line:

$ optimization solve critical_point_two_variables a=3 b=2 c=1 d=-4 e=6 f0=1
f_value = -20.5
x = 2.5
y = -5.5
type = local minimum
  by: Critical point of a two-variable quadratic and its classification  (grad f = 0 solved exactly; D = f_xx f_yy - f_xy^2 = 4ac - b^2 classifies it)
  a = 3
  b = 2
  c = 1
  d = -4
  e = 6
  f0 = 1
  assumes: 4ac - b^2 != 0 (an isolated critical point)

Python:

import optimization_ai999x as m

a = m.solve("critical_point_two_variables", a="3", b="2", c="1", d="-4", e="6", f0="1")
print(a.values or (a.value, a.unit))

Browser: open this capability on the model page and press Compute in your browser.

What this means: This calculation identifies the stationary point for the quadratic function defined by coefficients 3, 2, 1, -4, 6, and 1. Solving the gradient equal to zero shows that the critical point occurs at coordinates x equals 2.5 and y equals -5.5. The value of the function at this specific location is -20.5. Based on the second derivative test, which examines the curvature using the determinant of the Hessian matrix, the nature of this point is classified as a local minimum. This classification indicates whether the surface curves upward like a bowl, downward like a dome, or forms a saddle shape at this coordinate.

Largest rectangle for a given length of fencing

The dimensions and maximum area of a rectangle enclosed by a fence of length L, on four sides or on three sides with a wall or river as the fourth.

Command line:

$ optimization solve fence_max_area wall=False L=200
width = 50
length = 50
max_area = 2500
  by: Largest rectangle for a given length of fencing  (area as a function of one side, derivative set to zero)
  L = 200
  assumes: L > 0

Python:

import optimization_ai999x as m

a = m.solve("fence_max_area", wall="False", L="200")
print(a.values or (a.value, a.unit))

Browser: open this capability on the model page and press Compute in your browser.

What this means: This calculation finds the largest rectangular area possible for a given fencing length. Using the total fence amount 200, it determines the optimal side dimensions. The resulting maximum area is 2500, achieved when the perpendicular side is 50 and the parallel side is 50. This ensures the most efficient use of boundary materials.

Documentation

• Rust user guide: the command line, every capability with a command and what it printed: PDF · Markdown
• Python guide: every capability with a Python example: PDF · Markdown
• Desktop app & command line: downloads, install, ask and search, figures, offline use, checksums: https://www.ai999x.com/desktop-cli/
• Model page: compute in your browser, every capability explained: https://www.ai999x.com/models/optimization-ai999x/

Highlights

• Verified, not generated. Every answer is computed by a certified optimization capability and checked (substituted back, dimension-checked or compared with an independent computation) before it is returned.
• Zero wrong answers by design. Seven explicit outcomes (SOLVED, NEEDS_INPUT, AMBIGUOUS, INVALID_INPUT, OUT_OF_SCOPE, UNRECOGNIZED_FORMAT, INTERNAL_ERROR), never a silent guess. Every one of its 240 documented answers is reproduced exactly on every tested platform.
• Plain-English questions and a guaranteed typed interface. ask() reads documented question forms; solve(capability_id, **inputs) always solves a supported task with valid inputs.
• Explains itself. Steps, formulas in LaTeX, units, assumptions and the model used, ready for notebooks, tutoring apps and homework checkers.
• Native and in the browser. Every capability also computes natively in Rust: the desktop app and the optimization command line for Windows and Linux, and WebAssembly on the model page, no Python, no network.
• Small and portable. Pure Python (sympy, mpmath); identical answers on Windows and Linux (Python 3.10, 3.11, 3.12, 3.13, 3.14).
• Coverage: linear and integer programming, convex and nonlinear optimisation, assignment, transportation and shortest-route problems, scheduling, inventory, queueing, game theory and multi-criteria decisions.

Model overview

Property	Value
Package	optimization-ai999x 1.5.0 (import optimization_ai999x)
Type	Deterministic neuro-symbolic optimization engine (rule-based reading + certified symbolic capabilities)
Capabilities	72 certified procedures
Knowledge pack	version 1.0.1
Golden answers	240 documented answers, identical on every tested platform
Platforms tested	Windows (Python 3.13), Linux x86-64 (Python 3.10, 3.11, 3.12, 3.13, 3.14)
Runtime	CPU only, offline; no GPU, no network, no API key
License	Apache-2.0
Creator / engineer	Ibnul Jaif Farabi
Organization	AI999X

Python quick start

pip install optimization-ai999x
# or straight from this repository:
pip install https://huggingface.co/AI999X/optimization-ai999x/resolve/main/dist/optimization_ai999x-1.5.0-py3-none-any.whl

import optimization_ai999x as optimization

optimization.ask("Solve the transportation problem with supply [20, 30], demand [10, 25, 15] and cost matrix [[2, 3, 1], [5, 4, 8]].").show()
# -> SOLVED — min_cost = 150, shipments = [[5.0, 0.0, 15.0], [5.0, 25.0, 0.0]], plan_unique = yes, dual = [-3.0, 0.0, 5.0, 4.0, 4.0]  (transportation_problem)
optimization.ask("Find the shortest path from A to D in the undirected graph with edges A-B 4, A-C 2, B-C 1, B-D 5, C-D 8.").show()
# -> SOLVED — distance = 8, path = ['A', 'C', 'B', 'D'], path_unique = yes  (shortest_path)
optimization.solve("eoq", D=1200, S=50, H=3).show()
# -> SOLVED — Q = 200, total_cost = 600, orders_per_period = 6, cycle_time = 0.166667  (eoq)
optimization.ask("Find the AHP consistency ratio for the pairwise judgments [3, 5, 2].").show()
# -> SOLVED — CR = 0.003185, CI = 0.0018473, lambda_max = 3.00369, weights = [0.6483290138222366, 0.2296507940626371, 0.12202019211512621]  (ahp_consistency_ratio)
optimization.ask("Find the pure-strategy Nash equilibria of the bimatrix game with row payoffs [[3, 0], [5, 1]] and column payoffs [[3, 5], [0, 1]].").show()
# -> SOLVED — count = 1, equilibria = [[2.0, 2.0]]  (pure_nash_equilibria)
optimization.ask("Who created you?").message
# -> optimization-ai999x was created and engineered by Ibnul Jaif Farabi at AI999X.

