*Model 4*

If at the beginning of the period [0, *T*] uncontrollable parameters of the logistic process such as T = 400 days, r = 0.001, and µ = 25 units/day are known and the cost of delivery and the price increase during the period [0, *T*] with ρ*<sup>c</sup>* = 0.003, ρ*<sup>p</sup>* = 0.00075, then the growth pattern will be *<sup>c</sup>s*(*t*) = <sup>400</sup> <sup>∗</sup> 1.003*<sup>t</sup>* , *<sup>p</sup>*(*t*) = <sup>20</sup> <sup>∗</sup> 1.00075*<sup>t</sup>* , *t* ∈ [0, *T*] (α*<sup>c</sup>* = 3, α*<sup>p</sup>* = 0.75).

When using the EOQ model, excluding the increase in delivery costs and prices, the time between deliveries will be *t<sup>w</sup>* = 40 days.

The total purchase, delivery, and storage costs for 400 days are as follows:

$$TC\_w = (400 + 20 \ast 25 \ast 40) \ast 1.001^{400} + (400 \ast 1.003^{40} + 20 \ast 1.00075^{40} \ast 25 \ast 40)$$

$$(40) \ast 1.001^{360} + \dots + (400 \ast 1.003^{360} + 20 \ast 1.00075^{360} \ast 25 \ast 40) \ast 1.001^{40}$$

$$= 294083 \text{ EUR}$$

When applying model 4, the time between deliveries is found by Formula (32):

$$t\_{\rm so} = \frac{1.003^{100}}{1.00075^{100}} \frac{40}{\sqrt{1 - 0.75}} = 100 \text{ days}$$

The total purchase, delivery, and storage costs for 400 days are as follows:

$$\text{TC}\_0 = (400 + 20 \ast 25 \ast 100) \ast 1.001^{400} + (400 \ast 1.003^{100} + 20 \ast 1.00075^{100} \ast 25 \ast 1.001^{100})$$

$$(80) \ast 1.001^{300} + \dots + (400 \ast 1.003^{300} + 20 \ast 1.00075^{300} \ast 25 \ast 80) \ast 1.001^{100} = 1$$

$$290748 \text{ EUR}$$

The savings will be as follows:

$$
\Delta TC = 294083 - 290748 = 3335 \text{EUR}
$$

If the cost of delivery decreases and the price increases during the period [0, *T*] with ρ*<sup>c</sup>* = <sup>−</sup>0.0039, <sup>ρ</sup>*<sup>p</sup>* = 0.00075, the growth pattern has the form *<sup>c</sup>s*(*t*) = <sup>400</sup> <sup>∗</sup> 0.9961*<sup>t</sup>* , *<sup>p</sup>*(*t*) = <sup>20</sup> <sup>∗</sup> 1.00075*<sup>t</sup>* , *t* ∈ [0, *T*] (α*<sup>c</sup>* = −3.9, α*<sup>p</sup>* = 0.75).

When using the EOQ model, excluding the reduction in delivery costs and the increase in prices, the time between deliveries will be *t<sup>w</sup>* = 40 days.

The total purchase, delivery, and storage costs for 400 days are as follows:

$$TC\_w = (400 + 20 \ast 25 \ast 40) \ast 1.001^{400} + (400 \ast 0.9961^{40} + 20 \ast 1.00075^{40} \ast 25 \ast 1.001^{40})$$

$$(40) \ast 1.001^{360} + \dots + (400 \ast 0.9961^{360} + 20 \ast 1.00075^{360} \ast 25 \ast 40) \ast 1.001^{40} = 1$$

$$288180 \text{ EUR}$$

When applying model 4, the time between deliveries is found by Formula (32):

$$t\_{\rm so} = \frac{0.9961^{100}}{1.00075^{100}} \frac{40}{\sqrt{1 - 0.75}} = 50 \text{ days}$$

