Maximum-Profit Inventory Model with Generalized Deterioration Rate

Abstract

We developed a maximum profit inventory model with a generalized deterioration rate where the selling rate is dependent on the inventory level that is an extension of two published papers. A complete solution structure is provided to decide the optimal solution with reasonable conditions supported by numerical examples, and then we prove that the optimal solution is independent of the demand pattern. Numerical examples are provided to illustrate our findings. In a previously published paper, three examples had symmetric conditions to decide the local maximum solution. Our approach provides a reasonable explanation for this symmetric phenomenon. Our findings will help researchers develop new inventory models in the future.

1. Introduction

Hill [1] is a pioneer, creating an inventory model with ramp-type demands where the demand is separated into two phases. In the first phase, the demand is an increasing function, and then in the second phase, the demand is a constant function. There has been a trend of considering these kinds of inventory models with minimum-cost and maximum-profit problems. For example, we list several articles by Mandal and Pal [2] on decay products; Wu et al. [3] on the backlogging rate related to the waiting time; Wu and Ouyang [4] on two replenishment policies starting with a shortage or without a shortage; Wu [5] on the deteriorated items following the Weibull distribution; and Giri et al. [6] on three-parameter Weibull distribution. Wu and Ouyang [4] examined the cyclic deterioration of items. Deng [7], to improve upon the work of Wu et al. [3], showed the existence of an optimal solution. Manna and Chaudhuri [8] developed inventory models where the deterioration is dependent on time. Deng et al. [9] improved upon the work of Mandal and Pal [2] to prove that an optimal solution exists. Wu et al. [10] examined the maximum-profit problem in inventory models with stock-dependent selling rates and ramp-type demand. They constructed two inventory systems which were dependent on the changing point of the ramp-type demand, and then examined the optimal solution for each case. In Wu et al. [10], the demand is a ramp type, and Chuang [11] generalized it to arbitrary demand where the selling rate is stock-dependent, and then Chuang [11] derived a complete solution procedure. Several related papers include: Lin et al. [12] with net present value, Mahajan and Diatha [13] with continuous compounding, Alım and Beullens [14] with a flexible delivery option, Khakzad and Gholamian [15] with inspection and advanced payment, Khan et al. [16] with an all-units discount environment and price-sensitive demand, Rezagholifam et al. [17] with non-instantaneous deteriorating items and capacity constraint, Sarkar et al. [18] with the time value of money, and Tiwari et al. [19] with trade credit. These papers are worth mentioning to point out the current research trend for this kind of inventory system. Several related papers that studied inventory models with deteriorating items are worthwhile to mention. Padiyar et al. [20] solved an inventory model of decay products under a fuzzy environment and with shortages where the demand is related to the selling price. Acevedo-Chedid et al. [21] developed a four-level inventory model, consisting of the supplier, production plants, distribution, and retailers, to examine collaborative planning for decay items. Mishra et al. [22] solved a sustainable supply chain inventory system with carbon emissions and controllable deteriorating products in a greenhouse environment. Mashud et al. [23] constructed a sustainable inventory system for imperfect items with controllable emissions and decay products to reduce carbon dioxide through green technology investments. Mishra et al. [24] examined an EOQ inventory system where the demand is dependent on the selling price and inventory level with preservation technology investment, shortages, and a controllable deterioration rate. Tiwari et al. [25] developed a two-echelon inventory system with partial trade credits for decay products with an expiration date to provide solution algorithms and sensitivity analysis. Khan et al. [26] constructed inventory models of deteriorating items with an expiration date where the demand is related to the selling price. They presented a solution algorithm to find the optimal solution. Padiyar et al. [27] considered an integrated EPQ inventory system of producer and buyer, with an imperfect production process for decay items that have two different demands: the producer with exponential demand, and the buyer with triangular demand, where the production rate is related to the demand. We also point out the following two other important papers related to inventory systems. Yadav et al. [28] examined a two-level integrated inventory system of manufacturer and retailer where the lead time is crashable with a service-level constraint, and the setup cost has a learning and forgetting effect. Navarro et al. [29] studied a maximum-profit EPQ inventory model with a supply chain of three levels, manufacturers, distribution centers, and retailers, to consider the effect of the marketing effort on the demand under multiple-item environments. We provide

Table 1. Comparison of our present work and some related papers.

Paper	D	II	PDD	PC	SP	SA	S	SM	SDD	SC
[10]	X			X	X		X	X	X
[11]	X			X	X		X	X	X
[20]	X		X		X	X	X			X
[21]	X	X		X	X		X			X
[22]	X		X		X	X	X	X		X
[23]	X	X	X	X	X	X		X
[24]	X		X		X	X	X	X	X
[25]	X		X	X	X	X	X	X		X
[26]	X		X	X	X	X	X	X
[27]	X	X		X		X	X			X
[28]					X	X		X		X
[29]		X		X	X	X				X
ours	X			X	X	X	X	X	X

D deterioration; II imperfect items; PDD price-dependent demand; PC purchasing cost; SP selling price; SA sensitivity analysis; S shortages; SM solution method; SDD stock-dependent demand; SC supply chain.

below to compare our paper with the above-mentioned articles to indicate the current trend of research directions. Our findings will help researchers to realize maximum-profit inventory models.

Based on Table 1, the selling price and purchasing cost have been studied by many papers. It shows that maximum-profit models are more prevalent than minimum-cost models. Many models consider the deterioration cost, revealing that decision-makers pay attention to decay problems. Sensitivity analysis and shortages are the third and fourth most common issues in Table 1. Sensitivity analysis examines the influence of decision variables by the variation in constant parameters. Investigation of shortages can prevent stocking too many products to provide a balance between inventory and backorders. The solution method is another important research issue, showing that besides developing new inventory systems, researchers also focus on finding the optimal solution. Our paper reviews the above-mentioned six issues to indicate that it follows the main research direction.