The Python package also installs a Python command, optimization-ai999x "<question>"; for the native program use optimization (above). The outputs above were produced by running this exact code on optimization-ai999x 1.5.0.

How every answer is verified

1. Reading: the question is turned into typed quantities (value, unit, kind, role) and one asked quantity; every stated number must be used or the question is refused.
2. Planning: only certified capabilities may answer; a capability applies only under the conditions the wording states (for example a pendulum, a liquid at rest, boiling).
3. Computing: exact symbolic or high-precision numeric evaluation.
4. Checking: dimensions, substitution back into the defining equation, and agreement between independent readings; any disagreement becomes a refusal, never a number.

Capabilities are certified before they ship: each must reproduce hand-worked reference cases, reject a deliberately wrong variant (a control), and leave every benchmark answer unchanged.

Python worked examples, with their steps

Each question was put to optimization-ai999x; the answer is the one it returned, with the steps it took.

Transportation problem (minimum shipping cost) · Network optimisation

Solve the transportation problem with supply [20, 30], demand [10, 25, 15] and cost matrix [[2, 3, 1], [5, 4, 8]].

→ min_cost = 150, shipments = [[5, 0, 15], [5, 25, 0]], plan_unique = yes, dual = [-3, 0, 5, 4, 4]

Steps

• Transportation problem (minimum shipping cost): supply = [20, 30], demand = [10, 25, 15], cost = [2 3 1; 5 4 8] → min_cost = 150, shipments = [[5, 0, 15], [5, 25, 0]], plan_unique = yes, dual = [-3, 0, 5, 4, 4]

Shortest path (Dijkstra's algorithm) · Network optimisation

Find the shortest path from A to D in the undirected graph with edges A-B 4, A-C 2, B-C 1, B-D 5, C-D 8.

→ distance = 8, path = [A, C, B, D], path_unique = yes

Steps

• Shortest path (Dijkstra's algorithm): edges = A-B (4), A-C (2), B-C (1), B-D (5), C-D (8), source = A, target = D, directed = False → distance = 8, path = [A, C, B, D], path_unique = yes

Economic order quantity (EOQ) · Inventory management

m.solve("eoq", D=1200, S=50, H=3)

→ Q = 200, total_cost = 600, orders_per_period = 6, cycle_time = 0.1666666667

Steps

• Economic order quantity (EOQ): D = 1200, S = 50, H = 3 → Q = 200, total_cost = 600, orders_per_period = 6, cycle_time = 0.1666666667

AHP consistency ratio · Multi-criteria decision analysis

Find the AHP consistency ratio for the pairwise judgments [3, 5, 2].

→ CR = 0.003184998331, CI = 0.001847299032, lambda_max = 3.003694598, weights = [0.6483290138, 0.2296507941, 0.1220201921], acceptable = yes

Steps

• AHP consistency ratio: judgments = [3, 5, 2] → CR = 0.003184998331, CI = 0.001847299032, lambda_max = 3.003694598, weights = [0.6483290138, 0.2296507941, 0.1220201921], acceptable = yes

Pure-strategy Nash equilibria of a bimatrix game · Game theory

Find the pure-strategy Nash equilibria of the bimatrix game with row payoffs [[3, 0], [5, 1]] and column payoffs [[3, 5], [0, 1]].

→ count = 1, equilibria = [[2, 2]]

Steps

• Pure-strategy Nash equilibria of a bimatrix game: A = [3 0; 5 1], B = [3 5; 0 1] → count = 1, equilibria = [[2, 2]]

Closed cylinder of least surface area for a given volume · Calculus optimisation

A closed cylindrical can must hold 500 cm^3. Find the dimensions that minimize the surface area.

→ radius = 4.301270069, height = 8.602540138, min_area = 348.7342055

Steps

• Closed cylinder of least surface area for a given volume: V = 500 → radius = 4.301270069, height = 8.602540138, min_area = 348.7342055

Figures

Every point was computed by optimization-ai999x.

A linear programme: feasible region and optimal corner

Maximise 3x + 5y subject to x ≤ 4; 2y ≤ 12; 3x + 2y ≤ 18, x, y ≥ 0: the feasible region with the corner points the package returns ((0, 0), (0, 6), (2, 6), (4, 0), (4, 3)) and the objective line through the optimum; marked: x = 2, y = 6, objective 36.

Steepest descent zig-zagging to the minimum

Steepest descent with exact line search on f(x) = ½ xᵀQx, Q = [[2, 0], [0, 8]], from x₀ = (4, 1): six steps computed by the package, each perpendicular to the level set through its start (f = 20, 7.2, 2.592, 0.93312, 0.335923, 0.120932, 0.0435356); marked: x₆ = (0.186624, 0.046656).

Shortest path and minimum spanning tree

A weighted network (A-B 4, A-C 2, B-C 1, B-D 5, C-D 8, A-E 7, E-D 3, C-E 6): the shortest path from A to D the package finds, A → C → B → D of length 8, and its minimum spanning tree (B-C, A-C, E-D, B-D, total weight 11).

