β—€ Back to Note

Tony's Intertwingled Memex β€’ Reference Manual

Automated Manufacturing Cost Surface Optimization (Systems & Industrial Engineering) Solution

Formulas & Definitions

Problem Overview

We evaluate the operational expenditure surface C(x,y)=x2+3xy+2y2C(x, y) = x^2 + 3xy + 2y^2C(x,y)=x2+3xy+2y2 at operating setpoint P(4,2)P(4, 2)P(4,2). We compute the gradient βˆ‡C(4,2)\nabla C(4, 2)βˆ‡C(4,2), determine the parameter vector of steepest descent βˆ’βˆ‡C-\nabla Cβˆ’βˆ‡C, find the optimal unit vector uβƒ—opt\vec{u}_{\text{opt}}uopt​, calculate the sensitivity rate of cost reduction, and interpret the algorithm in terms of industrial gradient descent control.


Step-by-Step Mathematical Solution

1. Compute and Evaluate the Cost Gradient

State the Gradient Formula (defined in The Gradient Vector):

βˆ‡C(x,y)=βŸ¨βˆ‚Cβˆ‚x,βˆ‚Cβˆ‚y⟩\nabla C(x, y) = \left\langle \frac{\partial C}{\partial x}, \frac{\partial C}{\partial y} \right\rangleβˆ‡C(x,y)=βŸ¨βˆ‚xβˆ‚C​,βˆ‚yβˆ‚Cβ€‹βŸ©

Compute the partial derivatives:

Cx=βˆ‚βˆ‚x[x2+3xy+2y2]=2x+3yC_x = \frac{\partial}{\partial x}[x^2 + 3xy + 2y^2] = 2x + 3yCx​=βˆ‚xβˆ‚β€‹[x2+3xy+2y2]=2x+3y
Cy=βˆ‚βˆ‚y[x2+3xy+2y2]=3x+4yC_y = \frac{\partial}{\partial y}[x^2 + 3xy + 2y^2] = 3x + 4yCy​=βˆ‚yβˆ‚β€‹[x2+3xy+2y2]=3x+4y

Evaluate at the current operational setpoint P(4,2)P(4, 2)P(4,2):

Cx(4,2)=2(4)+3(2)=8+6=14Β $M/unitC_x(4, 2) = 2(4) + 3(2) = 8 + 6 = 14\text{ \$M/unit}Cx​(4,2)=2(4)+3(2)=8+6=14Β $M/unit
Cy(4,2)=3(4)+4(2)=12+8=20Β $M/unitC_y(4, 2) = 3(4) + 4(2) = 12 + 8 = 20\text{ \$M/unit}Cy​(4,2)=3(4)+4(2)=12+8=20Β $M/unit
βˆ‡C(4,2)=⟨14,20⟩ $M/unit\nabla C(4, 2) = \langle 14, 20 \rangle\text{ \$M/unit}βˆ‡C(4,2)=⟨14,20⟩ $M/unit

2. Vector of Steepest Cost Reduction (Gradient Descent)

To reduce costs most rapidly, adjustments must move along the direction of steepest descent, given by the negative gradient:

vβƒ—opt=βˆ’βˆ‡C(4,2)=βˆ’βŸ¨14,20⟩=βŸ¨βˆ’14,βˆ’20⟩\vec{v}_{\text{opt}} = -\nabla C(4, 2) = -\langle 14, 20 \rangle = \langle -14, -20 \ranglevopt​=βˆ’βˆ‡C(4,2)=βˆ’βŸ¨14,20⟩=βŸ¨βˆ’14,βˆ’20⟩

Normalize to obtain the unit vector direction uβƒ—opt\vec{u}_{\text{opt}}uopt​:

βˆ₯vβƒ—optβˆ₯=(βˆ’14)2+(βˆ’20)2=196+400=596=2149β‰ˆ24.413\|\vec{v}_{\text{opt}}\| = \sqrt{(-14)^2 + (-20)^2} = \sqrt{196 + 400} = \sqrt{596} = 2\sqrt{149} \approx 24.413βˆ₯vopt​βˆ₯=(βˆ’14)2+(βˆ’20)2​=196+400​=596​=2149β€‹β‰ˆ24.413
uβƒ—opt=βŸ¨βˆ’14,βˆ’20⟩2149=βŸ¨βˆ’7149,βˆ’10149βŸ©β‰ˆβŸ¨βˆ’0.573,βˆ’0.819⟩\vec{u}_{\text{opt}} = \frac{\langle -14, -20 \rangle}{2\sqrt{149}} = \left\langle -\frac{7}{\sqrt{149}}, -\frac{10}{\sqrt{149}} \right\rangle \approx \langle -0.573, -0.819 \rangleuopt​=2149β€‹βŸ¨βˆ’14,βˆ’20βŸ©β€‹=βŸ¨βˆ’149​7​,βˆ’149​10β€‹βŸ©β‰ˆβŸ¨βˆ’0.573,βˆ’0.819⟩

3. Localized Sensitivity Rate

The rate of cost reduction achieved per unit step in parameter space along uβƒ—opt\vec{u}_{\text{opt}}uopt​ is:

Duβƒ—optC(4,2)=βˆ’βˆ₯βˆ‡C(4,2)βˆ₯=βˆ’596=βˆ’2149β‰ˆβˆ’24.41Β $MΒ perΒ unitΒ stepD_{\vec{u}_{\text{opt}}} C(4, 2) = -\|\nabla C(4, 2)\| = -\sqrt{596} = -2\sqrt{149} \approx -24.41\text{ \$M per unit step}Duopt​​C(4,2)=βˆ’βˆ₯βˆ‡C(4,2)βˆ₯=βˆ’596​=βˆ’2149β€‹β‰ˆβˆ’24.41Β $MΒ perΒ unitΒ step

4. Machine Learning & Gradient Descent in Automation

  • Mathematical Routine: Automated factory controllers update setpoints via the iterative gradient descent update:
    Pk+1=Pkβˆ’Ξ±βˆ‡C(Pk)P_{k+1} = P_k - \alpha \nabla C(P_k)Pk+1​=Pkβ€‹βˆ’Ξ±βˆ‡C(Pk​)
    where Ξ±>0\alpha > 0Ξ±>0 is the learning rate or actuator gain coefficient.
  • Physical Implementation: The vector βŸ¨βˆ’14,βˆ’20⟩\langle -14, -20 \rangleβŸ¨βˆ’14,βˆ’20⟩ informs the programmable logic controller (PLC) that raw material feed rate (xxx) and conveyor velocity (yyy) must both be decreased simultaneously, with conveyor speed trimmed slightly more aggressively (ratio of 20:14=10:720:14 = 10:720:14=10:7). Taking steps along this negative gradient vector drives the manufacturing line down the steepest cost decline toward the global optimal operating minimum.