AE4 Module 3 - Linear Programming (Part 2)- IAN & CJ PDF

Preview

Summary

LECTURES ONLY...

Description

IAN Interpret and Navigate!

1. The Kalo Fertilizer Company makes a fertilizer using two chemicals that provide nitrogen, phosphate, and potassium. A pound of ingredient 1 contributes 10 ounces of nitrogen and 6 ounces of phosphate, while a pound of ingredient 2 contributes 2 ounces of nitrogen, 6 ounces of phosphate, and 1 ounce of potassium. Ingredient 1 costs $3 per pound, and ingredient 2 costs $5 per pound. The company wants to know how many pounds of each chemical ingredient to put into a bag of fertilizer to meet the minimum requirements of 20 ounces of nitrogen, 36 ounces of phosphate, and 2 ounces of potassium while minimizing cost. a. Formulate a linear programming model for this problem. b. Solve this model by using graphical analysis. OBJECTIVE FUNCTION Minimize Z = 3x + 5y Where Z = total cost 3x = cost of ingredient 1 5y = cost of ingredient 2 MODEL x = Ingredient 1 CONSTRAINT 10x Nitrogen 6x Phosphate Potassium Minimize Z = 3x + 5y Subject to 10x + 2y ≥ 20 6x + 6y ≥ 36 y≥ 2 x, y ≥ 0

y = Ingredient 2

Resource Available

2y 6y 1

20 36 2

Graphical Solution of Minimization Model 1. Plot the constraints of the Nitrogen 10x + 2y = 20 x – 20/ 10 y – 20/ 10 x=2 y=2 (2, 10)

Phosphate 6x + 6y = 36 x = 36/ 6 y = 36/6 x=6 y=6 (6, 6)

x=0

Potassium y= 2 y =2 (0,2)

y 10

9

8

7

6

5

4

3

2

1 0

7

8

9

10

11

x

8

9

10

11

x

2. Locate the point in the feasible solution y 10

A 10x + 2y = 20 9

8

7

6

5

B 6x + 6y = 36

4

3

C

y= 2

2

1 0

7

3. SOLUTION VALUES of x and y and once the optimal solution point have been found. 10x + 2y = 20 10x + 2y = 20 6x + 6y = 36 6x + 6y = 36 10(1) + 2y = 20 6(1) + 6y = 36 10 + 2y = 20 6 + 6y = 36 6(10x + 2y = 20)6 6y = 36-6 2y = 20 – 10 2(6x + 6y = 36)2 2y = 10 6y =30 60x + 12y = 120 y=5 y=5 12x + 12y = 72 48x + 0 = 48 x=1 (1, 5) 6x + 6y = 36 6x + 6y = 36 y=2 y=2 6(4) + 6y = 36 6x + 6y = 36 24 + 6y = 36 6(y = 2)6 6y = 36 -24 6x + 6y = 36 6y = 12 6y = 12 y=2 6x = 24 (4, 2) x =4 A= (0,10) Z = 3x + 5y Z = 3(0) + 5(10) Z = 50

B = (1,5) Z = 3x + 5y Z = 3(1) + 5(5) Z = 3 + 25 Z = 28 The Kalo Fertilizer Company should have 4 ingredient meet the minimum cost of Php 22.00

C= (4, 2) Z = 3x + 5y Z = 3(4) + 5(2) Z = 12 + 10 Z = 22 1 and 2 for ingredient 2 to

y 10

A 10x + 2y = 20

Z = 3x + 5y

9

8

7

6

5

B 6x + 6y = 36

4

3

C

y= 2

2

1 0

7

8

9

10

11

x

2. Copperfield Mining Company owns two mines, each of which produces three grades of ore — high, medium, and low. The company has a contract to supply a smelting company with at least 12 tons of high-grade ore, 8 tons of medium-grade ore, and 24 tons of lowgrade ore. Each mine produces a certain amount of each type of ore during each hour that it operates. Mine 1 produces 6 tons of high-grade ore, 2 tons of medium-grade ore, and 4 tons of low-grade ore per hour. Mine 2 produces 2, 2, and 12 tons, respectively, of high-, medium-, and low-grade ore per hour. It costs Copperfield $200 per hour to mine each ton of ore from mine 1, and it costs $160 per hour to mine each ton of ore from mine 2. The company wants to determine the number of hours it needs to operate each mine so that its contractual obligations can be met at the lowest cost. a. Formulate a linear programming model for this problem. b. Solve this model by using graphical analysis. LINEAR PROGRAMMING MODEL OBJECTIVE FUNCTION Minimize Z = 200x + 160y Where Z = total cost 200x = cost from mine 1 160y = cost from mine 2 MODEL CONSTRAINT High Medium Low

