Innovation of Economic Order Quantity (EOQ) Model for Deteriorating Items with Time-Linked Quadratic Demand Under Non-decreasing Shortages

Abstract

Demand for several variety of commodity depends on its nature like; price, corrosion and time. In most inventory models demand is considered constant or time-dependent. In this study, we set up a deterministic EOQ for weakening stuffs when demand is quadratic time linked. Shortages are permitted and totally backlogged. Mathematical representation is derived and then some constructive outcomes is framed to demonstrate most favorable answers. Numerical illustrations are incorporated to demonstrate the optimal clarification. Sensitivity analysis with effects of system constraints is provided to scrutinize the nature of representation. Mathematica 7.1 software is employed to get arithmetical results.

Introduction

In past few years, several investigations is available on inventory problems for failing goods. Majority of substances in nature suffer decay or spoilage over time. Green vegetables, fruits, radioactive substances, blood, alcohol etc. bear from exhausted by spoilage during storage. The foremost difficulty of any production association is how to manage and uphold the inventories of failing products. Demand for a particular item made by producer depends on price, quality and season etc.

Ghare and Schrader [1] were originally planned an EOQ model with exponentially decaying foodstuffs. Covert and Philip [2] analyzed a model with two-parameters Weibull distribution weakening. Philip [3] widespread and extended the model of Covert and Philip taking into description with three parameters Weibull distribution deterioration. Whitin [4] recognized the corrosion of fashion merchandise at the last part of agreed cargo space time. Shah and Jaiswal [5] addressed an arrange level EOQ model in favor of weakening objects. Aggarwal [6] and Dave and Patel [7] explored inventory models for item fading at a steady rate. Shah et al. [8] measured an inventory structure by means of failing commodity where demand rate is associated to advertising and selling charge. Goyal and Giri [9] addressed a comprehensive evaluation of deteriorating inventory. Some research effort linked to deteriorating objects has been done by Goyal [10], Raafat et al. [11], Wee [12], Mishra [13], Deb and Chaudhuri [14], Chung and Ting [15], Fujiwara [16], Chakraborti and Chaudhuri [17], Jalan and Chaudhuri [18], Wu [19], Dye et al. [20], Panda et al. [21], Widyadana et al. [22], Tripathi and Kumar [23], Tripathi and Pandey [24], Singh et al. [25] and Tripathi [26].

In traditional economic order quantity models demand rate is unspecified to be stable. However, demand rate for most of the products may not be always constant, it may be stock, time, selling price and advertising sensitive demand rate. Roy [27] published an EOQ model where decline rate is time comparative and demand associated by means of selling price. Tripathi [28] presented an inventory model for item whose demand is a decreasing function of selling price. Yang [29] established an EOQ model for stock-dependent demand and holding cost. Jaggi et al. [30] proposed a model under two stages of traffic credit strategy by guessing demand is linked with credit epoch presented by vendor to buyer via DCF approach.

Teng et al. [31] introduced an EOQ model under trade credit financing through escalating demand under tolerable delay in expenses. Hsieh et al. [32] widespread the demand for an EOQ model by means of fluctuating trade credits to an rising function of time. Skouri et al. [33] pointed out EOQ model for incline type demand in the midst of Weibull corrosion rate. Khanra et al. [34] established a model for a failing objects containing time-linked demand in which holdup in expenses are tolerable. Several researchers worked in this direction like Donaldson [35], Silver [36], Ritchie [37], Mitra et al. [38], Min et al. [39], Patra [40], Sahoo et al. [41] and others.

