TEXTILE MILL SCHEDULING case

Topics: Cost, Optimization, Profit maximization Pages: 2 (516 words) Published: October 21, 2014

Vi Le
MGM350C
Professor Rohit RampalCase: Textile Mill Scheduling
Let X3R = yards of fabric 3 on regular looms
X4R = yards of fabric 4 on regular looms
X5R = yards of fabric 5 on regular looms
X1D = yards of fabric 1 on dobbie looms
X2D = yards of fabric 2 on dobbie looms
X3D = yards of fabric 3 on dobbie looms
X4D = yards of fabric 4 on dobbie looms
X5D = yards of fabric 5 on dobbie looms
Y1 = yards of fabric 1 purchased
Y2 = yards of fabric 2 purchased
Y3 = yards of fabric 3 purchased
Y4 = yards of fabric 4 purchased
Y5 = yards of fabric 5 purchased
Profit Contribution per Yard:
Fabric = Manufactured (Variable cost –Selling Price)
Purchased = (Selling Price – Purchase Price)
Fabric Manufactured Purchased
1 0.99-0.66=0.33 0.99-0.80=0.19
2 0.86-0.55=0.31 0.86-0.70=0.16
3 1.10-0.49=0.61 1.10-0.60=0.50
4 1.24-0.51=0.73 1.24-0.70=0.54
5 0.70-.50=0.20 0.70-.70=0
Production Times in Hours per Yard:
Fabric Regular Dobbie1 1 / 4.63 = 0.21598
2 1 / 4.63 = 0.21598
3 1 / 5.23 = 0.1912 1 / 5.23 = 0.1912
4 1 / 5.23 = 0.1912 1 / 5.23 = 0.1912
5 1 / 4.17 = 0.2398 1 / 4.17 = 0.2398
Max Profit objective function
Max 0.61 X3R + 0.73X4R + 0.20X5R + 0.33X1D + 0.31X2D + 0.61X3D + 0.73X4D + 0.20X5D + 0.19Y1 + 0.16Y2 + 0.50Y3 + 0.54Y4 Dobbie Hours Available: 8 Looms x 30 days x 24 hours/day = 5760 Regular Hours Available: 30 Looms x 30 days x 24 hours/day = 21600 Loom Constraints:

Constraint 1: Regular Looms: 0.192X3R + 0.1912X4R + 0.2398X5R ≤ 21600 Constraint 2: Dobbie Looms: 0.21598X1D + 0.21598X2D + 0.1912X3D + 0.1912X4D + 0.2398X5D ≤ 5760 Demand Constraints:
Constraint 3: X1D + Y2 = 16500
Constraint 4: X2D + Y2 = 22000
Constraint 5: X3R + X3D + Y3 = 62000
Constraint 6: X4R + X4D + Y4 = 7500
Constraint 7: X5R + X5D + Y5 = 62000
According to the
Production/ purchased schedule (yards)
Fabric Regular DobbieLooms Looms Purchased
1 4469 11831
2 22000 3 2771...
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