x = Mine 1

y = Mine 2

Resource Available

6x 2x 4x

2y 2y 12y

12 8 24

Minimize Z = 200x + 160y Subject to 6x + 2y ≥ 12 2x + 2y ≥ 8 4x + 12y ≥ 24 x, y ≥ 0 GRAPHICAL SOLUTION OF MINIMIZATION MODEL 1. Plot the constraints of the High 6x + 2y = 12 x – 12/6 y – 12/2 x=2 y=6 (2, 6)

Medium 2x + 2y = 8 x = 8/ 2 y = 8/2 x=4 y=4 (4, 4)

Low 4x + 12y = 24 x=24/4 y = 24/12 x=6 y=2 (6,2)

y 6

5

4

3

2

1 01

2

3

4

5

6

7

8

9

10

11

x

2. Locate the point in the feasible solution y 6

5

4

A 6x + 2y = 12 2x + 2y = 8

B

3

C

2

D 4x+ 12y= 24 1 01

2

3

4

5

6

7

8

9

10

11

x

3. SOLUTION VALUES of x and y and once the optimal solution point have been found. 2x + 2y = 8 6x + 2y = 12 6x + 2y = 12 2(1) + 2y = 8 2x + 2y = 8 6(1) + 2y = 12 2 + 2y = 8 6x + 2y = 12 6 + 2y = 12 2y = 8-2 2y = 12 – 6 2x + 2y = 8 2y =6 2y = 6 4x + 0 = 4 y=3 x=1 y=3 B: (1, 3)

2x + 2y = 8 4x + 12y = 24 2(2x + 2y = 8)2 4x + 12y = 24 4x + 4y = 16 4x +12y = 24 8y = 8 y=1

2x + 2y = 8 2x + 2(1) = 8 2x + 2 = 8 2x = 8 – 2 2x = 6 x=3

4x + 12y = 24 4x + 12(1) = 24 4x + 12 = 24 4x = 24-12 4x = 12 x=3 C: (3, 1)

A: (0,6) B: (1, 3) C: (3, 1) Z = 200x + 160y Z = 200x + 160y Z = 200x + 160y Z = 200(0) + 160(6) Z = 200(1) + 160(3) Z = 200(3) + 160(1) Z = 0 + 960 Z = 200 + 480 Z = 600 + 160 Z = 680 Z = 960 Z = 760 D: (6, 0) Z = 200x + 160y Z = 200(6) + 160(0) Z = 1,200 + 0 Z = 1200 Copperfield Mining Company should have 1.5 number of hours to operate in mine 1 and 2 to meet the minimum cost of Php 540.00 6

A

6x + 2y = 12

5

4

Z = 200x + 160y

2x + 2y = 8

B

3

C

2

D 4x+ 12y= 24 1 01

2

3

4

5

6

7

8

9

10

11

x

COUNT JUAN Count your Score!

PROBLEMS PART I. The poultry farmer decided to make his own chicken scratch by combining alfalfa and corn in rail car quantities. A rail car of corn costs Php 400 and a rail car of alfalfa costs Php 200. The farmer's chickens have a minimum daily requirement of vitamin K (500 milligrams) and iron (400 milligrams), but it doesn't matter whether those elements come from corn, alfalfa, or some other grain. A unit of corn contains 150 milligrams of vitamin K and 75 milligrams of iron. A unit of alfalfa contains 250 milligrams of vitamin K and 50 milligrams of iron. a. Formulate the linear programming model for this situation. b. Determine the value of x and y that will maximize revenue. c. Find the optimal profit at the optimal solution. LINEAR PROGRAMMING MODEL OBJECTIVE FUNCTION Minimize Z = 400x + 200y Where Z = total cost 400x = cost from corn 200y = cost from alfalfa MODEL x = Corn y = Alfalfa CONSTRAINT 150x 250y Vitamin K Iron 75x 50y Minimize Z = 400x + 200y 150x + 250y ≥ 500 75x + 50y ≥ 400 x, y ≥ 0 GRAPHICAL SOLUTION OF MINIMIZATION MODEL 1. Plot the constraints of the Vitamin K 150x + 250y = 500 x – 500/150 y – 500/250 x = 3.33 y=2 (3.3, 2)