An optimal transportation plan

The optimal transportation plan the package computes for supplies [20, 30], demands [10, 25, 15] and unit costs [[2, 3, 1], [5, 4, 8]] (rows: sources, columns: destinations; left the unit costs, right the quantities shipped): minimum total cost 150.

A two-machine schedule in Johnson's order

A two-machine flow shop: processing times [5, 2, 6, 7] on machine 1 and [3, 4, 8, 2] on machine 2. The package's Johnson order is job 2 → job 3 → job 1 → job 4, with makespan 22; each job starts on machine 2 as soon as it leaves machine 1 and machine 2 is free.

Capabilities

72 certified procedures. A few of them:

• Procedures: 0/1 knapsack problem; 2 x 2 zero-sum game (saddle point or mixed-strategy formula); AHP consistency ratio; AHP priority vector (principal eigenvector); AHP weights by the geometric mean method; Absolute extrema of a polynomial on a closed interval; All-pairs shortest distances (Floyd-Warshall); Assignment problem (Hungarian method), maximum total value; Assignment problem (Hungarian method), minimum total cost; Bin packing (minimum number of bins); Break-even quantity; Chromatic number (minimum graph colouring); Closed cylinder of least surface area for a given volume; Critical path method (project duration and slack); Critical point of a two-variable quadratic and its classification; Discounted Markov decision process (optimal values and policy); Dynamic lot sizing (Wagner-Whitin); EOQ with planned backorders; Earliest due date rule (minimum maximum lateness); Economic order quantity (EOQ); Economic production quantity (EPQ); Equality-constrained quadratic program (Lagrange multipliers / KKT); Erlang B blocking probability (M/M/c/c loss system); Integer linear program (branch and bound) …

optimization_ai999x.templates() lists every question form ask() reads; the capability guide (PDF, every example executed) is included in this repository.

All 72 formulas

Show every formula, by topic

Calculus optimisation / geometry

Closed cylinder of least surface area for a given volume

$$r^{\ast }={\left(\frac{V}{2\pi }\right)}^{1\mathrm{/}3},h^{\ast }=2r^{\ast }r^*\; =\; \backslash left(\backslash frac\{V\}\{2\backslash pi\}\backslash right)^\{1/3\},\; \backslash quad\; h^*\; =\; 2r^*$$

Largest rectangle for a given length of fencing

$$A_{\max }=\frac{L^2}{16}\text{ (4 sides)},\frac{L^2}{8}\text{ (3 sides)}A\_\{\backslash max\}\; =\; \backslash frac\{L^2\}\{16\}\; \backslash text\{\; (4\; sides)\},\; \backslash quad\; \backslash frac\{L^2\}\{8\}\; \backslash text\{\; (3\; sides)\}$$

Open box of maximum volume from a rectangular sheet

$$x^{\ast }=\frac{(L+W)-\sqrt{L^2-LW+W^2}}{6}x^*\; =\; \backslash frac\{(L+W)\; -\; \backslash sqrt\{L^2\; -\; LW\; +\; W^2\}\}\{6\}$$

Calculus optimisation / one variable

Absolute extrema of a polynomial on a closed interval

Calculus optimisation / several variables

Critical point of a two-variable quadratic and its classification

$$D=f_{xx}{f}_{yy}-f_{xy}^2=4ac-b^{2}D\; =\; f\_\{xx\}\; f\_\{yy\}\; -\; f\_\{xy\}^2\; =\; 4ac\; -\; b^2$$

Cost-volume-profit analysis

Break-even quantity

$$Q_{BE}=\frac{F}{p-v}Q\_\{BE\}\; =\; \backslash frac\{F\}\{p\; -\; v\}$$

Dynamic programming / Markov decision processes

Discounted Markov decision process (optimal values and policy)

$$V^{\ast }(s)=\underset{a}{\max }(R(s,a)+\gamma \sum _{s^{\mathrm{\prime }}}P(s^{\mathrm{\prime }}\mid s,a)V^{\ast }(s^{\mathrm{\prime }}))V^*(s)\; =\; \backslash max\_a\; \backslash Big(\; R(s,a)\; +\; \backslash gamma\; \backslash sum\_\{s\text{'}\}\; P(s\text{'}\; \backslash mid\; s,a)\backslash ,\; V^*(s\text{'})\; \backslash Big)$$

Dynamic programming / inventory

Dynamic lot sizing (Wagner-Whitin)

$$C_t=\underset{1\le s\le t}{\min }(C_{s-1}+K+h\sum _{u=s}^t(u-s)d_u)C\_t\; =\; \backslash min\_\{1\; \backslash le\; s\; \backslash le\; t\}\; \backslash Big(\; C\_\{s-1\}\; +\; K\; +\; h\; \backslash sum\_\{u=s\}^\{t\}\; (u\; -\; s)\backslash ,\; d\_u\; \backslash Big)$$

Economic optimisation / monopoly pricing

Profit-maximising output with linear demand

$$q^{\ast }=\frac{a-c}{2b},p^{\ast }=\frac{a+c}{2}q^*\; =\; \backslash frac\{a\; -\; c\}\{2b\},\; \backslash quad\; p^*\; =\; \backslash frac\{a\; +\; c\}\{2\}$$

Economic optimisation / pricing

Revenue-maximising price with linear demand

$$p^{\ast }=\frac{\alpha }{2\beta },R^{\ast }=\frac{{\alpha }^2}{4\beta }p^*\; =\; \backslash frac\{\backslash alpha\}\{2\backslash beta\},\; \backslash quad\; R^*\; =\; \backslash frac\{\backslash alpha^2\}\{4\backslash beta\}$$

