Prescriptive Analytics

Prescriptive analytics represents the most advanced form of analytics, focusing on "what should we do" by providing actionable recommendations through optimization algorithms, simulation, and decision theory. In automotive applications, prescriptive analytics drives strategic decisions in pricing, inventory management, marketing allocation, and operational optimization.

Mathematical Foundation

Prescriptive analytics combines optimization theory, decision science, and simulation to recommend optimal actions:

\boxed{\mathbf{Data + Models + Constraints \rightarrow Optimal Decisions}}

Linear Programming

Standard Form

Linear programming solves optimization problems with linear objective functions and constraints:

\begin{aligned} \text{Maximize: } \quad & \mathbf{c}^T \mathbf{x} \\ \text{Subject to: } \quad & \mathbf{A}\mathbf{x} \leq \mathbf{b} \\ & \mathbf{x} \geq \mathbf{0} \end{aligned}

Where:

$\mathbf{c}$ is the objective function coefficient vector
$\mathbf{x}$ is the decision variable vector
$\mathbf{A}$ is the constraint coefficient matrix
$\mathbf{b}$ is the constraint bounds vector

Simplex Method

The simplex algorithm iteratively moves between vertices of the feasible region:

\mathbf{x}^{(k+1)} = \mathbf{x}^{(k)} + \alpha^{(k)} \mathbf{d}^{(k)}

Where $\mathbf{d}^{(k)}$ is the search direction and $\alpha^{(k)}$ is the step size.

Automotive Example: Production Planning Optimization

Business Context: An automotive manufacturer needs to optimize production across multiple plant locations to minimize cost while meeting demand.

Decision Variables:

$x_{ij}$ = Number of vehicles of type $i$ produced at plant $j$

Objective Function (Minimize Total Cost):

\text{Minimize: } Z = \sum_{i=1}^{m} \sum_{j=1}^{n} c_{ij} x_{ij}

Constraints:

Demand Satisfaction:

\sum_{j=1}^{n} x_{ij} \geq d_i \quad \forall i

Plant Capacity:

\sum_{i=1}^{m} x_{ij} \leq C_j \quad \forall j

Non-negativity:

x_{ij} \geq 0 \quad \forall i,j

Sample Problem:

Vehicles: Sedan (S), SUV (U), Truck (T)
Plants: Detroit (D), Atlanta (A), Los Angeles (L)
Production costs: $c_{SD} = 25,000$ , $c_{SA} = 24,000$ , $c_{SL} = 26,000$

Mathematical Formulation:

\text{Minimize: } 25,000x_{SD} + 24,000x_{SA} + 26,000x_{SL} + \ldots

Solution: Optimal production allocation minimizes total cost while satisfying all constraints.

Business Impact: 12% cost reduction through optimal plant utilization.

Integer Programming

Mathematical Formulation

Integer programming extends linear programming with integer constraints:

\begin{aligned} \text{Maximize: } \quad & \mathbf{c}^T \mathbf{x} \\ \text{Subject to: } \quad & \mathbf{A}\mathbf{x} \leq \mathbf{b} \\ & x_j \in \mathbb{Z}^+ \quad \forall j \in J \\ & x_j \geq 0 \quad \forall j \notin J \end{aligned}

Binary Integer Programming

For yes/no decisions:

x_j \in \{0, 1\} \quad \forall j

Automotive Example: Dealership Location Selection

Business Context: An automotive group wants to select optimal locations for new dealerships to maximize market coverage while staying within budget.

Decision Variables:

$x_i = 1$ if location $i$ is selected, 0 otherwise

Objective Function (Maximize Market Coverage):

\text{Maximize: } Z = \sum_{i=1}^{n} p_i x_i

Where $p_i$ is the market potential at location $i$ .

Budget Constraint:

\sum_{i=1}^{n} c_i x_i \leq B

Market Coverage Constraint:

\sum_{i \in S_j} x_i \geq 1 \quad \forall j

Where $S_j$ represents locations that can serve market segment $j$ .

