Mathematical model for one dimensional Cutting stock

1. Process Description

1.1 Slitting

Slitting is also a shearing process, but rather than making cuts at the end of a work piece like shearing, slitting is used to cut a wide coil of metal into a number of narrower coils as the main coil is moved through the slitter. During the slitting process, the metal coil passes lengthwise through the slitter's circular blades.

The slitting process characteristics include:

Being restricted to cutting relatively thin materials (0.001 to 0.125 in.), leaving left-over burrs on slit edges of the narrower coals, its ability to be used on both ferrous and nonferrous metals, Its categorization as a high production designed to control metal coil width.

The illustration that follows provides a two-dimensional look at a typical coil slitting process. Note how the metal work piece is drawn past the upper and a lower slitting blade, leaving two coils the same length as the original wide coils.

Slitting can be used equally well for both sheet or coil rolls.

Slitting blades are designed depending on the job required. The three critical determinants of the blade configuration include:

The work piece material thickness. The type of material to be slit. The tolerances that must be held while slitting.

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1.2 Process Conversion

Hot Rolling HR Coil Slitting Machine

Slitting Process Slit Coils

Tubes

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2. Problem Definition

2.1 Cutting Stock Problems Identification

As reported by Tokuyama and Nomuyuki (1981) of Sumitomo Metal Industries, Japan, the characteristics of the cutting stock problems in the iron and steel industries are as follows:

There are a variety of criteria such as maximizing yield and increasing efficiency. Cutting problems are usually accompanied by inventory stocking problems.

Practical algorithms that give near-optimal solutions in the real world have been developed. In their paper, Tokuyama and Nomuyuki discuss applications to one-dimensional cutting of large sections and two dimensional cutting of plates.

2.2 Cutting Stock Optimization

During slitting of hot and/or cold rolled coils for tube making purpose, it has observed that some amount of side trimmings generated which leads to loss of metal sheets which has further no use in prime product making. If we taking the consideration of this kind of loss in big project then the loss is approximately 23%.

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3. Formulation of Linear Programming

There are mainly four steps in the mathematical formulation of linear programming problem as a mathematical model. We will discuss formulation of those problems which involve only two variables.

Identify the decision variables and assign symbols. These decision variables are those quantities whose values we wish to determine.

Identify the set of constraints and express them as linear equations/in equations in terms of the decision variables. These constraints are the given conditions.

Identify the objective function and express it as a linear function of decision variables. It might take the form of maximizing profit or production or minimizing cost.

Add the non-negativity restrictions on the decision variables, as in the physical problems, negative values of decision variables have no valid interpretation.

Here we are basically try to make out the formulation of decision variable and the constrains for minimize the loss of metal during slitting process. In one dimensional cutting stock, we consider only that dimension which is fixed in nature or which is directly effect the result.

Let’s take a 20 inch width of hot rolled coil (sheet) and we have to cut the coil vertically. In this case the width of the coil is our central focus because the width of slit coil will be going to be the circumference of the tube. So that we have to make the formulation in such a way where we can cut the main coil as well as to minimize the leftover.

So let start with a one dimensional cutting stock problem and try to make the formulation.

3.1 Taking an example where one processing agency gets an order of slit coils of four types from a tube manufacturing company.

9 inch 511 nos. 8 inch 301 nos. 7 inch 263 nos. 6 inch 383 nos.

It is assuming that all the cut sheet of same length (because length of the coils may differ as per the weight). The problem is to cut the sheets to minimize the waste. (As per the given picture)

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20”

Variable length depending Upon the coil

If we cut two 9” slit then 2” will be the waste, similarly if we cut two 8” slit then 4” will be the waste and so on. So the first thing we have to do to try and find out; how many cutting patterns are possible? These patterns are made according to the main coil width and keep in mind that any of the combination should not crossed the coil width, which is going to slit.

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4.2 COMBINATION CHART

Slit coil width (in inch) Pattern

Pattern No. Combination of measurement [9” 8” 7” 6”] Waste (inch)

1. 2X9” = 18” [2 0 0 0] waste= 2”2. 2X8” = 16” [0 2 0 0] waste= 4”3. (2X7”)+(1X6”) =20” [0 0 2 1] waste= 04. 3X6” = 18” [0 0 0 3] waste= 2”5. (1X9”)+(1X8”) = 17” [1 1 0 0] waste= 3”6. (1X9”)+(1X7”) = 16” [1 0 1 0] waste= 4”7. (1X9”)+(1X6”) = 15” [1 0 0 1] waste= 5”8. (1X8”)+(1X7”) = 15” [0 1 1 0] waste= 5”9. (1X8”)+(2X6”) = 20” [0 1 0 2] waste= 010. (1X7”)+(2X6”) = 19” [0 0 1 2] waste= 1”

Let,

Xj = number of sheets cut using pattern j. Shaded portion is optimum, means no waste.

Constraints

2X1 + X5 + X6 + X7 >= 511 9” Sheets

2X2 + X5 + X8 + X9 >= 301 8” Sheets

2X3 + X6 + X8 + X10 >= 263 7” Sheets

X3+3X4+X7+2X9+2X10 >= 383 6” Sheets

Xj >= 0

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If we so the object function is to minimize the waste, so from the above chart we get the function as: -

Minimize (2X1+4X2+2X4+3X5+4X6+5X7+5X8+X10)

However if we make an assumption that excess sheet produce in every width (they are all inequalities because it may not be possible to get the exact number required), then the objective function become.

Minimize (2X1+4X2+2X4+3X5+4X6+5X7+5X8+X10) + 9(2X1 + X5 + X6 + X7 – 511) + 8(2X2 + X5 + X8 + X9 – 301) + 7(2X3 + X6 + X8 + X10 – 263) + 6(X3+3X4+X7+2X9+2X10 – 383)

This further reduces and we get,

Minimize (20X1 + 20X2 + 20X3 + 20X4 + 20X5 + 20X6 + 20X7 + 20X8 + 20X9 + 20X10) – 11146

Now this 11146 is a constant and we taken out from formulation, because it is not depends on variables. So this constant can be removed.

Similarly the 20 which is common multiplier of all variable, so this 20 also can be taken out. Now the objective function become

Minimize ∑Xj

So the problem of minimizing the waste now become minimizing the total number of cuts. If we assume that the excess material cut is also treated as waste, and then we can show that the cutting stock problem to minimize waste now becomes the minimizing total number of cuts.

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4. Reference

“A Survey of Mathematical Programming Application in Integrated Steel Plants” By, Goutam Dutta & Robert Fourer.Indian Institute of Management, Ahmadabad (INDIA)Northwestern University, Illinois (USA).

“Application of Cutting Stock Problem to a Construction Company” (A Case Study) By, Seda Alp, Gurdel Ertek & S Ilker Birbil.Sabanchi University, Istanbul (TURKEY)

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