Game theory / non-zero-sum games

Mixed-strategy Nash equilibrium of a 2 x 2 game

$$q=\frac{a_{22}-a_{12}}{a_{11}-a_{12}-a_{21}+a_{22}},p=\frac{b_{22}-b_{21}}{b_{11}-b_{12}-b_{21}+b_{22}}q\; =\; \backslash frac\{a\_\{22\}\; -\; a\_\{12\}\}\{a\_\{11\}\; -\; a\_\{12\}\; -\; a\_\{21\}\; +\; a\_\{22\}\},\backslash quad\; p\; =\; \backslash frac\{b\_\{22\}\; -\; b\_\{21\}\}\{b\_\{11\}\; -\; b\_\{12\}\; -\; b\_\{21\}\; +\; b\_\{22\}\}$$

Pure-strategy Nash equilibria of a bimatrix game

$$a_{rs}=\underset{k}{\max }{a}_{ks},b_{rs}=\underset{k}{\max }{b}_{rk}a\_\{rs\}\; =\; \backslash max\_k\; a\_\{ks\},\backslash quad\; b\_\{rs\}\; =\; \backslash max\_k\; b\_\{rk\}$$

Game theory / zero-sum games

2 x 2 zero-sum game (saddle point or mixed-strategy formula)

$$v=\frac{ad-bc}{a+d-b-c},p=\frac{d-c}{a+d-b-c},q=\frac{d-b}{a+d-b-c}v\; =\; \backslash frac\{ad\; -\; bc\}\{a\; +\; d\; -\; b\; -\; c\},\backslash quad\; p\; =\; \backslash frac\{d\; -\; c\}\{a\; +\; d\; -\; b\; -\; c\},\backslash quad\; q\; =\; \backslash frac\{d\; -\; b\}\{a\; +\; d\; -\; b\; -\; c\}$$

Two-person zero-sum game (value and optimal mixed strategies)

$$v=\underset{x\in {\mathrm{\Delta }}_m}{\max }\underset{j}{\min }\sum _i{x}_i{a}_{ij}=\underset{y\in {\mathrm{\Delta }}_n}{\min }\underset{i}{\max }\sum _j{a}_{ij}{y}_{j}v\; =\; \backslash max\_\{x\; \backslash in\; \backslash Delta\_m\}\; \backslash min\_\{j\}\; \backslash sum\_i\; x\_i\; a\_\{ij\}\; =\; \backslash min\_\{y\; \backslash in\; \backslash Delta\_n\}\; \backslash max\_i\; \backslash sum\_j\; a\_\{ij\}\; y\_j$$

Integer programming

Integer linear program (branch and bound)

$$\max \mathrm{/}\min c^{\mathrm{\top }}x\text{s.t. }Ax(\le ,\ge ,=)b,x\in {\mathbb{Z}}_{\ge 0}^n\backslash max\; /\; \backslash min\; \backslash ;\; c^\backslash top\; x\; \backslash quad\; \backslash text\{s.t.\; \}\; A\; x\; \backslash ;(\backslash le,\backslash ge,=)\backslash ;\; b,\backslash ;\; x\; \backslash in\; \backslash mathbb\{Z\}\_\{\backslash ge\; 0\}^n$$

Integer programming / covering

Minimum vertex cover (and maximum independent set)

$$\tau (G)=\min \{\mathrm{\mid }C\mathrm{\mid }:C\cap \{u,v\}\mathrm{\ne }\mathrm{\varnothing }\mathrm{\forall }uv\in E\},\alpha (G)=\mathrm{\mid }V\mathrm{\mid }-\tau (G)\backslash tau(G)\; =\; \backslash min\backslash \{\backslash vert\; C\backslash vert\; :\; C\; \backslash cap\; \backslash \{u,\; v\backslash \}\; \backslash neq\; \backslash emptyset\backslash \; \backslash forall\; uv\; \backslash in\; E\backslash \},\backslash quad\; \backslash alpha(G)\; =\; \backslash vert\; V\backslash vert\; -\; \backslash tau(G)$$

Minimum-cost set cover

$$\min \sum _i{c}_i{x}_i\text{ s.t. }\sum _{i\ni e}{x}_i\ge 1\mathrm{\forall }e,x\in \{0,1\}\backslash min\; \backslash sum\_i\; c\_i\; x\_i\backslash \; \backslash text\{s.t.\}\backslash \; \backslash sum\_\{i\; \backslash ni\; e\}\; x\_i\; \backslash ge\; 1\backslash \; \backslash forall\; e,\backslash \; x\; \backslash in\; \backslash \{0,1\backslash \}$$

Integer programming / graph colouring

Chromatic number (minimum graph colouring)

$$\chi (G)=\min \{k:\mathrm{\exists }c:V\to \{\mathrm{1..}k\},c(u)\mathrm{\ne }c(v)\mathrm{\forall }uv\in E\}\backslash chi(G)\; =\; \backslash min\backslash \{k\; :\; \backslash exists\backslash ,\; c:\; V\; \backslash to\; \backslash \{1..k\backslash \},\backslash \; c(u)\; \backslash neq\; c(v)\backslash \; \backslash forall\; uv\; \backslash in\; E\backslash \}$$

Integer programming / knapsack

0/1 knapsack problem

$$\max \sum _i{v}_i{x}_i\text{ s.t. }\sum _i{w}_i{x}_i\le W,x_i\in \{0,1\}\backslash max\; \backslash sum\_i\; v\_i\; x\_i\; \backslash \; \backslash text\{s.t.\}\backslash \; \backslash sum\_i\; w\_i\; x\_i\; \backslash le\; W,\backslash \; x\_i\; \backslash in\; \backslash \{0,1\backslash \}$$