2. Notation and Assumptions

To develop an inventory system where the demand is dependent on the inventory level, to be compatible with Wu et al. [10] and Chuang [11], we use their assumptions and notations, except the deterioration function is generated from a constant function to an arbitrary function. Moreover, we constructed an auxiliary function, $\mathrm{m}\left(t_{\mathrm{1}}\right)$. The notations are given below:

(1)

The lead time is zero because replenishments are instantaneous with an infinite replenishment rate.

(2)

$A$ is the ordering cost per order.

(3)

$c$ is the purchasing cost per unit.

(4)

$C_d$ is the deterioration cost per unit.

(5)

$C_h$ is the inventory holding cost per unit, per unit of time.

(6)

$C_s$ is the shortage cost per unit, per unit of time.

(7)

$s$ is the selling price per unit.

(8)

$t_1$ is the time when the inventory level drops to zero.

(9)

$t_1^{*}$ is the optimal solution for the variable $t_1$.

(10)

$T$ is the finite time horizon under consideration.

(11)

$\alpha$ is the portion of extra selling related to inventory level. We define those sales related to inventory level as “extra sells”. For a technical reason, we assume that $\alpha \left(s+C_d\right)<C_h$, which will be explained in Section 3.

(12)

$\mu$ is the time that the ramp-type demand changes from an increasing phase to a constant phase.

(13)

$\theta \left(t\right)$ is the deteriorated fraction of the on-hand inventory deterioration per unit of time.

(14)

$I\left(t\right)$ is the on-hand inventory level at the time $t$ over the ordering cycle $\left[0,T\right]$.

(15)

The shortage is allowed and fully backlogged.

(16)

Without referring to extra sales, the demand rate R(t) is a positive continuous function.

(17)

$D\left(t\right)$ is the consumption sale rate (the actual rate), which is stock-dependent with $D\left(t\right)=\{\begin{array}{c}R(t)+\alpha I(t) ,I\left(t\right)>0,\hfill \\ R(t) ,\hspace{0.17em}I(t)\le 0, \hfill \end{array}$ where $\alpha$ is a positive constant.

(18)

$Z\left(t_1\right)$ is the total profit that consists of sale revenue, order cost, purchase cost, holding cost, deterioration cost, and shortage cost.

(19)

$m\left(t_1\right)$ is an auxiliary function such that $m\left(t_1\right)=0$ and $\frac{\mathrm{d}}{\mathrm{d}{t}_1}\mathrm{Z}\left(t_1\right)=0$ have the same solution.

We recall that (a) $t_1$ is the time when the inventory level drops to zero, and (b) $\mu$ is the time when the ramp-type demand changes from an increasing phase to a constant phase. Depending on the relation between $t_1$ and $\mu$, Wu et al. [10] divided their model into two cases: Case 1, $t_1\ge \mu$, and Case 2, $t_1\le \mu$. The deterioration rate is a constant function in Wu et al. [10]. Hence, there were two objective functions in Wu et al. [10] with different domains. Chuang [11] studied the same inventory model as Wu et al. [10], and he also considered the deterioration cost. Chuang [11] did not write down the explicit formula for the ramp-type demand, such that there is only one objective function in Chuang [11]. Our paper extended the work of Wu et al. [10] and Chuang [11] considering a deterioration rate from a constant function to a non-negative function. We followed the solution procedure of Chuang [11]. We focus on (i) the demand function and (ii) the deterioration function, both in an implicit setting, such that there is only one objective function in our paper.

3. Our proposed Inventory Model

We recall the inventory systems with ramp-type demand. These kinds of inventories were first proposed by Hill [1], and then further investigated by Mandal and Pal [2], Deng et al. [9], Wu and Ouyang [4], and Wu et al. [10]. Replenishment occurs at the time $t=0$ when the inventory level begins at its maximum level. From $t=0$ to $t_1$, the inventory level decreases owing to deterioration and demand, $R\left(t\right)$. At $t_1$, the inventory level drops to zero, after which shortages occurred for the period $\left(t_1,T\right)$, and all of the unsatisfied demand during the shortage period $\left(t_1,T\right)$ is completely backlogged. Since the actual selling rate is dependent on the inventory level, the analytical model is described by the following two differential equations:

$$\frac{d}{dt}I(t)+\left(\theta \left(t\right)+\alpha \right) I(t)=-R(t),0<t<t_1$$

(1)

and

$$\frac{d}{dt}I(t)=-R(t),t_1<t<T$$

(2)

We try to solve the differential system of Equations (1) and (2), then

$$I\left(t\right)={\displaystyle \underset{t}{\overset{t_1}{\mathrm{\int }}}R\left(y\right)e^{{\displaystyle {\mathrm{\int }}_t^y\left(\theta \left(x\right)+\alpha \right)dx}}dy}$$

(3)

for $0\le t\le t_1$, and

$$I\left(t\right)={\displaystyle \underset{t}{\overset{t_1}{\mathrm{\int }}}R\left(x\right)dx}$$

(4)

for $t_1\le t\le T$.