In real life, demands for medicines, foodstuffs, vegetables etc. are generally lost throughout the storage phase, it means that items produced may have fraction of defective. Ghiami et al. [42] explored a two-level supply succession model in favor of fading inventory where vendor’s storehouse has inadequate capability. Jaggi et al. [43] offered an EOQ model in support of defective quality item to settle on best possible ordering strategies of a seller with allowed delay in expenditure for acceptable shortages. Ouyang and Chang [44] established a mathematical model to learn the most advantageous fabrication procedure for an economic production quantity (EPQ) scheme in the midst of unsatisfactory construction process under permitted delay in payment and complete backlogging. Wee et al. [45] projected an EPQ model for unfinished backorders. Bhunia et al. [46] presented an inventory model for single weakening commodity through two part warehouses containing dissimilar conserving services under shortages. Sana [47] recognized best possible selling price by means of time changeable weakening and fractional backlogging.

In this paper an EOQ model is presented for an item with quadratic time sensitive demand rate (i.e. D(t)= a + bt + ct²) with deterioration. If c = 0 then demand rate is linear, and if, b = c = 0, demand rate is invariable. Shortages are acceptable and linearly time dependent. The foremost purpose of the model is to settle on minimum total average cost.

The remaining component of manuscript is prepared as followed. In the next allotment 2, we afford assumption and notations used all over the study. In “Mathematical Formulation” section, mathematical formulation is obtained for deciding minimum total cost/unit time. The optimal solutions are provided in subdivision 4. In “Some Results” section, some significant results are provided based on optimal solution. Numerical illustrations are conversed in “Numerical Examples” section. The sensitivity investigation is provided in “Sensitivity Analysis and Managerial Implication” section. We establish conclusion and future research directives in “Conclusion and Future Investigation” section.

Assumptions and Notations

The following assumptions are made to build up the model:

• Replenishment rate in instantaneous.

• Demand rate D(t) is a quadratic time dependent during [0, t₁] and time dependent between [t₁, T].

• Shortages are tolerable and fully backlogged. Backlog rate is time induced.

• Time horizon is endless.

• Deterioration rate θ is invariable.

Notations are as follows:

q(t):

inventory level at instant ‘t’

A :

cost of placing order

h :

holding cost/unit time

θ :

worsening rate 0≤ θ <1

p :

unit cost of an item

s :

shortage cost/unit time

t ₁ :

time for inventory depletion due to corrosion and demand, at time t₁, shortages occurring just after t₁

T :

length of cycle time

SC, HC and DC:

shortage, holding and deterioration costs

Z(T, t₁):

total cost/unit time

Z(T*, t₁*):

optimal Z(T, t₁)

t₁* and T*:

optimal t₁ and T

Q ₁ :

positive order quantity

Q ₂ :

negative order quantity

Q :

order quantity

Q*:

optimal Q

D₁(t)= a + bt + ct²:

demand rate in [0, t₁] and a > b > c >0, b ≠ 0, c ≠ 0

D₂(t)= a + bt:

backlogged rate in [t₁, T] and a > b >0, b ≠ 0.

Mathematical Formulation

Inventory level q(t) at time ‘t’ during the time interval [0, T] is given by the following differential equations:

$$ \frac{dq(t)}{dt} + \theta q(t) = - \,D_{1} (t), $$

(1)

$$ \frac{dq(t)}{dt} = - \,D_{2} (t), $$

(2)

with the condition

$$ q\left( {t_{1} } \right) = 0. $$

(3)

Solution of (1) and (2) with (3) are specified in “Appendix A” (Fig. 1)

Fig. 1

Figure between times versus inventory level

Thus, maximum inventory level and shortage quantity during [0, T] are:

$$ Q_{1} = \frac{1}{\theta }\left[ {\left\{ {\left( {a + bt_{1} + ct_{1}^{2} } \right) - \left( \frac{{b + 2ct_{1} }}{\theta } \right) + \frac{2c}{{\theta^{2} }}} \right\}e^{{\theta t_{1} }} - \left( {a - \frac{b}{\theta } + \frac{2c}{{\theta^{2} }}} \right)} \right] $$

