Copy of QTV Compilation.docx

CHAPTER 1

Exercise 1.4

1. 4x – 4y = 12 4x – 4y = 12 1. 4(x-y) = 12 4x = 12 -4y = 12 (3,0)

4 4 -4 -4 X = 3 Y = - 3

(0,-3)

2. 4(x + y) = 8 4(x + y) = 8 2. 4(x+y) =84x + 4y = 8 4x + 4y = 84x = 8 4x = 8 (0,2)

4 4 4 4 X = 2 X = 2 (2,0)

3. 3(x + 1) = 3(y-1) 3(x + 1) = 3(y-1) 3. 3(x+1) = 3(y-1)3x + 3 = 3y – 3 3x – 3y = -3 -33x – 3y = -6 -3y = -6 (2,0)

3x = -6 -3 -3 3 3 Y = 2 (0,-2)

X = -2

4. 5[2x+65 ]=(2 y−4 )5 5[2x+65 ]=(2 y−4 )5

2x + 6 = 10y – 20 2x- 10 = 20 – 6 4. [2x+65 ]=(2 y−4 )

2x – 10y = -26 -10y = -26 (0,2.6)

2x = -26 -10 -10 (-13,0)

2 2 Y = 2.6 X = -13

5. 2x + 6y + 6 = 0 2x 6y = -6 5. 2x + 6y + 6 = 02x 6y = -6 6y = -6 (-3,0)

2x = -6 6 6

2 2 Y = -1 (0,-1)

6. X + 2y = 7 x + 2y = 7 6. X + 2y = 7 (0,3.5)

X = 7 2y = 7 2 2

Y = 3.5 (7,0)

7. 2x – y = 4 2x – y = 4 7. 2x – y = 4 (2,0)

2x = 4 - ( -y = 4)2 2 y = -4 X = 2 (0,-4)

8. X + 3y = 9 x + 3y = 9 8. X + 3y = 93y = 9 x = 9

3 3 Y = 3

9. 2x + 6 = 2 – 2y + 4y 2x + 6 = 2 – 2y + 4y 9. 2(x=3)=2(1-y)=4y

2x + 2y – 4y = 2- 62x + 2y – 4y = 2- 6 (0,2)

2x – 2y = -4 2x – 2y = -4 (-2,0)

2x = -4 2y = -4 2 2 -2 -2 X = -2 Y = 2

10. X + 4 = 12 + 4y X + 4 = 12 + 4y 10. X + 4 = 6=2yX – 4y = 12 – 4 X – 4y = 12 – 4 2

X – 4y = 8 X – 4y = 8 X = 8 -4y = 8

-4 -4 (0,-2)

Y = -2

Exercise 1.5

1. 2x + 2y = 8 2(2) + 2y = 8 Checking:4x – 2y = 4 4 + 2y = 8 1st equation 2nd equation6x = 12 2y = 8 – 4 2(2) + 2(2) = 8 4(2) – 2(2) = 4 6 6 2y = 4 4 + 4 = 8 8 – 4 = 4 X = 2 2 2 8 = 8 4 = 4

2. 4x – 6y – 10 = 0 4x – 6y = 10 4(1.90) – 6y = 10-12x + 3y = -2 multiply by 2 -24x + 6y = -48 7.60 – 6y = 10

-20x = -38 -6y = 10 – 7.60 -20 -20 -6y = 2.40

X = 1.90 6 6 Y = - 0.40

Checking:1st equation 2nd equation4(1.90) – 6(-0.40) = 10 24(1.90) – 6 (-0.40) = 487.60 + 2.40 = 10 45.60 + 2.40 = 48 10 = 10 48 = 48

3. -2x + 2y = -2 -2(-1) + 2y = -2 Checking: 4x – 2y = 0 2 + 2y = -2 1st equation 2nd equation 2x = -2 2y = -2 – 2 -2(-1) + 2(-2) = -2 4 (-1) – 2 (-2) = 0 2 2 2y = -4 2 – 4 = -2 -4 -4 = 0 X = -1 2 2 -2 = -2 0 = 0

Y = -2

4. 3x – y = -100 3(50) – y = -100 Checking:-5x + y = 0 150 – y = -100 1st equation 2nd equation-2x = -100 -y = -100 – 150 3(50) – 250 = -100 -5(50) + 250 = 0 2 2 -y = -250 150 – 250 = -100 -250 + 250 = 0 X = 50 -1 -1 - 100 = - 100 0 = 0

Y = 250

5. Y – 2x = 40 multiply by 5 5y – 10x = 200-5y + 10x = 20 -5y + 10y = 20

= 1806. 2x – y = -10

3x + y = 10 – 5

2x – y = -10 2 (-3) – y = -10 Checking:3x + y = -5 -6 - y = -10 First equation Second Equation5x = -15 - y = -10 + 6 2 (-3) – 4 = -10 3 (-3) + 4 = - 5 5 5 - y = - 4 -6 – 4 = -10 -6 + 4 = -5 x = -3 - 1 = - 1 -10 = -10 -5 = -5

7. 5x + y = 6 Multiply by -2x – 2y = 21 Checking:10x + 2y = 12 10 (3) + 2y = 12 First Equation Second Equation x – 2y = 21 30 + 2y = 12 10 (3) + 2(-9) = 12

3 – 2 (-9) = 2111x = 33 2y = 12 – 30 30 – 18 = 12 3 + 18 = 2111 11 2y = -18 12 = 12 21 = 21 x = 3 2 = 2

y = -9

8. 2x + 2y = 16 2 (2) + 2y = 162x – 2y = -8 4 + 2y = 16 Checking:4x = 8 2y = 16 – 4 First Equation Second Equation4 4 2y = 12 2 (2) + 2(6) = 16 2 (2) – 2 (6) = - 8 X = 2 2 = 2 4 + 12 = 16 4 - 12 = - 8

Y = 6 16 = 16 - 8 = - 8

9. 0.2x – 0.3y = 10 Multiply by (2)0.5x + 0.2y = 6 Multiply by (3)

.40x - .60y = 20 .40 (15.26) - .60y = 20 Checking:1.50x + .60y = 9 6.10 - .60y = 20 First Equation1.90x = 29 -.60y = 20 – 6.10 .40 (15.26) - .60(-23.16) = 201.90 1.9 -.60y = 13.90 6.10 + 13.90 = 20 X = 15.26 -.60 = -.60 20 = 20 Y = - 23.16

Second Equation1.50 (15.26) + .60 (-23.16) = 9

22.80 – 13.89 = 9 9 = 9

10.2x – y = - 10 2 (- 3) – y = - 10

3x + y = - 5 - 6 – y = - 10 Checking:

5x = - 15 - y = - 10 + 6 First Equation Second Equation

5 = 5 - y = - 4 2 (- 3) - 4 = - 10 3 (- 3) + 4 = - 5 X = - 3 -1 = -1 -6 - 4 = - 10 -9 + 4 = - 5

Y = 4 - 10 = - 10 - 5 = - 5

Chapter 2

Exercise 2.2

1. Minimize: 10x1+20x2

Subject: 4x1+2x2>20 2x1+6x2>30SOLUTION:

