13. Bandford Tile Company makes ceramic and porcelain tile for residential and c

Question

13. Bandford Tile Company makes ceramic and porcelain tile for residential and commercial use. They produce three different grades of tile (for walls, residential floor- ing, and commercial flooring), each of which requires different amounts of materials and production time, and generates different contribution to profit. The information below shows the percentage of materials needed for each grade and the profit per square foot. ental tea uses tea Californ GradeI Grade II Grade IlI Profit/square foot $2.50 $4.00 5.00 50% Qua Clay Silica Sand Feldspar 50% 5% 20% 25% 10% 15% 50% 15% 15% 30% 15% Prer Duk 25% 40% Each week, Sanford Tile receives raw material ship- ments and the operations manager must schedule the plant to efficiently use the materials to maximize profit- Net pro mium, irm's r Indian t and 16,4 solve a ability. Currently, inventory consists of 6,000 paunds clay, 3,000 ds of silica, 5,000 pounds of sand, and Gemand varies tor erent grades, marketing estimates that at most 8,000 square feet of Grade tile should be produced, and that at least 1,500 square feet of Grade I tiles are the pr inforn required. Each square foot of tile weighs approximately two pounds. Develop a linear optimization model to determine how many of each grade of tile the company should make next week to maximize profit contribution. a. b. Implement your model on a spreadsheet and find an optimal solution. Explain the sensitivity information for the objective coefficients. What happens if the profit on Grade Iis increased by $0.05? If an additional 500 pounds of feldspar is available, how will the optimal solution be affected? c. d. e. Suppose that 1,000 pounds of clay are foundit inferior quality. What should the company

Explanation / Answer

a)

Let g1, g2,g3 be the sq. foot of grade1, grade2 and grade3 tiles produced.

The model is then

Maximise the profit

2.5*g1+4*g2+5*g3

subject to

the inventory constraints of materials

0.5*2*g1+0.3*2*g2+0.25*2*g3 <=6000 (clay)

0.05*2*g1+0.15*2*g2+0.1*2*g3 <=3000 (silica)

0.2*2*g1+0.15*2*g2+0.15*2*g3 <=5000 (sand)

0.25*2*g1+0.4*2*g2+0.5*2*g3 <=8000 (feldspar)

g1 >=1500 (min sq. ft of grade1 tiles)

g3 <= 8000 (max sq. ft of grade 3 tiles)

g1, g2 and g3 being integers

b)

model implementation

optimal solution is

c)

This was the original solution

This is after the grade1 profit increased to 2.55 dollars / sq. ft

Earlier it was bound by 1500 minimum requirement and hence was contrained to minimum

Now it is much profitable to go beyond the minimum 1500 and hence broken that barrier.

d)

If additional 500 pounds of feldspare is available, then

Grade 1 Grade 2 Grade 3 Profit/ sq foot 2.5 4 5 Clay 0.5 0.3 0.25 6000 Silica 0.05 0.15 0.1 3000 Sand 0.2 0.15 0.15 5000 Feldspar 0.25 0.4 0.5 8000 sq ft 1500 8000 sq ft of tiles 1500 0 7250 pounds of tiles =+B9*2 =+C9*2 =+D9*2 Required Clay =+B$10*B3 =+C$10*C3 =+D$10*D3 =+SUM(B13:D13) Silica =+B$10*B4 =+C$10*C4 =+D$10*D4 =+SUM(B14:D14) Sand =+B$10*B5 =+C$10*C5 =+D$10*D5 =+SUM(B15:D15) Feldspar =+B$10*B6 =+C$10*C6 =+D$10*D6 =+SUM(B16:D16) Profit =+B9*B2 =+C9*C2 =+D9*D2 =+SUM(B18:D18)

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