A stochastic hybrid state model for optimizing hedging policies in manufacturing systems with randomly occurring defects

Abstract

The paper addresses the optimal production control problems for an unreliable manufacturing system that produces items that can be regarded as conforming or nonconforming. A new stochastic hybrid state Markovian model with three discrete states, also called modes is introduced. The first two, operational sound and operational defective are not directly observable, while the third mode, failure, is observable. Production of defective parts is respectively initiated and stopped at the random entrance times to and departure times from the defective operational mode. The intricate, piecewise-deterministic dynamics of the model are studied, and the associated Kolmogorov equations are developed under the suboptimal class of hedging policies. The behavior of the model is numerically investigated, optimized under hedging policies, and subsequently compared to that of a tractable extension of the two-mode Bielecki-Kumar single machine model, where both conforming and defective parts are simultaneously produced in the operational mode, while the ratio of produced non conforming to conforming parts remains fixed.

Appendices

Appendices

Derivation of Eqs. 7 and 12

The stock dynamics is defined by

$$ \begin{array}{rll} \frac{dx_1(t)}{dt} &=& -d\\ \frac{dx_2(t)}{dt} &=& k-\frac{x_2(t) \, d}{x_1(t)} \end{array} $$

therefore:

$$ \begin{array}{rll}\label{eq:A1} x_1(t)&=&x_1(0) -d \, t \\ \frac{dx_2(t)}{dt}&=&k-\frac{x_2(t) \, d}{x_1(0) -d \, t } \end{array} $$

(56)

becomes:

$$ \label{eq:A2} \frac{dx_2(t)}{dt}+\frac{x_2(t) \, d}{x_1(0) -d \, t }=k $$

by multiplying both sides of the equation by the same term, we get:

$$ \begin{array}{rll}\label{eq:A3} &&\exp\left(\int^{t}_{0}\frac{ d }{x_1(0) -d \, \tau } \, d\tau \right) \left(\frac{dx_2(t)}{dt}+\frac{x_2(t) \, d}{x_1(0) -d \, t }\right)=k\,\exp\left(\int^{t}_{0}\frac{ d }{x_1(0) -d \, \tau } \, d\tau \right) \\ \text{i.e.} \quad &&\frac{d}{dt}\left[ \exp\left(\int^{t}_{0}\frac{ d }{x_1(0) -d \, \tau } \, d\tau \right) \, x_2(t) \right] = k\,\exp\left(\int^{t}_{0}\frac{ d }{x_1(0) -d \, \tau } \, d\tau \right) \end{array} $$

(57)

and since

$$ \begin{array}{rll} \exp\left(\int^{t}_{0}\frac{ d \, d\tau}{x_1(0) -d \, \tau } \right)&=&\exp\left( -\left[\ln\left( x_1(0) -d \, \tau \right)\right]^{t}_{0}\right) \\ &=&\exp\left(\ln\left(\frac{ x_1(0)}{x_1(0) -d \,t} \right)\right) \\ &=&\frac{ x_1(0)}{x_1(0) -d \,t} \end{array} $$

Eq. 57 becomes:

$$ \begin{array}{rll}\label{eq:A4} \frac{d}{dt}\left[ \frac{ x_1(0)}{x_1(0) -d \,t} \, x_2(t) \right]&=&k\,\frac{ x_1(0)}{x_1(0) -d \,t}\\ \frac{ x_1(0)}{x_1(0) -d \,t} \, x_2(t) - x_2(0) &=&\frac{k\, x_1(0)}{d}\,\int^{t}_{0} \frac{ d \, d\tau}{x_1(0) -d \,\tau}\\ \frac{ x_1(0)}{x_1(t)} \, x_2(t) &=&x_2(0)\,+ \frac{k\, x_1(0)}{d}\, \ln\left(\frac{ x_1(0)}{x_1(t)}\right) \end{array} $$

We finally retrieve Eq. 7

$$ \begin{array}{rll}\label{eq:A5} x_1(t)&=&x_1(0) -d \, t \\ x_2(t)&=&x_1(t)\,\left(\frac{x_2(0)} {x_1(0)}+\frac {k}{d} \ln{\frac{x_1(0)} {x_1(t)}}\right) \end{array} $$

Equation 12 is a special case of Eq. 7 where the initial points are $x_1(0)=\frac{z\,d}{k}$ and $x_2(0)=\frac{z(k-d)}{k}$:

$$ \begin{array}{rll}\label{eq:A10} && x_2(t)= x_1(t)\,\left(\frac{k-d}{d}+\frac {k}{d} \ln\left({\frac{z\,d}{k}}\right)-\frac {k}{d} \ln({x_1(t)})\right) \\ && x_1(t)=\frac{z\,d}{k}- d\, t \end{array} $$

Derivation of Eq. 10

The stock dynamics in the region x ₁(t) + x ₂(t) = z is defined by:

$$ \frac{dx_1(t)}{dt} = \frac{x_2(t) \, d}{x_1(t)} \label{eq:A6} $$

(58)

$$ x_2(t) = z -x_1(t) \label{eq:A7} $$

(59)

Substituting Eq. 59 in Eq. 58, we obtain:

$$ \frac{dx_1(t)}{dt}= \frac{z-x_1(t) \, d}{x_1(t)} $$

Therefore

$$ \begin{array}{rll}\label{eq:A9} d &=& \frac{dx_1(t)}{dt} \left(\frac{x_1(t)}{z-x_1(t)}\right)\\ &=& \frac{dx_1(t)}{dt} \left(-1+\frac{z}{z-x_1(t)}\right)\\ &=& \frac{dx_1(t)}{dt} \left(-1+\frac{z}{z-x_1(t)}\right)\\ &=& -\frac{dx_1(t)}{dt}+\frac{z}{z-x_1(t)}\frac{dx_1(t)}{dt} \end{array} $$

(60)

Integrating Eq. 60 with respect to time,we have:

$$ \begin{array}{rll} d\,t &=& -(x_1(t)-x_1(0))- z \, \left[\ln(z-x_1(t))\right]_{0}^{t} \\ &=& -(x_1(t)-x_1(0))- z \, \ln\left(\frac{z-x_1(t)}{z-x_1(0)}\right) \end{array} $$

i.e.

$$ x_1(t)-x_1(0)+z\ln{\frac{z-x_1(t)}{z-x_1(0)}}+d\,t=0 $$

Hence, Equation 10 is retrieved:

$$ \begin{array}{rll} &&x_1(t)-x_1(0)+z\ln{\frac{z-x_1(t)}{z-x_1(0)}}+d\,t=0\\ &&x_2(t)-x_2(0)-z\ln{\frac{x_2(t)}{x_2(0)}}-d\,t=0 \end{array} $$