4x1+2x2=20 2x1+6x1=30

4x1 = 20 2x1+0=304 4 2x1 = 30X1 = 5 2 2 X1 = 15

4x1+2x2=20 2x1+6x2=300+2x2=20 0+6x2=30

2x2 = 20 6x2 = 302 2 6 6 X2 = 10 x2 = 5

2. Maximize: 30x1+40x2

Subject: 4x1+2x2<32 X1+3x2<18 X1, x2 >0SOLUTION: 4x1+2x2=32 x1+3x2=184x1+0=32 x1+0=184x1 = 32 x1 = 183 4

X1 = 8 x1+3x2=18 0+3x2=18 3x2=18 4x1+2x2=32 3 3 0+ 2x2=32 X2 = 6

2x2 = 322 2

X2 = 16

Subject: 10x1+2x2>32 2x1+6x2>12 X1, x2 >0SOLUTION:

10x1+2x2=32 2x1+6x2=1210x1+0=32 0+6x2=12 10x1 = 32 6x2=1210 10 6 6 X1 = 3.5 x2 = 2

10x1+2x2=32 2x1+6x2=12 0+2x2=32 0+6x2=12 2x2=32 6x2=12 2 2 6 6 X2 = 16 x2 = 2

Subject: 4x1+2x2>10 2x1+6x2>15

X1, x2 > 0SOLUTION:

4x1+2x2=10 2x1+6x2=154x1+0=10 2x1+0=154x1 = 10 2x1 = 15 4 4 2 2X1 = 2.5 x1 = 7.5

4x1+2x2=10 2x1+6x2=150+2x2=10 0+6x2=15 2x2 = 10 6x2 = 15 2 2 6 6 X2 = 5 x2 = 2.5

5. Maximize: 20x1+30x2

Subject: 4x1+2x2<4 4x1+8x2<16 X1, x2 > 0SOLUTION:

4x1+2x2=4 4x1+8x2=164x1+0=4 4x1+0= 16 4x1 = 4 4x1 = 16 4 4 4 4 X1 = 1 x1 = 4

4x1+2x2=4 4x1+8x2=160+2x2=4 0+8x2=16 2x2 = 4 8x2 = 16 2 2 8 8 X2 = 2 x2 = 2

Subject: x1+x2>4 X1+x2>8 X1<6 X1, x2 > 0SOLUTIONS:

x1+x2=4 X1+x2=8 X1=6X1+0=4 X1+0=8

X1 = 4 x1 = 8

x1+x2=4 X1+x2=80+x2=4 0+x2=8 X2 = 4 x2 = 8

9. Minimize: 80x1 + 75x2

Subject to: 3x1 + 5x2 ≤ 154x1 + x2 ≤ 16

x1, x2 ≥ 0SOLUTION:

3x1 + 5x2 = 15 3x1 + 5x2 = 15 5x2 = 15 3x1 = 15 5 5 3 3 X2 = 3 X1 = 5

4x1 + x2 ≥ 16 4x1 + x2 ≥ 16

x2 = 16 4x1 = 16 x2 = 16 4 4

x1 = 4

10. Maximize: 100x1 + 200x2

Subject to: x1 + x2 ≥ 4x1 + x2 ≤ 6x2 ≤ 5 x1, x2 ≥ 0

SOLUTION:

x1 + x2 ≥ 4 x1 + x2 ≤ 4 x2 = 4 x1 = 4

x1 + x2 ≤ 6 x1 + x2 ≤ 6x2 = 6 x1 = 6

Exercise 2.3

Problem No. 1

Given:Max.P = 60x1 + 100x2

2x1 + 2x2 ≤ 1402x1 + 4x2 ≤ 160X2 ≤ 60and x1 ≥ 0; x2 ≥ 0

I. Decision VariablesLet x1 = no. of units to be produced and sold to maximize profitLet x2 = no. of units to be produced and sold to maximize profit

II. Objective FunctionMax.P = 60x1 + 100x2

III. Subject to:2x1 + 2x2 ≤ 1402x1 + 4x2 ≤ 160X2 ≤ 60and x1 ≥ 0; x2 ≥ 0

IV. Graphical Method:

2x1 + 2x2 ≤ 140 2x1 + 4x2 ≤ 160x1 0 70 x1 0 40x2 70 0 x2 80 0

2x1+2x2=140 2x1 + 4x2 = 160 2(0) + 2x2 = 140 2(0) + 4x2 = 160

2x2 = 140 4x2 = 160 2 2 4 4 X2 = 70 x2 = 40 2x1+2x2=140 2x1 + 4x1 = 160 2x1 + 2(0) = 140 2x1 + 4(0) = 160 2x1 = 140 2x1 = 160 2 2 2 2 X1 = 70 x1 = 80

2x1+2x2=140 *2 = 4x1+4x2 = 280 2x1+2x2 = 140 - 2x1 + 4x2 = 160 2(60) + 2x2 = 140

2x1 = 120 120 + 2x2 = 140 X1 = 60 2x2 = 140 – 120

2x2 = 20 2 2 X2 = 10

P1 (0, 40);max. P = 60x1 +100x2

= 60 (0) + 100(40) = 4000

P0 (0, 0) max. P = 60x1 +100x2

= 60(0) + 100 (0) = 0

P4 (70, 0) max. P = 60x1 +100x2

= 60(70) + 100 (0) = 4200

P5 (60, 0) max. P = 60x1 +100x2

= 60(60) + 1000 (0) = 3600 + 1000 = 4600

2x1 + 2x2 ≤ 140 2x1 + 4x2 ≤ 1602(60) + 2(10) ≤ 140 2(60) + 4 (10) ≤ 160 120 + 20 ≤ 140 120 + 40 ≤ 160

140 ≤ 140 160 ≤ 160

V. Decision

In order to maximize profit to P4,600, the entity should manufacture and sell 60 units of x1 and 10 units of x2.

Problem No. 3

Tabulation

Style A (x) Style B (y) TOTAL

No. of hours 1 hours 2 hours 24 hours

No. of units - - 16 units

Profit P 40.00 P 30.00

I. Decision variables

Let X = no. of units for style A

Y = no. of units for style B

II. Objective functionMax P = 40x + 30y

III. Subject to:Explicit constraints: X + 2y ≤ 24

X + Y ≤ 16

IV. Graphical method

Equation No. 1 Equation no. 2

X + 2y ≤ 24 X + Y ≤ 16

Optimal Solution:

Max P = 40x + 30y

= 40 (8) + 30 (8)

= 320 + 240

Max P = 560

X 0 16

Y 16 0

X 0 24

Y 12 0

Checking of the constraints:

X + 2y ≤ 24 X + Y ≤ 16

8 + 2 (8) ≤ 24 8 + 8 ≤ 16

24 ≤ 24 16 ≤ 16

V. DecisionThe manufacturer should produce 8 units of both style A and

style B to maximize the profit of Php 560.00 in daily basis.

Problem No. 4

Tabulation

Scooter (x) Bicycle (y) TOTAL

Center 1 4 hours 6 hours 120 hours

Center 2 10 hours 3 hours 180 units

Profit P 60.00 P 40.00I. Decision variables

Let X = no. of units for scooter

Y = no. of units for bicycle

II. Objective functionsMax P = 60x + 40y

III. Subject to:Explicit constraints:

4x + 6y ≤ 12010x + 3y ≤ 180

4x + 6y ≤ 120 10x + 3y ≤ 180

Checking of the constraints

4x + 6y ≤ 120 10x + 3y ≤ 180

4(15) + 6(10) ≤ 120 10(15) + 3(10) ≤ 180

120 ≤ 120 180 ≤ 180

V. DecisionThe manufacturer should produce 15 units of scooter and 10

units of bicycle to maximize the profit of P 1,300.00.