The total purchase, delivery, and storage costs for 400 days are as follows:

$$\text{TC}\_0 = (400 + 20 \ast 25 \ast 50) \ast 1.001^{400} + (400 \ast 0.9961^{50} + 20 \ast 1.00075^{50} \ast 25 \ast 1.50)$$

$$(50) \ast 1.001^{350} + \dots + (400 \ast 0.9961^{350} + 20 \ast 1.00075^{350} \ast 25 \ast 50) \ast 1.001^{50} = 1$$

$$288015 \text{ EUR}$$

The savings will be as follows:

$$
\Delta TC = 288180 - 288015 = 165 \text{ EUR}
$$

If the cost of delivery increases and the price decreases during the period [0, *T*] with ρ*<sup>c</sup>* = 0.002, <sup>ρ</sup>*<sup>p</sup>* = <sup>−</sup>0.003, the growth pattern has the form *<sup>c</sup>s*(*t*) = <sup>400</sup> <sup>∗</sup> 1.002*<sup>t</sup>* , *<sup>p</sup>*(*t*) = <sup>20</sup> <sup>∗</sup> 0.997*<sup>t</sup>* , *t* ∈ [0, *T*] (α*<sup>c</sup>* = 2, α*<sup>p</sup>* = −3).

When using the EOQ model, excluding the increase in the cost of delivery and the decrease in the price, the time between deliveries will be *t<sup>w</sup>* = 40 days.

The total purchase, delivery, and storage costs for 400 days are as follows:

$$TC\_w = (400 + 20 \ast 25 \ast 40) \ast 1.001^{400} + (400 \ast 1.002^{40} + 20 \ast 0.997^{40} \ast 25 \ast 40) \ast 1.001^{360} + \dots + (400 \ast 1.002^{360} + 20 \ast 0.997^{360} \ast 25 \ast 40) \ast 1.001^{40} = 168249 \text{ EUR}$$
 
$$168249 \text{ EUR}$$

When applying model 4, the time between deliveries is found by Formula (32):

$$t\_{so} = \frac{1.002^{100}}{0.997^{100}} \frac{40}{\sqrt{1+3}} = 33 \text{ days}$$

The total purchase, delivery, and storage costs for 400 days are as follows:

$$T\mathbb{C}\_0 = (400 + 20 \ast 25 \ast 33) \ast 1.001^{400} + (400 \ast 1.002^{33} + 20 \ast 0.997^{33} \ast 25 \ast 33) \ast 1.001^{367} + \dots + (400 \ast 1.002^{367} + 20 \ast 0.997^{367} \ast 25 \ast 33) \ast 1.001^{33} = $$
 
$$167620 \text{ EUR}$$

The savings will be as follows:

$$
\Delta TC = 168249 - 167620 = 629 \text{ EUR}
$$

If uncontrollable parameters of the logistic process such as T = 400 days, r = 0.001, and µ = 25 units/day are known at the beginning of the period [0, *T*] and the cost of delivery and the price are reduced during the period [0, *T*] with ρ*<sup>c</sup>* = −0.001, ρ*<sup>p</sup>* = −0.003, then the growth pattern will be: *<sup>c</sup>s*(*t*) = <sup>400</sup> <sup>∗</sup> 0.999*<sup>t</sup>* , *<sup>p</sup>*(*t*) = <sup>20</sup> <sup>∗</sup> 0.997*<sup>t</sup>* , *t* ∈ [0, *T*] (α*<sup>c</sup>* = −1, α*<sup>p</sup>* = −3).

When using the EOQ model, excluding the reduction in delivery costs and prices, the time between deliveries will be *t<sup>w</sup>* = 40 days.

The total purchase, delivery and storage costs for 400 days are as follows:

$$TC\_w = (400 + 20 \ast 25 \ast 40) \ast 1.001^{400} + (400 \ast 0.999^{40} + 20 \ast 0.997^{40} \ast 25 \ast 40) \ast 1.001^{360} + \dots + (400 \ast 0.999^{360} + 20 \ast 0.997^{360} \ast 25 \ast 40) \ast 1.001^{40} = 165335 \text{ EUR}$$
 
$$165335 \text{ EUR}$$

When applying model 4, the time between deliveries is found by Formula (32):

$$t\_{\rm so} = \frac{0.999^{100}}{0.997^{100}} \frac{40}{\sqrt{1+3}} = 25 \text{ days}$$

The total purchase, delivery, and storage costs for 400 days are as follows:

$$T\mathbb{C}\_0 = (400 + 20 \ast 25 \ast 25) \ast 1.001^{400} + (400 \ast 0.999^{25} + 20 \ast 0.997^{25} \ast 25 \ast 25) \ast 1.001^{375} + \dots + (400 \ast 0.999^{375} + 20 \ast 0.997^{375} \ast 25 \ast 25) \ast 1.001^{25} = $$
 
$$163141 \text{ EUR}$$

The savings will be as follows:

$$
\Delta T \text{C} = 165335 - 163141 = 2194 \text{ EUR}
$$