(4)

and

$$ Q_{2} = \left( {T - t_{1} } \right)\left\{ {a + \frac{b}{2}\left( {T + t_{1} } \right)} \right\}\quad {\text{respectively}} $$

(5)

Order quantity

$$ Q \, = \, Q_{1} + \, Q_{2} = \frac{1}{\theta }\left[ {\left\{ {\left( {a + bt_{1} + ct_{1}^{2} } \right) - \left( \frac{{b + 2ct_{1} }}{\theta } \right) + \frac{2c}{{\theta^{2} }}} \right\}e^{{\theta t_{1} }} - \left( {a - \frac{b}{\theta } + \frac{2c}{{\theta^{2} }}} \right)} \right] + \left( {T - t_{1} } \right)\left\{ {a + \frac{b}{2}\left( {T + t_{1} } \right)} \right\} $$

(6)

$$ {\text{Cost of placing order}} = A $$

(7)

Number of deteriorated item during [0, t₁] is:

$$ \begin{aligned} & = Q_{1} - \int\limits_{0}^{{t_{1} }} {(a + bt + ct^{2} )dt} \\ & = \frac{1}{\theta }\left[ {\left\{ {(a + bt_{1} + ct_{1}^{2} ) - \left( \frac{{b + 2ct_{1} }}{\theta } \right) + \frac{2c}{{\theta^{2} }}} \right\}e^{{\theta t_{1} }} - \left( {a - \frac{b}{\theta } + \frac{2c}{{\theta^{2} }}} \right) - \theta t_{1} \left( {a + \frac{{bt_{1} }}{2} + \frac{{ct_{1}^{2} }}{3}} \right)} \right] \\ \end{aligned} $$

(8)

Cost of deterioration during [0, t₁] is

$$ CD = \frac{p}{\theta }\left[ {\left\{ {(a + bt_{1} + ct_{1}^{2} ) - \left( \frac{{b + 2ct_{1} }}{\theta } \right) + \frac{2c}{{\theta^{2} }}} \right\}e^{{\theta t_{1} }} - \left( {a - \frac{b}{\theta } + \frac{2c}{{\theta^{2} }}} \right) - \theta t_{1} \left( {a + \frac{{bt_{1} }}{2} + \frac{{ct_{1}^{2} }}{3}} \right)} \right] $$

(9)

Total cost/unit time is

$$ \begin{aligned} & Z(T,t_{1} ) = \frac{A + HC + CD + SC}{T} Z = \frac{A}{T} + \left( \frac{h + p\theta }{{\theta^{2} T}} \right)\\ &\quad \left[ {\left\{ \left( {a + bt_{1} + ct_{1}^{2} } \right) - \left( \frac{{b + 2ct_{1} }}{\theta } \right) + \frac{2c}{{\theta^{2} }} \right\}e^{{\theta t_{1} }} - \theta t_{1} \left( {a + \frac{{bt_{1} }}{2} + \frac{{ct_{1}^{2} }}{3}} \right) - \left( {a - \frac{b}{\theta } + \frac{2c}{{\theta^{2} }}} \right)} \right] \\ & \quad + \frac{{s(T - t_{1} )^{2} }}{6T}\left\{ {3a + b(T + 2t_{1} )} \right\}. \\ \end{aligned} $$

(10)

It is complicated to grip Eq. (10) in the present form. Truncated Taylor’s series estimate is used for exponential term to obtain closed form solution. From Eq. (10), we have

$$ Z = \frac{A}{T} + \left( \frac{h + p\theta }{\theta T} \right)t_{1}^{2} \left\{ \frac{{\theta \left( {a + bt_{1} + ct_{1}^{2} } \right)}}{2} - \frac{{ct_{1} }}{3} \right\} + \frac{{s(T - t_{1} )^{2} }}{6T}\left\{ {3a + b(T + 2t_{1} )} \right\}. $$

(11)

