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 Parameters: Cost I = 20, U = 16 Process Yield: I = 0.3t + 0.4d + 0.2c U = 0.4t + 0.2d + 0.3c Customer Demand: 2t +1.5d +0.5c Objective: min Cost
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.