In today’s competitive scenario, organizations face immense challenges for meeting the transitional consumer's demand, and maintaining the inventory plays a major role. Indulging the betterment of various promotional activities that yield a reputation for the concern. The classical economic order quantity by Harris
In the early 1990’s the Economic order quantity (EOQ) and Economic Production Quantity (EPQ) played a vital role in the operations management area, contrastingly it failed in meeting with the real-world challenges. Since these models assumed that the items received or produced are of a perfect quality which is highly challenging. In inventory management, EOQ plays a vital role in minimizing holding and ordering costs. Ford W. Harris (1915) developed the EOQ model but it was further extended and extensively applied by R.H. Wilson. Wang
Briefly, the input parameters of inventory models are often taken as crisp values due to variability in nature.H.J. Zimmerman
Considering these inputs, the non-linear programming problem is solved for total cost function using Lagrangian and Kuhn-Tucker method and concluded with a sensitivity analysis, which exhibits the variations between the fuzzy and crisp values.
Considering these inputs, non-linear programming is solved for total cost function using Lagrangian and Kuhn-Tucker method and done a sensitivity analysis between the slight variations between the fuzzy and crisp values.
Definition 1:
A fuzzy set à defined on R (∞,-∞), if the membership function of à is defined by
Definition 2:
A trapezoidal fuzzy number Ã=(a, b, c, d) with a membership function μà is defined by
Function principle
Suppose Ã=(x1, x2, x3, x4) and
The addition of
à ⊕ B̅ =(x1+y1, x2+y2, x3+y3, X4+y4)
Where x1, x2, x3, x4,y1, y2, y3, y4 are any real numbers.
(2) The multiplication of
à ⊕
Where Z1={x1y1, x1y4, x4y1, x4y4}, Z2 = ={x2y2, x2y3, x3y2, x3y4}, C1 = min Z1, C2 = min Z1, C1 = max Z2, C1 = max Z2.
If x1, x2, x3, x4,y1, y2, y3 and y4 are all zero positive real numbers then
à ⊗
(3) The subtraction of
Where also x1, x2, x3, x4,y1, y2, y3 and y4 are any real numbers.
(4) The division of
where y1, y2, y3 and y4 are positive real numbers. Also x1, x2, x3, x4,y1, y2, y3 and y4are nonzero positive numbers.
(5) For any ∝ ∈ R
Extension of the Lagrangian Method.
Solving a nonlinear programming problem by obtaining the optimum solution was discussed by Taha
Suppose if the problem is given as
Minimize y = f(x)
Sub to gi(x) ≥ 0, i = 1, 2, · · ·, m.
The constraints are non-negative say x ≥ 0 if included in the m constraints. Then the procedure of the extension of the Lagrangian method will involve the following steps.
Step 1:
Solve the unconstrained problem
Min y = f(x)
If the resulting optimum satisfies all the constraints, then stop since all the constraints are inessential. Or else set K = 1 and move to step 2.
Step 2:
Activate any K constraints (i.e., convert them into equalities) and optimize f(x) subject to the K active constraints by the Lagrangian method. If the resulting solution is feasible with respect to the remaining constraints, the steps have to be repeated. If all sets of active constraints taken K at a time are considered without confront a feasible solution, go to step 3.
Step 3:
If K = m, stop; there’s no feasible solution.
Otherwise set K = K + 1 and go to step 2.
Graded mean Integration Representation Method was introduced by Hsieh et al
In this paper trapezoidal fuzzy numbers is used as fuzzy parameters for the production inventory model. Let
The input parameters for the corresponding model are
K - Ordering cost
a - constant demand rate coefficient
b - price-dependent demand rate coefficient
P - selling price
Q - Order size
c - unit purchasing cost
g - constant holding cost coefficient
Let’s consider the total cost
The total cost per cycle is given by
Partially differentiating w.r.t Q,
Equating
By using trapezoidal numbers, fuzzified input parameters are as follows
The optimal order quantity
Partially differentiating w.r.t ‘Q’ and equating to zero,
Hence the optimal economic order quantity for crisp values is derived,
Applying the graded mean representation
Now partially differentiating w.r.t Q1, Q2, Q3, Q4 and equating to zero,
The above derived results depict that Q1 > Q2 > Q3 > Q4 failing to satisfy the constraints 0 ≤ Q1 ≤ Q2 ≤ Q3 ≤ Q4 . So, converting the inequality constraint Q2 - Q1 ≥ 0 into equality constraint Q2 - Q1 = 0 Optimizing P(TC(Q,P) subject to Q2 - Q1 = 0 by Lagrangian method.
From equations (1) and (2) the results are,
Since Q3 > Q4 which does not satisfy the constraint 0 ≤ Q1 ≤ Q2 ≤ Q3 ≤ Q4. Now converting the inequality constraints Q2 - Q1 ≥ 0, Q3 - Q2 ≥ 0 into equality constraints Q2 - Q1 = 0 and Q3 - Q2 = 0. Optimizing,
From equations (1’) , (2’) and (3’),
In the above-mentioned results Since Q1 > Q4 which does not satisfy the constraint 0 ≤ Q1 ≤ Q2 ≤ Q3 ≤ Q4. Converting the inequality constraints Q2 - Q1 ≥ 0, Q3 - Q2 ≥ 0 and Q4 - Q3 ≥ 0 into equality constraints Q2 - Q1 = 0, Q3 - Q2 = 0 and Q4 - Q3 = 0 . Optimizing,
After Partially differentiation (Appendix)
satisfies the required inequality constraints.
By applying Kuhn Tucker conditions, the total cost is minimized by finding the solution of Q1, Q2, Q3, Q4 with Q1, Q2, Q3, Q4
The above conditions simplify to the following 𝞴1, 𝞴2, 𝞴3, 𝞴4.
It is known that, Q1 > 0 then in
𝞴1Q1 = 0 arrive at 𝞴4 = 0 In a similar fashion, if 𝞴1= 𝞴2 = 𝞴3 = 0, Q4 ≤ Q3 ≤ Q2 ≤ Q1 does not satisfy 0 ≤ Q1 ≤ Q2 ≤ Q3 ≤ Q4. Therefore the conclusion is Q2 = Q1, Q3 = Q2 and Q4 = Q3
i.e., Q1 = Q2 = Q3 = Q4 = Q*
Let us consider an integrated inventory system having the following statistics with crisp parameters having following values
As mentioned earlier, the trapezoidal numbers
yielding the below results. Equation 1 was the optimal order quantity for crisp (equation 1) values and optimization to the EOQ is done by applying Lagrangian (equation 2) and Kuhn-Tucker (equation 3) methods under graded-mean defuzzification method.