Resource Available 500 400

Iron 75x + 50y = 400 x = 400/75 y = 400/50 x = 5.33 y=8 (5.3, 8)

y 10

9

y

01

2

3

4

5

6

7

8

9

10

11

x

2. Locate the point in the feasible solution y 10

9

A

150x + 250y = 500

B 01

2

3

4

5

75x + 50y = 400 6

7

8

9

10

11

x

3. SOLUTION VALUES of x and y and once the optimal solution point have been found. A: (0,8) C: (5.3, 0) Z = 400x + 200y Z = 400x + 200y Z = 400(0) + 200(8) Z = 400(5.3) + 200(0) Z = 0 + 1,600 Z = 2,120 + 0 Z = 1,600 Z = 2,120 The poultry farmer should have zero milligrams of vitamin k and 8 milligrams of iron to meet the minimum cost of Php 1,600.00

y 10

9

Z = 400x + 200y

A

150x + 250y = 500

B 01

2

3

4

5

75x + 50y = 400 6

7

8

9

10

11

x

PART II. The Munchies Cereal Company makes a cereal from several ingredients. Two of the ingredients, oats and rice, provide vitamins A and B. The company wants to know how many ounces of oats and rice it should include in each box of cereal to meet the minimum requirements of 48 milligrams of vitamin A and 12 milligrams of vitamin B while minimizing cost. An ounce of oats contributes 8 milligrams of vitamin A and 1 milligram of vitamin B, whereas an ounce of rice contributes 6 milligrams of A and 2 milligrams of B. An ounce of oats costs $0.05, and an ounce of rice costs $0.03. a. Formulate a linear programming model for this problem. b. Solve this model by using graphical analysis. LINEAR PROGRAMMING MODEL OBJECTIVE FUNCTION Minimize Z = 0.05x + 0.03y Where Z = total cost 0.05x = cost from oats 0.03y = cost from rice MODEL x = Oats CONSTRAINT Vitamin A 8x 1x Vitamin B Minimize Z = 0.05x + 0.03y Subject to 8x + 6y ≥ 48 x + 2y ≥ 12 x, y ≥ 0

y = Rice

Resource Available

6y 2y

48 12

GRAPHICAL SOLUTION OF MINIMIZATION MODEL 1. Plot the constraints of the Vitamin A 8x + 6y = 48 x – 48/8 y – 48/6 x=6 y=8 (6, 8)

Vitamin B x + 2y = 12 x = 12 y = 12/2 y=6 (12, 6)

y 10

9

8

7

6

5

4

3

2

1 01

2

3

4

5

6

7

8

9

10

11

12

x

2. Locate the point in the feasible solution y 10

9

A 8x + 6y = 48

8

7

6

B 5

4

x + 2y = 12

3

2

C 1 01

2

3

4

5

6

7

8

9

10

11

12

x

3. SOLUTION VALUES of x and y and once the optimal solution point have been found. 8x + 6y = 48 x + 2y = 12 8x + 6y = 48 3(x + 2y = 12)3 8x + 6y = 48 3x + 6y = 36 5x =12 x = 2.4

8x + 6y = 48 8(2.4) + 6y = 48 19.2 + 6y = 48 6y = 48 – 19.2 6y = 28.8 y = 4.8

x + 2y = 12 2.4 + 2y = 12 2y = 12-2.4 2y = 9.6 y = 4.8 B: (2.4, 4.8)

A: (0,8) B: (2.4, 4.8) C: (12, 0) Z = 0.05x + 0.03y Z = 0.05x + 0.03y Z = 0.05x + 0.03y Z = 0.05(0) + 0.03(8) Z = 0.05(2.4) + 0.03(4.8) Z = 0.05(12) + 0.03(0) Z = 0 + 0.24 Z = 0.12 + 0.144 Z = 0.6 + 0 Z = 0.26 Z = 0.6 Z = 0.24 The Munchies Cereal Company should have zero ounces of oats and 8 ounces of rice to meet the minimum cost of Php 0.24.

