Fuzzy Multi-objective Linear Programming Application in Product-Mix DecisionMaking Sani Susanto1, Neneng Tintin Rosmiyanti2, Pandian Vasant3, Arijit Bhattacharya4 Management Science Group, Department of Industrial Engineering, Parahyangan Catholic University, Bandung, Indonesia. 2 Department of Mathematics, Padjadjaran University, Jatinangor, Indonesia. 3 Electrical & Electronic Engineering Program, Universiti Teknologi Petronas, 31750 Tronoh, BSI, Perak DR, Malaysia. 4 The Patent Office, CP-2, Sector V, Kolkata – 700091, West Bengal, India. 1
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Abstract This paper focuses on optimizing product-mix problem using real-world data of a food processing industry. The extensively used LP model has been remodeled using fuzzy sets and applied to the realworld problem. The developed LP model is able to consider the fuzziness in the parameters and to tackle the presence of multiple objective functions. The degree of satisfaction of the product-mix decisionmaker in terms of (i) the total profit obtained and (ii) the waste resulted, is at least 70%, which is encouraging.
single objective in nature. Those two flaws are to be tackled in this paper developing a model that can adopt those two aspects. The developed model will be able to consider the fuzziness in the parameters and to tackle the presence of multiple objective functions [6]. A real-world application will be demonstrated to validate the developed model. This validation is based on a real research held in PT Campina Ice Cream Industry, a market leader in ice cream and frozen food industry located at Surabaya, East Java, Indonesia. A family comprising three ice cream products – viz., Didi Cup Chocolate-Vanilla, Bazooka Chocolate and Homepacks-Chocolate Fudge – is selected as a case study in product-mix problem. Here are the brand pictures of these three products:
1. Introduction One of the main goals of all enterprises is to gain maximum profit. In achieving this goal multiple criteria are to be considered, viz., financial systems, marketing management, human resources management and production management. This research concentrates on one important aspect of production management for enterprise producing family of products, i.e., the product-mix problem. The product-mix problem intends to determine the number of each product produced in order to achieve objectives of the enterprise, e.g., maximize profit, minimize wastes, considering limitations of the enterprise e.g., the availability of material, funds, space. Linear Programming (LP) is used extensively for product-mix problem. However, this technique assumes that all the parameters in the model should be certain [2,3,4,5]. Moreover, the LP technique is of
Figure 1. Didi Cup® Chocolate-Vanilla
Figure 2. Bazooka® Chocolate
Figure 3. Homepacks Chocolate Fudge®
the three types of product; d1 = waste resulted from the production of Didi Cup Chocolate Vanilla; d2 = waste resulted from the production of Didi Cup Bazooka Chocolate; and d3 = waste resulted from the production of Homepacks Chocolate Fudge Step 1.3 Determine the systems constraints as follows: o The amount of milk powder needed must not exceed its availability, which is:
2. Problem Formulation The steps to formulate the production management problem into fuzzy linear programming with multiple objectives are as follows: Step-1: Formulating the crisp mathematical model for the product-mix problem Step 1.1: Determine the decision variables There are three decision variables, they are: x1 = the amount of 1 litre package of Didi Cup Chocolate-Vanilla to be produced; x2 = the amount of 1-litre package of Didi Cup Bazooka Chocolate to be produced; and x3 = the amount of 1-litre package of Homepacks Chocolate Fudge to be produced.
3
∑a j =1
1j
o The amount of chocolate powder needed must not exceed its availability, i.e.,: 3
j =1
For the instant problem there are two objective functions. The first objective function (eqn. 1) is to maximize the total profit, i.e., the sum of all profit gained from the sale of each type of product:
z1 =
3
∑c j =1
j
xj
… (1)
where, z1 = total profit gained from the three types of product; c1 = profit contribution from Didi Cup Chocolate Vanilla (in Indonesian Rupiah); c2 = profit contribution from Didi Cup Bazooka Chocolate (in Indonesian Rupiah); and c3 = profit contribution from Homepacks Chocolate Fudge (in Indonesian Rupiah). The second objective function (eqn. 2) is to minimize the total waste resulted from the production of the three types of product:
z2 =
3
∑d j =1
j
xj
… (2)
where, z2 = total waste resulted from the production of
… (3)
where, a11 = the amount of milk powder needed to produce packages of 1 litre of Didi Cup ChocolateVanilla; a12 = the amount of milk powder needed to produce packages of 1 litre of Didi Cup Bazooka Chocolate; a13 = the amount of milk powder needed to produce packages of 1 litre of Homepacks Chocolate Fudge; and b1 = the amount of milk powder available to produce packages of 1 liter of the three product.