Unbounded knapsack problem

$$V(c)=\max (V(c-1),\underset{i:w_i\le c}{\max }V(c-w_i)+v_i)V(c)\; =\; \backslash max\backslash big(V(c-1),\backslash \; \backslash max\_\{i:\; w\_i\; \backslash le\; c\}\; V(c\; -\; w\_i)\; +\; v\_i\backslash big)$$

Integer programming / location

Uncapacitated facility location

$$\underset{S\mathrm{\ne }\mathrm{\varnothing }}{\min }\sum _{i\in S}{f}_i+\sum _j\underset{i\in S}{\min }{c}_{ij}\backslash min\_\{S\; \backslash neq\; \backslash emptyset\}\; \backslash sum\_\{i\; \backslash in\; S\}\; f\_i\; +\; \backslash sum\_j\; \backslash min\_\{i\; \backslash in\; S\}\; c\_\{ij\}$$

Integer programming / packing

Bin packing (minimum number of bins)

$$\min \mathrm{\#}\text{bins }\ge \lceil \sum _j{s}_j\mathrm{/}C\rceil \backslash min\; \backslash \#\backslash text\{bins\}\; \backslash \; \backslash ge\backslash \; \backslash left\backslash lceil\; \backslash sum\_j\; s\_j\; /\; C\; \backslash right\backslash rceil$$

Inventory management

EOQ with planned backorders

$$Q^{\ast }=\sqrt{\frac{2DS}{H}\cdot \frac{H+B}{B}},s^{\ast }=Q^{\ast }\frac{H}{H+B}Q^*\; =\; \backslash sqrt\{\backslash frac\{2DS\}\{H\}\backslash cdot\backslash frac\{H+B\}\{B\}\},\; \backslash quad\; s^*\; =\; Q^*\; \backslash frac\{H\}\{H+B\}$$

Economic order quantity (EOQ)

$$Q^{\ast }=\sqrt{\frac{2DS}{H}},T{C}^{\ast }=\sqrt{2DSH}Q^*\; =\; \backslash sqrt\{\backslash frac\{2\; D\; S\}\{H\}\},\; \backslash quad\; TC^*\; =\; \backslash sqrt\{2\; D\; S\; H\}$$

Economic production quantity (EPQ)

$$Q^{\ast }=\sqrt{\frac{2DS}{H(1-D\mathrm{/}P)}}Q^*\; =\; \backslash sqrt\{\backslash frac\{2DS\}\{H(1\; -\; D/P)\}\}$$

Order quantity with quantity discounts (all-units)

$$TC(Q)=D{p}_k+\frac{DS}{Q}+\frac{i{p}_{k}Q}{2}TC(Q)\; =\; D\; p\_k\; +\; \backslash frac\{D\; S\}\{Q\}\; +\; \backslash frac\{i\; p\_k\; Q\}\{2\}$$

Periodic-review order-up-to level

$$S=\bar{d}(R+L)+z_{SL}{\sigma }_d\sqrt{R+L}S\; =\; \backslash bar\; d\; (R\; +\; L)\; +\; z\_\{SL\}\backslash ,\backslash sigma\_d\backslash sqrt\{R\; +\; L\}$$

Reorder point with safety stock

$$ROP=\bar{d}L+z_{SL}{\sigma }_d\sqrt{L}ROP\; =\; \backslash bar\; d\; L\; +\; z\_\{SL\}\backslash ,\backslash sigma\_d\; \backslash sqrt\{L\}$$

Inventory management / single-period

Newsvendor order quantity (discrete demand)

$$Q^{\ast }=\min \{d:F(d)\ge \frac{p-c}{p-s}\}Q^*\; =\; \backslash min\backslash \{\; d\; :\; F(d)\; \backslash ge\; \backslash tfrac\{p\; -\; c\}\{p\; -\; s\}\; \backslash \}$$

Newsvendor order quantity (normal demand)

$$F(Q^{\ast })=\frac{p-c}{p-s},Q^{\ast }=\mu +z\sigma F(Q^*)\; =\; \backslash frac\{p\; -\; c\}\{p\; -\; s\},\; \backslash quad\; Q^*\; =\; \backslash mu\; +\; z\backslash ,\backslash sigma$$

Newsvendor order quantity (uniform demand)

$$Q^{\ast }=a+\frac{p-c}{p-s}(b-a)Q^*\; =\; a\; +\; \backslash frac\{p\; -\; c\}\{p\; -\; s\}(b\; -\; a)$$

Linear programming

Linear program (simplex method)

$$\max \mathrm{/}\min c^{\mathrm{\top }}x\text{s.t. }Ax(\le ,\ge ,=)b,x\ge 0\backslash max\; /\; \backslash min\; \backslash ;\; c^\backslash top\; x\; \backslash quad\; \backslash text\{s.t.\; \}\; A\; x\; \backslash ;(\backslash le,\; \backslash ge,\; =)\backslash ;\; b,\backslash ;\; x\; \backslash ge\; 0$$

Two-variable linear program (graphical corner-point method)

$$z^{\ast }=\underset{v\in \text{corners}}{\max }{c}^{\mathrm{\top }}vz^*\; =\; \backslash max\_\{v\; \backslash in\; \backslash text\{corners\}\}\; c^\backslash top\; v$$

Linear programming / sensitivity analysis

Shadow prices of a linear program

$$y^{\mathrm{\top }}=c_B^{\mathrm{\top }}{B}^{-1},\frac{\mathrm{\partial }{z}^{\ast }}{\mathrm{\partial }{b}_i}=y_{i}y^\backslash top\; =\; c\_B^\backslash top\; B^\{-1\},\; \backslash qquad\; \backslash frac\{\backslash partial\; z^*\}\{\backslash partial\; b\_i\}\; =\; y\_i$$