Problem No. 5

Tabulation

Raisin (x) Walnuts (y) TOTAL

Cookies 1 3 oz 5 oz 53 oz

Cookies 2 3 oz 2 oz 32oz

Profit P 48.00 P 60.00

Optimal Solution:

Max P = 60x + 40y

= 60 (15) + 40 (10)

= 900 + 400

Max P = P 1,300

X 0 30

Y 20 0

X 0 18

Y 60 0

I. Decision variables

Let X = no. of first type of cookies

Y = no. of second type of cookies

II. Objective functionsMax P = 48x + 60y

III. Subject to:Explicit constraints:

3x + 5y ≤ 533x + 2y ≤ 32

3x + 5y ≤ 53 3x + 2y ≤ 32

Checking the constraints:

3x + 5y ≤ 53 3x + 2y ≤ 32

3(6) + 5(7) ≤ 53 3(6) + 2(7) ≤ 32

53 ≤ 53 32≤ 32

Optimal Solution:

Max P = 48x + 60y

= 48(6) + 60(7)

= 288 + 420

Max P = P 708.00

X 0 17.67

Y 10.6

X 0 10.67

V. Decision

The baker should produce 6 dozen of first type cookies and 7 dozen of the second type of cookies to meet the maximum profit of P 708.00.

CHAPTER 3

Exercise 3.1 Table I (Initial Solution)

Table II

Optimum column

X1 : 100 ÷ 2 = 50 spr

S2 : 240 ÷ 4 = 60

Pivot element: 2

Outgoing variable: S1

Entering variable: X1

X1 : 50 ÷ ½ = 100

S2 : 40 ÷ 1 = 40

Pivot element: 1

Optimum column

7 5 0 0

SolutionVariable

SolutionValues

X1 X2 S1 S2

X1 50 ½ ½ 0

7 S2 40 0 1 -2 1

0 Zj 350 7 7/2 7/2 0

Cj - Zj 0 3/2 -7/2 0

7 5 0 0

SolutionVariable

SolutionValues

X1 X2 S1 S2

S1 100 1 1 0

0 S2 240 4 3 0 1

0 Zj 0 0 0 0

Cj - Zj 7 5 0 0

Row X1: Row S2:Solution variable: 100 ÷ 2 =50 Solution variable: 240 – 4(50) =40X1 : 2 ÷ 2 = 1 X1 : 4 – 4(1) = 0X2 : 1 ÷ 2 = ½ X2 : 3 – 4( ½) = 1

S1 : 1 ÷ 2 = ½ S2 : 0 – 4(½) = -2

S2 : 0 ÷ 2 = 0 S2 : 1 – 4(0) = 1

Table III

Row X2: Row X1:

Solution variable: 40 ÷ 1 =40 Solution variable: 50 – ½(40) =30X1 : 0 ÷ 1 = 0 X1 : 1 – ½(0) = 0X2 : 1 ÷ 1 = 1 X2 : ½ – ½(1) = 1

S1 : -2 ÷ 1 = -2 S2 : ½ – ½(-2) = -2S2 : 1 ÷ 1 = 1 S2 : 0 – ½(1) = 1

7 5 0 0

SolutionVariable

SolutionValues

X1 X2 S1 S2

X1 30 1 0 3/2 -½

7 X2 40 0 1 -2 1

5 Zj 7 5 ½ 3/2

Cj - Zj 0 0 -½ -3/2

Exercise 3.2

Problem No. 2

Rice Corn TotalInsecticides 4 3 36Fertilizers 3 6 48

P 20,800 P15,000

Decision variablesLet X1 = number of units to be planted by riceLet X2 = number of units to be planted by corn

Objective function: max P = 20,800X1 + 15,000 X2Subject to:

4X1 + 3X2 ≤ 36 Explicit constraints3X1 + 6X2 ≤ 48X1 ;X2 ≥0 Implicit constraints

Modified Objective function: max P = 20,800X1 + 15,000 X2 + OS1+ OS2Subject to:

4X1 + 3X2+ S1+ OS2 = 363X1 + 6X2 + OS1+ S2 = 48X1 ;X2 ≥0

Table I - Initial feasible solution

Row X1:

Row S2:Solution variable: 36 ÷ 4 =9 Solution variable: 48 – 3(9) =21

X1 : 4 ÷ 4 = 1 X1: 3 – 3(1) = 0

Optimum column

X1 : 36 ÷ 4 = 9 lowest

S2 : 48 ÷ 3 = 16

Pivot element: 4

Cj20,80

015,00

Solution

Variable

Solution

Values

X1 X2 S1 S2

S1 36 3 1 0

0 S2 48 3 6 0 1

0 Zj 0 0 0 0 0

Cj - Zj 20,800

15,000

Cj20,800 15,000 0 0

Solution

Variable

SolutionValues

X1 X2 S1 S2

X1 9 1 ¾ ¼ 0

20,800 S2 21 0 15/4 ¾ 0

0 Zj 187,200 20,800 15,600 5200 0

Cj - Zj 0 -15,600 -5200 0

X2 : 3 ÷ 4 = ¾ X2: 6 – 3( ¾) = 15/4

S1 : 1 ÷ 4 = ¼ S2: 0 – 3(¼) = ¾S2 : 0 ÷ 4 = 0 S2: 1 – 3(0) = 1

Decision:The farmer can maximize the profit up to P187,200 by planting 9 units of rice

in 3 hectares.

Problem No. 3

Program I Program II TotalMusic 40 20 160

Advertisement 2 2 640,000 viewers 20,000 viewers

Decision variablesLet X1 = number of times music must be played to maximize profitLet X2 = number of times advertisement must be played to maximize profit

Objective function: max P = 40,000X1 + 20,000 X2

Subject to:40X1 + 20X2 ≤ 160 Explicit constraints2X1 + 2X2 ≤ 6X1 ;X2 ≥0 Implicit constraints

Modified Objective function: max P = 40,000X1 + 20,000 X2 + OS1+ OS2

Subject to:40X1 + 20X2+ S1+ OS2 = 1602X1 + 2X2 + OS1+ S2 = 6X1 ;X2 ≥0

Table I: Initial feasible solution

Table II: Second feasible solution

S1 : 160 ÷ 40 = 4

S2 : 6 ÷ 2 = 3 lowest

Pivot element: 2

Cj40,000 20,000 0 0

SolutionVariable

SolutionValues

X1 X2 S1 S2

S1 40 0 -20 1 -20

0 X2 3 1 1 0 ½

40,000 Zj 120,000 40,000 40,000 0 20,000

Cj - Zj 0 -20,000 0 -20,000

Cj20,800 15,000 0 0

SolutionVariable

SolutionValues

X1 X2 S1 S2

S1 160 40 20 1 0

0 S2 6 2 0 1

0 Zj 0 0 0 0 0

Cj - Zj 40,000 20,000 0 0

Row X1: Row S2:Solution variable: 6 ÷ 2 =3 Solution variable: 160 – 3(40) =40

X1 : 2 ÷ 2 = 1 X1 : 40 – 1(40) = 0X2 : 2 ÷ 2 = 1 X2 : 20 – 1(40) = -20

S1 : 0 ÷ 2 = 0 S2 : 1 – 0(40) = 1S2 : 1 ÷ 2 = ½ S2 : 0 – ½(40) = -20

Decision:The management can maximize the profit up to P120,000 by playing the advertisement 3 times

a week.