Binary Constraint:

x_i \in \{0, 1\} \quad \forall i

Business Results: Optimal selection of 8 locations from 25 candidates, achieving 95% market coverage within budget.

Network Optimization

Minimum Cost Flow Problem

Network flow problems optimize flow through networks:

\begin{aligned} \text{Minimize: } \quad & \sum_{(i,j) \in A} c_{ij} x_{ij} \\ \text{Subject to: } \quad & \sum_{j:(i,j) \in A} x_{ij} - \sum_{j:(j,i) \in A} x_{ji} = b_i \quad \forall i \\ & 0 \leq x_{ij} \leq u_{ij} \quad \forall (i,j) \in A \end{aligned}

Where:

$c_{ij}$ is the cost per unit flow on arc $(i,j)$
$x_{ij}$ is the flow on arc $(i,j)$
$b_i$ is the supply/demand at node $i$
$u_{ij}$ is the capacity of arc $(i,j)$

Automotive Example: Supply Chain Optimization

Business Context: An automotive parts supplier optimizes distribution from factories to dealers to minimize transportation costs.

Network Structure:

Supply Nodes: 3 factories with production capacity
Demand Nodes: 50 dealerships with parts demand
Transshipment Nodes: 8 distribution centers

Mathematical Model:

Flow Conservation at each node:

\sum_{\text{in}} x_{ij} - \sum_{\text{out}} x_{jk} = b_j

Capacity Constraints:

x_{ij} \leq u_{ij} \quad \forall (i,j)

Cost Minimization:

\text{Minimize: } \sum_{(i,j)} c_{ij} \cdot d_{ij} \cdot x_{ij}

Where $d_{ij}$ is the distance between nodes $i$ and $j$ .

Business Impact: 18% reduction in transportation costs through optimal routing.

Dynamic Programming

Bellman Equation

Dynamic programming solves multi-stage decision problems:

V_t(s_t) = \max_{a_t} \left[ R_t(s_t, a_t) + \gamma V_{t+1}(s_{t+1}) \right]

Where:

$V_t(s_t)$ is the value function at state $s_t$ and time $t$
$R_t(s_t, a_t)$ is the immediate reward
$\gamma$ is the discount factor
$s_{t+1}$ is the next state given action $a_t$

Automotive Example: Inventory Management Optimization

Business Context: A dealership optimizes parts inventory ordering to minimize holding and stockout costs.

State Variables:

$s_t$ = Current inventory level at time $t$

Decision Variables:

$a_t$ = Order quantity at time $t$

Cost Function:

C_t(s_t, a_t) = c \cdot a_t + h \cdot (s_t + a_t) + p \cdot E[\max(D_t - (s_t + a_t), 0)]

Where:

$c$ = ordering cost per unit
$h$ = holding cost per unit
$p$ = penalty cost per unit shortage
$D_t$ = demand at time $t$

Bellman Equation:

V_t(s_t) = \min_{a_t \geq 0} \left[ C_t(s_t, a_t) + \gamma \sum_{d} P(D_{t+1} = d) V_{t+1}(s_t + a_t - d) \right]

Optimal Policy: $(s, S)$ policy where:

If inventory $s_t < s$ , order up to $S$
If inventory $s_t \geq s$ , don't order

Business Results: 25% reduction in total inventory costs while maintaining 98% service level.

Game Theory

Nash Equilibrium

In competitive environments, game theory provides optimal strategies:

Pure Strategy Nash Equilibrium: Strategy profile $(s_1^*, s_2^*, \ldots, s_n^*)$ where:

u_i(s_i^*, s_{-i}^*) \geq u_i(s_i, s_{-i}^*) \quad \forall s_i \in S_i, \forall i

Mixed Strategy Equilibrium

When no pure strategy equilibrium exists:

\sum_{s_i \in S_i} \sigma_i(s_i) u_i(s_i, s_{-i}) = u_i(\sigma_i, s_{-i})

Where $\sigma_i$ is the probability distribution over strategies.

Automotive Example: Competitive Pricing Strategy

Business Context: Two automotive dealerships compete on pricing for the same vehicle model.

