Some general properties for the newsboy problem with an extraordinary order

Abstract

In this paper, an extension of the standard newsboy problem is presented, involving an extraordinary order and a variable mixture of backorders and lost sales. The backlogged demand ratio is given by a nonincreasing function of the quantity of shortage. Some general properties for the expected cost are derived under weak assumptions about the backorder rate function. When the backorder rate is a linear function, some sufficient conditions for the global convexity of the expected cost are derived. A sufficient condition for the local concavity of this function is also provided. Numerical examples are presented to illustrate the theoretical results and a specific practical case is proposed and solved. Moreover, a sensitivity analysis of the optimal solution with respect to the parameters of the backorder rate function is included. Finally, some extensions of the proposed model are suggested as possible directions for future research.

Appendix

In this Appendix, all the proofs for the results obtained in this paper are included.

Proof of Lemma 1

If 0<Q<a, then x−Q>x−a and β(x−Q)≤β(x−a). Therefore, from (3), it follows that

$$\begin{aligned} T(Q) =&\int_{a}^{b} ( x-Q ) \bigl[ p+ ( \omega -p ) \beta (x-Q) \bigr] f(x)\,dx\\ >&\int_{a}^{b} ( x-a ) \bigl[ p+ ( \omega -p ) \beta (x-a) \bigr] f(x)\,dx=T(a). \end{aligned}$$

On the other hand, if Q>b then it is clear that

$$ T(Q)=h\int_{a}^{b}(Q-x)f(x)\,dx>h\int _{a}^{b}(b-x)f(x)\,dx=T(b) $$

and the proof is complete. □

Proof of Lemma 2

The following cases are considered:

• (a) M<b−a

  • (i) If Q∈(a,b−M), the expression (4) can be rewritten as

    $$\begin{aligned} T(Q) =&h\int_{a}^{Q}(Q-x)f(x)\,dx+p\int _{Q}^{b}(x-Q)f(x)\,dx \\ &{}+( \omega -p ) \int _{Q}^{Q+M} (x-Q)\beta (x-Q)f(x)\,dx. \end{aligned}$$

    With the given assumptions, the Leibniz rule for the derivation of an integral with variable limits of integration can be applied to obtain the formula (5).

  • (ii) If Q∈(b−M,b), T(Q) is given by (4) and, as in the previous case, applying the Leibniz rule the expression (6) is obtained.

• (b) M≥b−a. The same reasoning as in (ii) can now be applied.

The rest of the proof follows easily. □

Proof of Corollary 3

It is immediate from part (iii) of Lemma 2 and taking into account that f(b)=0 when b=∞. □

Proof of Corollary 4

First of all, $T_{+}^{\prime }(a)$ is examined. From Lemma 2(4), it follows that, if a<b−M, then

$$\begin{aligned} T_{+}^{\prime }(a) =&-p+(\omega -p) \biggl\{ -\int _{a}^{a+M} \bigl[ \beta (x-a)+(x-a)\beta ^{\prime }(x-a) \bigr] f(x)\,dx\\ &{}+M\beta _{-}(M)f(a+M) \biggr\} \\ \leq &-p+(p-\omega )\beta _{o}\int_{a}^{a+M}f(x) \,dx\\ &{}+(p-\omega )\int_{a}^{a+M}(x-a)\beta ^{\prime }(x-a)f(x)\,dx+0 \\ \leq &-p+(p-\omega )\beta _{o}<0, \end{aligned}$$

and, if a≥b−M, we have

$$\begin{aligned} T_{+}^{\prime }(a) =&-p+(\omega -p) \biggl\{ \int _{a}^{b}\bigl[-\beta (x-a)-(x-a)\beta ^{\prime }(x-a)\bigr]f(x)\,dx \biggr\} \\ \leq& -p+(p-\omega )\beta _{o}<0. \end{aligned}$$

On the other hand, applying Lemma 2(5), $T_{-}^{\prime }(b)=h>0$ is obtained.