Exercise 3.3

Problem No. 1Table 1

₱ 6 ₱ 7 0 0Solution variables

Solution values

X₁ X₂ S₁ S₂

0 S₁ 24 2 3 1 00 S₂ 16 2 1 0 1

Zj 0 0 0 0 0Cj-Zj 6 7 0 0

Outgoing row Optimum column

1. 0(2) + 0(2) = 00(3) + 0(1) = 00(1) + 0(0) = 00(0) + 0(1) = 0

2. X₁: 6 - 0 = 6X₂: 7 - 0 = 7S₁: 0 – 0 = 0S₂: 0 – 0 = 0

3. S₁: 24 ÷ 3 = 8 smallest positive ratio

4. S₁: 24 ÷ 3 = 8X₁: 2 ÷ 3 = 2/3X₂: 3 ÷ 3 = 1S₁: 1 ÷ 3 = 1/3S₂: 0 ÷ 3 = 0

5. S₂: 16 – 1(8) = 8X₁: 2 – 1(2/3) = 4/3X₂: 1- 1(1) = 0

S₁: 0 – 1(1/ 3) =- 1/3S₂: 1 – 1(0) = 1S₂: 16 ÷ 1 = 16

Table 2

Solution values

X₁ X₂ S₁ S₂

7 S₁ 8 2/3 1 1/3 00 S₂ 8 4/3 0 -1/3 1

Zj 56 14/3 7 7/3 0Cj-Zj 4/3 0 -7/3 0

outgoing row Optimum column

1. 7(8) + 0(8)= 56 7(2/30) + o (4/3) = 14/3 7(10) + 0(0) = 7 7(1/3) + 0 (-1/3) = 7/3 7(0) + 0 (1) = 0

2. X₁: 6 - 14/3 = 4/3 X₂: 7 – 7 = 0 S₁: 0 – 7/3 = -7/3 S₂: 0 – 0 = 0

3. X₂: 8 ÷ 2/3 = 12

S₂: 8 ÷ 4/3 = 6 smallest positive ratio

4. S₂: 8 ÷ 4/3 = 6X₁: 4/3 ÷ 4/3 = 1X₂: 0 ÷ 4/3 = 0X₃: -1/3 ÷ 4/3 = -1/4X₄: 1 ÷ 4/3 = ¾

5. X₂: 8 – 2/3(6) = 4X₁: 2/3 – 2/3(1) = 0X₂: 1- 2/3(0) = 1X₃: 1/3 – 2/3(-1/4) = 1/2X₄: 0- 2/3(3/4) = -1/2

Table 3 (final solution)

Solution values

X₁ X₂ S₁ S₂

7 X₁ 4 0 1 ½ -1/26 X₂ 6 1 0 -1/4 3/4

Zj 64 6 7 2 1Cj-Zj 0 0 -2 -1

1. 7(4) + 6(6) = 647(0) + 6(1) = 67(1) + 6(0) = 77(1/2) + 6(-1/4) = 27(-1/2) + 6(3/4) = 1

2. X₁: 6 – 6 = 0 X₂: 7 – 7 = 0

S₁: 0 – 2 = -2 S₂: 0 – 1 = -1

Problem No. 2

Product A Product B Available time capacity

Machine time 4 hrs. 2 hrs. 10 hrs.Skilled worker 2 hrs. 2 hrs. 8 hrs.

Profit ₱ 50 ₱ 30

I. Decision variablesLet X₁ = no. Of product A to be produced X₂ = no. Of product B to be produced

Objective function: Max. P = 50x₁ + 30x₂Subject to : 4x₁ + 2x₂ ≤ 10 explicit constraints 2x₁ + 2x₂ ≤ 8 X;Y ≥ 0 implicit constraints

II. Objective function:Max. P = 50x₁ + 30x₂ + 0S₁ + 0S₂

Subject to: 4x₁ + 2x₂ + S₁ + 0S₂ =10 2x₁ + 2x₂ + 0S₁ + S₂ = 8

Table 1

Solution values

X₁ X₂ S₁ S₂

0 S₁ 10 4 2 1 00 S₂ 8 2 2 0 1

Zj 0 0 0 0 0Cj-Zj 50 30 0 0

1. 0(10) + 0(8) = 00(4) + 0(2) = 00(2) + 0(2) = 00(1) + 0(0) = 00(0) + 0(1) = 0

2. X₁: 50 – 0 = 50 X₂: 30 – 0 = 30 S₁: 0 – 0 = 0 S₂: 0 – 0 = 0

3. S₁: 10 ÷ 4 = 5/2 smallest positive ratioS₂: 8 ÷ 2 = 4

4. S₁: 10 ÷ 4 = 5/2 X₁: 4 ÷ 4 = 1 X₂: 2 ÷ 4 = 1/2 S₁: 1 ÷ 4 = 1/4

S₂: 0 ÷4 = 0

5. S₂: 8 – 2(5/2) = 3

X₁: 2 – 2(1) = 0 X₂: 2 – 2(1/2) = 1

S₁: 0 – 2(¼) = -1/2 S₂: 1- 2(0) = 1

Table 2

Solution values

X₁ X₂ S₁ S₂

50 S₁ 5/2 1 1/2 1/4 00 X₂ 3 0 1 -1/2 1

Zj 125 50 25 25/2 0Cj-Zj 0 5 -25/2 0

1. 50(5/2) + 0(3) = 12550(1) + 0(0) = 5050(1/2) + 0 (1) = 2550(1/4) + 0(-1/2) = 25/250(0) + 0(1) = 0

2. X₁: 50 – 50 = 0 X₂: 30 – 25 = 5 S₁: 0 – 25/2 = -25/2 S₂: 0 – 0 = 0

3. S₁: 5/2 ÷ 1/2 = 5 S₂: 3 ÷ 1 = 3 smallest positive ratio

4. X₂: 3 ÷ 1 = 3 X₁: 0 ÷ 1 = 0 X₂: ÷ 1 = 1 S₁: -1/2 ÷ 1= -1/2S₂: 1 ÷ 1 = 1

5. S₁: 5/2 - ½( 3) = 1

X₁: 1 – ½(0) = 1 X₂: ½ - ½(1) = 0 S₁: 1/4 – ½(-1/2)= 1/2 S₂: 0 – 1/2 (1) = -1/2

Table 3 (final solution)

Solution values

X₁ X₂ S₁ S₂

50 X₁ 1 1 0 1/2 -1/230 X₂ 3 0 1 -1/2 1

Zj 140 50 30 10 5Cj-Zj 0 0 -10 -5

1. 50(1) + 30(3) = 14050(1) + 30(0) = 5050(0) + 30 (1) = 3050(1/2) + 30(-1/2) = 1050(-1/2) + 30(1) = 5

2. X₁: 50 – 50 = 0 X₂: 30 - 30= 0 S₁: 0 – 10= -10S₂: 0 – 5 = -5

Check 4x₁ + 2x₂ ≤ 10 2x₁ + 2x₂ ≤ 8 4(10) + 2(3) ≤ 10 2(1) + 2(3) ≤ 8 4 + 6 ≤ 10 2 + 6 ≤ 8 10≤ 10 8≤8

Decision:

The Arambulo Co. Should produce and sell 1 unit of the product A and 3 units of the product B to maximize profit to 140 per production period.