∑a
Step 1.2: Determine the objective functions
x j ≤ b1
2j
x j ≤ b2
… (4)
where, a21 = the amount of chocolate powder needed to produce 1 litre of Didi Cup Chocolate-Vanilla; a22 = the amount of chocolate powder needed to produce 1 litre of Didi Cup Bazooka Chocolate; a23 = the amount of chocolate powder needed to produce 1 litre of Homepacks Chocolate Fudge; and b2 = the amount of chocolate powder available to produce packages of 1 liter of the three product. o The amount of vegetable oil needed must not exceed its availability, that is: 3
∑a j =1
3j
x j ≤ b3
… (5)
where, a31 = the amount of vegetable oil needed to produce 1 litre of Didi Cup Chocolate-Vanilla; a32 = the amount of vegetable oil needed to produce 1 litre of Didi Cup Bazooka Chocolate; a33 = the amount of vegetable oil needed to produce 1 litre of Homepacks Chocolate Fudge; and b3 = the amount of vegetable oil available to produce packages of 1 liter of the three product. Thus, the complete crisp mathematical model for the product mix problem is as follows:
Maximize z1 = c1x1 + c2x2 + c3x3 Minimize z2 = d1x1 + d2x2 + d3x3 subject to: a11x1 + a12x2 + a13x3 ≤ b1 a21x1 + a22x2 + a23x3 ≤ b2 a31x1 + a32x2 + a33x3 ≤ b3 x1, x2, x3 ≥ 0
... (6) … (7) … (8) … (9) … (10) … (11)
Step-2: Developing the crisp mathematical model formulated in Step-1 to fuzzy mathematical model. Since there will be potential conflict between the achievement of the two objectives, some tolerances must be defined to resolve the problem. The tolerances permitted by the company are as follows: − Tolerance 1: at least 75% of the potential maximum profit must be achieved − Tolerance 2: the total waste must not exceed 30% more than the potential minimum waste To include these two tolerances following steps are considered: 1. Solve the problem to maximize and minimize the objective function (6) subject to the constraints (8)(11) and let the corresponding optimal objective function value be, respectively, z1’ and z1”. When the objective function z1 achieves the value of z1’ (or more) or z1” (or less), then to the degree of optimality of the value achieved by this function will be assigned the value, respectively, 1 or 0. When the objective function z1 = c1x1 + c2x2 + c3x3 achieves any values between z1” and z1’, then the following function will determine the degree of satisfaction to the value achieved by the function z1:
0, if d1 x1 + d 2 x2 + d 3 x3 ≥ z 2 " z "-( d x + d x + d x ) 2 1 1 2 2 3 3 µ z2 ( x1 , x2 , x3 ) = , if z 2 ' ≤ d1 x1 + d 2 x 2 + d 3 x3 ≤ z 2 " z 2 "- z 2 ' 1 if d1 x1 + d 2 x2 + d 3 x3 ≤ z 2 '
… (13) 3. The multi-objective optimization problem (1)-(6) is then modified to a multi-objective problem which maximizes degree of satisfaction of the decision-maker by quantifying the objective functions z1 and z2, as follows: Maximize µ z1 ( x1 , x 2 , x3 ) … (14) Maximize µ z 2 ( x1 , x 2 , x3 ) subject to the constraints (8) – (11). This multi-objective optimization problem is then modified to a single objective optimization problem by applying the maxi-min criterion: max α … (16) subject to constraints (3) – (6) and the following additional constraints: µ z1 ( x1 , x 2 , x3 ) ≥ α … (17)
µ z2 ( x1 , x 2 , x3 ) ≥ α
