Case Problem: Textile Mill Scheduling

Case Problem: Textile Mill Scheduling Assuming, X1 = Yards of fabric 1 purchased X2 = Yards of fabric 1 on dobbie looms X3 = Yards of fabric 2 purchased X4 = Yards of fabric 2 on dobbie looms X5 = Yards of fabric 3 purchased X61 = Yards of fabric 3 on dobbie looms X62 = Yards of fabric 3 on regular looms X7 = Yards of fabric 4 purchased X81 = Yards of fabric 4 on dobbie looms X82 = Yards of fabric 4 on regular looms X9 = Yards of fabric 5 purchased X101 = Yards of fabric 5 on dobbie looms X102 = Yards of fabric 5 on regular looms

Production Times in Hours per Yard: Fabric | Dobbie | Regular | 1 | 1 / 4.63 = 0.215983 | - | 2 | 1 / 4.63 = 0.215983 | - | 3 | 1 / 5.23 = 0.191205 | 1 / 5.23 = 0.191205 | 4 | 1 / 5.23 = 0.191205 | 1 / 5.23 = 0.191205 | 5 | 1 / 4.17 = 0.239808 | 1 / 4.17 = 0.239808 |

Profit Contribution per Yard: Fabric | Manufactured (Variable cost –Selling Price) | Purchased (Selling Price – Purchase Price) | 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 |

Max Profit objective function
Max -
P = .19X1 + 0.33X2 + 0.16X3 + 0.31X4 + 0.5X5 + 0.16X61 + 0.16X62 + 0.54X7 + 0.73X81 + 0.73X82 + 0X9 + 0.2X101 + 0.2X102

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

Demand Constraints:
Constraint1: X1 + X2 = 16500
Constraint2: X3 + X4 = 22000
Constraint3: X5 + X61 + Y62 = 62000
Constraint4: X7 + X81 + Y82 = 7500
Constraint5: X9 + X101 + Y102 = 62000

Loom Constraints:
Constraint6: Dobbie Looms: 0.21598X2 + 0.21598X4 + 0.1912X61 + 0.1912X81 + 0.2398X101 ≤ 5760
Constraint7: Regular Looms: 0.192X62 + 0.1912X82 + 0.2398X102 ≤ 21600

OPTIMAL SOLUTION:
Attached is Excel sheet for optimal solution with