*Literature Review*

There is a large volume of published studies concerning the inventory models with cash payments, credit payments, or advance payments, such as those that we have reviewed or cited below.

In these literatures, Taleizadeh et al. [2] established an economic order quantity (EOQ) model with partial backordering in which the supplier asks the retailer to pay a fraction of the purchasing cost in advance and allows them to divide the prepaymen<sup>t</sup> into multiple equal installments during a lead time. Taleizadeh [3] extended the inventory model of Taleizadeh et al. [2] to the cases of deteriorating items with and without shortages. Taleizadeh [4] used an advance-cash-payment plan to develop an EOQ model for an evaporating item with partial backordering for a real case study of a gasoline station. Zhang et al. [5] developed an inventory model under advance payment, which includes all payments in advance and partially advanced-partially delayed paymen<sup>t</sup> plans. Eck et al. [6] explored the role of cash-in-advance financing for export decisions in firms. Lashgari et al. [7] considered an EOQ model with hybrid partial payment, such as upstream partial prepaymen<sup>t</sup> and downstream partial delayed paymen<sup>t</sup> without shortage, with full backordering, or with partial backordering. Tavakoli and Taleizadeh [8] gave a lot-sizing model for decaying item for the retailer to pay all the purchasing cost in advance with no shortage or with full backordering shortage or partial lost sale. Heydari et al. [9] assumed the demand is stochastic and credit-dependent under the two-level trade credit, then they found the optimal ordering quantity and the length of the credit period. Feng and Chan [10] expanded the two-level trade credit to include joint pricing and production decisions for new products with pronounced learning-by-doing phenomenon. Much of the current research attention has been directed towards trade-credit inventory models for deteriorating items with its own expiration date. For example, Chen and Teng [11] established an inventory model for deteriorating items under two-level trade credit by discounted cash flow analysis in which the deterioration rate is non-decreasing over time and near 100 percent particularly close to its expiration date. Then, they demonstrated that the retailer's optimal credit period and cycle time not only exist, but are also unique. Mahata [12] discussed an EOQ model for deteriorating items having fixed lifetime under two-level trade credit. He showed that the retailer's optimal replenishment cycle time not only exists but is also

unique. Wu et al. [13] examined an inventory model with expiration date dependent deterioration under an advance-cash-credit paymen<sup>t</sup> scheme to find the optimal replenishment cycle time and the fraction of no shortages such that the total profit is maximized. Moreover, some related recent articles are those by, for example, Zia and Taleizadeh [14], Wu et al. [15], Chen et al. [16], Teng et al. [17], Diabat et al. [18], Feng et al. [19], Mahata and De [20], Tiwari et al. [21], Taleizadeh et al. [22], Li et al. [23,24], Taleizadeh [25], Krommyda et al. [26], Tsao et al. [27], Mashud et al. ([28,29]), AlArjani et al. [30], and Hou et al. [31].

There is a large volume of published studies describing the inventory models for non-instantaneous deteriorating items. Udayakumar and Geetha [32] considered time, value of money, and the effect of inflation to develop an economic-ordering policy for non-instantaneous deteriorating items over a finite time horizon in which the demand is a deterministic function of selling price and advertisement cost. They found the optimal length of replenishment and the optimal order quantity. Lashgari et al. [33] presented an EOQ model for non-instantaneous deteriorating items under an advance-delay paymen<sup>t</sup> when shortages are allowed in a partial form. They found the optimal order and shortage quantities to minimize the retailer's total inventory cost function. Udayakumar and Geetha [34] developed an EOQ model for non-instantaneous deteriorating items with capacity constraint under a trade credit policy. They found the optimal replenishment cycle time and order quantity to minimize the total inventory cost. Babangida and Baraya [35] showed an inventory model for non-instantaneous deteriorating items with two-phase demand rates, capacity constraint and complete backlogged under trade credit policy. They provided the necessary and sufficient conditions for the existence and uniqueness of solutions. Soni and Suthar [36] revealed an inventory model for non-instantaneous deteriorating items with partial backlogging; they considered that the demand rate has a negative and positive exponential effect of price and promotional effort, respectively, while the item is not in a state of deterioration and then found the joint optimal pricing and replenishment policy for the non-instantaneous deteriorating items. Cenk Çalı¸skan [37] deals with the inventory model for deteriorating items in which the opportunity cost is based on compound interest, and backorders are allowed. The article determines a near-optimal and intuitive closed-form solution, which is simple to the practitioners. Under a variety of practical conditions, some researchers have considered the above items, such as Tiwari et al. [38], Tsao [39], Geetha and Udayakumar [40], Jaggi et al. [41], Maihami et al. [42], Mashud et al. [43], Bounkhel et al. [44], and Udayakumar et al. [45].

Given that it is worthwhile studying the effect of defective items on inventory problems, numerous researchers, such as Khanna et al. [46], have developed inventory models for deteriorating imperfect quality items with allowable shortages and permissible delays in payments. Zhou et al. [47] found a synergy economic order quantity model, in which the concepts of imperfect quality, inspection error, and shortages with trade credit are considered. They found the annual profit function is concave and obtained the closed form optimal solution to the model. Datta [48] proposed a production-inventory model with defective items. The model incorporates additional investment opportunity on quality improvement for reducing the proportion of defective products. Taleizadeh et al. [49] developed an imperfect EPQ model with upstream trade credit periods linked to raw material order quantity and downstream trade credit periods. Pal and Mahapatra [50] developed an inventory model with imperfect products for a three-level supply chain, and three different ways of dealing with defective products were investigated in their model. Khakzad and Gholamian [51] investigated the effect of inspection time on the deterioration rate; they showed the convexity of the model and illustrated the uniqueness of the solution. Taleizadeh et al. [52] revealed an EOQ inventory model with imperfect items and partial backordering. They assumed a percent of the products in a lot is imperfect and the imperfect items are replenished by perfect ones at a higher cost. The objective is to obtain the optimal value of the length period and the percent of period duration in which the inventory level is positive. Some imperfect production models with trade credit have been studied, in recent years, by (among others) Wang et al. [53], Alamri

et al. [54], Palanivel and Uthayakumar [55], Aghili and Hoseinabadi [56], Tsao et al. [57,58], Khanna et al. [59], Liao et al. [60], Kazemi et al. [61], Liao et al. [62], Mashud et al. [63], and Srivastava et al. [64].

We remark, in passing to the next section, that the mathematical analytic solution procedures, which we have used in the mathematical analysis and discussions of the inventory and supply chain models considered in this article, are full of elaborate usages of the intricate techniques of calculus in determining the continuity, convexity, monotonicity (increasing or decreasing), and differentiability properties of the object functions (that is, the total optimal cost functions). We have stated our main results of this investigation in the form of five theorems (Theorems 1 to 5), which we have proved by appealing also to two Lemmas (Lemma 1 and Lemma 2). For the sake of brevity and compact presentation, the proof of Lemma 2 has been given in the Appendix A instead of the main text. It is quite natural to expect such a format and style in a mathematically oriented article. Furthermore, as we have already mentioned, for the accuracy, completeness, and safe applicability of the results and discussions presented in this article, the usage of the mathematical analytic solution procedures, which are based upon the elaborate and intricate techniques of calculus, is essential here.

### **2. Mathematical Formulation of the Supply Chain Model and Its Analysis**

Based on the above assumptions, the inventory level drops at the demand rate and the defective rate during the time interval [0, *td*]. Then, the inventory level drops to zero due to the demand and the deterioration with the expiration dates during the time interval [*td*, *<sup>T</sup>*]. Furthermore, the variations in the inventory level with respect to time *t* can be expressed below.

The differential equation representing the inventory status during the time interval *t* ∈ [0, *ts*], *ts* = *y x*, is given by

$$\frac{dI\_1(t)}{dt} = -D \tag{1}$$

where *t* is restricted, as in Equation (2). Under the condition *<sup>I</sup>*1(0) = *y*, by solving Equation (1), we obtain

$$I\_1(t) = y - Dt,\\ 0 < t \le t\_s \tag{2}$$

In the second interval [*ts*, *td*], the differential equation represents the inventory status:

$$\frac{dI\_2(t)}{dt} = -D\_\prime t\_\delta \le t \le t\_d \tag{3}$$

Under the condition *<sup>I</sup>*2(*td*)=(<sup>1</sup> − *p*)*y* − *Dtd*, Equation (3) yields

$$I\_2(t) = (1 - p)y - Dt, \ t\_\kappa \le t \le t\_d \tag{4}$$

During the third interval [*td*, *<sup>T</sup>*], the change in the inventory level is represented by the following differential equation:

$$\frac{dI\_3(t)}{dt} + \theta(t) \cdot I\_3(t) = -D\_\prime t\_d \le t \le T \tag{5}$$

Under the condition *<sup>I</sup>*3(*T*) = 0, the solution of Equation (5) is given by

$$I\_3(t) = D(1+m-t) \cdot \ln(\frac{1+m-t}{1+m-T}), t\_d \le t \le T \tag{6}$$

Making use of the continuity property of *<sup>I</sup>*2(*td*) = *<sup>I</sup>*3(*td*), it follows from Equations (4) and (6) that

$$(1 - p)y - Dt\_d = D(1 + m - t\_d) \ln(\frac{1 + m - t\_d}{1 + m - T}) \tag{7}$$

which implies that the order quantity is given by

$$y = \frac{D}{(1-p)}[t\_d + (1+m-t\_d) \cdot \ln(\frac{1+m-t\_d}{1+m-T})] \tag{8}$$

Substituting Equation (8) into Equations (2) and (4), we ge<sup>t</sup>

$$I\_1(t) = \frac{D}{(1-p)}[t\_d + (1+m-t\_d) \cdot \ln(\frac{1+m-t\_d}{1+m-T}) - (1-p)t] \tag{9}$$

and

$$I\_2(t) = D[t\_d + (1 + m - t\_d) \cdot \ln(\frac{1 + m - t\_d}{1 + m - T}) - t] \tag{10}$$

Additionally, this article focuses on *ts* ≤ *td*, so we have *T* ≤ *R*∗ if and only if *ts* ≤ *td*. Here,

$$R^\* = (1+m) - (1+m-t\_d) \cdot e^{-\frac{[(1-p)x-D]t\_d}{D(1+m-t\_d)}}\tag{11}$$

We now calculate the annual total relevant cost which results from the following components:

1. Order cost = *oT*

2. The holding cost (excluding interest charges) after receiving *y* units at time 0 is given below:

Case 1. When 0 < *T* < *td*, Theholding cost

$$\begin{cases} \frac{\eta}{T} \left[ \int\_{0}^{\frac{\eta}{2}} (y - Dt)dt + \int\_{\frac{\eta}{2}}^{T} D(T - t)dt \right] = \frac{hD}{(1 - p)T} \left[ \frac{(1 - p)T}{2} + \frac{pDT}{x(1 - p)} \right] \\ \text{Case 2. When } t\_{d} \le T \\ \text{The holding cost} = \frac{h}{T} \left\{ \int\_{0}^{\frac{\eta}{2}} I\_{1}(t)dt + \int\_{\frac{\eta}{2}}^{t\_{d}} I\_{2}(t)dt + \int\_{t\_{d}}^{T} I\_{3}(t)dt \right\} \\ \\ = \frac{h}{T} \left\{ \frac{D}{2} t\_{d}^{2} + D(1 + m - t\_{d}) \cdot t\_{d} \ln(\frac{1 + m - t\_{d}}{1 + m - T}) + \frac{pD^{2}}{x(1 - p)^{2}} \\ \times \left[ t\_{d} + (1 + m - t\_{d}) \cdot \ln(\frac{1 + m - t\_{d}}{1 + m - T}) \right]^{2} + \frac{D}{2} (1 + m - t\_{d})^{2} \cdot \ln(\frac{1 + m - t\_{d}}{1 + m - T}) \\ + \frac{D}{4} (1 + m - T)^{2} - \frac{D}{4} (1 + m - t\_{d})^{2} \right\} \end{cases}$$

3. The procurement cost per replenishment cycle is: Case 1. When 0 < *T* < *td* The procurement cost = *cyT* = *cD* (<sup>1</sup>−*p*) Case 2. When *td* ≤ *T* The procurement cost = *cyT* = *cD* (<sup>1</sup>−*p*)*<sup>T</sup>* [*td* + (<sup>1</sup>+*m* − *td*) · ln( 1+*m*−*td* 1+*m*−*T* )]


$$\eta = \frac{cD}{T} [t\_d + (1 + m - t\_d) \cdot \ln(\frac{1 + m - t\_d}{1 + m - T}) - T]$$

6. The interest charged for advance paymen<sup>t</sup> per replenishment cycle is

$$\begin{aligned} &\frac{cI\_kDT}{T}(\int\_0^N \mathfrak{a}dt) + \frac{cI\_kD}{T} \int\_N^{T+N} \mathfrak{a}(T+N-t)t^2 \\ &= \frac{\mathfrak{a}cI\_kDT}{T}(N+L) + \frac{\mathfrak{a}cI\_kDT^2}{2T} \end{aligned}$$

7. The interest charged for cash paymen<sup>t</sup> per replenishment cycle is

$$\frac{cI\_k D T}{T} (\int\_0^N \beta dt) + \frac{cI\_k D}{T} \int\_N^{T+N} \beta (T+N-t) dt = \frac{\beta c I\_k D T N}{T} + \frac{\beta c I\_k D T^2}{2T}$$

8. The interest charged for credit paymen<sup>t</sup> per replenishment is Case 1. When *N* ≤ *M* and *M* ≤ *T* ≤ *T* + *N* The interest charged for credit paymen<sup>t</sup>

$$=\frac{\pi cDI\_k}{T}\left\{\rho\left[\int\_M^{T+N}(T+N-t)dt\right]+(1-\rho)\left[\int\_M^T(T-t)dt\right]\right\}$$

$$=\frac{\pi cDI\_k}{2T}\left\{\rho(T+N-M)^2+(1-\rho)\left(T-M\right)^2\right\}$$

Case 2. When *N* ≤ *M* and *T* ≤ *M* ≤ *T* + *N* The interest charged for credit paymen<sup>t</sup>

$$=\frac{\tau cDI\_k}{T}\left[\rho\int\_M^{T+N}(T+N-t)dt\right]=\frac{\tau cDI\_k}{2T}\left[\rho(T+N-M)^2\right]$$

Case 3. When *N* ≤ *M* and *T* + *N* ≤ *M* The interest charged for credit paymen<sup>t</sup> is zero. Case 4. When *N* > *M* and *M* ≤ *T* The interest charged for credit paymen<sup>t</sup>

$$=\frac{\pi c D I\_k}{T} \left\{ \rho \left[ \int\_M^N T dt + \int\_N^{T+N} (T+N-t) dt \right] + (1-\rho) \left[ \int\_M^T (T-t) dt \right] \right\}$$

$$=\frac{\pi c D I\_k}{2T} \left\{ \rho \left[ T^2 + 2(N-M)T \right] + (1-\rho) \left( T-M \right)^2 \right\}$$

Case 5. When *N* > *M* and *M* ≥ *T* The interest charged for credit paymen<sup>t</sup>

$$\begin{array}{c} = \frac{\pi cDI\_k}{T} \left\{ \rho \left[ \int\_M^N Tdt + \int\_N^{T+N} (T+N-t)dt \right] \right\} \\ = \frac{\pi cDI\_k}{2T} \left\{ \rho \left[ T^2 + 2T(N-M) \right] \right\} \end{array}$$

9. The interest earned for credit paymen<sup>t</sup> per replenishment is Case 1. When *N* ≤ *M* and *M* ≤ *T* ≤ *T* + *N* The interest earned for credit paymen<sup>t</sup>

$$\theta = \frac{\text{tr}\, \text{DL}\_{\ell}}{\text{T}} \left\{ \rho \int\_{N}^{M} (t - N)dt + (1 - \rho) \int\_{0}^{M} tdt \right\} = \frac{\text{tr}\, \text{DL}\_{\ell}}{\text{T}T} \left[ \rho (M - N)^{2} + (1 - \rho) \int\_{0}^{M} Ndt \right]$$

Case 2. When *N* ≤ *M* and *T* ≤ *M* ≤ *T* + *N* The interest earned for credit paymen<sup>t</sup>

$$=\frac{\text{tr}\,DI\_{\varepsilon}}{T}\left\{\rho\int\_{N}^{M}(t-N)dt+(1-\rho)\left[\int\_{0}^{T}tdt+\int\_{T}^{M}Tdt\right]\right\}$$

$$=\frac{\pi \nu D I\_{\mathfrak{E}}}{2T} \left\{ \rho (M - N)^2 + (1 - \rho) \left[ T^2 + 2T(M - T) \right] \right\}$$

Case 3. When *N* ≤ *M* and *T* + *N* ≤ *M* The interest earned for credit paymen<sup>t</sup>

$$\rho\_1 = \frac{\tau \nu D I\_\varepsilon}{T} \left\{ \rho \left[ \int\_N^{T+N} (t - N) dt + \int\_{T+N}^M T dt \right] + (1 - \rho) \left[ \int\_0^T t dt + \int\_T^M T dt \right] \right\}$$

$$= \frac{\tau \nu D I\_\varepsilon}{2T} \left\{ \rho \left[ T^2 + 2T(M - T - N) \right] + (1 - \rho) \left[ T^2 + 2T(M - T) \right] \right\}$$

$$\text{where } M = M - M \text{ and } \tau \text{ in } \mathbb{T}$$

Case 4. When *N* > *M* and *M* ≤ *T* = *τνD Ie T* [(1 − *ρ*) *M*0 *tdt*] = *τνD Ie* 2*T* [(1 − *ρ*)*M*<sup>2</sup>] Case 5. When *N* > *M* and *N* ≥ *T* = *τνD Ie T* (1 − *ρ*)[ *T*0 *tdt* + *MT Tdt*] = *τνD Ie T* )(1 − *ρ*)[*T*<sup>2</sup> + 2*T*(*M* − *T*)]+

Finally, the total annual relevant cost *TC*(*T*) is obtained as follows:

*TC*(*T*) = ordering cost + stock holding cost (excluding interest charges) + procurement cost + screening cost + deterioration cost + interest charged – interest earned.

Furthermore, we obtain the following cases:

Case I. Suppose that *N* ≤ *M*

Case (I-1). Suppose that *td* < *M* − *N* < *M*

$$TC(T) = \begin{cases} TC\_1(T) & \text{if } \qquad 0 < T < t\_d \\ TC\_2(T) & \text{if } \quad t\_d \le T < M - N \\ TC\_3(T) & \text{if } \quad M - N \le T < M \\ TC\_4(T) & \text{if } \qquad M \le T \le R^\* \end{cases} \tag{12}$$

where

$$\begin{array}{ll} \text{TC}\_{1}(T) = & \{ \begin{array}{l} \mathfrak{f} + \frac{\mathsf{h}D}{\{1-\rho\}}[\frac{(1-\rho)T}{2} + \frac{\mathsf{p}DT}{x(1-\rho)}] + \frac{(c+\mathsf{s})D}{1-\rho} + \frac{c\mathsf{l}\_{1}\mathsf{d}T}{T}[a(N+L) + \mathsf{f}\ \mathsf{N}] \\ + \frac{c\mathsf{l}\_{1}\mathsf{d}T^{2}}{2T}(a+\mathsf{f}) - \frac{\mathsf{r}\mathsf{p}D\mathsf{l}\_{1}}{2T}\left\{\rho[T^{2}+2T(M-T-N)] + (1-\rho)[T^{2}+2T(M-T)]\right\} \end{array} \tag{13}$$

$$\begin{array}{lcl} \text{TC2}(T) & = \frac{\rho}{T} + \frac{k}{T} \Bigg\{ \frac{D}{2} t\_d^2 + D(1+m-t\_d) \cdot t\_d \cdot \ln\left(\frac{1+m-t\_d}{1+m-T}\right) + \frac{pD^2}{x(1-p)^4} \Big[ t\_d + (1+m-t\_d) \Big] \\ & \times \ln\left(\frac{1+m-t\_d}{1+m-T}\right) \big[^2 + \frac{D}{2}(1+m-t\_d)^2 \cdot \ln\left(\frac{1+m-t\_d}{1+m-T}\right) + \frac{D}{4}(1+m-T)^2 - \frac{D}{4}(1+m-t\_d)^2 \Big] \\ & + \frac{(c+s)D}{(1-p)T} \big[ t\_d + (1+m-t\_d) \cdot \ln\left(\frac{1+m-t\_d}{1+m-T}\right) \big] + \frac{cD}{T} \big[ t\_d + (1+m-t\_d) \cdot \ln\left(\frac{1+m-t\_d}{1+m-T}\right) - T \Big] \\ & + \frac{c\Delta DT}{T} \big[ a(N+L) + \beta N \big] + \frac{c\beta\_L \mathcal{D}^2}{2T} (a+\beta) - \frac{\pi v D L}{2T} \big[ \rho \big[ T^2 + 2T(M-T-N) \big] \\ & + (1-\rho) \big[ T^2 + 2T(M-T) \big] \end{array} \tag{14}$$