Multi-criteria decision analysis

TOPSIS ranking (closeness to the ideal solution)

C_i = \frac{D_i^-}{D_i^+ + D_i^-},\quad D_i^\pm = \Big\\vert w \odot r_i - v^\pm \Big\\vert _2

Weighted product model

$$P_i=\prod _j{n}_{ij}^{w_j}P\_i\; =\; \backslash prod\_j\; n\_\{ij\}^\{\backslash ,w\_j\}$$

Weighted sum model (simple additive weighting)

$$S_i=\sum _j{w}_j{n}_{ij},n_{ij}=\frac{x_{ij}}{\underset{i}{\max }{x}_{ij}}\text{ or }\frac{\underset{i}{\min }{x}_{ij}}{x_{ij}}S\_i\; =\; \backslash sum\_j\; w\_j\; n\_\{ij\},\backslash quad\; n\_\{ij\}\; =\; \backslash frac\{x\_\{ij\}\}\{\backslash max\_i\; x\_\{ij\}\}\backslash \; \backslash text\{or\}\backslash \; \backslash frac\{\backslash min\_i\; x\_\{ij\}\}\{x\_\{ij\}\}$$

Multi-criteria decision analysis / AHP

AHP consistency ratio

$$CI=\frac{{\lambda }_{\max }-n}{n-1},CR=\frac{CI}{R{I}_n}CI\; =\; \backslash frac\{\backslash lambda\_\{\backslash max\}\; -\; n\}\{n\; -\; 1\},\backslash quad\; CR\; =\; \backslash frac\{CI\}\{RI\_n\}$$

AHP priority vector (principal eigenvector)

$$Pw={\lambda }_{\max }w,\sum _i{w}_i=1P\; w\; =\; \backslash lambda\_\{\backslash max\}\; w,\backslash quad\; \backslash sum\_i\; w\_i\; =\; 1$$

AHP weights by the geometric mean method

$$w_i=\frac{(\prod _j{a}_{ij}{)}^{1\mathrm{/}n}}{\sum _k(\prod _j{a}_{kj}{)}^{1\mathrm{/}n}}w\_i\; =\; \backslash frac\{(\backslash prod\_j\; a\_\{ij\})^\{1/n\}\}\{\backslash sum\_k\; (\backslash prod\_j\; a\_\{kj\})^\{1/n\}\}$$

Network optimisation / assignment

Assignment problem (Hungarian method), maximum total value

$$\underset{\sigma }{\min }\sum _i{c}_{i\sigma (i)}=\max \{\sum _i{u}_i+\sum _j{v}_j:u_i+v_j\le c_{ij}\}\backslash min\_\{\backslash sigma\}\; \backslash sum\_i\; c\_\{i\backslash sigma(i)\}\; =\; \backslash max\; \backslash Big\backslash \{\; \backslash sum\_i\; u\_i\; +\; \backslash sum\_j\; v\_j\; :\; u\_i\; +\; v\_j\; \backslash le\; c\_\{ij\}\; \backslash Big\backslash \}$$

Assignment problem (Hungarian method), minimum total cost

$$\underset{\sigma }{\min }\sum _i{c}_{i\sigma (i)}=\max \{\sum _i{u}_i+\sum _j{v}_j:u_i+v_j\le c_{ij}\}\backslash min\_\{\backslash sigma\}\; \backslash sum\_i\; c\_\{i\backslash sigma(i)\}\; =\; \backslash max\; \backslash Big\backslash \{\; \backslash sum\_i\; u\_i\; +\; \backslash sum\_j\; v\_j\; :\; u\_i\; +\; v\_j\; \backslash le\; c\_\{ij\}\; \backslash Big\backslash \}$$

Network optimisation / flows

Maximum flow and minimum cut (Edmonds-Karp)

$$\max \mathrm{\mid }f\mathrm{\mid }=\underset{(S,\bar{S})}{\min }c(S,\bar{S})\backslash max\; \backslash vert\; f\backslash vert\; =\; \backslash min\_\{(S,\; \backslash bar\; S)\}\; c(S,\; \backslash bar\; S)$$

Minimum-cost flow (transshipment)

$$\min \sum c_{ij}{x}_{ij}\text{ s.t. }\sum _j{x}_{vj}-\sum _i{x}_{iv}=b_v,0\le x_{ij}\le u_{ij}\backslash min\; \backslash sum\; c\_\{ij\}\; x\_\{ij\}\backslash \; \backslash text\{s.t.\}\backslash \; \backslash sum\_j\; x\_\{vj\}\; -\; \backslash sum\_i\; x\_\{iv\}\; =\; b\_v,\backslash \; 0\; \backslash le\; x\_\{ij\}\; \backslash le\; u\_\{ij\}$$

Network optimisation / matching

Maximum bipartite matching

$$\nu (G)=\tau (G)\text{ for bipartite }G\backslash nu(G)\; =\; \backslash tau(G)\; \backslash text\{\; for\; bipartite\; \}\; G$$

Maximum-weight matching (general graph)

$$\underset{M\text{ matching}}{\max }\sum _{e\in M}{w}_e\backslash max\_\{M\; \backslash text\{\; matching\}\}\; \backslash sum\_\{e\; \backslash in\; M\}\; w\_e$$

Network optimisation / routing

Travelling salesman tour (exact, Held-Karp)

$$g(S,k)=\underset{m\in S\setminus \{k\}}{\min }(g(S\setminus \{k\},m)+d_{mk})g(S,\; k)\; =\; \backslash min\_\{m\; \backslash in\; S\; \backslash setminus\; \backslash \{k\backslash \}\}\; \backslash big(g(S\; \backslash setminus\; \backslash \{k\backslash \},\; m)\; +\; d\_\{mk\}\backslash big)$$