The purchased quantity at $t=0$ plus the backorder quantity at $t=T$ is the sum of $I\left(0\right)$ and $-I\left(T\right)$; then, the purchased quantity will be

$$I\left(0\right)-I\left(T\right)={\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}R\left(y\right)e^{{\displaystyle {\mathrm{\int }}_0^y\left(\theta \left(x\right)+\alpha \right)dx}}dy}+{\displaystyle \underset{t_1}{\overset{T}{\mathrm{\int }}}R\left(x\right)dx}$$

(5)

The holding cost during the period $\left[0,t_1\right]$ is evaluated by changing the order of integration

$$C_h{\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}I\left(t\right)dt}=C_h{\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}R\left(y\right){\displaystyle \underset{0}{\overset{y}{\mathrm{\int }}}{e}^{{\displaystyle {\mathrm{\int }}_t^y\left(\theta \left(x\right)+\alpha \right)dx}}}}dtdy$$

(6)

The number of deteriorated items during the period, $\left[0,t_1\right]$, is evaluated

$$I\left(0\right)-{\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}D\left(x\right)}dx={\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}R\left(y\right)\left(e^{{\displaystyle {\mathrm{\int }}_0^y\left(\theta \left(x\right)+\alpha \right)dx}}-1\right)dy}-\alpha {\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}R\left(y\right){\displaystyle \underset{0}{\overset{y}{\mathrm{\int }}}{e}^{{\displaystyle {\mathrm{\int }}_t^y\left(\theta \left(x\right)+\alpha \right)dx}}}}dtdy$$

(7)

The shortage cost during the period $\left[t_1,T\right]$ is computed by changing the order of integration

$$C_s{\displaystyle \underset{t_1}{\overset{T}{\mathrm{\int }}}-I\left(t\right)dt}=C_s{\displaystyle \underset{t_1}{\overset{T}{\mathrm{\int }}}\left(T-x\right)}R\left(x\right)dx$$

(8)

We evaluate the sale revenue per replenishment to derive

$$s\left({\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}D\left(t\right)dt-I\left(T\right)}\right)=s\alpha {\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}R\left(y\right){\displaystyle \underset{0}{\overset{y}{\mathrm{\int }}}{e}^{{\displaystyle {\mathrm{\int }}_t^y\left(\theta \left(x\right)+\alpha \right)dx}}}}dtdy+s{\displaystyle \underset{0}{\overset{T}{\mathrm{\int }}}R\left(x\right)}dx$$

(9)

We obtain the total profit per replenishment cycle to denote it as the sale revenue minus the total cost, where the total cost is consistent with deterioration cost, shortage cost, holding cost, purchasing cost, and ordering cost.

$$Z\left(t_1\right)=\frac{1}{T}\{s\alpha {\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}R\left(y\right){\displaystyle \underset{0}{\overset{y}{\mathrm{\int }}}{e}^{{\displaystyle {\mathrm{\int }}_t^y\left(\theta \left(x\right)+\alpha \right)dx}}}}dtdy+s{\displaystyle \underset{0}{\overset{T}{\mathrm{\int }}}R\left(x\right)}dx-A$$

$$\begin{array}{c}-c[{\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}R\left(y\right)e^{{\displaystyle {\mathrm{\int }}_0^y\left(\theta \left(x\right)+\alpha \right)dx}}dy}+{\displaystyle \underset{t_1}{\overset{T}{\mathrm{\int }}}R\left(x\right)dx}]-C_h{\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}R\left(y\right){\displaystyle \underset{0}{\overset{y}{\mathrm{\int }}}{e}^{{\displaystyle {\mathrm{\int }}_t^y\left(\theta \left(x\right)+\alpha \right)dx}}}}dtdy\\ -C_s{\displaystyle \underset{t_1}{\overset{T}{\mathrm{\int }}}\left(T-x\right)}R\left(x\right)dx-C_d[{\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}R\left(y\right)\left(e^{{\displaystyle {\mathrm{\int }}_0^y\left(\theta \left(x\right)+\alpha \right)dx}}-1\right)dy}\\ -\alpha {\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}R\left(y\right){\displaystyle \underset{0}{\overset{y}{\mathrm{\int }}}{e}^{{\displaystyle {\mathrm{\int }}_t^y\left(\theta \left(x\right)+\alpha \right)dx}}}}dtdy]\}\end{array}$$

(10)

Based on Equation (10), we derive

$$\begin{array}{c}\frac{d}{d{t}_1}Z\left(t_1\right)=\frac{R\left(t_1\right)}{T}\{\left(s\alpha -C_h+\alpha \hspace{0.17em}{C}_d\right){\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}{e}^{{\displaystyle {\mathrm{\int }}_t^{t_1}\left(\theta \left(x\right)+\alpha \right)dx}}dt}\\ -\left(c+C_d\right)\left(e^{{\displaystyle {\mathrm{\int }}_0^{t_1}\left(\theta \left(x\right)+\alpha \right)dx}}-1\right)+C_s\left(T-t_1\right)\}\end{array}$$

(11)

Based on Equation (11), we assume an auxiliary function, say $m\left(t_1\right)$, to simplify the later discussion, as follows

$$m\left(t_1\right)=\left(s\alpha -C_h+\alpha \hspace{0.17em}{C}_d\right){\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}{e}^{{\displaystyle {\mathrm{\int }}_t^{t_1}\left(\theta \left(x\right)+\alpha \right)dx}}dt}-\left(c+C_d\right)\left(e^{{\displaystyle {\mathrm{\int }}_0^{t_1}\left(\theta \left(x\right)+\alpha \right)dx}}-1\right)+C_s\left(T-t_1\right)$$

(12)

such that

$$\frac{d}{d{t}_1}Z\left(t_1\right)=\frac{R\left(t_1\right)}{T}m\left(t_1\right)$$

(13)

for $0\le t_1\le T.$ Hence, $\frac{d}{d{t}_1}Z\left(t_1\right)$ and $m\left(t_1\right)$ have the same roots.

