Problem Overview
We evaluate the operational expenditure surface C ( x , y ) = x 2 + 3 x y + 2 y 2 C(x, y) = x^2 + 3xy + 2y^2 C ( x , y ) = x 2 + 3 x y + 2 y 2 P ( 4 , 2 ) P(4, 2) P ( 4 , 2 ) β C ( 4 , 2 ) \nabla C(4, 2) β C ( 4 , 2 ) β β C -\nabla C β β C u β opt \vec{u}_{\text{opt}} u opt β
Step-by-Step Mathematical Solution
1. Compute and Evaluate the Cost Gradient
State the Gradient Formula (defined in The Gradient Vector ):
β C ( x , y ) = β¨ β C β x , β C β y β© \nabla C(x, y) = \left\langle \frac{\partial C}{\partial x}, \frac{\partial C}{\partial y} \right\rangle β C ( x , y ) = β¨ β x β C β , β y β C β β© Compute the partial derivatives:
C x = β β x [ x 2 + 3 x y + 2 y 2 ] = 2 x + 3 y C_x = \frac{\partial}{\partial x}[x^2 + 3xy + 2y^2] = 2x + 3y C x β = β x β β [ x 2 + 3 x y + 2 y 2 ] = 2 x + 3 y C y = β β y [ x 2 + 3 x y + 2 y 2 ] = 3 x + 4 y C_y = \frac{\partial}{\partial y}[x^2 + 3xy + 2y^2] = 3x + 4y C y β = β y β β [ x 2 + 3 x y + 2 y 2 ] = 3 x + 4 y Evaluate at the current operational setpoint P ( 4 , 2 ) P(4, 2) P ( 4 , 2 )
C x ( 4 , 2 ) = 2 ( 4 ) + 3 ( 2 ) = 8 + 6 = 14 Β $M/unit C_x(4, 2) = 2(4) + 3(2) = 8 + 6 = 14\text{ \$M/unit} C x β ( 4 , 2 ) = 2 ( 4 ) + 3 ( 2 ) = 8 + 6 = 14 Β $M/unit C y ( 4 , 2 ) = 3 ( 4 ) + 4 ( 2 ) = 12 + 8 = 20 Β $M/unit C_y(4, 2) = 3(4) + 4(2) = 12 + 8 = 20\text{ \$M/unit} C y β ( 4 , 2 ) = 3 ( 4 ) + 4 ( 2 ) = 12 + 8 = 20 Β $M/unit β C ( 4 , 2 ) = β¨ 14 , 20 β© Β $M/unit \nabla C(4, 2) = \langle 14, 20 \rangle\text{ \$M/unit} β C ( 4 , 2 ) = β¨ 14 , 20 β© Β $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:
v β opt = β β C ( 4 , 2 ) = β β¨ 14 , 20 β© = β¨ β 14 , β 20 β© \vec{v}_{\text{opt}} = -\nabla C(4, 2) = -\langle 14, 20 \rangle = \langle -14, -20 \rangle v opt β = β β C ( 4 , 2 ) = β β¨ 14 , 20 β© = β¨ β 14 , β 20 β© Normalize to obtain the unit vector direction u β opt \vec{u}_{\text{opt}} u opt β
β₯ v β opt β₯ = ( β 14 ) 2 + ( β 20 ) 2 = 196 + 400 = 596 = 2 149 β 24.413 \|\vec{v}_{\text{opt}}\| = \sqrt{(-14)^2 + (-20)^2} = \sqrt{196 + 400} = \sqrt{596} = 2\sqrt{149} \approx 24.413 β₯ v opt β β₯ = ( β 14 ) 2 + ( β 20 ) 2 β = 196 + 400 β = 596 β = 2 149 β β 24.413 u β opt = β¨ β 14 , β 20 β© 2 149 = β¨ β 7 149 , β 10 149 β© β β¨ β 0.573 , β 0.819 β© \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 \rangle u opt β = 2 149 β β¨ β 14 , β 20 β© β = β¨ β 149 β 7 β , β 149 β 10 β β© β β¨ β 0.573 , β 0.819 β© 3. Localized Sensitivity Rate
The rate of cost reduction achieved per unit step in parameter space along u β opt \vec{u}_{\text{opt}} u opt β
D u β opt C ( 4 , 2 ) = β β₯ β C ( 4 , 2 ) β₯ = β 596 = β 2 149 β β 24.41 Β $MΒ perΒ unitΒ step D_{\vec{u}_{\text{opt}}} C(4, 2) = -\|\nabla C(4, 2)\| = -\sqrt{596} = -2\sqrt{149} \approx -24.41\text{ \$M per unit step} D u opt β β C ( 4 , 2 ) = β β₯β C ( 4 , 2 ) β₯ = β 596 β = β 2 149 β β β 24.41 Β $MΒ perΒ unitΒ step 4. Machine Learning & Gradient Descent in Automation
Mathematical Routine: Automated factory controllers update setpoints via the iterative gradient descent update:
P k + 1 = P k β Ξ± β C ( P k ) P_{k+1} = P_k - \alpha \nabla C(P_k) P k + 1 β = P k β β Ξ± β C ( P k β ) Ξ± > 0 \alpha > 0 Ξ± > 0 Physical Implementation: The vector β¨ β 14 , β 20 β© \langle -14, -20 \rangle β¨ β 14 , β 20 β© x x x y y y 20 : 14 = 10 : 7 20:14 = 10:7 20 : 14 = 10 : 7