$$\begin{array}{lcl} \text{TC}\_{3}(T) & = \frac{\varphi}{\Upsilon} + \frac{k}{\Upsilon} \Big\{ \frac{D}{2} t\_{d}^{2} + D(1 + m - t\_{d}) \cdot t\_{d} \cdot \ln\left(\frac{1 + m - t\_{d}}{1 + m - T}\right) + \frac{pD^{2}}{\mathbf{x}(1 - p)^{2}} [t\_{d} + (1 + m - t\_{d}) \\ & \qquad \times \ln\left(\frac{1 + m - t\_{d}}{1 + m - T}\right) \big[ \begin{matrix} +\frac{D}{2}(1 + m - t\_{d}) \cdot \ln\left(\frac{1 + m - t\_{d}}{1 + m - T}\right) + \frac{D}{4}(1 + m - T)^{2} - \frac{D}{4}(1 + m - t\_{d})^{2} \Big{\\ m\frac{d}{2}\left(1 + m - T\right)} + \frac{1}{2}(1 + m - T) \Big{]} - \frac{cD}{T}(1 + m - t\_{d}) \cdot \ln\left(\frac{1 + m - t\_{d}}{1 + m - T}\right) - T\Big{]} \\ & + \frac{k\_{d}DT}{\Gamma} [\alpha(N + L) + \not{p}\text{N}] + \frac{c\_{d}D\Gamma^{2}}{2\Gamma} (\alpha + \not{p}) + \frac{\pi Dk\_{d}}{2\Gamma} [\rho(T + N - M)^{2}] \\ & - \frac{\operatorname{Tr}DL}{2\Gamma} \Big{(}\rho(M - N)^{2} + (1 - \rho)[\Gamma^{2} + 2T(M - T)]\Big{)} \end{array} \tag{15}$$

$$\begin{array}{rcl} \mathsf{TC}\_{4}(T) & = & \frac{a}{\mathsf{T}} + \frac{b}{\mathsf{T}} \Big\{ \frac{D}{2} t\_{d}^{2} + D(1+m-t\_{d}) \cdot t\_{d} \cdot \ln\left(\frac{1+m-t\_{d}}{1+m-t}\right) + \frac{pD^{2}}{\mathsf{x}(1-p)^{2}} \Big[ t\_{d} + (1+m-t\_{d}) \\ & \qquad \times \ln\left(\frac{1+m-t\_{d}}{1+m-t}\right) \Big[ t^{2} + \frac{D}{2}(1+m-t\_{d})^{2} \cdot \ln\left(\frac{1+m-t\_{d}}{1+m-t}\right) + \frac{D}{\mathsf{x}}(1+m-T)^{2} - \frac{D}{4}(1+m-t\_{d})^{2} \Big] \\ & + \frac{(c+d)D}{(1-p)^{T}} [t\_{d} + (1+m-t\_{d}) \cdot \ln\left(\frac{1+m-t\_{d}}{1+m-t}\right)] + \frac{cD}{\mathsf{T}} [t\_{d} + (1+m-t\_{d}) \cdot \ln\left(\frac{1+m-t\_{d}}{1+m-t}\right) - T] \\ & + \frac{c\underline{b}\cdot\mathsf{T}\mathcal{I}}{\mathsf{T}} [a(N+L) + \mathcal{f}\mathcal{N}] + \frac{c\underline{b}\cdot\mathsf{T}\mathcal{I}^{2}}{2\mathsf{T}} (a+\mathcal{f}) + \frac{\mathsf{r}\epsilon\mathcal{D}l\_{0}}{2\mathsf{T}} [\rho(T+N-M)^{2} \\ & + (1-\rho)(T-M)^{2}\Big] - \frac{\mathsf{r}\epsilon\mathcal{D}L}{2T} [\rho(M-N)^{2} + (1-\rho)M^{2}] \end{array} \tag{16}$$

Since *TC*1(*td*) = *TC*2(*td*), *TC*2(*<sup>M</sup>* − *N*) = *TC*3(*<sup>M</sup>* − *N*) and *TC*3(*M*) = *TC*4(*M*), *TC*(*T*) is continuous and well-defined on *T* > 0.

Case (I-2). Suppose that *M* − *N* < *td* < *M*

$$TC(T) = \begin{cases} TC\_1(T) & \text{if } \quad 0 < T < M - N \\ TC\_5(T) & \text{if } \quad M - N \le T < t\_d \\ TC\_3(T) & \text{if } \quad t\_d \le T < M \\ TC\_4(T) & \text{if } \quad M \le T \le R^\* \end{cases} \tag{17}$$

where

$$\begin{array}{l} \text{TC}\_{5}(T) \quad = \frac{\varrho}{T} + \frac{hD}{(1-p)} \big[ \frac{(1-p)T}{2} + \frac{pDT}{x(1-p)} \big] + \frac{(c+s)D}{1-p} + \frac{cI\_{k}DT}{T} \big[ a(N+L) + \beta N \big] \\ \quad + \frac{cI\_{k}DT^{2}}{2T} (a+\beta) + \frac{\text{tr}\,\text{D}I\_{k}}{2T} \big[ \rho(T+N-M)^{2} \big] - \frac{\text{tr}\,\text{D}I\_{k}}{2T} \big[ \rho(M-N)^{2} \\ \quad + (1-\rho)[T^{2}+2T(M-T)] \big] \end{array} \tag{18}$$

Since *TC*1(*<sup>M</sup>* − *N*) = *TC*5(*<sup>M</sup>* − *<sup>N</sup>*), *TC*5(*td*) = *TC*3(*td*) and *TC*3(*M*) = *TC*4(*M*), *TC*(*T*) is continuous and well-defined on *T* > 0. Case (I-3). Suppose that *M* − *N* < *M* < *td*

$$TC(T) = \begin{cases} TC\_1(T) & \text{if } \quad 0 < T < M - N \\ TC\_5(T) & \text{if } \quad M - N \le T < M \\ TC\_6(T) & \text{if } \quad M \le T < t\_d \\ TC\_4(T) & \text{if } \quad t\_d \le T \le R^\* \end{cases} \tag{19}$$

where

$$\begin{array}{ll} \text{TC6}(T) &= \frac{\varrho}{T} + \frac{hD}{\left(1-p\right)} \left[ \frac{(1-p)T}{2} + \frac{pDT}{\mathbf{x}(1-p)} \right] + \frac{(c+s)D}{1-p} + \frac{cI\_kDT}{I} \left[ \mathbf{a}(N+L) + \beta N \right] \\ &+ \frac{cI\_kDT^2}{2T} (\mathbf{a} + \boldsymbol{\beta}) + \frac{\tau cDI\_k}{2T} \left[ \boldsymbol{\rho}(T+N-M)^2 + (1-\boldsymbol{\rho})(T-M) \right]^2 \\ &- \frac{\tau vDI\_\ell}{2T} \left[ \boldsymbol{\rho}(M-N)^2 + (1-\boldsymbol{\rho})M^2 \right] \end{array} \tag{20}$$

Since *TC*1(*<sup>M</sup>* − *N*) = *TC*5(*<sup>M</sup>* − *<sup>N</sup>*), *TC*5(*M*) = *TC*6(*M*) and *TC*6(*td*) = *TC*4(*td*), *TC*(*T*) is continuous and well-defined on *T* > 0.

Case II. Suppose that *N* > *M*

Case (II-1). Suppose that *td* < *M*

$$TC(T) = \begin{cases} \begin{array}{l} T\mathcal{C}\_7(T) \quad \text{if} \quad 0 < T < t\_d\\ T\mathcal{C}\_8(T) \quad \text{if} \quad t\_d \le T < M\\\ T\mathcal{C}\_9(T) \quad \text{if} \quad M \le T \le R^\* \end{array} \tag{21}$$

where

$$\begin{array}{ll} \text{TC}\_{7}(T) &= \frac{\varrho}{T} + \frac{hD}{(1-p)} [\frac{(1-p)T}{2} + \frac{pDT}{x(1-p)}] + \frac{(c+s)D}{1-p} + \frac{cI\_{k}DT}{T} [a(N+L) + \beta N] \\ &+ \frac{cI\_{k}DT^{2}}{2T}(a+\beta) + \frac{\tau cDI\_{k}}{2T} \{\rho[T^{2}+2T(N-M)]\} \\ &- \frac{\tau cDI\_{r}}{2T} \{(1-\rho)[T^{2}+2T(M-T)]\} \end{array} \tag{22}$$

$$\begin{array}{lcl} \text{TCs}(T) &= \frac{\mu}{T} + \frac{k}{T} \left\{ \frac{\mu}{2} t\_d^2 + D(1 + m - t\_d) \cdot t\_d \cdot \ln\left(\frac{1 + m - t\_d}{1 + m - T}\right) + \frac{pD^2}{\mathbf{x}(1 - p)^2} [t\_d + (1 + m - t\_d) \\ & \qquad \times \ln\left(\frac{1 + m - t\_d}{1 + m - T}\right)]^2 + \frac{\mu}{2} (1 + m - t\_d)^2 \cdot \ln\left(\frac{1 + m - t\_d}{1 + m - T}\right) + \frac{\mu}{2} (1 + m - T)^2 - \frac{D}{\mathbf{x}} (1 + m - t\_d)^2 \right\} \\ & \qquad + \frac{(\varepsilon + \nu)D}{(1 - p)^2} [t\_d + (1 + m - t\_d) \cdot \ln\left(\frac{1 + m - t\_d}{1 + m - T}\right)] + \frac{\varepsilon D}{2} [t\_d + (1 + m - t\_d) \cdot \ln\left(\frac{1 + m - t\_d}{1 + m - T}\right) - T] \\ & \qquad + \frac{\varepsilon L\_d T}{2} [\mathbf{a}(N + L) + \beta N] + \frac{\varepsilon L\_d T^2}{2T} (\mathbf{a} + \beta) + \frac{\varepsilon c D l\_k}{2T} \{\rho \left[ T^2 + 2T(N - M) \right] \} \\ & \qquad - \frac{\tau c D l\_k}{2T} \{(1 - \rho) \left[ T^2 + 2T(M - T) \right] \} \end{array} \tag{23}$$

