Inventory Systems

Abstract

In traditional inventory systems the typical decisions involve the determination of the level of service to be provided, the frequency of or time between replenishments R, the order up to level S which determines how much to order, and the reorder point s which determines when in the cycle the order is placed. Depending on the conditions and assumptions of the inventory system, these values can be either parameters or variables and this has created many different types of inventory control policies. An inventory control policy is considered “optimal” if it minimizes the long-range or average total system cost. While this long-range average cost differs significantly from industry to industry and from product to product, 25% of the product value is often used as a first order approximation of the average cost for holding a product in inventory for 1 year.

Appendices

Appendix A: Expected Lost Sales for Discrete Distribution Function

Several alternative methods exist for the derivation of the expected lost sales for a discrete cumulative demand distribution. The approach taken in the following derivation approximates the continuous demand distribution by a single value in each interval. In an attempt to make this appendix self-contained several of the formulas shown previously in the chapter are repeated here. The intervals are assumed to be sorted by increasing cumulative distribution values, i.e., it is assumed that the cumulative distribution is given for N intervals for with as upper bounds Q ₁, Q ₂, … Q _N and that F(Q ₁) £ F(Q ₂) £ … £ F(Q _N ). To simplify the notation, the cumulative distribution function at the lower bound of the first interval will be denoted by F(Q ₀) and its value is by definition equal to zero.

For the first interval, the computation of the expected lost sales is given by the following two expressions.

$$ \begin{array}{l} ns\left( {Q_0 } \right) = \bar d \\ns\left( {Q_1 } \right) = \bar d - Q_1\\\end{array} $$

(10.95)

Recall that for a continuous demand distribution the expected number of lost sales is equal to the following.

$$ ns\left( s \right) = E\left[ {NS\left( s \right)} \right] = \int_s^\infty{(x - s)f\left( x \right)dx}$$

(10.96)

The expected lost sales at the upper bounds of the intervals can then be computed as follows.

$$ \begin{aligned} ns\left( {Q_k } \right) &= \int\limits_{Q_k }^\infty{\left( {x - Q_k } \right)f\left( x \right)} dx = \int\limits_{Q_k }^\infty{xf\left( x \right)} dx - Q_k \int\limits_{Q_k }^\infty{f\left( x \right)} dx \\& = \sum\limits_{j = 0}^{N - 1 - k} {\left( {\int\limits_{Q_{k + j} }^{Q_{k + j + 1} } {xf\left( x \right)} dx} \right)}- Q_k \left( {1 - F\left( {Q_k } \right)} \right) \\\end{aligned} $$

(10.97)

The integral in this expression calculates the expected demand in one interval. It is assumed that the only possible values of the demand in each interval is equal to the upper bound of that interval. The overall expected demand and the expected demand in a interval are then computed by the following two expressions. Recall that for discrete cumulative distribution functions, the critical ratio identifies the interval and the optimal value of the inventory is set to the upper bound of that interval.

$$ \bar d = \sum\limits_{j = 1}^N {Q_j\cdot } \left( {F\left( {Q_j } \right) - F\left( {Q_{j - 1} } \right)} \right) $$

(10.98)

$$ \int\limits_{Q_{k + j} }^{Q_{k + j + 1} } {xf\left( x \right)} dx = Q_{k + j + 1} \left( {F\left( {Q_{k + j + 1} } \right) - F\left( {Q_{k + j} } \right)} \right) $$

(10.99)

The expected lost sales can then be expressed in quantities that only depend on interval values.

$$ ns\left( {Q_k } \right) = \sum\limits_{j = 0}^{N - 1 - k} {\left( {Q_{k + j + 1}\cdot \left( {F\left( {Q_{k + j + 1} } \right) - F\left( {Q_{k + j} } \right)} \right)} \right)}- Q_k \left( {1 - F\left( {Q_k } \right)} \right) $$

(10.100)

Next the difference between two values of the expected lost sales for adjacent interval end points is computed.

$$ \begin{array}{l} ns\left( {Q_{k + 1} } \right) - ns\left( {Q_k } \right) \\\quad\quad\quad= \sum\limits_{j = 0}^{N - 2 - k} {\left( {Q_{k + j + 2}\cdot \left( {F\left( {Q_{k + j + 2} } \right) - F\left( {Q_{k + j + 1} } \right)} \right)} \right)}- Q_{k + 1} \left( {1 - F\left( {Q_{k + 1} } \right)} \right) \\ \quad\quad\quad\quad- \sum\limits_{j = 0}^{N - 1 - k} {\left( {Q_{k + j + 1}\cdot \left( {F\left( {Q_{k + j + 1} } \right) - F\left( {Q_{k + j} } \right)} \right)} \right)}+ Q_k \left( {1 - F\left( {Q_k } \right)} \right) \\\end{array} $$

(10.101)

The two summations in the above equation contain the same terms except that the second summation contains one extra term corresponding to the interval [Q _{k + j} , Q _{k + j +1}]. The expression can thus be simplified to the following.

