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Optimisation Methods Liverpool Hope University. Optimisation. Example.
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Example Hope Fabrics Ltd. runs a textile manufacturing industry in the northern England. The industry uses cotton as its primary raw material. The cotton are sourced from India and the United States of America (USA); and processed cotton mainly yield three textile products: terrycloth (for the manufacture of high absorbent bath towels and robes), denim (for the manufacture of blue jean), and chambray (for the manufacture of shirts). The cotton imported from the above sources yield different product mixes. Each bale of Indian cotton yields 0.3 bale of terrycloth, 0.4 bale of denim, and 0.2 bale of chambray. Each bale of cotton from the USA yields 0.4 bale of terrycloth, 0.2 bale of denim and 0.3 bale of chambray. The remaining 10% is lost to processing. The cotton also differ in cost and availability. Hope Fabrics can purchase up to 9000 bales per day from India at £200 per bale. Up to 6000 bales per day of American cotton are available at the lower cost of £160 per bale. Hope Fabrics contracts with its customers’ require it to produce 2000 bales per day of terrycloth,1500 bales per day of denim and 500 bales per day of chambray. Develop an optimization model to satisfy these requirements at minimum cost.
Solution Problem Description • Cotton are sourced from India and USA • Customer Demand: 2000 terrycloth, 1500 denim, and 500 chambray (in bales) • India Number of Bales Purchased: ? Cost Per Bale: £200 Maximum Supply Available: 9 Processed Bale: 0.3t, 0.4d, 0.2c • USA Number of Bales Purchased: ? Cost Per Bale: £160 Maximum Supply Available: 6 Processed Bale: 0.4t, 0.2d, 0.3c
Visual Description Cost Bale of terrycloth Bale of denim Bale of chambray Design Variables: , Parameters: Cost I = 20, U = 16 Objective: min Cost Process Yield: I = 0.3t + 0.4d + 0.2c U = 0.4t + 0.2d + 0.3c Constraints: Bale of terrycloth Bale of denim Bale of chambray Customer Demand: 2t +1.5d +0.5c
Mathematical Format The general form of a mathematical program or (single objective) optimisation model is: min or max Subject to: where are given functions of decision variables ; and are specified constant parameters.
Hope Fabrics Ltd. Model min s.t. where,, Design Variables? Variable Type Constraints? Main Constraints? Objective?
Simultaneous Equations • (1) • (2) multiplying Eq. 2 by 2; we obtain: • (3) subtracting Eq. 3 from Eq. 2; we obtain: or thus,
contd… The design constraint of is therefore, Substituting the value (vector) of in all the main constraints, we obtain: =
contd… The design constraint on is thus the negative elements in the column vectors are discarded. The feasible bound for the design variables are therefore: Using these values, we evaluate the optimisation problem objective function i.e. min
contd… We obtain the vector: Now, because this is a cost minimisation problem, our solution is the lowest objective function value – which in this case is 96. This corresponds to a design variable optimum of and Thus, Hope Textile Ltd should purchase 2000 and 3500 bales of cotton from India and the United States respectively.
Graphical Solution Simple optimisation problems such as this can also be solved graphically. This involves plotting feasible 2D designs that satisfy the inequality constraints. For each of the inequality constraint, one design variable is set at zero, the other is solved, and vice-versa. This is used to plot the lines representing the constraints. The feasible direction is determined by arbitrarily choosing a point on any side of the line, and checking if it is a feasible point.
feasible Design The meeting point of 2 or more lines is considered the optimal point.
Optimal Solution The optimal solution as solved mathematically and graphically are: Therefore, considering the constraints in this problem; and the need to satisfy customers’ demand, Hope Textile Ltd must purchase 2000 and 3500 bales of cotton from India and the United States respectively.
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