From Equation (12), it yields

$$\begin{array}{c}{m}^{\prime }\left(t_1\right)=\left(s\alpha -C_h+\alpha \hspace{0.17em}{C}_d\right)\left(1+\left(\theta \left(t_1\right)+\alpha \right){\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}{e}^{{\displaystyle {\mathrm{\int }}_t^{t_1}\left(\theta \left(x\right)+\alpha \right)dx}}dt}\right)\\ -\left(c+C_d\right)\left(\theta \left(t_1\right)+\alpha \right)e^{{\displaystyle {\mathrm{\int }}_0^{t_1}\left(\theta \left(x\right)+\alpha \right)dx}}-C_s\end{array}$$

(14)

We rewrite Equation (14) as

$$\begin{array}{c}{m}^{\prime }\left(t_1\right)=\alpha \left(s+C_d\right)-C_h-C_s\\ +\left(\theta \left(t_1\right)+\alpha \right)\left\{\left(s\alpha +\alpha C_d-C_h\right){\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}{e}^{{\displaystyle {\mathrm{\int }}_t^{t_1}\left(\theta \left(x\right)+\alpha \right)dx}}dt}-\left(c+C_d\right)e^{{\displaystyle {\mathrm{\int }}_0^{t_1}\left(\theta \left(x\right)+\alpha \right)dx}}\right\}\end{array}$$

(15)

We also compute

$$m\left(0\right)=C_{s}T>0$$

(16)

to

$$m\left(T\right)=\left(\alpha \left(s+C_d\right)-C_h\right){\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}{e}^{{\displaystyle {\mathrm{\int }}_t^{t_1}\left(\theta \left(x\right)+\alpha \right)dx}}dt}-\left(c+C_d\right)\left(e^{{\displaystyle {\mathrm{\int }}_0^{t_1}\left(\theta \left(x\right)+\alpha \right)dx}}-1\right)<0$$

(17)

From Equations (15) and (17), we notice that without giving an explicit expression of the deterioration function $\theta \left(t\right)$, researchers cannot compute those integrations related to $\frac{d}{dt}m\left(t\right)$. The alternative approach is to consider numerical support to face the monotonic problem of $\frac{d}{dt}m\left(t\right)$ and $m\left(T\right)<0$.

We can say that the most feasible way to prove that $m\left(t_1\right)=0$ has a unique solution for $0<t_1<T$ is to show that $m\left(t_1\right)$ is a strictly decreasing function. Hence, we try to find conditions to guarantee $\frac{d}{d{t}_1}m\left(t_1\right)<0$ and $m\left(T\right)<0$.

We check the coefficients in Equations (15) and (17) to find that

$$\alpha \left(s+C_d\right)-C_h<0$$

(18)

and

$$\alpha \left(s+C_d\right)-C_h-C_s<0$$

(19)

Owing to the inequality in Equation (18), the inequality in Equation (19) is implied, so we will try to verify that the inequality in Equation (18) is valid.

Consequently, we can derive our desired result of the existence and uniqueness of only one root for $m\left(t_1\right)$, which is the optimal solution for $Z\left(t_1\right)$, which is the unique solution for $\frac{d}{d{t}_1}Z\left(t_1\right)=0$.

Since our proposed model is new, in the literature, there are no inventory models that have considered those parameters altogether. Hence, we have to break our condition into two parts:

$$\alpha s<0.5C_h$$

(20)

and

$$\alpha C_d<0.5C_h$$

(21)

The reason we separate Equation (18) into two parts is explained below.

We recall that in the numerical examples of Wu et al. [10] and Chuang [11], $\alpha =0.01$, $s=20$ $C_h=3$, such that

$$\alpha s=0.2<0.5C_h=1.5$$

(22)

Therefore, our assertion of Equation (21) is supported by the numerical examples of Wu et al. [10] and Chuang [11].

On the other hand, Wu et al. [10] did not include the deterioration cost per unit, $C_d$, in their inventory models. Chuang [11] has already developed an inventory model with the deterioration cost. However, Chuang [11] did not inform us of his deterioration cost. In comparison with Wu et al. [10], in the numerical examples of Chuang [11], he still assumed that $C_d$ = 0. Hence, we must refer to another paper, Lin [30], and his numerical example, $C_d=3$ and $C_h=10$, to imply that

$$\alpha \hspace{0.17em}{C}_d=0.03<0.5C_h=5$$

(23)

Thus, our claim of Equation (22) is confirmed by the numerical example of Lin [30].

Hence, our extra condition of Equation (18) is supported by the numerical examples of Wu et al. [10], Lin [30], and Chuang [11].

Since exponential functions are positive, we imply that those two integrations of exponential functions are also positive. The two coefficients before integrations are negative. We combine our observations to derive

$$m^{\prime }\left(t_1\right)<0$$

(24)

We show that $m\left(t_1\right)$ is a decreasing function from $m\left(0\right)>0$ to $m\left(T\right)<0$.

Hence, there is a unique point, say $t_1^{*}$, that is satisfying $m\left(t_1^{*}\right)=0$. We recall that $m\left(t_1\right)$ and $Z^{\prime }\left(t_1\right)$ have the same sign, such that

$$Z^{\prime }\left(t_1\right)>0, \mathrm{for} 0<t_1<t_1^{*}$$

(25)

and

$$Z^{\prime }\left(t_1\right)<0, \mathrm{for} t_1^{*}<t_1<T$$

(26)

We prove that $t_1^{*}$ is the optimal solution for our maximum-profit inventory model. Equation (12) reveals that the demand rate, $R\left(t_1\right)$, is not included in the expression of $m\left(t_1\right)$, so the optimal solution is independent of the demand rate. This finding is consistent with the results of Lin [30] and Hung [31], which are minimum-cost inventory models, that is, the optimal solution is independent of the demand pattern.

4. Numerical Examples

We recall numerical examples in Chuang [11] with the following data: the setup (the ordering) cost $A=50$, the holding cost $C_{\mathrm{h}}=3$, the shortage cost $C_{\mathrm{s}}=5$, the purchasing cost $c=15$, the actual sale rate $D_0=400$, the selling price $s=20$, the planning horizon $T=1$, the ratio of extra selling related to the inventory level $\alpha =0.01$, and the deterioration rate $\theta =0.05$. In Chuang [11], his inventory model did not include the deterioration cost. Hence, we refer to Lin [30] to adopt $C_{\mathrm{d}}=3$. To check the influence of the variation for parameters, we examine a detailed sensitivity analysis to alter the values from the decreasing percentages of 30%, 20%, and 10%, to increasing percentages of 10%, 20%, and 30% for the setup cost A, the holding cost $C_h$, the shortage cost $C_s$, the purchasing cost c, the selling price s, the planning horizon T, the ratio of extra selling related to the inventory level $\alpha$, and the deterioration function $\theta$. We list the results for variation in the setup cost in

Table 2. Variation in the setup cost A.