$$\begin{array}{l} s\left( {Q_{k + 1} } \right) - ns\left( {Q_k } \right) \\\quad= - Q_{k + 1} \cdot \left( {F\left( {Q_{k + 1} } \right) - F\left( {Q_k } \right)} \right) - Q_{k + 1} \left( {1 - F\left( {Q_{k + 1} } \right)} \right) + Q_k \left( {1 - F\left( {Q_k } \right)} \right) \\\quad = - \left( {Q_{k + 1} - Q_k } \right)\left( {1 - F\left( {Q_k } \right)} \right) \end{array}$$

(10.102)

This establishes the recursive relationship for the expected lost sales between subsequent intervals.

$$ ns\left( {Q_{k + 1} } \right) = ns\left( {Q_k } \right) - \left( {Q_{k + 1}- Q_k } \right)\left( {1 - F\left( {Q_k } \right)} \right)\quad for\;k = 1 \ldots N - 1 $$

(10.103)

Appendix B: Optimal (s, Q) Policy for Service Level 2

In an attempt to make this appendix self-contained, several of the formulas shown previously in the chapter are repeated here. The objective is to minimize the total, long-range inventory cost subject to the fill rate service constraint. The total inventory cost includes the ordering cost and the holding costs for the cycle and the safety inventory. The optimization problem is thus:

$$ \min \quad TC = \frac{D}{Q}oc + hc\frac{Q}{2} + hc \cdot \left( {s - d \cdot lt} \right) $$

(10.104)

$$ s.t.\quad \frac{{ns\left( s \right)}}{Q} \le \left( {1 - \beta } \right) $$

(10.105)

$$ Q \ge 0 $$

This is a nonlinear optimization problem with a single inequality constraint. Given that both the objective function and the constraint are convex in function of Q, the Karush–Kuhn–Tucker (KTT) conditions are sufficient conditions for a nonnegative Q ^* to be the global optimum. The optimization problem can be written in the following standard form.

$$ \begin{array}{l} \min \quad f\left( x \right) \\s.t.\quad g\left( x \right) \le 0 \\\end{array} $$

(10.106)

The sufficient KKT conditions for a feasible x ^* to be the global optimum are then as follows, where ∇f(x) denotes the gradient of the function f(x). The gradient has to be taken with respect to the two variables Q and s.

$$ \nabla f\left( {x^* } \right) + \mu\cdot \nabla g\left( {x^* } \right) = 0 $$

(10.107)

$$ \mu\cdot g\left( {x^* } \right) = 0 $$

(10.108)

$$ \mu\ge 0 $$

(10.109)

For the inventory problem, the sufficient KKT conditions can then be written as follows.

$$ \nabla \left( {\frac{D}{{Q^* }}oc + hc\frac{{Q^* }}{2} + hc \cdot \left( {s^*- d \cdot lt} \right)} \right) + \mu \nabla \left( {\frac{{ns\left( {s^* } \right)}}{{Q^* }} - \left( {1 - \beta } \right)} \right) = 0 $$

(10.110)

$$ \mu \left( {\frac{{ns\left( {s^* } \right)}}{{Q^* }} - \left( {1 - \beta } \right)} \right) = 0 $$

(10.111)

$$ \mu\ge 0 $$

(10.112)

If m equals zero, then the EOQ quantity provides the optimal solution to this problem and the safety inventory level is equal to zero. This can be interpreted as the case where the inventory level is computed using the EOQ formula while ignoring the service level requirement and this inventory level satisfies the service level. In other words, ordering enough inventory to cover the expected demand during the lead time satisfies the required service level.

$$ \begin{array}{l} Q^*= \sqrt {\displaystyle \frac{{2 \cdot D \cdot oc}}{{hc}}}\\s^*= d \cdot lt \\\end{array} $$

(10.113)

The remainder of the discussion treats the case where m is strictly positive. In this case, the following equality must be satisfied.

$$ ns\left( {s^* } \right) = Q^* \left( {1 - \beta } \right) $$

(10.114)

First we focus on the computation of the gradient constraint with respect to the reorder point s. This yields the following equality.

$$ hc + \frac{{\mu ^* }}{{Q^* }}\,\frac{{dns\left( {s^* } \right)}}{{ds}} = 0 $$

(10.115)

Recall that ns(s) is the expected number of units short per cycle, or

$$ ns\left( s \right) = E\left[ {NS\left( s \right)} \right] = \int_s^\infty{\left( {x - s} \right)f\left( x \right)dx}$$

(10.116)

The derivative of ns(s) with respect to the reorder point s can be computed with Leibniz’s rule.