$$\begin{array}{lcl} \text{TCy}(T) &= \frac{\rho}{T} + \frac{h}{T} \Bigg\{ \frac{D}{2} t\_d^2 + D(1+m-t\_d) \cdot t\_d \cdot \ln\left(\frac{1+m-t\_d}{1+m-T}\right) + \frac{pD^2}{x(1-p)^2} \Big[ t\_d + (1+m-t\_d) \Big] \\ &\times \ln\left(\frac{1+m-t\_d}{1+m-T}\right) \big[^2 + \frac{D}{2}(1+m-t\_d)^2 \cdot \ln\left(\frac{1+m-t\_d}{1+m-T}\right) + \frac{D}{2}(1+m-T)^2 - \frac{D}{4}(1+m-t\_d)^2 \Big] \\ &+ \frac{(c+s)D}{(1-p)T} \big[ t\_d + (1+m-t\_d) \cdot \ln\left(\frac{1+m-t\_d}{1+m-T}\right) \big] + \frac{cD}{T} \big[ t\_d + (1+m-t\_d) \cdot \ln\left(\frac{1+m-t\_d}{1+m-T}\right) - T \Big] \\ &+ \frac{c\_1 D T}{T} \big[ a(N+L) + \beta' N \right] + \frac{c\_2 D T^2}{2T} (a+\beta) + \frac{vcD L}{2T} \big[ \rho \big[ T^2 + 2T(N-M) \big] \\ &+ (1-\rho) \left( T - M \right)^2 \Big] - \frac{\tau\_2 D L}{2T} (1-\rho) M^2 \end{array} \tag{24}$$

Since *TC*7(*td*) = *TC*8(*td*) and *TC*8(*M*) = *TC*9(*M*), *TC*(*T*) are continuous and welldefined on *T* > 0.

Case (II-2). Suppose that *td* ≥ *M*

$$TC(T) = \begin{cases} \begin{array}{c} TC\_7(T) \quad \text{if} \quad 0 < T < M \\ TC\_{10}(T) \quad \text{if} \quad M \le T < t\_d \\\ TC\_9(T) \quad \text{if} \quad t\_d \le T \le R^\* \end{cases} \tag{25}$$

where

$$\begin{array}{l} \text{TC}\_{10}(T) & = \frac{\rho}{T} + \frac{hD}{(1-p)}[\frac{(1-p)T}{2} + \frac{pDT}{x(1-p)}] + \frac{(\varepsilon+s)D}{1-p} + \frac{cI\_kDT}{T}[a(N+L) + \beta N] \\ & + \frac{cI\_kDT^2}{2T}(a+\beta) + \frac{\tau cDl\_k}{2T}\left\{\rho\left[T^2 + 2T(N-M)\right] + (1-\rho)\left(T-M\right)^2\right\} \\ & - \frac{\tau vDl\_k}{2T}(1-\rho)M^2 \end{array} \tag{26}$$

Since *TC*7(*M*) = *TC*10(*M*) and *TC*10(*td*) = *TC*9(*td*), *TC*(*T*) is continuous and welldefined on *T* > 0.

### **3. The Convexity and Monotonicity Properties of** *TCi***(***T***)**

**(***i* **= 1, 2, 3, 4, 5, 6, 7, 8, 9, 10)**

In this section, we continue the derivations described in above section and adopt Equations (13)–(16), (18), (20), (22)–(24) and (26) to find the first-order and the second-order derivatives of the annual total relevant costs *TCi*(*T*) with respect to *T* in order to obtain the convexity properties as follows:

$$TC\_1'(T) = \frac{1}{T^2} \left\{-o + \frac{hD}{2} [1 + \frac{2pD}{x(1-p)^2}]T^2 + \frac{cI\_kD}{2}(a+\beta)T^2 + \frac{\pi \nu D I\_t}{2}T^2\right\} \tag{27}$$

$$TC''\_1(T) = \frac{2o}{T^3} > 0\tag{28}$$

$$\begin{cases} TC\_2'(T) &= \frac{1}{T^2} \left\{-o + h \cdot G(T) + [\frac{(2-p)c + s}{1-p}] \cdot [D(1+m-t\_d) \cdot \frac{T}{1+m-T} - Dt\_d \\ &- D(1+m-t\_d) \cdot \ln(\frac{1+m-t\_d}{1+m-T})] + \frac{cl\_d D}{2}(a+\beta)T^2 + \frac{\pi vDL\_t}{2}T^2 \right\} \end{cases} \tag{29}$$

$$\begin{cases} TC'^{\prime}(T) &= \frac{1}{T^{5}} \left\{ 2o + h \cdot H(T) + \left[ \frac{(2-p)c + s}{1-p} \right] \cdot \left[ D(1+m-t\_d) \cdot \frac{(1+m)T}{(1+m-T)^2} \right. \\ & \quad - 3D(1+m-t\_d) \frac{T}{1+m-T} + 2Dt\_d + 2D(1+m-t\_d) \cdot \ln\left(\frac{1+m-t\_d}{1+m-T}\right) \right] \end{cases} \tag{30}$$

$$\begin{cases} TC\_3(T) &= \frac{1}{T^2} \left\{-o + h \cdot G(T) + [\frac{(2-\rho)c+s}{1-\rho}] \cdot [D(1+m-t\_d) \cdot \frac{T}{1+m-T} - Dt\_d \\ &- D(1+m-t\_d) \cdot \ln(\frac{1+m-t\_d}{1+m-I})] + \frac{cL\_dD}{2}(a+\rho)T^2 + \frac{rcD\_b}{2} \left\{\rho \left[T^2 - (N-M)^2\right] \right\} \\ &+ \frac{rcD\_d}{2} [\rho(M-N)^2 + (1-\rho)T^2] \right\} \end{cases} \tag{31}$$

$$\begin{cases} TC\_3''(T) = \frac{1}{T^5} \Bigg\{ 2o + h \cdot H(T) + \left[ \frac{(2-p)c + s}{1-p} \right] \cdot \left[ D(1+m-t\_d) \cdot \frac{(1+m)T}{\left(1+m-T\right)^2} \right. \\ \qquad - 3D(1+m-t\_d) \frac{T}{1+m-T} + 2Dt\_d + 2D(1+m-t\_d) \cdot \ln\left(\frac{1+m-t\_d}{1+m-T}\right) \Bigr] \\ \qquad + \tau c D I\_k \wp \left( N-M \right)^2 - \tau v D I\_\ell \wp \left( M-N \right)^2 \Bigr] \end{cases} \tag{32}$$

$$\begin{split} TC\_4'(T) &= \begin{array}{c} \frac{1}{T^2} \left\{-o + h \cdot G(T) + \left[\frac{(2-p)c+s}{1-p}\right] \cdot \left[D(1+m-t\_d) \cdot \frac{T}{1+m-T} - Dt\_d - \right. \\ -D(1+m-t\_d) \cdot \ln\left(\frac{1+m-t\_d}{1+m-T}\right) \right] + \frac{cI\_bD}{2}(a+\beta)T^2 + \frac{rcDl\_l}{2} \left\{\rho\left[T^2 - \left(N-M\right)^2\right] \\ + \left(1-\rho\right)\left(T^2 - M^2\right) \right\} + \frac{rcDl\_t}{2} \left[\rho\left(M-N\right)^2 + \left(1-\rho\right)M^2\right] \end{array} \tag{33}$$

$$\begin{array}{lcl} \text{TC\\_4}(T) &= \frac{1}{T^4} \left\{ 2\nu + h \cdot H(T) + \left[ \frac{(2-\rho)c + s}{1-\rho} \right] \cdot \left[ D(1+m-t\_d) \cdot \frac{(1+m)T}{\left(1+m-T\right)^2} \right. \\ &\left. -3D(1+m-t\_d) \frac{T}{1+m-T} + 2Dt\_d + 2D(1+m-t\_d) \cdot \ln\left(\frac{1+m-t\_d}{1+m-T}\right) \right] \\ &+ \pi c D I\_k[\rho \left( N-M \right)^2 + \left(1-\rho\right)M^2] - \pi v D I\_\varepsilon[\rho \left( M-N \right)^2 + \left(1-\rho\right)M^2] \right\} \end{array} \tag{34}$$

$$\begin{split} TC\_5'(T) = \frac{1}{T^2} \left\{ -\sigma + \frac{bL}{2} [1 + \frac{2pD}{x(1-p)^2}] T^2 + \frac{c\_1lD}{2} (a+\rho)T^2 + \frac{rcDL\_l}{2} \left\{ \rho \left[ T^2 - (N-M)^2 \right] \right\} \\ \quad + \frac{rcDL\_l}{2} [\rho (M-N)^2 + (1-\rho)T^2] \right\} \end{split} \tag{35}$$

$$TC'\_5(T) = \frac{1}{T^5} \left\{ 2\rho + \text{tr} D I\_k \rho \left( N - M \right)^2 - \text{tr} v D I\_\varepsilon \rho \left( M - N \right)^2 \right\} \tag{36}$$

$$\begin{cases} TC\_6'(T) = \frac{1}{T^2} \left\{ -\rho + \frac{hD}{2} [1 + \frac{2pD}{x(1-p)^2}] T^2 + \frac{c\underline{h}D}{2} (a+\rho) T^2 + \frac{vcD\underline{h}}{2} \left\{ \rho \left[ T^2 - (N-M)^2 \right] \\ \quad + (1-\rho)(T^2 - M^2) \right\} + \frac{vvD L}{2} [\rho (M-N)^2 + (1-\rho)M^2] \right\} \end{cases} \tag{37}$$

$$TC^{\tau}\_{\delta}(T) = \frac{1}{T^{2}} \left\{ 2\sigma + \text{tr} D I\_{k} [\rho \left( N - M \right)^{2} + (1 - \rho)M^{2} ] - \text{tr} v D I\_{t} [\rho (M - N)^{2} + (1 - \rho)M^{2}] \right\} \tag{38}$$

$$\begin{cases} TC\_7(T) &= \frac{1}{T^2} \left\{-o + \frac{hD}{2} [1 + \frac{2pD}{x(1-p)^2}]T^2 + \frac{cI\_bD}{2}(a+\beta)T^2 + \frac{\tau cDI\_k}{2}\rho T^2 \\ &+ \frac{\tau vD I\_c}{2}(1-\rho)T^2 \right\} \end{cases} \tag{39}$$