A	−30%	−20%	−10%	base	+10%	+20%	+30%
$t_1^{*}$	0.5608	0.5608	0.5608	0.5608	0.5608	0.5608	0.5608

.

Our findings in Table 2 show that the optimal solution of $t_1^{*}$ is independent of the setup cost. Our findings for variation in the holding cost are listed in

Table 3. Variation in the holding cost $C_h$.

$C_h$	−30%	−20%	−10%	base	+10%	+20%	+30%
$t_1^{*}$	0.6245	0.6017	0.5805	0.5608	0.5424	0.5252	0.5090

.

Our results in Table 3 show that the optimal solution of $t_1^{*}$ has a negative relationship with the holding cost. Our results for variation in the shortage cost are presented in

Table 4. Variation in the shortage cost $C_s$.

$C_s$	−30%	−20%	−10%	base	+10%	+20%	+30%
$t_1^{*}$	0.4726	0.5058	0.5349	0.5608	0.5840	0.6048	0.6236

.

Our findings in Table 4 show the optimal solution $t_1^{*}$ has a positive relationship with the shortage cost. Our findings for variation in the purchasing cost are shown in

Table 5. Variation in the purchasing cost c.

$c$	−30%	−20%	−10%	base	+10%	+20%	+30%
$t_1^{*}$	0.5785	0.5725	0.5666	0.5608	0.5552	0.5497	0.5442

.

Our results in Table 5 they assert that the optimal solution $t_1^{*}$ has a negative relationship with the purchasing cost. Our results for variation in the selling price are listed in

Table 6. Variation in the selling price s.

$s$	−30%	−20%	−10%	base	+10%	+20%	+30%
$t_1^{*}$	0.5570	0.5583	0.5596	0.5608	0.5621	0.5634	0.5647

.

Our results in Table 6 denote that the optimal solution $t_1^{*}$ has a positive relationship with the selling price. Our findings for variation in the planning horizon are expressed in

Table 7. Variation in the planning horizon T.

T	−30%	−20%	−10%	base	+10%	+20%	+30%
$t_1^{*}$	0.3934	0.4493	0.5051	0.5608	0.6164	0.6720	0.7275

.

Our findings in Table 7 imply that the optimal solution $t_1^{*}$ has a positive relationship with the planning horizon. Our results for variation in the ratio of extra selling are denoted in

Table 8. Variation in the ratio of extra selling $\alpha$.

$\theta$	−30%	−20%	−10%	base	+10%	+20%	+30%
$t_1^{*}$	0.5601	0.5603	0.5606	0.5608	0.5611	0.5613	0.5616

.

Our results in Table 8 claim that the optimal solution $t_1^{*}$ has a positive relationship with the ratio of extra selling. Our findings for the results for variation in the deterioration function are shown in

Table 9. Variation in the deterioration function $\theta$.

$\theta$	−30%	−20%	−10%	base	+10%	+20%	+30%
$t_1^{*}$	0.5796	0.5732	0.5669	0.5608	0.5548	0.5489	0.5432

.

Our findings in Table 9 point out that the optimal solution $t_1^{*}$ has a negative relationship with the deterioration function. Our results for variation in the deterioration cost are expressed in

Table 10. Variation in the deterioration cost $C_{\mathrm{d}}$.

$C_{\mathrm{d}}$	−30.	−20%	−10%	base	+10%	+20%	+30%
$t_1^{*}$	0.5637	0.5627	0.5618	0.5608	0.5599	0.5589	0.5580

.

Our results in Table 10 claim that the optimal solution $t_1^{*}$ has a negative relation with the deterioration cost.

5. Application of Our Approach

When $\theta \left(t\right)$ is degenerated to a constant function, denoted as $\theta$, we can rewrite Equation (12) as

$$m\left(t_1\right)=\left[\frac{\alpha \left(s-c\right)-C_h-\theta \left(c+C_d\right)}{\theta +\alpha }\right]\left(e^{\left(\theta +\alpha \right)\hspace{0.17em}{t}_1}-1\right)+C_s\left(T-t_1\right)$$

(27)

which is the same result as the criterion of Chuang [11] for the optimal solution, revealing that Chuang [11] is a special case of our model.

To be comparable with Wu et al. [10], we consider the same numerical examples as their Example 1,$A=50$, $C_{\mathrm{h}}=3$, $C_{\mathrm{s}}=5$, $D_0=400$, $T=1$, $c=15$, $s=20$, $\alpha =0.01$, and $\theta =0.05$.

Wu et al. [10] did not consider the deterioration cost. In comparing our work with that of Wu et al. [10], we suppose $C_{\mathrm{d}}=0$. in the following discussion.

Wu et al. [10] studied a ramp-type demand as

$$D\left(t\right)=\{\begin{array}{c}{D}_{0}t, 0\le t\le \mu ,\\ D_0\mu , \mu \le t\le T.\end{array}$$

(28)

Depending on the relationship between $\mu$ and $t_1$, Wu et al. [10] partitioned their model into two cases: Case A: $t_1\ge \mu$, and Case B: $\mu \le t_1$. Consequently, they constructed their profit function ${\mathrm{Z}}_1\left(t_1\right)$ and ${\mathrm{Z}}_2\left(t_1\right)$ for Cases A, and B, respectively.