Network optimisation / shortest routes

All-pairs shortest distances (Floyd-Warshall)

$$d_{ij}\leftarrow \min (d_{ij},d_{ik}+d_{kj})d\_\{ij\}\; \backslash leftarrow\; \backslash min(d\_\{ij\},\backslash ,\; d\_\{ik\}\; +\; d\_\{kj\})$$

Shortest path (Dijkstra's algorithm)

$$d(v)=\underset{(u,v)}{\min }(d(u)+w_{uv})d(v)\; =\; \backslash min\_\{(u,v)\}\; \backslash big(d(u)\; +\; w\_\{uv\}\backslash big)$$

Shortest path with negative weights (Bellman-Ford)

$$d(v)\le d(u)+w_{uv}\mathrm{\forall }(u,v)d(v)\; \backslash le\; d(u)\; +\; w\_\{uv\}\; \backslash \; \backslash forall\; (u,v)$$

Network optimisation / spanning trees

Minimum spanning tree (Kruskal's algorithm)

$$\underset{T\text{ spanning tree}}{\min }\sum _{e\in T}{w}_e\backslash min\_\{T\; \backslash text\{\; spanning\; tree\}\}\; \backslash sum\_\{e\; \backslash in\; T\}\; w\_e$$

Network optimisation / transportation

Transportation problem (minimum shipping cost)

$$\min \sum _{ij}{c}_{ij}{x}_{ij}\text{ s.t. }\sum _j{x}_{ij}\le s_i,\sum _i{x}_{ij}=d_j,x\ge 0\backslash min\; \backslash sum\_\{ij\}\; c\_\{ij\}\; x\_\{ij\}\backslash \; \backslash text\{s.t.\}\backslash \; \backslash sum\_j\; x\_\{ij\}\; \backslash le\; s\_i,\backslash \; \backslash sum\_i\; x\_\{ij\}\; =\; d\_j,\backslash \; x\; \backslash ge\; 0$$

Nonlinear optimisation / descent methods

Steepest descent step with exact line search (quadratic)

$$\alpha =\frac{g^{\mathrm{\top }}g}{g^{\mathrm{\top }}Qg},x_1=x_0-\alpha g\backslash alpha\; =\; \backslash frac\{g^\backslash top\; g\}\{g^\backslash top\; Q\; g\},\backslash quad\; x\_1\; =\; x\_0\; -\; \backslash alpha\; g$$

Nonlinear optimisation / quadratic

Equality-constrained quadratic program (Lagrange multipliers / KKT)

Unconstrained quadratic minimisation

$$x^{\ast }=-Q^{-1}c,f^{\ast }=\frac{1}{2}{c}^{\mathrm{\top }}{x}^{\ast }x^*\; =\; -Q^\{-1\}\; c,\backslash quad\; f^*\; =\; \backslash tfrac12\; c^\backslash top\; x^*$$

Project scheduling

Critical path method (project duration and slack)

$$E{S}_k=\underset{j\to k}{\max }(E{S}_j+d_j),L{S}_k=\underset{k\to s}{\min }L{S}_s-d_k,{\text{slack}}_k=L{S}_k-E{S}_{k}ES\_k\; =\; \backslash max\_\{j\; \backslash to\; k\}(ES\_j\; +\; d\_j),\backslash quad\; LS\_k\; =\; \backslash min\_\{k\; \backslash to\; s\}\; LS\_s\; -\; d\_k,\backslash quad\; \backslash text\{slack\}\_k\; =\; LS\_k\; -\; ES\_k$$

PERT project duration and completion probability

$$t_e=\frac{a+4m+b}{6},{\sigma }^2={\left(\frac{b-a}{6}\right)}^2,P(T\le d)=\mathrm{\Phi }\text{\hspace{0.17em}⁣}\left(\frac{d-\mu }{\sigma }\right)t\_e\; =\; \backslash frac\{a\; +\; 4m\; +\; b\}\{6\},\backslash ;\; \backslash sigma^2\; =\; \backslash left(\backslash frac\{b\; -\; a\}\{6\}\backslash right)^2,\backslash ;\; P(T\; \backslash le\; d)\; =\; \backslash Phi\backslash !\backslash left(\backslash frac\{d\; -\; \backslash mu\}\{\backslash sigma\}\backslash right)$$

Queueing theory

M/G/1 queue (Pollaczek-Khinchine formula)

$$L_q=\frac{{\lambda }^2{\sigma }_S^2+{\rho }^2}{2(1-\rho )}L\_q\; =\; \backslash frac\{\backslash lambda^2\; \backslash sigma\_S^2\; +\; \backslash rho^2\}\{2(1-\backslash rho)\}$$

M/M/1 queue

$$L=\frac{\rho }{1-\rho },W=\frac{1}{\mu -\lambda },W_q=\frac{\rho }{\mu -\lambda }L\; =\; \backslash frac\{\backslash rho\}\{1-\backslash rho\},\backslash ;\; W\; =\; \backslash frac\{1\}\{\backslash mu\; -\; \backslash lambda\},\backslash ;\; W\_q\; =\; \backslash frac\{\backslash rho\}\{\backslash mu-\backslash lambda\}$$

M/M/c queue (Erlang C)

$$C(c,a)=\frac{B(c,a)}{1-\rho (1-B(c,a))},L_q=C\frac{\rho }{1-\rho }C(c,\; a)\; =\; \backslash frac\{B(c,a)\}\{1\; -\; \backslash rho(1\; -\; B(c,a))\},\; \backslash quad\; L\_q\; =\; C\backslash ,\backslash frac\{\backslash rho\}\{1-\backslash rho\}$$