Payoff Matrix (Daily Profit in $000s):

	Competitor Low	Competitor High
Low	(8, 8)	(12, 4)
High	(4, 12)	(10, 10)

Nash Equilibrium Analysis:

Pure Strategy: (High, High) with payoffs (10, 10) Mixed Strategy: Each player plays High with probability $p$

Indifference Condition:

8p + 4(1-p) = 12p + 10(1-p)

Solving: $p = 0.6$

Expected Payoff:

E[\pi] = 0.6 \times 10 + 0.4 \times 8 = 9.2

Business Strategy: Implement high pricing 60% of the time for optimal expected profit.

Simulation and Monte Carlo Methods

Monte Carlo Simulation

For complex systems with uncertainty:

E[f(X)] \approx \frac{1}{n} \sum_{i=1}^{n} f(X_i)

Where $X_i$ are random samples from the distribution of $X$ .

Confidence Interval:

\bar{f} \pm z_{\alpha/2} \frac{s}{\sqrt{n}}

Automotive Example: Service Center Capacity Planning

Business Context: A service center uses simulation to determine optimal staffing levels considering uncertain arrival patterns and service times.

Model Parameters:

Arrival Process: Poisson with rate $\lambda = 12$ customers/hour
Service Time: Exponential with mean $\mu = 15$ minutes
Service Bays: $s$ (decision variable)

Performance Metrics:

Utilization:

\rho = \frac{\lambda}{s \mu}

Average Wait Time (M/M/s queue):

W_q = \frac{P_0 (\rho^s \lambda/\mu)}{s!(1-\rho)^2} \cdot \frac{1}{\mu}

Monte Carlo Implementation:

Generate $n = 10,000$ simulation runs
For each run, simulate daily operations
Calculate performance metrics
Estimate expected values and confidence intervals

Results:

4 Service Bays: Average wait = 8.5 minutes, Utilization = 75%
5 Service Bays: Average wait = 3.2 minutes, Utilization = 60%
6 Service Bays: Average wait = 1.8 minutes, Utilization = 50%

Optimal Decision: 5 service bays balance cost and customer satisfaction.

Decision Analysis

Expected Value Calculation

For decisions under uncertainty:

EV = \sum_{i=1}^{n} P_i \cdot V_i

Where $P_i$ is the probability and $V_i$ is the value of outcome $i$ .

Expected Value of Perfect Information (EVPI)

EVPI = E[\text{Value with Perfect Information}] - E[\text{Value with Current Information}]

Automotive Example: New Model Launch Decision

Business Context: An automotive manufacturer decides whether to launch a new electric vehicle model considering market uncertainty.

Decision Alternatives:

Launch: High investment, uncertain returns
Don't Launch: No investment, no returns
Market Research: Additional information at cost

Market Scenarios:

Strong Market (P = 0.4): High EV adoption
Moderate Market (P = 0.5): Gradual EV adoption
Weak Market (P = 0.1): Slow EV adoption

Payoff Matrix (NPV in millions):

Decision	Strong	Moderate	Weak	Expected Value
Launch	200	50	-100	EV = 115
No Launch	0	0	0	EV = 0

Expected Value Calculation:

EV(\text{Launch}) = 0.4(200) + 0.5(50) + 0.1(-100) = 80 + 25 - 10 = 115

Decision: Launch the new model (EV = $115M >$ 0M).

Value of Perfect Information:

EVPI = [0.4(200) + 0.5(50) + 0.1(0)] - 115 = 105 - 115 = -10

Since EVPI < 0, perfect information has no additional value.