For Case A, Wu et al. [10] obtained

$$\frac{\alpha \left(s-c\right)-\left(C_{\mathrm{h}}+\theta c\right)}{\theta +\alpha }\left[{\mathrm{e}}^{\left(\theta +\alpha \right)t_1}-1\right]+C_{\mathrm{s}}\left(T-t_1\right)=0$$

(29)

to decide their optimal solution for $\mu \le t_1\le T$.

For Case B, Wu et al. [10] derived the same result as Equation (29) to decide their optimal solution for $0\le t_1\le \mu$.

Wu et al. [10] did not explain why their results for Cases A and B are identical under two different domains. They examined three examples for their inventory model with (i) Example 1, $\mu =0.4$, (ii) Example 2, $\mu =0.5953$, and (iii) Example 3, $\mu =0.6$.

For Example 1, Wu et al. [10] claimed that the local maximum solution for ${\mathrm{Z}}_1\left(t_1\right)$ is $\mu$ and the local maximum solution for ${\mathrm{Z}}_2\left(t_1\right)$ is $t_1^{*}=0.5953$, with

$${\mathrm{Z}}_2\left(t_1^{*}=0.5953\right)>{\mathrm{Z}}_2\left(\mu =0.4\right)={\mathrm{Z}}_1\left(\mu =0.4\right)$$

(30)

For Example 2, Wu et al. [10] found that the local maximum solution for ${\mathrm{Z}}_1\left(t_1\right)$ is $\mu$ and the local maximum solution for ${\mathrm{Z}}_2\left(t_1\right)$ is $\mu$, with

$${\mathrm{Z}}_2\left(\mu =0.5953\right)={\mathrm{Z}}_1\left(\mu =0.5953\right)$$

(31)

For Example 3, Wu et al. [10] found that the local maximum solution for ${\mathrm{Z}}_1\left(t_1\right)$ is $t_1^{*}=0.5953$ and the local maximum solution for ${\mathrm{Z}}_2\left(t_1\right)$ is $\mu$, with

$${\mathrm{Z}}_2\left(\mu =0.6\right)={\mathrm{Z}}_1\left(\mu =0.6\right)<{\mathrm{Z}}_1\left(t_1^{*}=0.5953\right)$$

(32)

Wu et al. [10] were not aware that their three examples had the same global maximum point, $t_1^{*}=0.5953$. Hence, Wu et al. [10] did not know that the maximum solution for these kinds of inventory models is independent of the demand type such that there are two divisions: the first one is to construct ${\mathrm{Z}}_1\left(t_1\right)$ and ${\mathrm{Z}}_2\left(t_1\right)$ for Cases A and B, respectively, concerning the relation between $t_1$ and $\mu$, which is unnecessary. The second one is to develop three examples, based on the relation $t_1^{*}$ and $\mu$, which is redundant. Hence, applying our approach, researchers can solve this kind of inventory model without those lengthy derivations proposed by Wu et al. [10].

At last, but not least, Wu et al. [10] and Chuang [11] mentioned that

$$t_1^{*}=0.5953$$

(33)

In the following, we list four related values to clearly indicate the values of $\mathrm{m}\left(t_1\right)$ decreasing from positive to negative. Hence, our solution is the right result.

On the other hand, we derive that $\mathrm{m}\left(0.570511\right)=9.922\times {10}^{-6}$, $\mathrm{m}\left(0.570512\right)=1.093\times {10}^{-6}$, $\mathrm{m}\left(0.570513\right)=-7.736\times {10}^{-6}$, and $\mathrm{m}\left(0.5953\right)=-0.219$ to indicate that the maximum solution should be revised as

$$t_1^{*}=0.570512$$

(34)

to show that the work of Wu et al. [10] and Chuang [11] needs revisions.

We can also apply our analytical approach to consider the deterioration rate function as a linear function such that $\mathsf{\theta }\left(t\right)=\mathrm{a}+\mathrm{b}t$, and then rewrite Equation (12) as follows:

$$\begin{array}{c}m\left(t_1\right)=\left(s\alpha -C_h+\alpha \hspace{0.17em}{C}_d\right){\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}{e}^{{\displaystyle {\mathrm{\int }}_t^{t_1}\left(a+\alpha +b\hspace{0.17em}x\right)dx}}dt}-\left(c+C_d\right)\left(e^{{\displaystyle {\mathrm{\int }}_0^{t_1}\left(a+\alpha +b\hspace{0.17em}x\right)dx}}-1\right)+C_s\left(T-t_1\right)\\ =\left(s\alpha -C_h+\alpha \hspace{0.17em}{C}_d\right)e^{\left(a+\alpha \right)t_1+\left(b\hspace{0.17em}{t}_1^2/2\right)}{\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}{e}^{-\left(a+\alpha \right)t-\left(b\hspace{0.17em}{t}^2/2\right)}dt}\\ -\left(c+C_d\right)\left(e^{\left(a+\alpha \right)t_1+\left(b\hspace{0.17em}{t}_1^2/2\right)}-1\right)+C_s\left(T-t_1\right)\end{array}$$

(35)

Under the condition of $\mathsf{\theta }\left(t\right)=\mathrm{a}+\mathrm{b}t$, solving $m\left(t_1\right)=0$ will be a challenging problem because, in the integration of Equation (35), the following quadratic form denoted as E, with

$$\mathrm{E}={\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}{e}^{-\left(a+\alpha \right)t-\left(b\hspace{0.17em}{t}^2/2\right)}dt}$$

(36)

appeared.

For a linear term in the exponential function, researchers can derive that

$${\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}{e}^{-\left(a+\alpha \right)t}dt}=\frac{1}{a+\alpha }\left[1-e^{-\left(a+\alpha \right)\hspace{0.17em}{t}_1}\right]$$

(37)

to indicate that when the deterioration function is a constant term as $\mathsf{\theta }\left(t\right)=\mathrm{a}$, then the integration can be directly solved as Equation (37).