M/M/c/K queue (finite capacity)

Machine-repair (finite-source) queue

$${\pi }_n\propto \frac{N!}{(N-n)!}\frac{(\lambda \mathrm{/}\mu {)}^n}{n!\text{ or }c!c^{n-c}}\backslash pi\_n\; \backslash propto\; \backslash frac\{N!\}\{(N-n)!\}\backslash ,\backslash frac\{(\backslash lambda/\backslash mu)^n\}\{n!\backslash \; \backslash text\{or\}\backslash \; c!\backslash ,c^\{n-c\}\}$$

Queueing theory / staffing

Minimum number of servers for a service target (Erlang C)

$$c^{\ast }=\min \{c>\lambda \mathrm{/}\mu :C(c,\lambda \mathrm{/}\mu )\le \alpha \}c^*\; =\; \backslash min\backslash \{c\; >\; \backslash lambda/\backslash mu\; :\; C(c,\; \backslash lambda/\backslash mu)\; \backslash le\; \backslash alpha\backslash \}$$

Queueing theory / teletraffic

Erlang B blocking probability (M/M/c/c loss system)

$$B(c,A)=\frac{A^c\mathrm{/}c!}{\sum _{k=0}^c{A}^k\mathrm{/}k!}B(c,\; A)\; =\; \backslash frac\{A^c\; /\; c!\}\{\backslash sum\_\{k=0\}^\{c\}\; A^k\; /\; k!\}$$

Revenue management

Littlewood's rule (two-fare protection level)

$$P(D_1>y^{\ast })=\frac{p_2}{p_1},y^{\ast }=\mu +z\sigma P(D\_1\; >\; y^*)\; =\; \backslash frac\{p\_2\}\{p\_1\},\; \backslash quad\; y^*\; =\; \backslash mu\; +\; z\backslash ,\backslash sigma$$

Scheduling / flow shop

Johnson's rule (two-machine flow shop)

$$\min C_{\max }\text{ for }F2vert{C}_{\max }\backslash min\; C\_\{\backslash max\}\; \backslash text\{\; for\; \}\; F2\; \backslash \backslash vert\; C\_\{\backslash max\}$$

Scheduling / parallel machines

Minimum makespan on identical parallel machines

$$\underset{\text{assignment}}{\min }\underset{k}{\max }\sum _{j\to k}{p}_j\backslash min\_\{\backslash text\{assignment\}\}\; \backslash max\_k\; \backslash sum\_\{j\; \backslash to\; k\}\; p\_j$$

Scheduling / single machine

Earliest due date rule (minimum maximum lateness)

$$\min L_{\max }=\min \underset{j}{\max }(C_j-d_j):d_{[1]}\le d_{[2]}\le \dots \backslash min\; L\_\{\backslash max\}\; =\; \backslash min\; \backslash max\_j\; (C\_j\; -\; d\_j):\backslash ;\; d\_\{[1]\}\; \backslash le\; d\_\{[2]\}\; \backslash le\; \backslash dots$$

Moore-Hodgson algorithm (minimum number of tardy jobs)

$$\min \sum _j{U}_j\backslash min\; \backslash sum\_j\; U\_j$$

Shortest processing time rule (minimum mean flow time)

$$\min \frac{1}{n}\sum _j{C}_j:p_{[1]}\le p_{[2]}\le \dots \backslash min\; \backslash tfrac\{1\}\{n\}\backslash sum\_j\; C\_j:\backslash ;\; p\_\{[1]\}\; \backslash le\; p\_\{[2]\}\; \backslash le\; \backslash dots$$

Weighted shortest processing time (Smith's rule)

$$\min \sum _j{w}_j{C}_j:\frac{p_{[1]}}{w_{[1]}}\le \frac{p_{[2]}}{w_{[2]}}\le \dots \backslash min\; \backslash sum\_j\; w\_j\; C\_j:\backslash ;\; \backslash frac\{p\_\{[1]\}\}\{w\_\{[1]\}\}\; \backslash le\; \backslash frac\{p\_\{[2]\}\}\{w\_\{[2]\}\}\; \backslash le\; \backslash dots$$

Unconstrained optimisation / one variable

Optimum of a quadratic function

$$x^{\ast }=-\frac{b}{2a},f^{\ast }=c-\frac{b^2}{4a}x^*\; =\; -\backslash frac\{b\}\{2a\},\; \backslash quad\; f^*\; =\; c\; -\; \backslash frac\{b^2\}\{4a\}$$

Out of scope

• Open-ended explanation, essay writing or tutoring dialogue: it is a solver, not a chatbot.
• Questions outside its certified optimization capabilities, and any question whose stated numbers it cannot all use: it refuses with the reason instead.
• High-stakes engineering, medical or legal decisions without independent review.

Limitations

• It answers what its certified capabilities cover (linear and integer programming, convex and nonlinear optimisation, assignment, transportation and shortest-route problems, scheduling, inventory, queueing, game theory and multi-criteria decisions); anything else is refused with a reason.
• ask() reads documented question forms and common phrasings; unusual wording may be refused; use solve() for guaranteed results.
• Results are as correct as the certified relations and their stated conditions; a certified relation is a checked model, not a claim about every physical situation.

Citation

@software{optimization_ai999x_1_5_0,
  title  = {optimization-ai999x: an offline, verified optimization solver},
  author = {Farabi, Ibnul Jaif}, organization = {AI999X},
  year   = {2026}, version = {1.5.0},
  url    = {https://huggingface.co/AI999X/optimization-ai999x}
}

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Created and engineered by Ibnul Jaif Farabi · AI999X · ai999x.com