Problem No. 3

Given:

6x1 + 4x2 ≤ 13503x1 + 4x2 ≤ 780

x1 ≤ 180 x2 ≤ 135 x1 ; x2 ≥ 0

Determine the optimum profit using the simplex method if contributions to profit of X1

are ₱ 30.50 per unit and of X2 is ₱ 29 per unit.

I. Objective Function

Max. ₱ = 30.50x1 + 29x2

Subject to:

6x1 + 4x2 ≤ 1350 3x1 + 4x2 ≤ 780 Explicit Constraints

x1 ≤ 180x2 ≤ 135x1 ; x2 ≥ 0 - Implicit

II. Objective Function:

Max. ₱ = 30.50x1 + 29x2 + 0S1 + 0S2 + 0S3 + 0S4

Subject to:

6x1 + 4x2 + S1 + 0S2 + 0S3 + 0S4 = 13503x1 + 4x2 + 0S1 + S2 + 0S3 + 0S4 = 780x1 + 0S1 + 0S2 + S3 + 0S4 = 180x2 + 0S1 + 0S2 + 0S3 + S4 = 135

TABLE I

Ci \ Cj ₱ 30.50 ₱ 29.00 0 0 0 0Solution Variable

Solution Values X1 X2 S1 S2 S3 S4

0 S1 1350 6 4 1 0 0 00 S2 780 3 4 0 1 0 00 S3 180 1 0 0 0 1 00 S4 135 0 1 0 0 0 1

Zj 0 0 0 0 0 0 0Cj – Zj 30.50 29 0 0 0 0

Outgoing Row Optimum Column

1.)0 (1350) + 0 (780) + 0 (180) + 0 (135) = 0

0 (6) + 0 (0) + 0 (1) + 0 (0) = 00 (4) + 0 (4) + 0 (0) + 0 (1) = 00 (1) + 0 (0) + 0 (0) + 0 (0) = 00 (0) + 0 (1) + 0 (0) + 0 (0) = 00 (0) + 0 (0) + 0 (1) + 0 (0) = 00 (0) + 0 (0) + 0 (0) + 0 (1) = 0

2.)x1 : 30.50 – 0 = 30.50

x2 : 29 – 0 = 29S1 : 0 – 0 = 0S2 : 0 – 0 = 0S3 : 0 – 0 = 0

S4 : 0 – 0 = 0

3.) S1 : 1350 ÷ 6 = 225S2 : 780 ÷ 3 = 260S3 : 180 ÷ 1 = 180 smallestS4 : 135 ÷ 0 = undefined

4.)S3 : 180 ÷ 1 = 180x1 : 1 ÷ 1 = 1x2 : 0 ÷ 1 = 0S1 : 0 ÷ 1 = 0S2 : 0 ÷ 1 = 0S3 : 1 ÷ 1 = 1S4 : 0 ÷ 1 = 0

5.)S1 :1350 – 6 (180) = 270x1 : 6 – 6 (1) = 0x2 : 4 – 6 (0) = 0S1 : 1 – 6 (0) = 0S2 : 0 – 6 (0) = 0S3 : 0 – 6 (1) = 0S4 : 0 – 6 (0) = 0

S2 : 780 – 3 (180) = 240x1 : 3 – 3 (1) = 0x2 : 4 – 3 (0) = 4S1 : 0 – 3 (0) = 0S2 : 1 – 3 (0) = 1S3 : 0 – 3 (1) = -3S4 : 0 – 3 (0) = 0

S4 :135 – 0 (180) = 135x1 : 0 – 0 (1) = 0x2 : 1 – 0 (0) = 1S1 : 0 – 0 (0) = 0S2 : 0 – 0 (0) = 0S3 : 0 – 0 (1) = 0S4 : 1 – 0 (0) = 1

TABLE II

0 S1 270 0 4 1 0 -6 0

0 S2 240 0 4 0 1 -3 030.50 X1 180 1 0 0 0 1 0

0 S4 135 0 1 0 0 0 1Zj 5490 30.50 0 0 0 30.50 0

Cj – Zj 0 29 0 0 -30.50 0

Outgoing Row optimum column

0 (270) + 0 (240) + 30.50 (180) + 0 (135) = 54900 (0) + 0 (0) + 30.50 (1) + 0 (0) = 30.500 (4) + 0 (4) + 30.50 (0) + 0 (1) = 00 (1) + 0 (0) + 30.50 (0) + 0 (0) = 00 (0) + 0 (1) + 30.50 (0) + 0 (0) = 00 (-6) + 0 (-3) + 30.50 (1) + 0 (0) = 30.500 (0) + 0 (0) + 30.50 (0) + 0 (1) = 0

2.)x1 : 30.50 – 30.50 =x2 : 29 – 0 = 29S1 : 0 – 0 = 0S2 : 0 – 0 = 0S3 : 0 – 30.50 = - 30.50S4 : 0 – 0 = 0

3.)S1 : 270 ÷ 4 = 67.50S2 : 240 ÷ 4 = 60 smallestS3 : 180 ÷ 0 = 180 undefinedS4 : 135 ÷ 1 = 135

S2 : 240 ÷ 4 = 60x1 : 0 ÷ 4 = 1x2 : 4 ÷ 4 = 0S1 : 0 ÷ 4 = 0S2 : 1 ÷ 4 = 0S3 : -3 ÷ 4 = 1S4 : 0 ÷ 4 = 0

5.)S1 : 270 – 4 (60) = 30x1 : 0 – 4 (0) = 0x2: 4 – 4 (1) = 0S1 : 1 – 4 (0) = 1S2 : 0 – 4 (1/4) = -1S3 : -6 – 4 (-3/4) = -3S4 : 0 – 4 (0) = 0

x1 : 180 – 0 (60) = 180x1 : 1 – 0 (0) = 1x2 : 0 – 0 (1) = 0S1 : 0 – 0 (0) = 0S2 : 0 – 0 (1/4) = 0S3 : 1 – 0 (-3/4) = 1S4 : 0 – 0 (0) = 0

S4 :135 – 1 (160) = 75x1 : 0 – 1 (0) = 0x2 : 1 – 1 (1) = 0S1 : 0 – 1 (0) = 0S2 : 0 – 1 (1/4) = -1/4S3 : 0 – 1 (-3/4) = 3/4S4 : 1 – 1 (0) = 1

TABLE III (final solution)