2. Solve the problem to minimize and maximize the objective function (7) subject to the constraints (8)(11) and let the corresponding optimal objective function value be, respectively, z2’ and z2”. When the objective function z2 achieves the value of z2’ (or less) or z2” (or more), then to the degree of optimality of the value achieved by this function will be assigned the value, respectively, 0 or 1. When the objective function z2 = d1x1 + d2x2 + d3x3 achieves any values between z2’ and z2”, then the following function will determine the degree of satisfaction to the value achieved by the function z2:
… (18)
c1x1 + c2x2 + c3x3 ≥ (0.75)( z1’) (Tolerance 1) … (19) d1x1 + d2x2 + d3x3 ≤ (0.30)( z2”) (Tolerance 2) … (20)
3. Results To solve the real product mix problem formulated in the previous section, the values of the parameters ci in (6), di in (7), and aij must be first obtained from the company (i , j = 1, 2, 3). The required values are obtained already from the
0, if c1 x1 + c 2 x 2 + c3 x3 ≤ z 1 " company as described in the Table 1. (c1 x1 + c 2 x 2 + c3 x3 ) - z1 " , if z 1 " ≤ c1 x1 + c 2 x 2 + c3 x3 ≤ z 1 ' µ z1 ( x1 , x 2 , x3 ) = z1 "- z1 ' Thus, the complete crisp mathematical 1 if c1 x1 + c 2 x 2 + c3 x3 ≥ z 1 ' the product-mix problem is as follows:
… (12)
… (15)
Maximize z1 = 5000x1 + 6250x2 + 4650x3 Minimize z2 = 4x1 + 2x2 + 3x3 subject to 2x1 + 3x2 + 6x3 ≤ 127 2x1 + 3x2 + 2x3 ≤ 52 1x1 + 2x2 + 3x3 ≤ 56 x1, x2, x3 ≥ 0
model for … (1”) … (2”) … (3”) … (4”) … (5”) … (6’’)
The values of z1’ and z1” are z1’= 130 000 and z1”= 0. Therefore, the constraint (19) is of the form: 5000x1 + 6250x2 + 4650x3 ≥ (0.75)(130000)= 97500 Similarly, the values of z2’ and z2” are z2’= 0 and z2”= 104 and the constraint (20) is of the form: 4x1 + 2x2 + 3x3 ≤ (0.30)(104) = 31.2 The final mathematical modelling to be solved
A thorough computation with MATLAB® software establishes the fact that if the above solution for the optimal product-mix is followed then the degree of satisfaction of the product-mix decision maker in terms of (i) the total profit obtained and (ii) the waste resulted, will be at least 70%, i.e., the optimal α will be 0.70.
will be as follows:
max α
subject to 2x1 + 3x2 + 6x3 ≤ 127 2x1 + 3x2 + 2x3 ≤ 52 1x1 + 2x2 + 3x3 ≤ 56 5000x1 + 6250x2 + 4650x3 ≥ 97 500 4x1 + 2x2 + 3x3 ≤ 31.2 µ z1 ( x1 , x 2 , x3 ) ≥ α , or
References
1 (5000 x1 + 6250 x2 + 4650 x3 ) ≥ α 130000 µ z2 ( x1 , x 2 , x3 ) ≥ α or 1 (104 - (4 x1 + 2 x 2 + 3x3 )) ≥ α 104 x1, x2, x3 ≥ 0.
4. Discussion and Conclusion The optimal solution is: x1 = 0 (this indicates that the company does not produce any 1 litre package of Didi Cup ChocolateVanilla); x2 = 15.6 (this indicates that the company produces 15.6 units of 1 litre package of Didi Cup Bazooka Chocolate to be produced); and x3 = 0 (this indicates that the firm does not produce any 1 litre package of Homepacks Chocolate Fudge).
Product 1. Didi Cup ChocolateVanilla 2. Bazooka Chocolate 3. Homepacks Chocolate Fudge Availability
Milk Powder (kgs) 2
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Table 1: Research data Ingredients per litre of product Chocolate Vegetable Oil Powder (kgs) (kgs) 2 1
Profit (Indonesian Rupiah) 5000
Waste (litres) 4
3
3
2
6250
2
6
2
3
4650
3
127
52
56
