O.R Case study

VYAS ,APOORV AGARWAL,AASTHA POKHARNA, ANURAG

OPERATION RESEARCH PROJECT SCOTSVILLE TEXTILE MILL

Scottsville Textile Mill Case Study: The Scottsville Textile Mill produces five different fabrics. Each fabric can be woven on one or more of the mill’s 38 looms. The sales department’s forecast of demand for the next month is shown in Table 4.16, along with data on the selling price per yard, variable cost per yard, and purchase price per yard. The mill operates 24 hours a day and is scheduled for 30 days during the coming month. The mill has two types of looms: dobbie and regular. The dobbie looms are more versatile and can be used for all 5 fabrics. The regular looms can produce only 3 of the fabrics. The mill has a total of 38 looms: 8 are dobbie and 30 are regular. The rate of production for each fabric on each type of loom is given in table 9.17. The time required to change over from producing one fabric to another is negligible and does not have to be considered. The Scottsville Textile Mill satisfies all demand with either its own fabric or fabric purchased from another mill. Fabrics that cannot be woven at the Scottsville Mill because of limited loom capacity will be purchased from another mill. The purchase price of each fabric is also shown in Table 9.16.

Monthly Demand, Selling Price, Variable cost, and purchase price data for Scotsville Textile Mill Fabrics Fabric

Demand (yards)

Selling Price ($/yard)

1

16,500

0.99

Variable Cost ($/yard) 0.66

2 3 4 5

22,000 62,000 7,500 62,000

0.86 1.10 1.24 0.70

0.55 0.49 0.51 0.50

Dobbie 4.63 4.63 5.23 5.23 4.17

0.80 0.70 0.60 0.70 0.70

Loom Production Rates for the Scotsville Textile Mill Fabric 1 2 3 4 5

Purchase Price ($/yard)

Regular 5.23 5.23 4.17

Managerial Report Develop a model that can be used to schedule production for the Scottsville Textile Mill, and at the same time, determine how many yards of each fabric must be purchased from another mill. Include a discussion and analysis of the following items in your report: 1. The final production schedule and loom assignments for each fabric. 2. The projected total contribution to profit. 3. A discussion of the value of additional loom time. (The mill is considering purchasing a ninth dobbie loom. What is your estimate of the monthly profit contribution of this additional loom?) 4. A discussion of the objective coefficients’ ranges. 5. A discussion of how the objective of minimizing total costs would provide a different model than the objective of maximizing total profit contribution. (How would the interpretation of the objective coefficients’ ranges differ for these two models?)

SolutionsVariable for the Solution-

X1: Yards of fabric 1 on dobbie loom X2: Yards of fabric 2 on dobbie loom X3: Yards of fabric 3 on dobbie loom X4: Yards of fabric 4 on dobbie loom X5: Yards of fabric 5 on dobbie loom X6: Yards of fabric 3 on regular loom X7: Yards of fabric 4 on regular loom X8: Yards of fabric 5 on regular loom X9: Yards of fabric 1 purchased X10: Yards of fabric 2 purchased X11: Yards of fabric 3 purchased X12: Yards of fabric 4 purchased X13: Yards of fabric 5 purchased

Time taken for producing a yard-

Fabric 1 2 3 4 5

Dobbie 1 / 4.63 = 1 / 4.63 = 1 / 5.23 = 1 / 5.23 = 1 / 4.17 =

0.215983 0.215983 0.191205 0.191205 0.239808

Regular 1 / 5.23 = 0.191205 1 / 5.23 = 0.191205 1 / 4.17 = 0.239808

Total no. of loom hours availableDobbie Loom = 8 looms * 24 hrs per day * 30 days = 5760 hrs. Regular Loom = 30 looms * 24 hrs per day * 30 days = 21,600 hrs.