$$TC^{\prime} \tau(T) = \frac{2\rho}{T^3} > 0\tag{40}$$

$$\begin{cases} TC\_8'(T) &= \frac{1}{T^2} \left\{-\sigma + h \cdot G(T) + [\frac{(2-p)c+s}{1-p}][D(1+m-t\_d)\frac{T}{1+m-T} \\ &- Dt\_d - D(1+m-t\_d)\cdot \ln(\frac{1+m-t\_d}{1+m-T})] + \frac{cL\_bD}{2}(\alpha+\beta)T^2 + \frac{\text{tr}Dl\_k}{2}\rho T^2 \\ &+ \frac{\text{tr}Dl\_k}{2}(1-\rho)T^2 \right\} \end{cases} \tag{41}$$

$$\begin{cases} TC's(T) &= \frac{1}{T^3} \left\{ 2o + h \cdot H(T) + [\frac{(2-p)c+s}{1-p}][D(1+m-t\_d)\frac{(1+m)T}{(1+m-T)^2} \\ &- 3D(1+m-t\_d)\frac{T}{1+m-T} + 2Dt\_d + 2D(1+m-t\_d)\cdot \ln(\frac{1+m-t\_d}{1+m-T})] \right\} \end{cases} \tag{42}$$

$$\begin{array}{ll} TC\_{\theta}(T) &= \frac{1}{T^{2}} \left\{-o + h \cdot G(T) + [\frac{(2-p)c+s}{1-p}][D(1+m-t\_{d})\frac{T}{1+m-l} \\ &- Dt\_{d} - D(1+m-t\_{d}) \cdot \ln(\frac{1+m-t}{1+m-l})] + \frac{cI\_{b}D}{2}(a+\beta)T^{2} + \frac{vcDl\_{b}}{2}[T^{2}-(1-\rho)M^{2}] \\ &+ \frac{x\chi QL\_{b}}{2}(1-\rho)M^{2} \right\} \end{array} \tag{43}$$

$$\begin{array}{ll} TC''g(T) &= \frac{1}{T^3} \Biggl\{ 2o + h \cdot H(T) + [\frac{(2-p)c + s}{1-p}][D(1+m-t\_d)\frac{(1+m)T}{(1+m-T)^2} \\ &- 3D(1+m-t\_d)\frac{T}{1+m-T} + 2Dt\_d + 2D(1+m-t\_d)\cdot \ln\left(\frac{1+m-t\_d}{1+m-T}\right) \Biggr\} \\ &+ \tau c D I\_k (1-\rho)M^2 - \tau v D I\_\ell (1-\rho)M^2 \Biggr\} \end{array} \tag{44}$$

$$\begin{array}{rcl} TC' \, \_{10}(T) &=& \frac{1}{T^2} \left\{ -\sigma + \frac{hD}{2} [1 + \frac{2pD}{x(1-p)^2}] T^2 + \frac{c l\_1 D}{2} (a+\beta) T^2 + \frac{vc D l\_1}{2} [T^2 - (1-\rho)M^2] \\ &+ \frac{vc D l\_\epsilon}{2} (1-\rho)M^2 \right\} \end{array} \tag{45}$$

$$TC'\_{10}(T) = \frac{1}{T^5} \left\{ 2\rho + \text{\textpi c}D I\_k(1-\rho)M^2 - \text{\textpi c}D I\_\ell(1-\rho)M^2 \right\} \tag{46}$$

where

$$\begin{array}{l} G(T) &= D(1+m-t\_d) \cdot t\_d \frac{T}{1+m-T} + \frac{2pD^2}{x(1-p)^2} (1+m-t\_d)[t\_d + (1+m-t\_d) \\ &\times \ln\left(\frac{1+m-t\_d}{1+m-T}\right)] \cdot \frac{T}{1+m-T} + \frac{D}{2}(1+m-t\_d)^2 \cdot \frac{T}{1+m-T} - \frac{D}{2}(1+m-T)T \\ &- \frac{D}{2}t\_d^2 - D(1+m-t\_d) \cdot t\_d \cdot \ln\left(\frac{1+m-t\_d}{1+m-T}\right) - \frac{pD^2}{x(1-p)^2} [t\_d + (1+m-t\_d) \cdot \ln\left(\frac{1+m-t\_d}{1+m-T}\right)]^2 \\ &- \frac{D}{2}(1+m-t\_d)^2 \cdot \ln\left(\frac{1+m-t\_d}{1+m-T}\right) - \frac{D}{4}(1+m-T)^2 + \frac{D}{4}(1+m-t\_d)^2 \end{array} \tag{47}$$

$$\begin{array}{l} H(T) &= D(1+m-t\_d)t\_d \frac{T^2}{(1+m-T)^2} + \frac{D}{2}(1+m-t\_d)^2 \frac{T^2}{(1+m-T)^2} + \frac{D}{2}T^2 \\ &- 2D(1+m-t\_d) \cdot t\_d \frac{T}{1+m-T} - D(1+m-t\_d)^2 \frac{T}{1+m-T} + D(1+m-T)T \\ &+ Dt\_d^2 + 2D(1+m-t\_d)t\_d \cdot \ln\left(\frac{1+m-t\_d}{1+m-T}\right) + D(1+m-t\_d)^2 \ln\left(\frac{1+m-t\_d}{1+m-T}\right) \\ &+ \frac{D}{2}(1+m-T)^2 - \frac{D}{2}(1+m-t\_d)^2 + \frac{2pD^2}{x(1-p)^2}(1+m-t\_d)^2 \frac{T^2}{(1+m-T)^2} \\ &+ \frac{2pD^2}{x(1-p)^2} [(1+m-t\_d)]t\_d + (1+m-t\_d) \cdot \ln\left(\frac{1+m-t\_d}{1+m-T}\right)] \frac{T^2}{(1+m-T)^2} \\ &- \frac{4pD^2}{x(1-p)^2} (1+m-t\_d)[t\_d + (1+m-t\_d) \cdot \ln\left(\frac{1+m-t\_d}{1+m-T}\right)] \frac{T}{1+m-T} \\ &+ \frac{2pD^2}{x(1-p)^2} [t\_d + (1+m-t\_d) \cdot \ln\left(\frac{1+m-t\_d}{1+m-T}\right)] \end{array} \tag{48}$$

Obviously, it is shown that *TC*1(*T*) and *TC*7(*T*) are convex functions on (0, <sup>∞</sup>), respectively.

Now, we let

$$\mathcal{W}\_1 = 2o - \tau \upsilon I\_\ell D\rho \left(M - N\right)^2\tag{49}$$

$$\mathcal{W}\_2 = 2o - \text{tr}\upsilon I\_c D \left[ \rho (M - N)^2 + (1 - \rho) M^2 \right] \tag{50}$$

and

$$\mathcal{W}\_3 = 2\rho - \pi v I\_c D (1 - \rho) M^2 \tag{51}$$

Then, we have the following convexity results.

**Lemma 1.** *Each of the following assertions holds true:*


**Remark 1.** *In our proof of Lemma 1, we need the assertions of Lemma 2.*

**Lemma 2.** *Each of the following assertions holds true:*

(A) *G*(*T*) > 0 *if T* ≥ *td*

$$\text{(B)}\quad D(1+m-t\_d)\frac{T}{1+m-T} - Dt\_d - D(1+m-t\_d) \cdot \ln(\frac{1+m-t\_d}{1+m-T}) > 0 \text{ if } T \ge t\_d$$

(C) *H*(*T*) > 0 *if T* ≥ *td*

$$\begin{array}{ll} \text{(D)} & D(1+m-t\_d) \frac{(1+m)T}{(1+m-T)^2} - 3D(1+m-t\_d) \frac{T}{1+m-T} \\ & + 2Dt\_d + 2D(1+m-t\_d) \cdot \ln(\frac{1+m-t\_d}{1+m-T}) > 0 \end{array} \quad \text{if} \quad T \ge t\_d.$$

**Remark 2.** *The proof of Lemma 2 is given in the Appendix A.*

### **Proof of Lemma 1.**


$$TC'\_{\dot{i}}(T) > \frac{1}{T^3} \left\{ 2o - \pi \upsilon D I\_t \rho (M - N)^2 \right\} > 0, \quad i = 3 \quad and \ 5$$

$$TC''\_{j}(T) > \frac{1}{T^{\mathfrak{J}}} \left\{ 2o - \text{tr} vDI\_{\mathfrak{e}}[\rho(M - N) + (1 - \rho)M^2] \right\} > 0 \,, \quad j = 4 \quad and \tag{6}$$

$$TC'\_{k}(T) > \frac{1}{T^{5}} \left\{ 2\rho - \text{tr}\boldsymbol{\nu} \boldsymbol{DI}\_{\boldsymbol{\varepsilon}}[(1-\rho)\boldsymbol{M}^{2}] \right\} > 0 \quad k=9 \quad \text{and} \quad 10 \tag{52}$$

The above results imply that *TCl*(*T*) (*l* = 3, 4, 5, 6, 9, 10) is convex on [*td*, <sup>∞</sup>).

(C) When *W*1 < 0, from Equations (31) and (33) and Lemma 2(A) and 2(B), we have the following results: 

*TCi*(*T*) > 12*T*2 −2*<sup>o</sup>* + *τvDIeρ*(*<sup>M</sup>* − *N*)<sup>2</sup> > 0 , *i* = 3 *and* 5 Furthermore, *TCi*(*T*) (*i* = 3, 5) is increasing on [*Td*, <sup>∞</sup>).

(D) When *W*2 < 0, from Equations (35) and (37) and Lemma 2(A) and 2(B), we have the following results:

$$TC\_j'(T) > \frac{1}{2T^2} \left\{-2\rho + \tau vDI\_c[\rho(M-N)^2 + (1-\rho)M^2] \right\} > 0, \quad j=4 \quad and \quad 6 \le j \le \frac{1}{2}$$

Furthermore, *TCj*(*T*) ( *j* = 4, 6) is increasing on [*td*, <sup>∞</sup>).

(E) When *W*3 < 0, from Equations (43) and (45) and Lemma 2(A) and 2(B), we have the following results:

$$TC\_k'(T) > \frac{1}{2T^2} \left\{-2\rho + \tau v D I\_\varepsilon (1 - \rho) M^2 \right\} > 0 \quad k = 9 \quad and \quad 10$$

Furthermore, *TCk*(*T*) (*k* = 9, 10) is increasing on [*td*, <sup>∞</sup>). This completes the proof of Lemma 1. -

### **4. The Main Theorems for Optimal Replenishment Cycle Time** *T∗* **of** *TC***(***T***)**

In this section, we apply the convexity and monotonicity properties in order to develop efficient decision rules for the optimal replenishment cycle time *T*∗ of *TC*(*T*).

