Case Problem: Textile Mill Scheduling

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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 given constraints (for both profit maximization and cost minimization models) Sheet – textile mill solution

There are 7 sheets in the same excel file –
For Profit maximization model -
1. Profit Limit Report
2. Profit Answer Report
3. Profit Sensitivity Report

For Cost minimization model -
4. Cost Limit Report
5. Cost Answer Report
6. Cost Sensitivity Report

Data and solution –
7. Data
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Discussion of Case Study
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Question 1
The final production schedule and loom assignment for each fabric Fabric1 purchase| 11831.2|
Fabric1 dobbie| 4668.8|
Fabric2 purchase| 0|
Fabric2 dobbie| 22000|
Fabric3 purchase| 62000|
Fabric3 dobbie| 0|
Fabric3 regular| 0|
Fabric4 purchase| 0|
Fabric4 dobbie| 0|
Fabric4 regular| 7500|
Fabric5 purchase| 0|
Fabric5 dobbie| 0|
Fabric5 regular| 62000|

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Fabric produced by Dobbie loom
Fabric1 dobbie| 4668.8|
Fabric2 dobbie| 22000|
Fabric3 dobbie| 0|
Fabric4 dobbie| 0|
Fabric5 dobbie| 0|
| |
Production hours of dobbie loom = 4666.8/4.63 + 22000/4.63 = 5760 hours

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Fabric production by regular loom
Fabric3 regular| 0|
Fabric4 regular| 7500|
Fabric5 regular| 62000|

Production hours of regular loom = 7500/5.23 + 62000/4.17 = 16302.13993

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Question 2
Projected total contribution to profit
Fabric1 purchase| 11540.7|
Fabric1 dobbie| 0|
Fabric2 purchase| 6820|
Fabric2 dobbie| 31000|
Fabric3 purchase| 0|
Fabric3 dobbie| 0|
Fabric3 regular| 0|
Fabric4 purchase| 0|
Fabric4 dobbie| 5475|
Fabric4 regular| 0|
Fabric5 purchase| 0|
Fabric5...
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