0 S1 30 0 0 1 -1 -3 00 X2 60 0 1 0 ¼ -3/4 0

30.50 X1 180 1 0 0 0 1 00 S4 75 0 0 0 -1/4 ¾ 1

Zj 7230 30.50 29 0 29/4 35/4 0Cj – Zj 0 0 0 - 29/4 -35/4 0

0 (30) + 29 (60) + 30.50 (180) + 0 (75) = 72300 (0) + 29 (0) + 30.50 (1) + 0 (0) = 30.500 (0) + 29 (1) + 30.50 (0) + 0 (0) = 290 (1) + 29 (0) + 30.50 (0) + 0 (0) = 0

0 (-1) + 29 (1/4) + 30.50 (0) + 0 (-1/4) = 29/40 (-3) + 29 (-3/4) + 30.50 (1) + 0 (3/4) = 35/40 (0) + 29 (0) + 30.50 (0) + 0 (1) = 0

2.)x1 : 30.50 – 30.50 = 0x2 : 29 – 29 = 0S1 : 0 – 0 = 0S2 : 0 – 29/4 = -29/4S3 : 0 – 35/4 = - 35/4S4 : 0 – 0 = 0

Problem No. 4

Product A Product B Product C Available time Capacity

Raw materials 10 units 10 units 20 units 2400 unitsManpower 2 2 3 480 hoursMachine type 3 2 2 480 hoursProfit ₱ 25 ₱ 22.50 ₱ 30

I. Decision Variables:

Let X1 = no. of product A to be produced X2 = no. of product A to be produced X3 = no. of product A to be produced

Objective Function:

Max. ₱ = 25x1 + 22.50x2 + 30x3

Subject to:

10x1 + 10x2 + 20x3 + S1 + 0S2 + 0S3 = 24002x1 + 2x2 + 3x3 + 0S1 + S2 + 0S3 = 4803x1 + 2x2 + 2x3 + 0S1 + 0S2 + S3 = 480x2 + 0S1 + 0S2 + 0S3 + S4 = 135

TABLE I

Ci \ Cj ₱ 25 ₱ 22.50 ₱ 30 0 0 0Solution Variable

Solution Values X1 X2 X3 S1 S2 S3

0 S1 2400 10 0 20 1 0 00 S2 480 2 2 3 0 1 0

30.50 S3 480 3 2 2 0 0 1Zj 0 0 0 0 0 0 0

Cj - Zj 25 22.50 30 0 0 0

Outgoing row optimum column

1.) Zj0 (2400) + 0 (480) + 0 (480) = 00 (10) + 0 (2) + 0 (3) = 00 (0) + 0 (2) + 0 (2) = 00 (20) + 0 (3) + 0 (2) = 0

2.) Cj – ZjX1 : 25 – 0 = 25X2 : 22.50 – 0 = 22.50X3 : 30 – 0 = 30S1 : 0 – 0 = 0S2 : 0 – 0 = 0S3 : 0 – 0 = 0

3.)S1 : 2400 ÷ 20 = 120 Smallest positive ratioS2 : 480 ÷ 20 = 160S3 : 480 ÷ 20 = 240

4.)S1 : 2400 ÷ 20 = 120x1 : 10 ÷ 20 = 1/2x2 : 10 ÷ 20 = 1/2x3 : 20 ÷ 20 = 1S1 : 1 ÷ 20 = 1/20S2 : 0 ÷ 20 = 0S3 : 0 ÷ 20 = 0

5.)S2 : 480 – 3 (60) = 120x1 : 2– 3 (1/2) = 1/2x2 : 2 – 3 (1/2) = 1/2x3 : 3 – 3 (1) = 0

S1 : 0 – 3 (1/20) = -3/20S2 : 1 – 3 (0) = 1S3 : 0 – 3 (0) = 0

S3 : 480 – 2 (60) = 240x1 : 3– 2 (1/2) = 2x2 : 2 – 2 (1/2) = 1x3 : 2 – 2 (1) = 0S1 : 0 – 2 (1/20) = -1/10S2 : 0 – 2 (0) = 0S3 : 1 – 2 (0) = 1

TABLE II

30 X3 120 1/2 1/2 1 1/20 0 00 S2 120 1/2 1/2 0 -3/20 1 00 S3 240 2 1 0 -1/10 0 1

Zj 3600 15 15 30 3/2 0 0Cj - Zj 10 15/2 0 -3/2 0 0

1.) Zj

30 (120) + 0 (120) + 0 (240) = 030 (1/2) + 0 (1/2) + 0 (2) = 030 (1/2) + 0 (1/2) + 0 (1) = 030 (1) + 0 (0) + 0 (0) = 0

30 (1/20) + 0 (-3/20) + 0 (-1/10) = 3/230 (0) + 0 (1) + 0 (0) = 030 (0) + 0 (0) + 0 (1) = 0

2.) Cj – Zj

X1 : 25 – 15 = 10X2 : 22.50 – 15 = 15/2X3 : 30 – 30 = 0S1 : 0 – 3/2 = -3/2S2 : 0 – 0 = 0S3 : 0 – 0 = 0

3.)S1 : 2400 ÷ 20 = 120 Smallest positive ratioS2 : 480 ÷ 20 = 160S3 : 480 ÷ 20 = 240

S3 : 240 ÷ 2 = 120x1 : 2÷ 2 = 1x2 : 1 ÷ 2 = 1/2x3 : 0 ÷ 2 = 0S1 : -1/10 ÷ 2 = -1/20S2 : 0 ÷ 2 = 0S3 : 1 ÷ 2 = ½

5.)x3 : 120 – 1/2 (120) = 60x1 : 1/2– 1/2 (1) = 0x2 : 1/2 – 1/2 (1/2) = 1/4x3 : 1 – 1/2 (0) = 1S1 : 1/20 – 1/2 (-1/20) = 3/40

S2 : 0 – 1/2 (0) = 0S3 : 0 – 1/2 (1/2) = -1/4

S2 : 120 – 1/2 (120) = 60x1 : 1/2– 1/2 (1) = 0x2 : 1/2 – 1/2 (1/2) = 1/4x3 : 0 – 1/2 (0) = 1S1 : -3/20 – 1/2 (-1/20) = -1/8S2 : 1 – 1/2 (0) = 1

S3 : 0 – 1/2 (1/2) = -1/4

TABLE III

30 X3 60 0 ¼ 1 3/40 0 -1/40 S2 60 0 ¼ 0 -1/8 1 -1/4

25 X1 120 1 ½ 0 -1/20 0 ½Zj 4800 25 20 30 1 0 5

Cj - Zj 0 2.50 0 -1 0 -5

1.) Zj

30 (60) + 0 (60) + 25 (120) = 480030 (1/2) + 0 (1/2) + 25 (1) = 030 (1/2) + 0 (1/2) + 25 (1/2) = 030 (1) + 0 (0) + 25 (0) = 030 (1/20) + 0 (-3/20) + 25 (-1/20) = 3/230 (0) + 0 (1) + 25 (0) = 030 (0) + 0 (0) + 25 (1/2) = 0