*4.1. Decision Rule of the Optimal Replenishment Cycle Time T*∗ *When N* ≤ *M*

4.1.1. Decision Rule of the Optimal Replenishment Cycle Time *T*∗ When *td* < *M* − *N* < *M* From Equation (12), we have

− *N*

<

*M*

$$TC(T) = \begin{cases} TC\_1(T) & \text{if } \quad 0 < T < t\_d \\ \begin{array}{l} TC\_2(T) & \text{if } \quad t\_d \le T < M - \\ TC\_3(T) & \text{if } \quad M - N \le T < T < T \\ TC\_4(T) & \text{if } \quad M \le T < R \end{array} \end{cases}$$

*TC*4(*T*) *if M* ≤ *T* < *R*∗

All *TCi*(*T*) (*i* = 1, 2, 3, 4) and *TC*(*T*) are defined on *T* > 0. From Equations (27), (29), (31) and (33), we have

$$TC\_1'(t\_d) = TC\_2(t\_d) = \frac{\Delta\_1}{t\_d^2}$$

$$TC\_2'(M - N) = TC\_3(M - N) = \frac{\Delta\_2}{\left(M - N\right)^2}$$

$$TC\_3'(M) = TC\_4(M) = \frac{\Delta\_3}{M^2}$$

and

$$TC\_4'(R^\*) = \frac{\Delta^\*}{R^{\*2}}$$

where

$$\Delta\_1 = -o + \frac{hD}{2} [1 + \frac{2pD}{x(1-p)^2}] t\_d^2 + \frac{cI\_k D}{2} (a+\beta) t\_d^2 + \frac{\pi \nu D I\_\varepsilon}{2} t\_d^2 \tag{53}$$

$$\begin{array}{rcl} \Delta\_{2} &=& -o + h \cdot \mathbb{G}(M-N) + \left[\frac{(2-p)c+s}{1-p}\right] \cdot \left[D(1+m-t\_{d}) \cdot \frac{(M-N)}{1+m-(M-N)} - Dt\_{d} \\ & -D(1+m-t\_{d}) \cdot \ln\left(\frac{1+m-t\_{d}}{1+m-(M-N)}\right)\right] + \frac{cl\_{D}D}{2}(a+\beta)\left(M-N\right)^{2} + \frac{\text{tr}\,\text{D}\!L}{2}(M-N)^{2} \end{array} \tag{54}$$

$$\begin{array}{ll} \Delta\_{3} &=& -o + h \cdot G(M) + \left[\frac{(2-p)c+s}{1-p}\right] \cdot \left[D(1+m-t\_{d}) \cdot \frac{M}{1+m-M} - Dt\_{d} \\ & -D(1+m-t\_{d}) \cdot \ln\left(\frac{1+m-t\_{d}}{1+m-M}\right)\right] + \frac{c\|\rho\|}{2}(a+\beta)M^{2} + \frac{vcDl\_{k}}{2}\left\{\rho \cdot \left[M^{2} - (N-M)^{2}\right]\right\} \\ & + \frac{\text{tr}\underline{D}L}{2}\left[\rho(M-N)^{2} + (1-\rho)M^{2}\right] \end{array} \tag{55}$$

$$\begin{array}{rcl}\Delta^\* &=& -\boldsymbol{\rho} + \boldsymbol{h} \cdot \boldsymbol{G}(\boldsymbol{R}^\*) + \left[\frac{(2-\rho)\boldsymbol{c} + \boldsymbol{s}}{1-\rho}\right] \cdot \left[D\left(1+\boldsymbol{m}-\boldsymbol{t}\_d\right) \cdot \frac{\boldsymbol{R}^\*}{1+\boldsymbol{m}-\boldsymbol{R}^\*} - \boldsymbol{D}\boldsymbol{t}\_d \\ & -\boldsymbol{D}\left(1+\boldsymbol{m}-\boldsymbol{t}\_d\right) \cdot \ln\left(\frac{1+\boldsymbol{m}-\boldsymbol{t}\_d}{1+\boldsymbol{m}-\boldsymbol{R}^\*}\right)\right] + \frac{\boldsymbol{c}\boldsymbol{l}\_d\boldsymbol{D}}{2}\left(\boldsymbol{a}+\boldsymbol{\beta}\right)\boldsymbol{R}^\* + \frac{\boldsymbol{v}\boldsymbol{c}\boldsymbol{D}\boldsymbol{l}\_k}{2}\left\{\rho\left[\boldsymbol{R}^{\*2}-\left(\boldsymbol{N}-\boldsymbol{M}\right)^2\right] \\ & +\left(1-\rho\right)\left(\boldsymbol{R}^{\*2}-\boldsymbol{M}^2\right)\right\} + \frac{\boldsymbol{v}\boldsymbol{c}\boldsymbol{D}\boldsymbol{l}\_k}{2}\left[\rho\left(\boldsymbol{M}-\boldsymbol{N}\right)^2+\left(1-\rho\right)\boldsymbol{M}^2\right] \end{array} \tag{56}$$

From the above results, we have Δ1 < Δ2. In addition, if *W*1 ≥ 0, then Δ1 < Δ2 < Δ3 < Δ<sup>∗</sup>. Otherwise, if *W*1 < 0, we obtain 0 < Δ2 < Δ3 < Δ<sup>∗</sup>. From the above discussions, the following results are achieved.

**Theorem 1.** *Suppose that td* < *M* − *N* < *M. Then, each of the following results holds true:*

	- (A) *If* Δ1 < 0*,* Δ2 < 0*,* Δ3 < 0 *and* Δ∗ < 0*, then TC*(*T*∗) = *TC*4(*R*<sup>∗</sup>). (B) *If* Δ1 < 0*,* Δ2 < 0*,* Δ3 < 0 *and* Δ∗ ≥ 0*, then TC*(*T*∗) = *TC*<sup>4</sup>*<sup>T</sup>*<sup>∗</sup><sup>4</sup> *.* (C) *If* Δ1 < 0*,* Δ2 < 0*,* Δ3 ≥ 0 *and* Δ∗ ≥ 0*, then TC*(*T*∗) = *TC*3(*T*<sup>∗</sup>3 )*.* (D) *If* Δ1 < 0*,* Δ2 ≥ 0*,* Δ3 ≥ 0 *and* Δ∗ ≥ 0*, then TC*(*T*∗) = *TC*2(*T*<sup>∗</sup>2 )*.* (E) *If* Δ1 ≥ 0*,* Δ2 ≥ 0*,* Δ3 ≥ 0 *and* Δ∗ ≥ 0*, then TC*(*T*∗) = *TC*<sup>1</sup>*<sup>T</sup>*<sup>∗</sup><sup>1</sup> *.*

(A) *If* Δ1 < 0*,* Δ2 ≥ 0*,* Δ3 ≥ 0 *and* Δ∗ ≥ 0*, then TC*(*T*∗) = *TC*2(*T*<sup>∗</sup>2 )*.* (B) *If* Δ1≥ 0*,* Δ2≥ 0*,* Δ3≥ 0 *and* Δ∗ ≥ 0*, then TC*(*T*∗) = *TC*<sup>1</sup>*<sup>T</sup>*<sup>∗</sup><sup>1</sup> *.*

**Proof.** The proof of Theorem 1 follows immediately from the above discussions. -

4.1.2. Decision Rule of the Optimal Replenishment Cycle Time *T*∗ When *M* − *N* < *td* < *M* From Equation (17), we have

$$TC(T) = \begin{cases} TC\_1(T) & if \quad 0 < T < M - N \\ TC\_5(T) & if \quad M - N \le T < t\_d \\ TC\_3(T) & if \quad t\_d \le T < M \\ TC\_4(T) & if \quad M \le T \le R^\* \end{cases}$$

Herein, *TC*5(*T*) is defined on *T* > 0 as well. Equations (27), (31) and (35) imply that

$$TC\_1'(M - N) = TC\_5'(M - N) = \frac{\Delta\_4}{\left(M - N\right)^2}$$

and

$$TC'\_5(t\_d) = TC'\_3(t\_d) = \frac{\Delta \varepsilon}{t\_d^2}$$

where

$$\Delta\_{4} = -\sigma + \frac{hD}{2} [1 + \frac{2pD}{x(1-p)^{2}}] (M-N)^{2} + \frac{c\beta\_{h}D}{2} (\sigma + \beta)(M-N)^{2} + \frac{\pi xDL}{2} (M-N)^{2} \tag{57}$$

and

$$\begin{cases} \Delta\_5 = -o + \frac{hD}{2} [1 + \frac{2pD}{x(1-p)^2}] t\_d^2 + \frac{\varepsilon I\_k D}{2} (a + \beta) t\_d^2 + \frac{\tau cDI\_k}{2} \left\{ \rho \left[ t\_d^2 - (N-M)^2 \right] \right\} \\ \quad + \frac{\tau cDI\_\varepsilon}{2} [\rho (M-N)^2 + (1-\rho) t\_d^2] \end{cases} \tag{58}$$

From the above results, we have these three situations: one is Δ4 < Δ5 < Δ3 < Δ∗ if *W*2 ≥ 0, another is Δ4 < Δ5 < Δ3 < Δ∗ and 0 < Δ3 < Δ∗ if *W*2 < 0 and *W*1 ≥ 0, and the other is 0 < Δ4 < Δ5 < Δ3 < Δ∗ if *W*1 < 0. Furthermore, we have the following results.