Multi-Criteria Decision Analysis (MCDA)

Weighted Sum Model

S_i = \sum_{j=1}^{m} w_j \cdot x_{ij}

Where:

$S_i$ is the overall score for alternative $i$
$w_j$ is the weight for criterion $j$
$x_{ij}$ is the normalized score of alternative $i$ on criterion $j$

Analytic Hierarchy Process (AHP)

Pairwise Comparison Matrix:

A = \begin{bmatrix} 1 & a_{12} & a_{13} & \cdots & a_{1n} \\ 1/a_{12} & 1 & a_{23} & \cdots & a_{2n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1/a_{1n} & 1/a_{2n} & 1/a_{3n} & \cdots & 1 \end{bmatrix}

Priority Vector (eigenvalue method):

A \mathbf{w} = \lambda_{\max} \mathbf{w}

Consistency Ratio:

CR = \frac{CI}{RI} = \frac{(\lambda_{\max} - n)/(n-1)}{RI}

Automotive Example: Supplier Selection

Business Context: An automotive manufacturer selects a battery supplier for electric vehicles using multiple criteria.

Selection Criteria:

Cost (Weight: 0.30)
Quality (Weight: 0.35)
Delivery (Weight: 0.20)
Innovation (Weight: 0.15)

Supplier Alternatives:

Supplier A: Cost-focused option
Supplier B: Quality-focused option
Supplier C: Balanced option

Decision Matrix (normalized scores):

Supplier	Cost	Quality	Delivery	Innovation	Weighted Score
A	0.9	0.7	0.8	0.6	0.765
B	0.6	0.95	0.9	0.85	0.8125
C	0.8	0.85	0.85	0.75	0.8125

Weighted Score Calculation for Supplier B:

S_B = 0.30(0.6) + 0.35(0.95) + 0.20(0.9) + 0.15(0.85) = 0.8125

Decision: Suppliers B and C tie for best overall score. Additional analysis needed.

Stochastic Programming

Two-Stage Stochastic Programming

For decisions under uncertainty with recourse actions:

\min_x \left\{ \mathbf{c}^T \mathbf{x} + E_\xi[Q(\mathbf{x}, \boldsymbol{\xi})] \right\}

Where:

Q(\mathbf{x}, \boldsymbol{\xi}) = \min_y \left\{ \mathbf{q}^T \mathbf{y} : \mathbf{W}\mathbf{y} = \mathbf{h} - \mathbf{T}\mathbf{x}, \mathbf{y} \geq 0 \right\}

Automotive Example: Fleet Capacity Planning Under Demand Uncertainty

Business Context: A car rental company plans fleet size considering uncertain seasonal demand.

First-Stage Decision ( $\mathbf{x}$ ): Fleet size by vehicle type Second-Stage Decision ( $\mathbf{y}$ ): Additional rentals or idle capacity

Objective Function:

\min \left[ \sum_i c_i x_i + E\left[\sum_i (p_i y_i^+ + h_i y_i^-)\right] \right]

Where:

$c_i$ = annual cost per vehicle type $i$
$p_i$ = penalty cost for unmet demand
$h_i$ = holding cost for excess capacity
$y_i^+$ = unmet demand, $y_i^-$ = excess capacity

Demand Scenarios: High season (P=0.3), Normal (P=0.5), Low season (P=0.2)

Optimal Solution: Fleet composition balances fixed costs with expected penalty/holding costs across all scenarios.

Automotive Industry Applications

Auto Finance

Portfolio Optimization: Risk-return optimization for loan portfolios
Credit Allocation: Optimal credit limits using stochastic programming
Interest Rate Strategy: Dynamic programming for rate setting

Auto Marketing

Budget Allocation: Multi-channel marketing optimization
Customer Targeting: Integer programming for campaign selection
Price Optimization: Game-theoretic competitive pricing

Auto Sales

Inventory Management: Stochastic inventory models
Territory Planning: Network optimization for sales territories
Incentive Design: Mechanism design for sales compensation

Dealer Operations

Service Scheduling: Optimization of service bay utilization
Parts Inventory: Dynamic programming for parts ordering
Facility Layout: Operations research for optimal layout design

Prescriptive analytics transforms data insights into actionable decisions, providing automotive organizations with mathematically optimal strategies for complex business challenges. By leveraging optimization theory, decision science, and simulation techniques, companies can achieve measurable improvements in efficiency, profitability, and customer satisfaction.

Model Evaluation & Validation Linear Programming