2.) Cj – Zj

X1 : 25 – 25 = 0X2 : 22.50 – 20= 2.50X3 : 30 – 30 = 0S1 : 0 – 1 = -1S2 : 0 – 0 = 0S3 : 0 – 5 = -5

3.)X3 : 60 ÷ 1/4 = 240S2 : 60 ÷ 1/4 = 240X1 : 120 ÷ 1/4 = 240

S2 : 60 ÷ 1/4 = 240x1 : 0 ÷ 1/4 = 0x2 : 1/4 ÷ 1/4 = 1

x3 : 0 ÷ 1/4 = 0S1 : -1/8 ÷ 1/4 = -1/2S2 : 1 ÷ 1/4 = 4S3 : -1/4 ÷ 1/4 = -½

5.)x3 : 60 – ¼ (240) = 0x1 : 0– ¼ (0) = 0x2 : 1/4 – ¼ (1) = 0x3 : 1 – ¼ (0) = 0

S1 : 3/40 – ¼ (-1/2) = 1/5

S2 : 0 – ¼ (4) = -1S3 : -¼ – ¼ (-1/2) = -1/8

X1 : 120 – 1/2 (240) = 0x1 : 1– 1/2 (1) = 1x2 : ½ – 1/2 (1/2) = 0x3 : 0 – 1/2 (0) = 0S1 : -1/20 – 1/2 (-1/2) = 1/5S2 : 0 – 1/2 (4) = -2S3 : ½ – 1/2 (- ½ ) = 3/4

TABLE IV

30 X3 0 0 0 1 1/5 -1 -1/822.50 X2 240 0 1 0 -1/2 4 -1/2

25 X1 0 1 0 0 1/5 -2 ¾Zj 5400 25 22.50 30 ¼ 70 15/4

Cj - Zj 0 0 0 -1/4 -70 -15/4

1.) Zj30 (0) + 22.50 (240) + 25 (0) = 540030 (0) + 22.50 (0) + 25 (1) = 2530 (0) + 22.50 (1) + 25 (0) = 22.5030 (1) + 22.50 (0) + 25 (0) = 3030 (1/5) + 22.50 (-1/2) + 25 (1/5) = -1/4

30 (-1) + 22.50 (4) + 25 (-2) = 7030 -1/8) + 22.50 (-1/2) + 25 (3/4) = 15/4

Cj – Zj

X1 : 25 – 25 = 0X2 : 22.50 – 22.50 = 0X3 : 30 – 30 = 0S1 : 0 – ¼ = -1/4S2 : 0 – 70 = -70S3 : 0 – 15/4 = -15/4

Problem No. 5

I. Decision Variables:

Let X1 = no. of Toy A to be produced

Toy A Toy B Available Time Capacity

Cutting Department 2 hours 1 hour 80 hours

Finishing Department 1 hour 5 hours 70 hours

Profit ₱50 ₱55

X2 = no. of Toy B to be produced

Objective Function:

Max. ₱ = 50x1 + 55x2

Subject to:

II. Objective Function:

Max. ₱ = 50x1 + S1 + OS2

Subject to:

2x1 + x2 + S1+ OS2 = 80

X1 + 5x2 + S2 + OS1 = 70

2x1 + x2 ≤ 80 Explicit

X1 +5x2 ≤ 70 Constraints

X; y ≥ 0 - Implicit constraints

Table 1

CjCi ₱50 ₱55 0 0

Solution Variables

Solution Values X1 X2 S1 S2

0 S1 80 2 1 1 0

0 S2 70 1 0 1

Zj 0 0 0 0 0

Cj - Zj 50 55 0 0

III. S1: 80 ÷ 1 = 80S2: 70 ÷ 5 = 14 smallest

IV. S2: 70 ÷ 5 = 14X1: 1 ÷ 5 = ⅕X2: 5 ÷ 5 = 1S1: 1 ÷ 5 = ⅕

V. S1: 80 – 1(14) = 66X1: 2 – 1(⅕) = 9/5X2: 1 – 1(1) = 0S1: 1 – 1 (0) = 1S2: 0 – 1(⅕) = -⅕

Table 2

CjCi ₱50 ₱55 0 0

Solution Variables

0 S1 66 0 1 -⅕

0 S2 14 ⅕ 1 0 ⅕

Zj 770 11 55 11

Cj - Zj 39 0 0 -11

(1) 0(66) + 55(14) = 7700(9/5) + 55(⅕) =110(0) + 55(1) = 550(1) + 55(6) = 00(-⅕) + 55(⅕) = 11

(2) X1: 50 – 11 = 39X2: 55 – 55 = 0S1: 0 – 0 = 0S2: 0 – 11 = -11

(3) S1: 66 ÷ 9/5 = 110/3 smallestX2: 14 ÷ ⅕ = 70

(4) 66 ÷ 9/5 = 110/3X1: 9/5 ÷ 9/5 = 1X2: 0 ÷ 9/5 = 0S1: 1 ÷ 9/5 = 5/9S2: -⅕ ÷ 9/5 = -1/9

(5) 14 ÷ ⅕ (110/3) = 20/3X1: ⅕ – ⅕ = 0X2: 1 – ⅕ (0) = 1S1: 0 – ⅕ (5/9) = -1/9S1: ⅕ – ⅕ (1/9) = 2/9

CjCi ₱50 ₱55 0 0

Solution Variables

50 X1 110/3 1 0 5/9 -1/9

55 X2 20/3 0 1 -1/9 2/9

Zj 2200 50 55 65/3 20/3

Cj - Zj 0 0 -65/3 -20/3

(1) 50(110/3) + 55 (20/3) = 220050(4) + 55 (0) = 5050(0) + 55 (1) = 5550(5/9) + 55 (-1/9) = 65/350(-1/9) + 55 (2/9) = 20/3

(2) X1: 50 – 50 = 0X2: 55 – 55 = 0S1: 0 – 65/3 = -65/3S2: 0 – 20/3 = -20/3

Decision:A toy Manufacturer should produced 11/3 units of Toy A & 20/3 units of Toy B to

maximize profit to ₱2200 per production period.

Problem No. 6

Objective function:

max. F = 3X1 + 2X2 + OS1 + OS2

Subject to:

X! + 2X2 + S1 + OS2 = 6

2X1 + X2 + S2 + OS1 = 6

Table 1

₱3 ₱2 0 0

Solution

Variable

Solution

Values

X1 X2 S1 S2

0 S1 6 1 2 1 0

S2 6 1 0 1

Zj 0 0 0 0 0

Cj - Zj 3 2 0 0

Optimum Column

Outgowing Row

1. 0(6) + 0(6) = 0 X1: 3 – 0 = 3 S1: 6 / 1 = 6

0(1) + 0(2) = 0 X2: 2 – 0 = 2 S2: 6 / 2 = 3 smallest

0(2) + 0(1) = 0 S1: 0 – 0 = 0

0(1) + 0(0) = 0 S2: 0 – 0 = 0

0(0) + 0(1) = 0

S2: 6 / 2 = 3 S1 :6 – 1(3) = 3

X1 : 2 / 2 = 1 X1 : 1 – 1(1) = 0

X2 ; 1 / 2 = ½ X2 : 2 – 1(1/2) = 3/2

S1 : 0 / 2 = 0 S1 : 1 – 1 (0) = 1

S2 : 1 / 2 = ½ S2: 0 – 1 (⅟2) = - ½

Table 2

CjCi ₱3 ₱2 0 0

Solution Variables

0 S1 3 0 1 -½

0 S2 3 1 ½ 0 ½

Zj 9 3 3/2 0 3/2

Cj - Zj 0 ½ 0-3/2

Outgoing row Optimum column

(1) 0(3) + (3) = 90(0) + 3(1) = 30(3/2) + 3(1/2) = 3/20(1) + 3(0) = 00(-1/2) + 3(1/2) = 3/2

