Hi everyone, I have a linear programming/optimization/cutting stock problem. The

Question

Hi everyone,

I have a linear programming/optimization/cutting stock problem. There are so many constraints and variables in this problem, and I can't solve it. I hope someone can help me with it. Thank you very much!

A company sells various pieces of lumber. They sell pieces of lengths: 3 ft, 5 ft, 7 ft, and 11 ft. The company makes these pieces by cutting raws of length 20 ft. The company has to fill customer demand of: 70 pieces of length 3, 45 pieces of length 5, 80 pieces of length 7, and 35 pieces of length 11. The company wants to minimize waste.

List all possible cutting patterns.

Formulate this problem as a linear programming.

Give an interpretation of the dual problem.

Solve the problem (either by hand or by software) by finding the solution that minimizes waste.

Describe the solution in terms of the cuts (and number of cuts) required.

Solve the dual problem, and give an interpretation of the solution to the dual problem.

Explanation / Answer

They sell pieces of lengths: 3 ft, 5 ft, 7 ft, and 11 ft. The company makes these pieces by cutting raws of length 20 ft. The company has to fill customer demand of: 70 pieces of length 3, 45 pieces of length 5, 80 pieces of length 7, and 35 pieces of length 11.

70 x 3 =210

45x5 = 225

80x7 = 560

35x11 = 385

Let X, Y, Z, W represent the number of pieces in each type.

Then X, Y, Z, W >0

Wood used = 3X+5Y+7Z+11W

Our objective is to minimize A, the wastage during the process.

Wood used Z = 20 x n where n is the number of 20 ft pieces.

Thus A = 20n - (3X+5Y+7Z+11W), which represents the wastage wood while curring.

The objective is to minimise this waste.

When x, y, z, w = 70,45,80,35, we have A = 20n-(1380)

A 20 ft wood can be cut into 4 pieces of 5 ft each, or 1 piece for 11 ft and 3 pieces of 3 ft each, or 2 pieces of 7 and 2 pieces of 3.

Thus we get the patterns of each piece as

4y or w+3x, or 2z+2x.

This ensures the maximum utilisation of the 20 ft wood subject to

(x, y, z, w) = (70,45,80,35)

This is the linear model.

The objective is to minimize the wastages.

Total feet required = 210+225+560+385 = 1380.

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