MARY Grace Panganiban - Quatech Activity no PDF

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Ba ta Techniques in Management OM 308 – Quantitative ng as St at e U ni ve rsi ty ACTIVITY NO. 2

NAME: MARY GRACE S. PANGANIBAN SR CODE: 18-56222

SECTION: OM 3201

Solve the following completely in any file form (jpeg, word, Excel, pdf) 1.

Objective Function

Maximize Z = 60X1 + 50X2

Subject to

Assembly

4X1 + 10X2  100 hours

Inspection

2X1 + 1X2  22 hours

Storage

3X1 + 3X2  39 cubic feet

X1, X2 ≥ 0

Assembly X1 X2 0 25

Inspection X1 X2

10 0

0 11

22 0

Storage X1 X2 0 23

13 0

Assembly

Inspection

Storage

4X1 + 10X2  100 4X1 + 10X2= 100

2X1 + 1X2  22 2X1 + 1X2= 22

3X1 + 3X2  39 3X1 + 3X2 = 39

4X1 + 10(0) = 100 X1= 25

2X1 + 1(0) = 22 X1=

3X1 + 3(0) = 39 X1= 13

4(0) + 10X2= 100

11 2(0) + 1X2= 22

3(0) + 3X2= 39

X2 = 10

X = 22

(5, 8) 2

(0, 10)

4

(9, 4)

X2 = 13

5

1 3

(0, 0)

(11, 0)

Corner Points: 1. (0, 0) 2. (0, 10) 3. (11, 0) 4. (5, 8) 5. (9, 4)

Maximize

Ba ta ng a S a e U ni ve

3 (4X1 + 10X2 = 100)

3 (2X1 + 1X2 = 22)

-4 (3X1 + 3X2 = 39) 12X1 + 30X2 = 300

-2 (3X1 + 3X2 = 39) 6X1 + 3X2 = 66

-12X1 - 12X2 = -156 18 X2 = 144

-6X1 - 6X2 = -78

X2 = 8

-3 X2 = -12 X2 = 4

-12X1 - 12X2 = -156 -12X1 – 12(8) = -156

-6X1 - 6X2 = -78 -6X1 – 6(4) = -78

-12X1 – 96 = -156 -12X1 = -156 + 96

-6X1 - 24 = -78 -6X1 = -78 + 24

-12X1 = -60 X =5

-6X1 = -54 X =9

Z = 60X1 + 50X2 Points (0, 0) (0, 10) (11, 0) (5, 8) (9, 4)

ty

Z Z = 60(0) + 50(0) Z = 60(0) + 50(10) Z = 60(11) + 50(0) Z = 60(5) + 50(8) Z = 60(9) + 50(4)