Determination of Optimal Solution

Differentiating (11) partially w.r.t. T and t₁ twice, we obtain

$$ \frac{\partial Z}{\partial T} = - \frac{A}{{T^{2} }} - \left( \frac{h + p\theta }{{\theta T^{2} }} \right)t_{1}^{2} \left\{ \frac{{\theta \left( {a + bt_{1} + ct_{1}^{2} } \right)}}{2} - \frac{{ct_{1} }}{3} \right\} + \frac{{s(T - t_{1} )}}{{6T^{2} }}\left\{ {3a(T + t_{1} ) + 2b\left( {T^{2} + Tt_{1} + t_{1}^{2} } \right)} \right\} $$

(12)

$$ \frac{\partial Z}{{\partial t_{1} }} = \left( \frac{h + p\theta }{\theta T} \right)t_{1} \left\{ \frac{\theta }{2}\left( {2a + 3bt_{1} + 4ct_{1}^{2} } \right) - ct_{1} \right\} - \frac{{s\left( {T - t_{1} } \right)}}{T}\left( {a + bt_{1} } \right) $$

(13)

$$ \frac{{\partial^{2} Z}}{{\partial T^{2} }} = \frac{2A}{{T^{3} }} + 2\left( \frac{h + p\theta }{{\theta T^{3} }} \right)t_{1}^{2} \left\{ \frac{\theta }{2}\left( {a + bt_{1} + ct_{1}^{2} } \right) - \frac{{ct_{1} }}{3} \right\} + \frac{s}{{3T^{3} }}\left( {bT^{3} + 2bt_{1}^{3} + 3at_{1}^{2} } \right) $$

(14)

$$ \frac{{\partial^{2} Z}}{{\partial t_{1}^{2} }} = \left( \frac{h + p\theta }{\theta T} \right)\left\{ {\theta \left( {a + 3bt_{1} + 6ct_{1}^{2} } \right) - 2ct_{1} } \right\} + \frac{s}{T}\left( {a + 2bt_{1} - bT} \right) $$

(15)

$$ \frac{{\partial^{2} Z}}{{\partial T\partial t_{1} }} = - \frac{{t_{1} }}{{2T^{2} }}\left[ {\left( {\frac{h + p\theta }{\theta }} \right)\left\{ {\theta \left( {2a + 3bt_{1} + 4ct_{1}^{2} } \right) - 2ct_{1} } \right\} + 2s\left( {a + bt_{1} } \right)} \right] < 0. $$

(16)

Our aim is to acquire least total cost/unit time. On putting $\frac{\partial Z}{\partial T} = 0\,{\text{and}}\,\frac{\partial Z}{{\partial t_{1} }} = 0;$ solving simultaneously, we get at t₁* and T*, which minimizes Z, provided $\frac{{\partial^{2} Z}}{{\partial T^{2} }} > 0, \, \frac{{\partial^{2} Z}}{{\partial t_{1}^{2} }} > 0;$ and $\left( {\frac{{\partial^{2} Z}}{{\partial T^{2} }}} \right)\left( {\frac{{\partial^{2} Z}}{{\partial t_{1}^{2} }}} \right) - \left( {\frac{{\partial^{2} Z}}{{\partial T\partial t_{1} }}} \right)^{2} > 0$ (See “Appendix C” section), at t₁* and T*. Putting $\frac{\partial Z}{\partial T} = 0 \,$ and $\frac{\partial Z}{{\partial t_{1} }} = 0$, we get

$$ 6A\theta + \left( {h + p\theta } \right)t_{1}^{2} \left\{ {3\theta \left( {a + bt_{1} + ct_{1}^{2} } \right) - 2ct_{1} } \right\} - s\theta (T - t_{1} )\left\{ {3a(T + t_{1} ) + 2b\left( {T^{2} + Tt_{1} + t_{1}^{2} } \right)} \right\} = 0 $$

(17)

and

$$ \left( {h + p\theta } \right)t_{1} \left\{ {\theta \left( {2a + 3bt_{1} + 4ct_{1}^{2} } \right) - 2ct_{1} } \right\} - 2s\theta \left( {T - t_{1} } \right)\left( {a + bt_{1} } \right) = 0. $$

(18)

Solving Eqs. (17) and (18) concurrently, we get T* and t₁*.