When we extend our inventory model for the deterioration from a constant term to a linear function as $\mathsf{\theta }\left(t\right)=\mathrm{a}+\mathrm{b}t$, the integration problem of Equation (36) cannot be directly solved to an exact result.

Therefore, for a quadratic form in the exponential function of Equation (36), we change the expression of Equation (36) to

$${\displaystyle \underset{0}{\overset{t_1}{\mathrm{\int }}}{e}^{-\left(a+\alpha \right)t-\left(b\hspace{0.17em}{t}^2/2\right)}dt}=\sqrt{\frac{2}{\mathrm{b}}}{\mathrm{e}}^{{\left(\mathrm{a}+\mathsf{\alpha }\right)}^2/2\mathrm{b}}{{\displaystyle \mathrm{\int }}}_{\left(\mathrm{a}+\mathsf{\alpha }\right)/\sqrt{2\mathrm{b}}}^{\left(\sqrt{\mathrm{b}/2}\right)\left(t_1+\left(\left(\mathrm{a}+\mathsf{\alpha }\right)/\mathrm{b}\right)\right)}{\mathrm{e}}^{-{\mathrm{y}}^2}\mathrm{dy}$$

(38)

with

$$\mathrm{y}=\sqrt{\frac{\mathrm{b}}{2}}\left(t+\frac{\mathrm{a}+\mathsf{\alpha }}{\mathrm{b}}\right)$$

(39)

and

$$\frac{-\mathrm{b}}{2}{t}^2-\left(\mathrm{a}+\mathsf{\alpha }\right)t=-{\left(\sqrt{\frac{\mathrm{b}}{2}}\left(t+\frac{\mathrm{a}+\mathsf{\alpha }}{\mathrm{b}}\right)\right)}^2+\frac{{\left(\mathrm{a}+\mathsf{\alpha }\right)}^2}{2\mathrm{b}}$$

(40)

For the improper integration of ${{\displaystyle \mathrm{\int }}}_0^{\infty }{\mathrm{e}}^{-{\mathrm{y}}^2\mathrm{dy}}$, researchers know that

$${{\displaystyle \mathrm{\int }}}_0^{\infty }{\mathrm{e}}^{-{\mathrm{y}}^2\mathrm{dy}}=\frac{\sqrt{\mathsf{\pi }}}{2}$$

(41)

However, for another definite integration, for example, that denoted as F, with

$$\mathrm{F}={{\displaystyle \mathrm{\int }}}_3^5{\mathrm{e}}^{-{\mathrm{y}}^2\mathrm{dy}},$$

(42)

researchers cannot find the exact solution for Equation (42). To the best of our knowledge, many practitioners have applied Taylor’s series expansion of the exponential function to derive an infinite series for the definite integration of Equation (42)

$$\begin{array}{c}{{\displaystyle \mathrm{\int }}}_3^5{\mathrm{e}}^{-{\mathrm{y}}^2\mathrm{dy}}={{\displaystyle \mathrm{\int }}}_3^5{\displaystyle {\displaystyle \sum }_{\mathrm{k}=0}^{\infty }}\frac{{\left(-{\mathrm{y}}^2\right)}^{\mathrm{k}}}{\mathrm{k}!}\mathrm{dy}\\ ={\displaystyle {\displaystyle \sum }_{\mathrm{k}=0}^{\infty }}\frac{{\left(-1\right)}^{\mathrm{k}}}{\mathrm{k}!}\left(\frac{5^{2\mathrm{k}+1}-3^{2\mathrm{k}+1}}{2\mathrm{k}+1}\right)\end{array}$$

(43)

Depending on the accuracy, practitioners selected a number, denoted as “m”, to accept that

$${{\displaystyle \mathrm{\int }}}_3^5{\mathrm{e}}^{-{\mathrm{y}}^2\mathrm{dy}}\approx {\displaystyle \sum }_{\mathrm{k}=0}^{\mathrm{m}}\frac{{\left(-1\right)}^{\mathrm{k}}}{\mathrm{k}!}\left(\frac{5^{2\mathrm{k}+1}-3^{2\mathrm{k}+1}}{2\mathrm{k}+1}\right)$$

(44)

where the estimation error is less than $\frac{5^{2\mathrm{m}+3}-3^{2\mathrm{m}+3}}{\left(\mathrm{m}+1\right)! \left(2\mathrm{m}+3\right)}$, because the infinite series on the right-hand side of Equation (44) is an alternative series.

The second way to solve the value of Equation (42) is to consider the well-known fact that finite integration is the limit of the Riemann sum. It shows that

$${{\displaystyle \mathrm{\int }}}_3^5{\mathrm{e}}^{-{\mathrm{y}}^2\mathrm{dy}}=\underset{\mathrm{n}\to \infty }{\lim }{\displaystyle \sum }_{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{e}}^{-{\left(3+\frac{2\mathrm{k}}{\mathrm{n}}\right)}^2}\frac{2}{\mathrm{n}}.$$

(45)

Based on the above discussion, in the beginning, we will adopt the second approach to solve $\mathrm{m}\left(t_1\right)=0$, when $\mathsf{\theta }\left(t\right)=\mathrm{a}+\mathrm{b}t$.

We construct an example for a linear deterioration function with $\mathrm{a}=0.05$ and $\mathrm{b}=0.1$ to extend the deterioration function to a linear form such that $A=50$, $C_{\mathrm{d}}=3$, $C_{\mathrm{h}}=3$, $C_{\mathrm{s}}=5$, $D_0=400$, $T=1$, $c=15$, $s=20$, $\alpha =0.01$, $\theta =0.05$, and $\mathsf{\theta }\left(t\right)=0.05+0.1t$.