**Theorem 2.** *Suppose that M* − *N* < *td* < *M Then, each of the following results holds true:*

	- (A) *If* Δ4 < 0*,* Δ5 < 0*,* Δ3 < 0 *and* Δ∗ < 0*, then TC*(*T*∗) = *TC*4(*R*<sup>∗</sup>)*.*
	- (B) *If* Δ4 < 0*,* Δ5 < 0*,* Δ3 < 0 *and* Δ∗ ≥ 0*, then TC*(*T*∗) = *TC*4(*T*<sup>∗</sup>4)*.*
	- (C) *If* Δ4 < 0*,* Δ5 < 0*,* Δ3 ≥ 0 *and* Δ∗ ≥ 0*, then TC*(*T*∗) = *TC*3(*T*<sup>∗</sup>3)*.*
	- (D) *If* Δ4< 0*,* Δ5≥ 0*,* Δ3≥ 0 *and* Δ∗ ≥ 0*, then TC*(*T*∗) = *TC*5(*T*<sup>∗</sup>5)*.*
	- (E) *If* Δ4≥ 0*,* Δ5 ≥ 0*,* Δ3 ≥ 0 *and* Δ∗ ≥ 0*, then TC*(*T*∗) = *TC*1(*T*<sup>∗</sup>1)*.*

(A) *If* Δ4 < 0*,* Δ5 < 0 *and* Δ3 ≥ 0*, then TC*(*T*∗) = *TC*3(*T*<sup>∗</sup>3)*.*

	- (A) *If* Δ4 ≥ 0*,* Δ5 ≥ 0 *and* Δ3 ≥ 0*, then TC*(*T*∗) = *TC*1(*T*<sup>∗</sup>1)*.*

**Proof.** The proof of Theorem 2 follows immediately from the above discussions. -

4.1.3. The Decision Rule of the Optimal Replenishment Cycle Time *T*∗ When *M* − *N* < *M* < *td*

$$TC(T) = \begin{cases} TC\_1(T) & \text{if } \quad 0 < T < M - N \\ TC\_5(T) & \text{if } \quad M - N \le T < M \\ TC\_6(T) & \text{if } \quad M \le T < t\_d \\ TC\_4(T) & \text{if } \quad t\_d \le T \le R^\* \end{cases}$$

Likewise, *TC*6(*T*) is defined on *T* > 0, Equations (33), (35) and (37) imply that

$$TC'\_5(M) = TC'\_6(M) = \frac{\Delta\_6}{M^2}$$

and

$$TC\_6'(t\_d) = TC\_4'(t\_d) = \frac{\Delta \tau}{t\_d^2}$$

where

$$\begin{array}{ll} \Lambda\_{6} &= -\sigma + \frac{hD}{2} [1 + \frac{2\nu D}{x(1-\rho)^{2}}] M^{2} + \frac{\epsilon hD}{2} (a+\rho) M^{2} + \frac{\text{v}\tau D\chi}{2} \left\{ \rho \left[M^{2} - (N-M)^{2}\right] \right\} \\ &+ \frac{\text{v}\tau D\mathcal{L}}{2} [\rho (M-N)^{2} + (1-\rho)M^{2}] \end{array} \tag{59}$$

and

$$\begin{array}{rcl} \Delta\_{7} &=& -o + \frac{hD}{2} [1 + \frac{2pD}{x(1-p)^{2}}] t\_{d}^{2} + \frac{cI\_{k}D}{2} (a+\beta) t\_{d}^{2} + \frac{\tau cD I\_{k}}{2} \left\{ \rho \left[ t\_{d}^{2} - (N-M)^{2} \right] \\ &+ (1-\rho) \left( t\_{d}^{2} - M^{2} \right) \right\} + \frac{\tau vD I\_{k}}{2} [\rho (M-N)^{2} + (1-\rho)M^{2}] \end{array} \tag{60}$$

In addition, there are three situations to occur here: one is Δ4 < Δ6 < Δ7 < Δ∗ when *W*2 ≥ 0, another is Δ4 < Δ6 and 0 < Δ6 < Δ7 < Δ∗ when *W*2 < 0 and *W*1 ≥ 0, and the other is 0 < Δ4 < Δ6 < Δ7 < Δ∗ when *W*1 < 0. Furthermore, we have the following results.

**Theorem 3.** *Suppose that M* − *N* < *M* < *td. Then, each of the the following results holds true:* (I) *If W*2 ≥ 0*, then*

(A) *If* Δ4 < 0*,* Δ6 < 0*,* Δ7 < 0 *and* Δ∗ < 0*, then TC*(*T*∗) = *TC*4(*R*<sup>∗</sup>)*.*

	- (A) *If* Δ4 < 0*,* Δ6 ≥ 0*,* Δ7 ≥ 0 *and* Δ∗ ≥ 0*, then TC*(*T*∗) = *TC*5(*T*<sup>∗</sup> 5
	- (B) *If* Δ4 ≥ 0*,* Δ6 ≥ 0*,* Δ7 ≥ 0 *and* Δ∗ ≥ 0*, then TC*(*T*∗) = *TC*1(*T*<sup>∗</sup> 1 )*.*

 )*.*

(III) *If W*1 < 0*, then* Δ4 > 0 *and TC*(*T*∗) = *TC*1(*T*<sup>∗</sup> 1)*.*

**Proof.** The proof of Theorem 3 follows immediately from the above discussions. -


$$TC(T) = \begin{cases} \begin{array}{cc} TC\tau(T) & if \quad 0 < T < t\_d\\ \begin{array}{cc} TC\_8(T) & if \quad t\_d \le T < M\\ TC\_9(T) & if \quad M \le T \le R^\* \end{array} \end{cases} $$

All *T* <sup>C</sup>*v*(*T*) (*v* = 7, 8, 9) are defined on *T* > 0. From Equations (39), (41), and (43), we have

$$TC\_7'(t\_d) = TC\_8'(t\_d) = \frac{\Delta\_8}{t\_d^2}$$

$$TC\_8'(M) = TC\_9'(M) = \frac{\Delta\_9}{M^2}$$

$$TC\_9'(R^\*) = \frac{\Delta^{\*\*}}{R^{\*2}}$$

and

$$\text{where}$$

$$\Delta \mathbf{s} = -o + \frac{hD}{2} [1 + \frac{2pD}{x(1-p)^2}] t\_d^2 + \frac{cI\_k D}{2} (a + \beta) t\_d^2 + \frac{\pi cDI\_k}{2} \rho t\_d^2 + \frac{\pi vDI\_\varepsilon}{2} (1-\rho) t\_d^2 \tag{61}$$

$$\begin{array}{rcl} \Delta \rho &=& -o + h \cdot G(M) + [\frac{(2-p)c+s}{1-p}][D(1+m-t\_d)\frac{M}{1+m-M} - \text{Dt}\_d \\ & - D(1+m-t\_d) \cdot \ln(\frac{1+m-t\_d}{1+m-M})] + \frac{cI\_kD}{2}(a+\beta)M^2 + \frac{\text{tr}Dl\_k}{2}\rho M^2 \\ & + \frac{\text{tr}Dl\_r}{2}(1-\rho)M^2 \end{array} \tag{62}$$

and

$$\begin{array}{lcl}\Delta^{\*\*} &= -o + h \cdot G(R^\*) + [\frac{(2-p)c + s}{1-p}][D(1+m-t\_d)\frac{R^\*}{1+m-K^\*} - Dt\_d \\ &- D(1+m-t\_d) \cdot \ln(\frac{1+m-t\_d}{1+m-R^\*})] + \frac{cI\_bD}{2}(a+\beta)R^{\*2} + \frac{rcDl\_k}{2}[R^{\*2} - (1-\rho)M^2] \\ &+ \frac{rcDl\_k}{2}(1-\rho)M^2 \end{array} \tag{63}$$

Additionally, if *W*3 ≥ 0, then Δ8 < Δ9 < Δ∗∗. Otherwise, if *W*3 < 0, we have Δ8 < Δ9 and 0 < Δ9 < Δ∗∗. Furthermore, we have the following Theorem.

**Theorem 4.** *Suppose that td* < *M Then, each of the following results holds true:*

	- (A) *If* Δ8 < 0*,* Δ9 < 0 *and* Δ∗∗ < 0*, then TC*(*T*∗) = *TC*9(*R*<sup>∗</sup>)*.*
	- (B) *If* Δ8 < 0*,* Δ9 < 0 *and* Δ∗∗ ≥ 0*, then TC*(*T*∗) = *TC*9(*T*<sup>∗</sup> 9 )*.*
	- (C) *If* Δ8 < 0*,* Δ9 ≥ 0 *and* Δ∗∗ ≥ 0*, then TC*(*T*∗) = *TC*8(*T*<sup>∗</sup> 8 )*.*
	- (D) *If* Δ8 ≥ 0*,* Δ9 ≥ 0 *and* Δ∗∗ ≥ 0*, then TC*(*T*∗) = *TC*7(*T*<sup>∗</sup> 7 )*.*
	- (A) *If* Δ8 < 0*,* Δ9 ≥ 0 *and* Δ∗∗ ≥ 0*, then TC*(*T*∗) = *TC*8(*T*<sup>∗</sup> 8)*.*
	- (B) *If* Δ8 ≥ 0*,* Δ9 ≥ 0 *and* Δ∗∗ ≥ 0*, then TC*(*T*∗) = *TC*7(*T*<sup>∗</sup> 7)*.*

**Proof.** The proof of Theorem 4 follows immediately from the above discussions. -

4.2.2. Decision Rule of the Optimal Replenishment Cycle Time *T*∗ When *td* ≥ *M* From Equations (25), we have

$$TC(T) = \begin{cases} \begin{array}{cc} TC\_7(T) & \text{if } \quad 0 < T < M\\ TC\_{10}(T) & \text{if } \quad M \le T < t\_d\\ TC\_9(T) & \text{if } \quad t\_d \le T \le R^\* \end{array} \end{cases}$$

Herein, *TC*10(*T*) is defined on *T* > 0. From Equations (39), (43), and (45), we have

$$TC'\_{\mathcal{T}}(M) = TC\_{10}(M) = \frac{\Delta\_{10}}{M^2}$$

and

$$TC'\_{10}(t\_d) = TC'\_{9}(t\_d) = \frac{\Delta\_{11}}{t\_d^2}$$

where

$$\Delta\_{10} = -\sigma + \frac{hD}{2} [1 + \frac{2pD}{\mathfrak{x}(1-p)^2}] M^2 + \frac{cI\_k D}{2} (\alpha + \beta) M^2 + \frac{\pi cDI\_k}{2} \rho M^2 + \frac{\pi vDI\_\varepsilon}{2} (1-\rho) M^2 \tag{64}$$

and

$$\begin{cases} \Lambda\_{11} &= -o + \frac{hD}{2} [1 + \frac{2pD}{x(1-p)^2}] t\_d^2 + \frac{cI\_k D}{2} (a+\beta) t\_d^2 + \frac{\tau cDI\_k}{2} [t\_d^2 - (1-\rho)M^2] \\ &+ \frac{\tau vDI\_\ell}{2} (1-\rho)M^2 \end{cases} \tag{65}$$

Likewise, if *W*3 ≥ 0, then Δ10 < Δ11 < Δ∗∗. Otherwise, if *W*3 < 0, we have 0 < Δ10 < Δ11 < Δ∗∗. From above arguments, we have the following theorem.