$$ \begin{aligned} \frac{d}{{dy}}\int_{l\left( y \right)}^{u\left( y \right)} {h\left( {x,y} \right)} dx &= \int_{l\left( y \right)}^{u\left( y \right)} {\left( {\frac{{\partial h\left( {x,y} \right)}}{{\partial y}}} \right)dx}+ h\left( {u\left( y \right),y} \right)\frac{{du\left( y \right)}}{{dy}} \\& \quad - h\left( {l\left( y \right),y} \right)\frac{{dl\left( y \right)}}{{dy}} \end{aligned}$$

(10.117)

This yields the following substitutions:

$$ \begin{array}{l} y = s \\h\left( {x,y} \right) = \left( {x - s} \right)f\left( x \right) \\u\left( y \right) = \infty ,\,\displaystyle\frac{{du\left( y \right)}}{{dy}} = 0 \\l\left( y \right) = s,\,\displaystyle\frac{{dl\left( y \right)}}{{dy}} = 1,\,h\left( {l\left( y \right),y} \right) = \left( {s - s} \right)f\left( s \right) = 0 \\ \displaystyle\frac{{\partial h\left( {x,y} \right)}}{{\partial y}} = \displaystyle\frac{{\partial \left( {x - s} \right)}}{{\partial s}}f\left( x \right) + \left( {x - s} \right)\displaystyle\frac{{\partial f\left( x \right)}}{{\partial s}} =- f\left( x \right) \\\end{array} $$

$$ \frac{{dns\left( s \right)}}{{ds}} = \int_s^\infty{ - f\left( x \right)dx}= \left. { - F\left( x \right)} \right|_s^\infty =- \left( {1 - F\left( s \right)} \right) $$

(10.118)

Substituting this in the gradient equation (10.115) yields

$$ hc + \frac{{\mu ^* }}{{Q^* }}\left( { - \left( {1 - F\left( {s^* } \right)} \right)} \right) = 0 $$

(10.119)

which can be rewritten as

$$ F\left( {s^* } \right) = 1 - \frac{{hc \cdot Q^* }}{{\mu ^* }} $$

The same equation can also be rewritten to determine the optimal value of μ ^*.

$$ \mu ^*= \frac{{hc \cdot Q^* }}{{1 - F\left( {s^* } \right)}} $$

(10.120)

Next we focus on the computation of the gradient constraint with respect to the reorder quantity Q. This yields the following equality.

$$- \frac{{oc \cdot D}}{{Q^{*2} }} + \frac{{hc}}{2} + \mu ^* \left[ { - \frac{{ns\left( {s^* } \right)}}{{Q^{*2} }}} \right] = 0 $$

(10.121)

Multiplying all terms by ${{2Q^{*2} } \mathord{\left/ {\vphantom {{2Q^{*2} } {hc}}} \right. \kern-\nulldelimiterspace} {hc}}$ and substituting for μ ^* with (10.120) yields the following quadratic equation

$$ \frac{{ - 2D \cdot oc}}{{hc}} + Q^{*2}- \frac{{2 \cdot ns\left( {s^* } \right)}}{{1 - F\left( {s^* } \right)}}Q^*= 0 $$

(10.122)

The general quadratic equation

$$ ax^2+ bx + c = 0 $$

(10.123)

has the following two roots, which are two real numbers provided the discriminant, which is the expression under the square root sign in this equation, is positive.

$$ x^*= \frac{{ - b \pm \sqrt {b^2- 4ac} }}{{2a}} $$

(10.124)

Since for the quadratic equation given in (10.122) a is equal to 1 and c is negative, the discriminant for this equation is always positive. For the quadratic equation for the reorder quantity only the root value computed with the positive sign yields a root with a positive value and is thus feasible. The optimal reorder quantity is given by.

$$ Q^*= \frac{{ns\left( {s^* } \right)}}{{1 - F\left( {s^* } \right)}} + \sqrt {\left( {\frac{{ns\left( {s^* } \right)}}{{1 - F\left( {s^* } \right)}}} \right)^2+ \frac{{2D \cdot oc}}{{hc}}}$$

(10.125)

For a normal demand distribution, ns(s ^*) can be written in function of the unit loss function, which yields the following equation for z ^*.

$$ z^*= L^{ - 1} \left( {\frac{{Q^* \left( {1 - \beta } \right)}}{{s_{dlt} }}} \right) $$

(10.126)

$$ s^*= d \cdot lt + z^*\cdot s_{dlt}= d \cdot lt + s_{dlt}\cdot L^{ - 1} \left( {\frac{{Q^* \left( {1 - \beta } \right)}}{{s_{dlt} }}} \right) $$

(10.127)

The optimal values of ns(s ^*) and F(s ^*) can then be determined from s ^*.

The Eqs. (10.125) and (10.127) provide two equations for the optimal reorder quantity and the optimal reorder point. Because the variables are mutually dependent in the equations, they have to be solved for iteratively through the method of cyclical coordinates. Finally, the optimal value of μ ^* can be computed with (10.120). Based on (10.71) and (10.120), the optimal value of μ ^* can thus be interpreted as the penalty for having a shortage equal to the total annual demand.

$$ \mu ^*= sc \cdot D $$

(10.128)