y 10

Z = 0.05x + 0.03y

9

A 8x + 6y = 48

8

7

6

B

5

4

x + 2y = 12

3

2

C 1 01

2

3

4

5

6

7

8

9

10

11

12

x

PART III. Universal Claims Processors processes insurance claims for large national insurance companies. Most claim processing is done by a large pool of computer operators, some of whom are permanent and some of whom are temporary. A permanent operator can process 16 claims per day, whereas a temporary operator can process 12 per day, and on average the company processes at least 450 claims each day. The company has 40 computer workstations. A permanent operator generates about 0.5 claim with errors each day, whereas a temporary operator averages about 1.4 defective claims per day. The company wants to limit claims with errors to 25 per day. A permanent operator is paid $64 per day, and a temporary operator is paid $42 per day. The company wants to determine the number of permanent and temporary operators to hire in order to minimize costs. a. Formulate a linear programming model for this problem. b. Solve this model by using graphical analysis. LINEAR PROGRAMMING MODEL OBJECTIVE FUNCTION Minimize Z = 64x + 42y Where Z = total cost 64x= cost from permanent operators 42y = cost from temporary operators MODEL CONSTRAINT Claims Workstations Errors

x = Permanent 16x x 0.5x

y = Temporary

Resource Available

12y y 1.4y

450 40 25

Minimize Z = 64x + 42y Subject to 16x + 12y ≥ 450 x + y ≤ 40 0.5x + 1.4y ≤ 25 x, y ≥ 0 GRAPHICAL SOLUTION OF MINIMIZATION MODEL 1. Plot the constraints of the Claims 16x + 12y = 450 x – 450/16 y – 450/12 x = 28.125 y = 37.5 (28.125, 37.5)

Workstations x + y = 40 x = 40 y = 40

(40, 40)

Errors 0.5x + 1.4y = 25 x – 25/0.5 y – 25/1.4 x = 50 y = 17.857 y = 17. 86 (50, 17.86)

y 45

40

35

30

25

20

15

10

5

1 01

5

10

15

45

50

55

x

2. Locate the point in the feasible solution y 45

16x + 12y = 4

40

35

30

25

20

15

10

C 0.5x + 1.4y = 25

5

1 01

D 5

10

15

45

50

55

x

3. SOLUTION VALUES of x and y and once the optimal solution point have been found. 16x + 12y = 450 0.5x + 1.4y = 25 16x 12y = 450 32(0.5x + 1.4y = 25)32 16x + 12y = 450 16x + 44.8y = 800 32.8y = 350 y = 10.671 x + y = 40 0.5x + 1.4y = 25 1.4(x + y = 40)1.4 0.5x + 1.4y = 25 1.4x + 1.4y = 56 0.5x + 1.4y = 25 0.9x =31 x = 34.44 x = 34.4

16x + 12y = 450 0.5x + 1.4y = 25 16x + 12(10.67) = 450 0.5x + 1.4(10.67) = 25 16x + 128.052 = 450 0.5x + 14.94 = 25 16x = 450 – 128.052 0.5x = 25 – 14.94 16x = 321.948 0.5x = 10.06 x = 20.122 x = 20.122 B: (20.122, 10.671) x + y = 40 34.4 + y = 40 y= 40 – 34.4 y = 5.556 y = 5.6

0.5x + 1.4y = 25 0.5(34.444) + 1.4y = 25 0.5(34.444) + 1.4y = 25 17.222 + 1.4y = 25 1.4y = 25 – 17.2 1.4y = 7.778 y = 5.556 y = 5.6

C: (34.444, 5.556) A: (28.125, 0) C: (34.444, 5.556) B: (20.122, 10.671) Z = 64x + 42y Z = 64x + 42y Z = 64x + 42y Z = 64(28.125) + 42(0) Z = 64(20.122) + 42(10.671) Z = 64(34.444) + 42(5.55) Z = 1,800 + 0 Z = 2,204.416 + 233.352 Z = 1,287.808 + 448.182 Z = 1,800 Z = 2,437.768 Z = 1,735.99 Universal Claims Processors should hire 20 permanent operators and 11 temporary minimum cost of Php 1,735.99. y 45