**Theorem 5.** *Suppose that td* ≥ *M Then, each of the following results holds true:*

(I) *If W*3 ≥ 0*, then* (A) *If* Δ10 < 0*,* Δ11 < 0 *and* Δ∗∗ < 0*, then TC*(*T*∗) = *TC*9(*R*<sup>∗</sup>)*.* (B) *If* Δ10 < 0*,* Δ11 < 0 *and* Δ∗∗ ≥ 0*, then TC*(*T*∗) = *TC*9(*T*<sup>∗</sup>9 )*.* (C) *If* Δ10 < 0*,* Δ11 ≥ 0 *and* Δ∗∗ ≥ 0*, then TC*(*T*∗) = *TC*10(*T*<sup>∗</sup>10)*.* (D) *If* Δ10 ≥ 0*,* Δ11 ≥ 0 *and* Δ∗∗ ≥ 0*, then TC*(*T*∗) = *TC*7(*T*<sup>∗</sup>7 )*.* (II) *If W*3< 0*, then* Δ10≥ 0 *and TC*(*T*∗) = *TC*7(*T*<sup>∗</sup>7)*.*

**Proof:** The proof follows immediately from the above discussions. -

### **5. Illustrative Numerical Examples**

In this section, we will provide numerical examples to illustrate the theoretical results. We assume that the maximum lifetime of the deteriorating items is 2 years (*m* = 2). The computed results are shown in Tables 1–9.


**Table 1.** The optimal replenishment policy used Theorem 1(I).

**Table 2.** The optimal replenishment policy used Theorem 1(II).


**Table 3.** The optimal replenishment policy used Theorem 2(I).


**Table 4.** The optimal replenishment policy used Theorem 2(II) and (III).



**Table 5.** The optimal replenishment policy used Theorem 3(I).

**Table 6.** The optimal replenishment policy used Theorem 3(II) and (III).


**Table 7.** The optimal replenishment policy used Theorem 4(I).


**Table 8.** The optimal replenishment policy used Theorem 4(II).


**Table 9.** The optimal replenishment policy used Theorem 5(I) and 5(II).


**Example 1.** *h* = 1.5, *c* = 3, *ν* = 4, *D* = 100, *x* = 300, *p* = 0.01, *s* = 10, *td* = 0.008, *N* = 0.01 year, *M* = 0.02 year, *L* = 0.3, *Ik*= \$0.15/\$/year, *Ie* = \$0.12/\$/year, *τ* = 0.3, *α* = 0.3, *β* = 0.4 and *ρ* = 0.3, Theorem 1(I) (if *W*1 ≥ 0) is applied to obtain optimal solution The computed result is shown in Table 1.

**Example 2.** *o* = 0.0002, *c* = 3, *ν* = 4, *s* = 10, *N* = 0.01 year, *M* = 0.02 year, *L* = 0.3, *Ie* = \$ 0.12/\$/year, *τ* = 0.3, *α* = 0.3, *β* = 0.4 and *ρ* = 0.3, Theorem 1(II) (if *W*1 < 0) is applied to obtain optimal solution. The computed result is shown in Table 2.

**Example 3.** *c* = 3, *ν* = 4, *D* = 95, *x* = 300, *p* = 0.01, *s* = 10, *td* = 0.012, *N* = 0.01 year, *M* = 0.02 year, *L* = 0.3, *Ik* = \$0.15/\$/year, *Ie* = \$ 0.12/\$/year, *τ* = 0.3, *α* = 0.3, *β* = 0.4 and *ρ*= 0.3, Theorem 2(I) (if *W*2 ≥ 0) is applied to obtain optimal solution. The computed result is shown in Table 3.

**Example 4.** *p* = 0.01, *s* = 10, *L* = 0.3, *τ* = 0.3, *α* = 0.3, *β* = 0.4 and *ρ* = 0.3, Theorem 2(II) (if *W*2 < 0 and *W*1 ≥ 0) and Theorem 2(III) (if *W*1 < 0) are applied to obtain optimal solution. The computed result is shown in Table 4.

**Example 5.** *c* = 3, *ν* = 4, *D* = 300, *x* = 1000, *p* = 0.001, *s* = 10, *td* = 0.025, *N* = 0.01 year, *M* = 0.02 year, *L* = 0.3, *Ik* = \$0.13/\$/year, *Ie* = \$ 0.12/\$/year, *τ* = 0.3, *α* = 0.3, *β* = 0.4 and *ρ* = 0.3, Theorem 3(I) (if *W*2 ≥ 0) is applied to obtain optimal solution. The computed result is shown in Table 5.

**Example 6.** *c* = 3, *ν* = 4, *x* = 1000, *p* = 0.001, *s* = 10, *td* = 0.025, *M* = 0.02 year, *L* = 0.3, *Ik* = \$0.13/\$/year, *Ie* = \$ 0.12/\$/year, *τ* = 0.3, *α* = 0.3, *β* = 0.4 and *ρ* = 0.3, Theorem 3(II) (if *W*2 < 0 and *W*1 ≥ 0) and Theorem 3(III) (if *W*1 < 0) are applied to obtain optimal solution. The computed result is shown in Table 6.

**Example 7.** *c* = 3, *ν* = 4, *x* = 1000, *p* = 0.001, *s* = 10, *td* = 0.01, *N* = 0.03 year, *M* = 0.02 year, *L* = 0.3, *Ik* = \$0.13/\$/year, *Ie* = \$ 0.12/\$/year, *τ* = 0.3, *α* = 0.3, *β* = 0.4 and *ρ* = 0.3, Theorem 4(I) (if *W*3 ≥ 0) is applied to obtain optimal solution. The computed result is shown in Table 7.

**Example 8.** *o* = 0.01, *c* = 3, *ν* = 4, *D* = 250, *x* = 1000, *p* = 0.001, *s* = 10, *td* = 0.01, *N* = 0.04 year, *M* = 0.03 year, *L* = 0.3, *Ik* = \$0.13/\$/year, *Ie* = \$ 0.12/\$/year, *τ* = 0.3, *α* = 0.3, *β* = 0.4 and *ρ* = 0.3, Theorem 4(II) (if *W*3 < 0) is applied to obtain optimal solution. The computed result is shown in Table 8.

**Example 9.** *h* = 1, *c* = 3, *ν* = 4, *D* = 250, *x* = 1000, *p* = 0.001, *s* = 10, = 0.03, *N* = 0.04 year, *M* = 0.02 year, *L* = 0.3, *Ik* = \$0.13/\$/year, *Ie* = \$ 0.12/\$/year, *τ* = 0.3, *α* = 0.3, *β* = 0.4 and *ρ* = 0.3, Theorem 5(I) (if *W*3 ≥ 0) and Theorem 5(II) (if *W*3 < 0) are applied to obtain optimal solution. The computed result is shown in Table 9.

Additionally, in Tables 1–9, *R*∗ and Δ∗ is defined as Equations (11) and (56), respectively. *T*∗ is the optimal cycle time so that *TC*(*T*∗) is the minimum.

### **6. Conclusions and Potential Directions for Further Research**

In our present investigation, we have established a sustainable inventory system in which the retailer sells the non-instantaneous deteriorating item that is fully deteriorated close to its expiry date and has imperfect quality such as those in seasonal products, food products, electronic components, and others. In order to manage the quality of the items, an inspection will occur during the state in which there is no deterioration. On the other hand, the supplier demands the retailer a distinct paymen<sup>t</sup> scheme, such as partial prepaymen<sup>t</sup> or cash and trade credit; in turn, the retailer grants customers partial cash and trade credit.

We have observed that some of the optimization methods lack mathematical rigor, and some of them are based on intuitive arguments, which result in the solution procedures being questionable from the logical viewpoints of mathematical analysis, such as those in the earlier works by Chang et al. (2004), Ouyang et al. (2006), and Cheng and Wang (2009). They ignored explorations of interrelations of functional behaviors of the total cost function to locate the optimal solution, so those shortcomings will naturally influence the implementation of their considered inventory model. Essentially, in order to explore the functional behaviors (such as continuity, monotonicity (increasing and decreasing) properties, differentiability, etc.) of the object functions (that is, the total cost functions), one can and should apply the mathematically accurate and reliable solution procedures. Moreover, if the object function (that is, the total cost functions) are convex, it is easier to find the optimal solution by using the convexity property. Consequently, the discussion

of the convexity of the total annual cost function is one of the main research topics of this article.

The object function (that is, the total cost function) of this article is a piecewise continuous function; it is increasing or decreasing in its own domain and we discuss the continuity of the object function, especially at its extreme point(s). Furthermore, the main purpose of this article has been to provide accurate and reliable mathematical analytic solution procedures for different scenarios by studying the convexity of the total annual cost function and the functional behaviors of the object function. For the proposed models, the convexity of the object functions has been proved and the closed-form optimal solution has been derived. Numerical examples, which illustrate the behavior of proposed models and the applied solution method, have been considered; a retailer, using the model obtained in this article, can effectively determine the optimal replenishment cycle.

Finally, the limitation of this article is that we have concentrated upon the inventory system without shortage, which can affect the supply chain from the producer to the retailer. Furthermore, this model has the potential to be extended to incorporate inflation and quantity discount effects, different demand forms such as credit-linked promotiondependent demand, and other issues under the system with shortages. Additionally, this article has considered the deterministic situation, so considering the stochastic situation, such as stochastic demand, can be another future research direction on the subject of this article.

**Author Contributions:** Conceptualization, J.-J.L., S.-D.L. and K.-J.C.; methodology, J.-J.L., S.-D.L. and H.M.S.; software, J.-J.L. and K.-N.H.; validation, J.-J.L., S.-D.L. and H.M.S.; formal analysis, J.-J.L., S.-D.L. and H.M.S.; investigation, J.-J.L., K.-N.H. and S.-D.L.; resources, J.-J.L., K.-J.C. and S.-D.L.; data curation, K.-N.H., S.-D.L. and S.-F.L.; writing—original draft preparation, J.-J.L., K.-J.C., S.-D.L. and S.-F.L.; writing—review and editing, J.-J.L. and S.-D.L.; visualization, J.-J.L. and S.-F.L.; supervision, J.-J.L., S.-D.L. and H.M.S.; project administration, J.-J.L., K.-J.C., S.-D.L. and H.M.S.; funding acquisition, Not Applicable. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research received no external funding.

**Institutional Review Board Statement:** Not applicable.

**Informed Consent Statement:** Not applicable.

**Data Availability Statement:** Not applicable.

**Conflicts of Interest:** The authors declare no conflict of interest.
