This research
article is structured into the following eight sections.: -
Notations and
Assumptions used for calculation in systematic way displayed in Section 2.
The
mathematical formulation of our model, along with various types of costs, is
derived in section 3. Additionally, the solution is obtained. In section 4 Two
numerical examples with specific values are employed to validate the proposed
model. The sensitivity of our problem is demonstrated section 5 through a table
created by varying the values of parameters used in this model. The section 6
presents graphs of the minimized cost function concerning various parameters,
along with their corresponding observations. Additionally, straightforward
graphs for sensitivity analysis are included. Next in the section 7 a visual
conclusion is provided. Section 8 covered novelty of article and Section 9
covered conclusion of our model, with relevance to its future importance.
2. Assumption
and Notations
This inventory
model has considered the following assumptions and notations.
2.1 Assumptions
The assumption
of this inventory models is as follow: -
1. The demand
rate is known, quadratic and continuous.
2. Two rates of
production are considered.
3. Rate
of production is known and constant.
4. The
replenishment occurs instantaneous i.e., replenishment rate is infinite.
5. The
production rate is always greater the demand rate.
6. The items is
single product; it does not interact with any other inventory items
7. Lead time is
assumed to be zero.
8. It is
assumed that no repair or replacement of the deteriorative items t during take
place
during a given cycle.
9. Holding cost
is constant.
2.2 Notations
(1) Sc –
Ordering Cost / Set up Cost per unit per order.
(2) κ -
Production Rate (units per unit time) during time interval [0, t1]
aκ - Production Rate (units per unit time) during time interval [t1,
t2], where a > 0
(3) CP -
Purchase Cost per unit.
(4) DC –
Cost of deteriorated items.
(5) λ-
The Rate of deterioration, where 0 < λ < 1
(6) HC =
Unit Holding Cost per unit per unit of time.
(7) t3 -
Production Cycle length (Total Cycle time)
(8) D(t) - The
demand rate, where c1, c2 and c3
> 0
c1=initial
demand rate (units per unit time), c2=initial rate of change in
demand., c3 =accretion of demand rate.
(9) I(t) -
Inventory level at time t.
(10) I1(t)
- Inventory level at time t during time interval [0, t1]
(11) I2(t)
- Inventory level at time t during time interval [t1, t2]
(12) I3(t)
- Inventory level at time t during time interval [t2, t3]
(13) t3 -
Time at which inventory level reaches to zero.
(14) IM1 -Maximum
inventory level at [0, t1]
(15) IM2 -Maximum
inventory level at [t1, t2]
(16) Q* - Order
quantity during the cycle length t3
(17) TAC* -
Total Cost / cycle time
(18) w.r.t. –
with respect to
Leveraging the
assumptions and notations, we endeavor to formulate a mathematical inventory
model aimed at attaining an optimal solution for the Economic Production
Quantity (EPQ).
3. Mathematical formulations of the proposed Model
In our model, the initial inventory level is 0 at time t=0. After a time period at time t=t1 the inventory level reaches IM1. After some time at time t=t2 the inventory level is IM2. Subsequently, with the impact of deterioration and demand the inventory level reaches to zero at time t=t3
Figure 1: Proposed inventory model with Two Level of Production
3.1 Mathematical Model
3.3 Cost Calculation of Proposed Model
To calculate cost of the proposed model we take different types of costs, which are follows:
(i) Ordering Cost / Set Up Cost (O.C.) Sc (12)
(ii) The order size during the period [0, t3] = Q*= IM2
4. Solution Algorithm: -
The solution algorithm of our proposed model is given below.
Step 1. Calculate different type of Costs using Equation (12) to (17).
Step 2. Calculate Total average inventory using equation (18).
Step 3. Find first derivative of TAC w.r.t. t3. And after find first derivative calculate
positive value of t3. We find the value of t3 is 8.16
t3=8.16
Step 5. Employing the mathematical software MATLAB (R2021a) and putting the
value of ɑ= 0.90, c1 = 50, c2= 30, c3 =40, κ=500 per unit time, t1=1, t2=5, λ=0.95,
Cp=20 per unit, Sc=100 per unit, HC=100 per unit time, DC=50 per unit in
equation of TAC.
Step 6.and after solving we find the value of TAC is 5789.6
Step 7. Change values of different parameters with rate of +20%, +10%, -10%, -20% and
find different results. From these results we construct sensitivity analysis.
Step 8. We get desired results.