From the above conversation, we find following fallouts.

Some Results

Result 1

T* is growing function of t₁*.

Proof

Differentiating (17) and (18) w.r.t. t₁, we get

$$ \frac{dT}{{dt_{1} }} = \frac{{t_{1} \left[ {(h + p\theta )\left\{ {\theta (2a + 3bt_{1} + 4ct_{1}^{2} ) - 2ct_{1} } \right\} + 2s\theta (a + bt_{1} )} \right]}}{2s\theta T(a + bT)} > 0. $$

(19)

and

$$ \frac{dT}{{dt_{1} }} = \frac{{(h + p\theta )\left\{ {\theta \left( {a + 3bt_{1} + 6ct_{1}^{2} } \right) - 2ct_{1} } \right\} + s\theta (a + 2bt_{1} )}}{{s\theta (a + bt_{1} )}} > 0. $$

(20)

From (19) and (20) we see that $\frac{dT}{{dt_{1} }} > 0$. Therefore, optimal T is rising function of t₁.□

Result 2

Z*(T*, t₁*) is convex function of T* and t₁*.

Proof

From Eqs. (14)–(16), it obvious that $\, \left( {\frac{{\partial^{2} Z}}{{\partial T^{2} }}} \right)\left( {\frac{{\partial^{2} Z}}{{\partial t_{1}^{2} }}} \right) - \left( {\frac{{\partial^{2} Z}}{{\partial T\partial t_{1} }}} \right)^{2} > 0 \,$, and $\frac{{\partial^{2} Z}}{{\partial T^{2} }} > 0 \, ,\frac{{\partial^{2} Z}}{{\partial t_{1}^{2} }} > 0$. Thus, optimal total cost/unit time Z*(T*, t₁*) is convex function of T* and t₁*.□

Numerical Examples

Numerical examples are making available to exhibit the outcomes of model developed in this study with the following data:

Numerical Example 1

Consider the value of a = 1500 units/year, b = 150 units/year, c = 15 units/year, A = $200/order, h = $2.0/year, s = $5/year, p = $20/unit and θ = 0.01. Putting theses in Eqs. (17) and (18) and solving, we have t₁* = 0.405452 year and T* = 0.518376 year and corresponding Q* = 799.002 units and Z(T*, t₁*) = $886.005.

Numerical Example 2

Let us consider numerical value of a = 2000, b = 200, c = 20, A = $200, h = $2.0, s = $5, p = $20 t and θ = 0.01 in appropriate units. On substitution these values in (17) and (18) and solve simultaneously, we get t₁* = 0.323048 year and T*= 0.423219 year, corresponding Q* = 865.428 units and Z(T*, t₁*) = $1039.09 (Figs. 2, 3).

Fig. 2

Graph between T and Z

Fig. 3

Graph between t₁ and Z

Sensitivity Analysis and Managerial Implication

In this section we discuss, the outcome of distinction of constraints a, b, c, θ, s, A, and h, on Q* and total cost Z* (T, t₁*). Sensitivity investigation will be very supportive in pronouncement making to scrutinize effect in modify of these differences. Using the identical data as in numerical Example 1, sensitivity investigation of unlike parameters has been completed. We learn outcome of discrepancies in a single factor maintaining additional scheme parameters equal on optimal solutions.

From the above Table 1 following results are made:

Table 1 Variation of T*, Q* and Z*(t₁*, T*) with a, b, c, θ, s, A, and h

• (i) Rise of ‘a’, results diminish of t₁ and T and enlarge in Q* and Z*.

• (ii) Raise in ‘c’ leads increase in t₁*, T*, Q* and decrease in Z*).