16x + 12y = 4

40

= 0.05x + 0.03y

35

30

25

20

15

10

C

0.5x + 1.4y = 25

5

1 01

D 5

10

15

45

50

55

x

D: (40, 0) Z = 64x + 42y Z = 64(40) + 42(0) Z = 2,560 + 0 Z = 2,560 operators to meet the

PART IV. The Elixer Drug Company produces a drug from two ingredients. Each ingredient contains the same three antibiotics, in different proportions. One gram of ingredient 1 contributes 3 units, and 1 gram of ingredient 2 contributes 1 unit of antibiotic 1; the drug requires 6 units. At least 4 units of antibiotic 2 are required, and the ingredients contribute 1 unit each per gram. At least 12 units of antibiotic 3 are required; a gram of ingredient 1 contributes 2 units, and a gram of ingredient 2 contributes 6 units. The cost for a gram of ingredient 1 is $80, and the cost for a gram of ingredient 2 is $50. The company wants to formulate a linear programming model to determine the number of grams of each ingredient that must go into the drug in order to meet the antibiotic requirements at the minimum cost. a. Formulate a linear programming model for this problem. b. Solve this model by using graphical analysis. LINEAR PROGRAMMING MODEL OBJECTIVE FUNCTION Minimize Z = 80x + 50y Where Z = total cost 80x= cost from Ingredient 1 50y = cost from Ingredient 2 MODEL CONSTRAINT Antibiotic 1 Antibiotic 2 Antibiotic 3

x = Ingredient 1 3x x 2x

y = Ingredient 2

Resource Available

1y y 6y

6 4 12

Minimize Z = 80x + 50y Subject to 3x + y ≥ 6 x+y≥4 2x + 6y ≥ 12 x, y ≥ 0 GRAPHICAL SOLUTION OF MINIMIZATION MODEL 1. Plot the constraints of the Antibiotic 1 3x + y = 6 x – 6/3 y–6 x=2 (2, 6)

x=4

Antibiotic 2 x+y=4 y=4 (4, 4)

Antibiotic 3 2x + 6y = 12 x – 12/2 y – 12/6 x=6 y=2 (6, 2)

y 10

9

8

7

6

5

4

3

1 01

2

3

4

5

7

8

9

10

11

12

x

2. Locate the point in the feasible solution y 10

9

8

3x + y = 6

7

A 6

5

4

B x+y=4

3

C 1 01

2

3

2x + 6y = 12

D 4

5

7

8

9

10

11

12

x

3. SOLUTION VALUES of x and y and once the optimal solution point have been found. 3x + y = 6 x+y=4 3x + y = 6 x+y=4 2x = 2 x=1

3x + y = 6 3(1) + y = 6 3+y=6 y=6-3 y=3

x+y=4 2x + 6y = 12 2(x + y = 4)2 2x + 6y = 12 2x + 2y = 8 2x + 6y = 12 4y = 4 y=1

x+y=4 x+1=4 x=4-1 x=3

x+y=4 1+y=4 y=4-1 y=3 B: (1, 3) 2x + 6y = 12 2x + 6(1) = 12 2x + 6 = 12 2x = 12 - 6 2x = 6 x=3 C: (3, 1)

A: (0, 6) C: (3, 1) D: (6, 0) B: (1, 3) Z = 80x + 50y Z = 80x + 50y Z = 80x + 50y Z = 80x + 50y Z = 80(0) + 50(6) Z = 80(3) + 50(1) Z = 80(6) + 50(0) Z = 80(1) + 50(3) Z = 0 + 300 Z = 240 + 50 Z = 480 + 0 Z = 80 + 150 Z = 300 Z = 290 Z = 480 Z = 230 The Elixer Drug Company should produce a gram of ingredient 1 and 3 grams of ingredient 2 to meet the minimum cost of Php 230.00 y 10

9

8

Z = 80x + 50y

3x + y = 6

7

A 6

5

4

B x+y=4

3

C 1 01

2

3

D 4

5

2x + 6y = 12 7

8

9

10

11

12

x

Did you learn well this week? Write your score here.

Score:

Feedbacks from the Professor: __________________________________________________...