For a given value of $t_1$, we uniformly partition the integration interval $\left[0,t_1\right]$ into twenty subintervals, and then we estimate $m\left(t_1\right)$ using the following approximation:

$$\begin{array}{c}\mathrm{m}\left(t_1\right)\approx =\left(s\alpha -C_h+\alpha \hspace{0.17em}{C}_d\right)e^{\left(a+\alpha \right)t_1+\left(b\hspace{0.17em}{t}_1^2/2\right)}\frac{t_1}{20}{\displaystyle \sum }_{\mathrm{k}=1}^{20}{\mathrm{e}}^{-\left(\mathrm{a}+\mathsf{\alpha }\right)\frac{\mathrm{k}}{20}{t}_1-\frac{\mathrm{b}}{2}{\left(\frac{\mathrm{k}}{20}{t}_1\right)}^2}\\ -\left(c+C_d\right)\left(e^{\left(a+\alpha \right)t_1+\left(b\hspace{0.17em}{t}_1^2/2\right)}-1\right)+C_s\left(T-t_1\right)\end{array}$$

(46)

We begin to search for the optimal solution, $t_1^{*}$, which is the solution of $\mathrm{m}\left(t_1\right)=0$, with the knowledge that $\mathrm{m}\left(0\right)>0$, $\mathrm{m}\left(T=1\right)<0$, and $\mathrm{m}\left(t_1\right)$ is a decreasing function.

Many methods can locate the value of the optimal solution, $t_1^{*}$. In the following, we will use a direct numerical method to list values of $\mathrm{m}\left(t_1\right)$ that will clearly reveal that $\mathrm{m}\left(t_1\right)$ is a decreasing function from a positive value to a negative value, and then we can find the solution of $\mathrm{m}\left(t_1\right)=0$.

For $t_1=0,0.1,\dots ,1$, we list the estimation result of $\mathrm{m}\left(t_1\right)$ in the following

Table 11. The estimation result for $\mathrm{m}\left(t_1\right)$ with $t_1=0,0.1,\dots ,1$.

$t_1$	0	0.1	0.2	0.3	0.4	0.5
$\mathrm{m}\left(t_1\right)$	5.000	4.105	3.188	2.250	1.288	0.302
$t_1$	0.6	0.7	0.8	0.9	1
$\mathrm{m}\left(t_1\right)$	−0.709	−1.747	−2.814	−3.910	−5.038

.

Based on Table 11, we know that the optimal solution, $t_1^{*}$, satisfies

$$0.5<t_1^{*}<0.6$$

(47)

Hence, for $t_1=0.50, 0.51,\dots , 0.6$, we list the estimation result of $\mathrm{m}\left(t_1\right)$ in

Table 12. The estimation result for $\mathrm{m}\left(t_1\right)$, with $t_1=0.50, 0.51,\dots , 0.6$.

$t_1$	0.50	0.51	0.52	0.53	0.54	0.55
$\mathrm{m}\left(t_1\right)$	0.302	0.202	0.102	$1.518\times {10}^{-3}$	−0.099	−0.200
$t_1$	0.56	0.57	0.58	0.59	0.60
$\mathrm{m}\left(t_1\right)$	−0.302	−0.403	−0.505	−0.607	−0.709

.

Based on Table 12, we know that the optimal solution, $t_1^{*}$, satisfies

$$0.53<t_1^{*}<0.54$$

(48)

Hence, for $t_1=0.530, 0.531,\dots , 0.540$, we list the estimation result of $\mathrm{m}\left(t_1\right)$ in

Table 13. The estimation result for $\mathrm{m}\left(t_1\right)$, with $t_1=0.530, 0.531,\dots , 0.540$.

$t_1$	0.530	0.531	0.532	0.533	0.534	0.535
$\mathrm{m}\left(t_1\right)$	$1.518\times {10}^{-3}$	$-8.54\times {10}^{-3}$	−0.019	−0.029	−0.039	−0.049
$t_1$	0.536	0.537	0.538	0.539	0.540
$\mathrm{m}\left(t_1\right)$	−0.059	−0.069	−0.079	−0.089	−0.099

.

Based on Table 13, we know that the optimal solution, $t_1^{*}$, satisfies

$$0.530<t_1^{*}<0.531$$

(49)

Hence, for $t_1=0.5300, 0.5301,\dots , 0.5310$, we list the estimation result of $\mathrm{m}\left(t_1\right)$ in

Table 14. The estimation result for $\mathrm{m}\left(t_1\right)$, with $t_1=0.5300, 0.5301,\dots , 0.5310$.

$t_1$	0.5300	0.5301	0.5302	0.5303	0.5304	0.5305
$\mathrm{m}\left(t_1\right)$	$1.518\times {10}^{-3}$	$5.11\times {10}^{-4}$	$-4.95\times {10}^{-4}$	$-1.50\times {10}^{-3}$	$-2.50\times {10}^{-3}$	$-3.51\times {10}^{-3}$
$t_1$	0.5306	0.5307	0.5308	0.5309	0.5310
$\mathrm{m}\left(t_1\right)$	$-4.52\times {10}^{-3}$	$-5.53\times {10}^{-3}$	$-6.53\times {10}^{-3}$	$-7.54\times {10}^{-3}$	$-8.54\times {10}^{-3}$

.

Based on Table 14, we know that the optimal solution, $t_1^{*}$, satisfies

$$0.5301<t_1^{*}<0.5302$$

(50)

We compare

$$\mathrm{m}\left(0.5301\right)=5.113\times {10}^{-4}$$

(51)

and

$$\mathrm{m}\left(0.5302\right)=-4.949\times {10}^{-4},$$

(52)

to select

$$t_1^{*}=0.5302$$

(53)

6. Conclusions

In this paper, we constructed a maximum-profit inventory model to show that the optimal solution is independent of the demand such that the deterioration rate is generalized to the general expression. We show that without writing the implicit expression of demand function and deterioration function, through our approach, complicated solution procedures with several different objective functions can be avoided. Our paper is not only an extension of the work of Wu et al. [10] and Chuang [11] but also provides a complete solution structure to find the optimal solution.