• (iii) Augment in b, θ, p and h will lead decrease in t₁*, T*, Q* and augment in optimal total cost.

• (iv) Amplify in s and A leads enlarge in t₁*, T*, Q* and Z*(t₁*, T*).

  (Please see “Appendix B” section).

Managerial Implication

The quadratic demand is rare in nature. This type of demand is often happens during earth quake, storm etc. In case of any trade deal, if initial demand increases order quantity and total cost increases. It means that initial demand, order quantity and total cost move in the same direction. In addition shortages cost, cost of placing order, order quantity and total cost shift in the same mode.

Conclusion and Future Investigation

In this paper, a deterministic EOQ model is established for fading objects by means of quadratic time induced demand. In case of seasonal foodstuffs demand rises suddenly to utmost in the middle season and then go down to a least abruptly as season ends. Quadratic time linked demand [i.e. D(t)= a + bt + ct², a > 0, b ≠ 0, c ≠ 0] is a best demonstration of these different types of demand. If b > 0, c > 0, the demand rate is the growing function of time. This is said to be accelerated expansion in demand, i.e. $\frac{{d^{2} D}}{{dt^{2} }} = 2c > 0$.If b > 0, c < 0, demand rate is said to be slow down growth in demand. If, b < 0, c < 0, demand rate is retarded decline for every time. The functional form of demand pattern D(t)= a + bt + ct², depending on the sign of b and c. Shortages are acceptable and linearly time dependent. Due to uncertainties, stock out is inescapable for many business connections. Mathematical model has been developed to formulate most advantageous answer. Next, numerical illustrations are given to authenticate the consequences. Sensitivity study with the dissimilarity of dissimilar scheme parameters on optimal solution has been also deliberated. We have obtained numerous administrative phenomena. For instance (i) a advanced value of ‘a’ and ‘A’ caused a higher order quantity and total cost, (ii) a superior value of θ, s, p and h caused lesser order quantity and higher total cost, (iii) a elevated value of ‘c’ caused lower order quantity and total cost and (iv) advanced value of ‘b’ caused elevated values of total cost and slightly decrease in order quantity.

At hand model provides quite a lot of probable extensions. For illustration, we may extend model for stochastic demand as well as Weibull deterioration rate. Another possible extension would be to regard as holding cost to be time or price sensitive

Appendices

Appendix A

$$ \begin{aligned}q(t) &= \frac{1}{\theta }\left[ \left\{ {\left( {a + bt_{1} + ct_{1}^{2} } \right) - \left( {\frac{{b + 2ct_{1} }}{\theta }} \right) + \frac{2c}{{\theta^{2} }}} \right\}e^{{\theta (t_{1} - t)}} \right.\\ &\quad - \left.\left\{ {\left( {a + bt + ct^{2} } \right) - \left( {\frac{b + 2ct}{\theta }} \right) + \frac{2c}{{\theta^{2} }}} \right\} \right],\quad 0 \, \le \, t \, \le \, t_{1}\end{aligned} $$

(21)

$$ I(t) = - \left\{ {a\left( {t - t_{1} } \right) + \frac{b}{2}\left( {t^{2} - t_{1}^{2} } \right)} \right\},t_{1} \le \, t \, \le \, T $$

(22)

The holding cost HC during [0, T₁] is

$$ \begin{aligned} HC & = h\int\limits_{0}^{{t_{1} }} {q(t)dt} = \frac{h}{\theta }\left[ \left\{ {\left( {a + bt_{1} + ct_{1}^{2} } \right) - \left( {\frac{{b + 2ct_{1} }}{\theta }} \right) + \frac{2c}{{\theta^{2} }}} \right\}\left( {\frac{{e^{{\theta t_{1} }} - 1}}{\theta }} \right)\right.\\ &-\left. t_{1} \left\{ \left( {a + \frac{{bt_{1} }}{2} + \frac{{ct_{1}^{2} }}{3}} \right) - \left( {\frac{{b + ct_{1} }}{\theta }} \right) + \frac{2c}{{\theta^{2} }} \right\} \right] \\ \end{aligned} $$