As a consequence, the function T(Q) attains its global minimum within the open interval (a,b). □

Proof of Lemma 5

It may be proved by reductio ad absurdum. Assume that L=lim _y↓0{−yβ′(y)}>0 (because β(y) is a non-increasing function). Then, for a given ε, with 0<ε<L/2, we could find δ∈(0,M) such that if y∈(0,δ), then $-\beta ^{\prime }(y)\,{>}\,\frac{\epsilon }{y}$. Therefore, $\int_{0}^{\delta }\,{-}\,\beta ^{\prime }(y)\,dy\,{>}\,\int_{0}^{\delta }\frac{\epsilon }{y}\,dy\,{=}\,\epsilon [ \ln \delta \,{-}\,\lim_{y\downarrow 0}\ln y]\,{=}\,\infty$, which is impossible because $\int_{0}^{\delta }-\beta ^{\prime }(y)\,dy= -\beta (y) \vert _{0}^{\delta }=\beta _{+}(0)-\beta _{-}(\delta )= \beta _{o}-\beta (\delta )<\infty$. □

Proof of Lemma 6

It is clear that the assumptions required here on the functions β(y) and f(x) involve the assumptions about these functions given in Lemma 2. The rest of the proof follows from the application of the Leibniz rule to (5) and (6), and next, using the result given in Lemma 5. □

Proof of Theorem 7

If Q∈(a,b) and Q≠b−M, from Lemma 6 and the given conditions, it is verified that T′′(Q)≥[h+p+β _o (ω−p)]f(Q). Therefore, taking into account that

$$ h+p+\beta _{o}(\omega -p)=h+\beta _{o}\omega +(1-\beta _{o})p>0, $$

(14)

we obtain T′′(Q)≥0. This inequality, together with expression (7), proves the convexity of T(Q) on (a,b).

Moreover, if f(x)>0 for all x∈(a,b), then (14) implies the strict convexity of T(Q). □

Proof of Lemma 8

The following cases are considered:

• (a) M<b−a

  • (i) From Lemma 2, if Q∈(a,b−M), now (5) becomes

    $$\begin{aligned} T^{\prime }(Q) =&hF(Q)+p\bigl[F(Q)-1\bigr]\\ &{}+(\omega -p) \biggl\{ \int _{Q}^{Q+M}\bigl[-\beta (x-Q)-(x-Q)\beta ^{\prime }(x-Q)\bigr]f(x)\,dx \biggr\} \end{aligned}$$

    and

    $$\begin{aligned} T^{\prime \prime }(Q) =&\bigl[h+p+\beta _{o}(\omega -p)\bigr]f(Q)\\ &{}+( \omega -p) \biggl\{ \int_{Q}^{Q+M}\bigl[2\beta ^{\prime }(x-Q)+(x-Q)\beta ^{\prime \prime }(x-Q)\bigr]f(x)\,dx \biggr\} \\ &{}+(p-\omega )M\beta _{-}^{\prime }(M)f(Q+M) \\ =&\bigl[h+p+\beta _{o}(\omega -p)\bigr]f(Q)\\ &{}+\frac{\beta _{o}(p-\omega )}{M} \bigl \{ 2\bigl[F(Q+M)-F(Q)\bigr]-Mf(Q+M) \bigr\}. \end{aligned}$$

  • (ii) On the other hand, if Q∈(b−M,b), the expression for T′(Q) remains as in Eq. (6). Hence,

    $$\begin{aligned} T^{\prime \prime }(Q) =&\bigl[h+p+\beta _{o}(\omega -p)\bigr]f(Q)\\ &{}+( \omega -p) \biggl\{ \int_{Q}^{b}\bigl[2\beta ^{\prime }(x-Q)+(x-Q)\beta ^{\prime \prime }(x-Q)\bigr]f(x)\,dx \biggr\} \\ =&\bigl[h+(1-\beta _{o})p+\beta _{o}\omega \bigr]f(Q)+ \frac{2\beta _{o}(p-\omega )}{M}\bigl[1-F(Q)\bigr]>0. \end{aligned}$$

• (b) If M≥b−a, the same reasoning as in (ii) can be applied.

The rest of the proof follows easily. □

Proof of Theorem 9

It follows directly from Lemma 8. □

Proof of Theorem 10

The proof follows easily taking into account that $T_{+}^{\prime \prime }(a)$ is obtained through the expression (11) if M<b−a, and through the expression (12) when M≥b−a. Thus, if M<b−a we have $T_{+}^{\prime \prime }(a)=[h+p+\beta _{o}(\omega -p)]f(a)+\frac{\beta _{o}(p-\omega )}{M} [ 2F(a+M)-Mf(a+M) ]$. On the other hand, if M≥b−a, then $T_{+}^{\prime \prime }(a)=[h+(1-\beta _{o})p+\beta _{o}\omega ]f(a)+\frac{2\beta _{o}(p-\omega )}{M}>0$. □