Total $0 $500 $660 $700 $740

Isoprofit

Z = 60X1 + 50X2

Z = 60X1 + 50X2

Z = 60X1 + 50X2

300 = 60X1 + 50X2 300 = 60(0) + 50X2

500 = 60X1 + 50X2 500 = 60(0) + 50X2

740 = 60X1 + 50X2 740 = 60(0) + 50X2

6 = X2 300 = 60X1 + 50(0) 5 = X1

10 = X2 500 = 60X1 + 50(0) 8.3 = X1

14.8 = X2 740 = 60X1 + 50(0) 12.3 = X1

Slack Variables

Ba ta ng as

Z = 60X1 + 50X2 + S1 + S2 + S3 (0, 10) X1 = 0,

X1, X2 ≥ 0

S1 + S2 + S3 ≥ 0

X2 = 10, S1 = 0, S2 = 12, S3 = 9

4X1 + 10X2 + S1 = 100 4(0) + 10(10) + S1 = 100 S1 = 100 – 100 S1= 0

at e U

2X1 + 1X2 + S2 = 22 2(0) + 1(10) + S2 = 22 S2 = 22 – 10 S2= 12

3X1 + 3X2 + S3 = 39 3(0) + 3(10) + S3 = 39 S3 = 39 – 30 S3= 9

(11, 0) X1 = 11, X2 = 0, S1 = 56, S2 = 0, S3 = 6 4X1 + 10X2 + S1 = 100 4(11) + 10(0) + S1 = 100 S1 = 100 – 44 S1= 56

(5, 8) X1 = 5,

2X1 + 1X2 + S2 = 22 2(11) + 1(0) + S2 = 22 S2 = 22 – 22 S2= 0

3X1 + 3X2 + S3 = 39 3(11) + 3(0) + S3 = 39 S3 = 39 – 33 S3= 6

ve rsi

X2 = 8, S1 = 0, S2 = 4, S3 = 0

4X1 + 10X2 + S1 = 100 4(5) + 10(8) + S1 = 100 S1 = 100 – 100 S1= 0

2X1 + 1X2 + S2 = 22 2(5) + 1(8) + S2 = 22 S2 = 22 – 18 S2= 4

3X1 + 3X2 + S3 = 39 3(5) + 3(8) + S3 = 39 S3 = 39 – 39 S3= 0

(9, 4) X1 = 0,

X2 = 10, S1 =24, S2 = 0, S3 = 0

4X1 + 10X2 + S1 = 100 4(9) + 10(4) + S1 = 100 S1 = 100 – 76 S1= 24

2.

2X1 + 1X2 + S2 = 22 2(9) + 1(4) + S2 = 22 S2 = 22 – 22 S2= 0

3X1 + 3X2 + S3 = 39 3(9) + 3(4) + S3 = 39 S3 = 39 – 39 S3= 0

Suppose 240 acres of land available. Profit: $40/acres corn; $30/acre oats. Have 320 hours of labour available. Corn takes 2 hours of labour per acre of land; however, Oats requires 1 hour of labour per acre of land

Problem: How many acres of each should be planted to maximize profit? (1) Formulate the problem. (2) Solve the problem using graphical method (corner solution) and Iso profit method.

Ba ta ng as St at e U

Let say: X1 = # of acres of corn X2 = # of acres of oats Maximization: Z = 40X1 + 30X2 Subject to: X1 + X2  240 2X1 + X2  320 X1 , X2 ≥ 0

X1 + X2  240 X1 + X2 = 240 X1 + (0) = 240 X1= 240 (0) + X2= 240 X2 = 240 Corner Points:

2

4

(160, 0)

1 3

2X1 + X2  320 2X1 + X2= 320

2X1 + (0) = 320 X1= 160

X1

X2

0 240

240 0

X1

X2

0 160

320 0

2(0) + X2= 320 X2 = 320

ty

-2(X1 + X2 = 240) 2X1 + X2= 320

1. (0, 0) 2. (0, 240) 3. (160, 0) 4. (80, 160)

-2X1 - 2X2 = -480 X2 = 160 2X1 + 160 = 320 2X1 = 160 X1= 80

Maximize Z = 40X1 + 30X2 Points (0, 0) (0, 240) (160, 0) (80, 160)

Z Z = 40(0) + 30(0) Z = 40(0) + 30(240) Z = 40(160) + 30(0) Z = 40(80) + 30(160)

Total $0 $7200 $6400 $8000

Isoprofit

Z = 40X1 + 30X2

as St at e U i e i ty

6000 = 40X1 + 30X2 6000 = 40(0) + 30X2 200 = X2

6000 = 40X1 + 30(0) 150 = X1

(160, 0)

Z = 40X1 + 30X2

Z = 40X1 + 30X2

7200 = 40X1 + 30X2 7200 = 40(0) + 30X2

8000 = 40X1 + 30X2 8000 = 40(0) + 30X2

240 = X2 7200 = 40X1 + 30(0) 180 = X1

266.67 = X2 8000 = 40X1 + 30(0) 200 = X1

Slack Variables Z = 40X1 + 30X2 + S1 + S2

X1, X2 ≥ 0

(0, 240)

X1 + X2 + S1 = 240

2X1 + 1X2 + S2 = 320

S1 + S2 ≥ 0

(0) + (240) + S1 = 240 S1 = 240 – 240 S1= 0

2(0) + 1(240) + S2 = 320 S2 = 320 – 240 S2= 80...