4.1 Numerical Example number 1
Should we consider a numerical illustration and employ select parameter values inherent to our inventory model: -
ɑ= 0.90, c1 = 50, c2= 30, c3 =40, κ=500 per unit time, t1=1, t2=5, λ=0.95, Cp=20 per unit, Sc=100 per unit, HC=100 per unit time, DC=50 per unit.
Subsequently, we insert these specific values into equations numbered (15) and (18). Employing the mathematical software MATLAB (R2021a), we tackle this problem, yielding the following optimal values as our resultant outcomes:-
t3* | Q* | TAC* |
8.16 | 152.2484 | 5789.6 |
4.2 Numerical Example number 2Should we consider a numerical illustration and employ select parameter values inherent to our inventory model: -
ɑ= 5, c1 = 30, c2= 10, c3 =10, κ=400 per unit time, t1=1.5, t2=3.5, λ=0.55, Cp=25 per unit, Sc=120 per unit, HC=0.5 per unit time, DC=10 per unit.
Subsequently, we insert these specific values into equations numbered (15) and (18). Employing the mathematical software MATLAB (R2021a), we tackle this problem, yielding the following optimal values as our resultant outcomes:-
5 Sensitivity analysisFor sensitivity analysis of this Model, we change values of parameters one by one and announce the effects on t3*, Q* and TAC*. Rate of changes (in percentage) in values of parameters are taken -20 %, -10%, +10% and +20%.
Parameter | Change in Parameter | t3* | Q* | TAC* |
| -20% | 8.11 | 121.9 | 1820.2 |
ɑ =0.9 | -10% | 8.13 | 137.02 | 3807.5 |
| 10% | 8.18 | 167.43 | 7766.5 |
| 20% | 8.20 | 182.69 | 9738.5 |
C1=50 | -20% | 8.49 | 161.64 | 2485.7 |
| -10% | 8.31 | 156.94 | 4396.8 |
| 10% | 8.02 | 147.55 | 6939.1 |
| 20% | 7.90 | 142.85 | 7896.8 |
| -20% | 7.94 | 174.77 | 10851 |
C2 = 30 | -10% | 8.05 | 163.49 | 8383.7 |
| 10% | 8.27 | 141.00 | 3050.5 |
| 20% | 8.40 | 129.75 | 143.46 |
C3=10 | -20% | 8.17 | 183.83 | 8902.1 |
-10% | 8.16 | 168.04 | 7346 |
10% | 8.15 | 136.45 | 4232.9 |
20% | 8.15 | 120.66 | 2675.9 |
κ=500 | -20% | 8.08 | 58.33 | - |
-10% | 8.12 | 105.28 | 2244.8 |
10% | 8.19 | 199.20 | 9318 |
20% | 8.23 | 246.16 | 12831 |
| -20% | 8.15 | 152.24 | 5695.4 |
t1 = 1 | -10% | 8.15 | 152.24 | 5741 |
| 10% | 8.16 | 152.24 | 5840.9 |
| 20% | 8.16 | 152.24 | 5894.5 |
| -20% | 6.92 | 240.01 | 11406 |
t2=5 | -10% | 7.54 | 199.19 | 8864.5 |
| 10% | 8.76 | 99.69 | 2225.3 |
| 20% | 9.36 | 41.87 | - |
| -20% | 9.08 | 224.49 | - |
λ =0.95 | -10% | 8.71 | 194.5 | 1771.3 |
| 10% | 8.00 | 141.21 | 6849.2 |
| 20% | 6.72 | 63.24 | 3687 |
| -20% | 8.15 | 152.24 | 5714.9 |
CP = 20 | -10% | 8.15 | 152.24 | 5752.3 |
| 10% | 8.16 | 152.24 | 5826.9 |
| 20% | 8.16 | 152.24 | 5864.2 |
| -20% | 8.16 | 152.24 | 5787.1 |
SC=100 | -10% | 8.16 | 152.24 | 5788.4 |
| 10% | 8.16 | 152.24 | 5790.8 |
| 20% | 8.16 | 152.24 | 5792 |
HC=100 | -20% -10% 10% 20% | 8.16 8.16 8.15 8.15 | 152.24 152.24 152.24 152.24 | 5056.8 5423.2 6156 6522.3 |
DC=50 | -20% -10% 10% 20% | 8.16 8.16 8.15 8.15 | 152.24 152.24 152.24 152.24 | 5441.5 5615.5 5963.6 6137.6 |