Calculation of profit / contribution per yard –

Fabric When Manufactured (Selling Price – Variable cost) 1 0.99-0.66=0.33 2 0.86-0.55=0.31 3 1.10-0.49=0.61 4 1.24-0.51=0.73 5 0.70-.50=0.20

When Purchased (Selling Price – Purchase Price) 0.99-0.80=0.19 0.86-0.70=0.16

1.10-0.60=0.50 1.24-0.70=0.54 0.70-.70=0

Max. Z = 0.33x1 + 0.31x2 + 0.61x3 + 0.73x4 + 0.20x5 + 0.61x6 + 0.73x7 + 0.20x8 + 0.19x9 + 0.16x10 + 0.50x11 + 0.54x12 + 0.00x13 Subject to, Constraint1: X1+X9=16500 (demand for fabric 1) Constraint2: X2+X10=22000 (demand for fabric 2) Constraint3: X3+X6+X11=62000 (demand for fabric 3) Constraint4: X4+X7+X12=7500 (demand for fabric 4) Constraint5: X5+X8+X13=62000 (demand for fabric 5) Constraint6: Available Dobbie Loom Hrs.: 0.216X1+0.216X2+0.191X3+0.19X4+0.24X5 ≤ 5760 Constraint7: Available Regular Loom Hrs.: 0.19X6+0.19X7+0.24X8 ≤ 21600 Where, X1, X2, X3, X4, X5, X6, X7, X8, X9, X10, X11, X12, X13 ≥ 0

LINGO OUTPUT Global optimal solution found. Objective value: Infeasibilities: Total solver iterations: Elapsed runtime seconds:

62548.86 0.000000 6 0.03

Model Class:

LP

Total variables: Nonlinear variables: Integer variables:

13 0 0

Total constraints: Nonlinear constraints:

8 0

Total nonzeros: Nonlinear nonzeros:

33 0

Variable X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 Row 1 2 3 4 5 6 7 8

Value 4666.667 22000.00 0.000000 0.000000 0.000000 27868.42 7500.000 62000.00 11833.33 0.000000 34131.58 0.000000 0.000000 Slack or Surplus 62548.86 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000

Reduced Cost 0.000000 0.000000 0.1379630E-01 0.1314815E-01 0.1660819E-01 0.000000 0.000000 0.000000 0.000000 0.1000000E-01 0.000000 0.8000000E-01 0.6105263E-01 Dual Price 1.000000 0.1900000 0.1700000 0.5000000 0.6200000 0.6105263E-01 0.6481481 0.5789474

Ans. 1 X1=4666.667, X2=22000, X3=0000, X4=0000, X5=0000, X6=27868, X7=7500, X8=62000, X9=11833, X10=0000, X11=34131, X12=0000, X13=0000

Ans. 2 Optimal solution = 62,548.86

Ans. 3 If a new loom is added the total no. of working hours will be increased by 1 * 24 hrs. * 30 days = 720 hrs. the new total no. of hours is = 5760+720 = 6,480 hrs. The range is (4756, 8316). Since the new number of hours is within the range the optimal solution is same. So the new objective function is 62548.86+.648*(6480-5760)=63015.42

Ans.4 Ranges in which the basis is unchanged: Objective Coefficient Ranges:

Variable X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13

Current Allowable Allowable Coefficient Increase Decrease 0.3300000 0.1000000E-01 0.1494737E-01 0.3100000 INFINITY 0.1000000E-01 0.6100000 0.1379630E-01 INFINITY 0.7300000 0.1314815E-01 INFINITY 0.2000000 0.1660819E-01 INFINITY 0.6100000 0.1314815E-01 0.1100000 0.7300000 INFINITY 0.1314815E-01 0.2000000 INFINITY 0.1660819E-01 0.1900000 0.1494737E-01 0.1000000E-01 0.1600000 0.1000000E-01 INFINITY 0.5000000 0.1100000 0.1314815E-01 0.5400000 0.8000000E-01 INFINITY 0.000000 0.6105263E-01 INFINITY

Righthand Side Ranges:

Row 2 3 4 5 6 7 8

Current RHS 16500.00 22000.00 62000.00 7500.000 62000.00 5760.000 21600.00

Allowable Increase INFINITY 4666.667 INFINITY 27868.42 22062.50 2556.000 6485.000

Allowable Decrease 11833.33 11833.33 34131.58 7500.000 27020.83 1008.000 5295.000

Ans.5 The optimal solution will be same as the profit is with in the range. For max profit coefficient means profit contribution. For min cost, we assume that the cost per unit is fixed.when the cost is in the range the optimal solution will be same.Coefficient means cost per unit For both max profit and min cost their mean production schedule is the difference between them,which is the meaning of coefficient.