Question

STAR Co. provides paper to smaller companies whose volumes are not large enough to warrant dealing directly with the paper mill. STAR receives 100-feet-wide paper rolls from the mill and cuts the rolls into smaller rolls of widths 12, 15, and 30 feet. The demands for these widths vary from week to week. The following cutting patterns have been established:

	Number of:
Pattern	12ft.	15ft.	30ft.	Trim Loss
1	0	4	1	10 ft.
2	4	3	0	7 ft.
3	8	0	0	4 ft.
4	2	1	2	1 ft.
5	2	3	1	1 ft.

Trim loss is the leftover paper from a pattern (e.g., for pattern 4, 2(12) + 1(15) + 2(30) = 99 feet used resulting in 100-99 = 1 foot of trim loss). Orders in hand for the coming week are 5,670 12-foot rolls, 1,680 15-foot rolls, and 3,350 30-foot rolls. Any of the three types of rolls produced in excess of the orders in hand will be sold on the open market at the selling price. No inventory is held.

Optimal Solution:

(a)	Formulate an integer programming model that will determine how many 100-foot rolls to cut into each of the five patterns in order to minimize trim loss. If your answer is zero enter “0” and if the constant is "1" it must be entered in the box.
	Min x1 + x2 + x3 + x4 + x5 s.t. x1 + x2 + x3 + x4 + x5 - Select your answer -≤≥=Item 11 12-foot rolls x1 + x2 + x3 + x4 + x5 - Select your answer -≤≥=Item 18 15-foot rolls x1 + x2 + x3 + x4 + x5 - Select your answer -≤≥=Item 25 30-foot rolls x1, x2, x3, x4, x5 are integers
(b)	Solve the model formulated in part a. What is the minimal amount of trim loss?
	Trim Loss:   feet
	How many of each pattern should be used and how many of each type of roll will be sold on the open market? If your answer is zero enter “0”.
	Pattern Number Rolls Used 1 2 3 4 5

Answer 1

a. Integer Linear Programming Model:

Decision Variables: Let X1, X2, X3, X4, X5 be the number of 100 ft rolls required from each of the patterns 1 to 5.

Objective Function: Minimise the total trim loss.

a 100 ft pattern 1 roll has 10ft trim loss and fo X1 number of rolls the trim loss would be 10X1. Similarly, the trim loss for the X quantities of all other pattern are expressed. The total of these values is the total trim loss. The objective is to minimize the total trim loss. Hence it can be expressed as:

Min 10X1+7X2+4X3+1X4+1X5

s.t Constraints:

0X1+4X2+8X3+2X4+2X5 >= 5670

4X1+3X2+0X3+1X4+3X5 >= 1680

1X1+0X2+0X3+2X4+1X5 >= 3350

the number of rolls in each pattern and width is given. For example, 4 rolls in pattern 2 and 12 ft width means that 4 rolls with 12ft width can be cut from a single 100ft roll. We need X2 number of 100 ft rolls of pattern 2. So, the requirement for 12ft rolls in pattern 2 would be 4X2.

Using the same logic, the requirement for each pattern and width is expressed in terms of X as mentioned above.

The requirement received for next week is 5670 for 12ft rolls and the excess produced would be sold in the open market at selling price. Hence the total requirement of 12 ft rolls should be such that it would meet the 5670 requirements. It can be exactly 5670 or more than this.

So the expression is

0X1+4X2+8X3+2X4+2X5 >= 5670

using the same logic the other two constraints can also be written as mentioned above.

As this is an integer programming, X1 to X5 are considered as integers. (no fractions)

b. Solving the integer programming using excel solver:

Spreadsheet model along with formulae:

Solver Parameters:

Solution:

Minimal amount of trim loss = 2836 feet (objective value)

Rolls used: (greenhighlighted cells in the above screenshot with heading no. of rolls)

Pattern Number	Rolles Used
1	0
2	0
3	290
4	1674
5	2

The number of rolls sold in the open market : (unused constraints)

Type of Roll	Sold in open market
12ft	2
15ft	0
30ft	0

In the above screenshot the LHS of constraints provides the data for utilization of resources.

For the 12ft roll, the requirement was 5670 but the use was 5672. As the excess rolls are sold in open market, these excess 2 units would be sold in the open market.

This value is 0 for the rest two rolls as the constraint is fully satisfied.