(23)

The shortage cost during [t₁, T] is given by

$$ SC = s\int\limits_{{t_{1} }}^{T} { - q(t)dt = } \frac{{s\left( {T - t_{1} } \right)^{2} }}{6}\left\{ {3a + b\left( {T + 2t_{1} } \right)} \right\} $$

(24)

Appendix B

See Fig. 4.

Fig. 4

Figures between system parameters a, b, c, s, p, A and total cost Z

Appendix C

Let

$$ \begin{aligned} \alpha & = \frac{h + p\theta }{\theta },\quad \beta = \left\{ {\frac{\theta }{2}\left( {a + bt_{1} + ct_{1}^{2} } \right) - \frac{{ct_{1} }}{3}} \right\},\quad \gamma = s\left( {bT^{3} + 3at^{2} + 2bt_{1}^{3} } \right), \\ A_{1} & = \left\{ {\theta \left( {a + 3bt_{1} + 6ct_{1}^{2} } \right) - 2ct_{1} } \right\},\quad A_{2} = s\left( {a + 2bt_{1} - bT} \right) \\ A_{3} & = \left\{ {\frac{\theta }{2}\left( {2a + 3bt_{1} + 4ct_{1}^{2} } \right) - ct_{1} } \right\}t_{1} ,\quad {\text{and}}\quad A_{4} = st_{1} \left( {a + bt_{1} } \right). \\ \end{aligned} $$

Then

$$ \begin{aligned} & \left( \frac{{\partial^{2} Z}}{{\partial T^{2} }} \right)\left( \frac{{\partial^{2} Z}}{{\partial t_{1}^{2} }} \right) - \left( \frac{{\partial^{2} Z}}{{\partial T\partial t_{1} }} \right)^{2} = \frac{1}{{T^{4} }}\left[ {\left( {2A + 2\alpha \beta t_{1}^{2} A_{1} + \frac{\gamma }{3}} \right)\left( {\alpha A_{1} + A_{1} } \right) - \left( {\alpha A_{3} + A_{4} } \right)^{2} } \right] \\ & \quad = \frac{1}{{T^{4} }}\left[ {2AA_{2} + 2\alpha \left( {AA_{1} + \beta t_{1}^{2} A_{2} - A_{3} A_{4} } \right) + \frac{{\alpha \gamma A_{1} }}{3} + \left( \frac{\gamma }{3}A_{2} - A_{4}^{2} \right) + \alpha^{2} \left( {2\beta t_{1}^{2} A_{1} - A_{3}^{2} } \right)} \right]. \end{aligned} $$

Consider

$$ \left( {\frac{\gamma }{3}A_{2} - A_{4}^{2} } \right) = \frac{{bs^{2} }}{3}\left\{ {T^{3} \left( {a - bT} \right) + t_{1}^{2} \left( {2at_{1} + bt_{1}^{2} - 3aT} \right) + 2bTt_{1} \left( {T^{2} - 2t_{1}^{2} } \right)} \right\} > 0. $$

Similarly, we can show that, $2\alpha \left( {AA_{1} + \beta t_{1}^{2} A_{2} - A_{3} A_{4} } \right)\,{\text{and}}\,\alpha^{2} \left( {2\beta t_{1}^{2} A_{1} - A_{3}^{2} } \right)$ are positive.

Hence

$$ \left( {\frac{{\partial^{2} Z}}{{\partial T^{2} }}} \right)\left( {\frac{{\partial^{2} Z}}{{\partial t_{1}^{2} }}} \right) - \left( {\frac{{\partial^{2} Z}}{{\partial T\partial t_{1} }}} \right)^{2} > 0. $$