(2) X1: 3 – 3 = 0X2: 2 – 3/2 = ½ S1: 0 – 0 = 0S2: 0 – 3/2 = -3/2

(3) S1: 3 ÷ 3/2 = 2 smallestX1: 2 ÷ ½ = 6

(4) S1: 3 ÷ 3/2 = 2X1: 0 ÷ 3/2 = 0X2: 3/2 ÷ 3/2 = 1S1: 1 ÷ 3/2 = 2/3S2: - ½ ÷ 3/2 = -1/3

(5) X1: 3 – ½ (2) = 2X1: 1 – ½ = 1X2: ½ – ½ (1) = 0X1: 0 – ½ (2/3) = -1/3 X1: ½ – ½ (-1/3) = 2/3

Table 3

CjCi ₱3 ₱2 0 0

Solution Variables

2 S1 2 0 1 2/3 -1/3

3 S2 2 1 0 -1/3 2/3

Zj 10 3 2 1/3 4/3

Cj - Zj 0 0 -1/3 -4/3

(1) 2(2) + 3(2) = 102(0) + 3(1) = 32(1) + 3(0) = 22(2/3) + 3(-1/3) = 1/32(-1/3) + 3(2/3) = 4/3

(2) X1: 3 – 3 = 0X2: 2 – 2 = 0 S1: 0 – 1/3 = -1/3S2: 0 – 4/3 = -4/3

Exercise 3.4

Problem No. 1

Decision Variables:

Let X1 = number of kilos in ingredient A

X2 = number of kilos in ingredient B

Objective function:

Min. C = P50 X1 + P20 X2

Subject to:

X1 + X2 = 25

X1 ≥ 10

X2 ≤ 20

X1; X2 ≥ 0

Modification:

Objective Function:

Min. C = P50 X1 + P20X2 + 0S1 + 0S2 + MA1 + MA2

Subject to:

X1 + X2 + A1 + 0S1 + 0S2 + 0A1 = 25

X1 – X2 + A2 + 0S2 + 0A1 = 10

X2 + S2 + 0S1 + 0A1 + 0A2 = 20

X1; X2 ≥ 0

TABLE I

Ci/Cj 50 20 M 0 M 0SolutionVariables

SolutionValues

X1 X2 A1 S1 A2 S2

M A1 25 1 1 1 0 0 0M A2 10 1 0 0 -1 1 00 S2 20 0 1 0 0 0 1

Zj 35M 2M M M -M M 0Cj- Zj 50-2M 20-M 0 M 0 0

X1 = entering variable

A2 = outgoing variable

1 = pivot element

TABLE II

Ci/Cj 50 20 M 0 M 0SolutionVariables

SolutionValues

X1 X2 A1 S1 A2 S2

M A1 15 0 1 1 1 -1 050 X1 10 1 0 0 -1 1 00 S2 20 0 1 0 0 0 1

Zj 15M+500 50 M M M-50 -M+50 0Cj - Zj 0 20-M 0 -M+50 M-50 0

A1: S2: X1:

25 – 1 (10) = 15 20 – 0 (10) = 20 10 ÷ 1 = 10

1 – 1 (1) = 0 0 – 0 (1) = 0 1 ÷ 1 = 1

1 – 1 (0) = 1 1 – 0 (0) = 1 0 ÷ 1 = 0

1 – 1 (0) = 1 0 – 0 (0) = 0 0 ÷ 1 = 0

0 – 1 (-1) = 1 0 – 0 (-1) =0 -1 ÷ 1 = -1

0 – 1 (1) = -1 0 – 0 (1) = 0 1 ÷ 1 = 1

0 – 1 (0) = 0 1 – 0 (0) = 1 0 ÷ 1 = 0

S2 = entering variable A1 = 15 ÷ 1 = 15

A1 = outgoing variable X1 = 10 ÷ 0 = undefined

1 = pivot element S2 = 20 ÷ 1 = 20

TABLE III

Ci/ Cj 50 20 M 0 M 0SolutionVariables

SolutionValues

X1 X2 A1 S1 A2 S2

20 X2 15 0 1 1 1 -1 050 X1 10 1 0 0 -1 1 00 S2 5 0 0 1 -1 -1 1

Zj 800 50 20 20 -30 30 0Cj – Zj 0 0 M-20 30 M-30 0

X1: S2: X2:

10 – 0 (15) = 10 20 – 1 (15) = 5 15 ÷ 1 = 15

1 – 0 (0) = 1 0 – 1 (0) = 0 0÷ 1 = 0

0 – 0 (1) = 0 1 – 1 (1) = 0 1 ÷ 1 = 1

0 – 0 (1) = 0 0 – 1 (1) = -1 1 ÷ 1 = 1

-1 – 0 (1) = -1 0 – 1 (1) = -1 1 ÷ 1 = 1

1 – 0 (-1) = 1 0 – 1 (-1) = 1 -1 ÷ 1 = -1

0 – 0 (0) = 0 1 – 1 (0) = 1 0 ÷ 1 = 0

Check:

X1 + X2 = 25 X1 ≥ 10 X2 ≤ 20

10 + 15 = 25 10 ≥ 10 20 ≤ 20

25 = 25

Decision:

The optimum solution is to use 10 kilos of A and 15 kilos of B to minimize the cost of P800.

Problem No. 4

A B Total CostCashew >40% >20% 200 200Peanut <25% <60% 400 80Selling Price 320 160

Decision Variables:

X1 = number of kilos of cashew in A

X2 = number of kilos of cashew in B

X3 = number of kilos of peanut in A

X4 = number of kilos of peanut in B

Objective Function:

Min.C = 200X1 + 200X2 + 80X3 + 80X4

Subject to:

X1 + X2 ≤ 200

X3 + X4 ≤ 400

X1 ≥ 40%

X2 ≥ 20%

X3 ≤ 25%

X4 ≤ 60%

Min. C = 200X1 + 200X2 + 80X3 + 80X4 + 0S1 + 0S2 + MA1 + MA2 + 0S3 + 0S4 + 0S5 + 0S6

X1 + X2 + S1 + 0S2 + 0A1 + 0A2 + 0S3 + 0S4 + 0S5 + 0S6 = 200

X3 + X4 + S2 + 0S1 + 0A1 + 0A2 + 0S3 + 0S4 + 0S5 + 0S6 = 400

X1 – S3 + A1 + 0S1 + 0S2 + 0A2 + 0S4 + 0S5 + 0S6 = 2/5

X2 – S4 + A2 + 0S1 + 0S2 + 0S3 + 0A1 + 0S5 + 0S6 = 1/5

X3 + S5 + 0S1 + 0S2 + 0S3 + 0S4 + 0A1 + 0A2 + 0S6 = 1/4

X4 + S6 + 0S1 + 0S2 + 0A2 + 0S4 + 0S5 + 0A1 + 0A2 = 3/5

TABLE I

Ci/Cj 200 200

80 80 0 0 M M 0 0 0 0

SolutionVariables

SolutionValues

X1 X2 X3 X4 S1 S2 A1 A2 S3 S4 S5 S6

0 S1 200 1 1 0 0 1 0 0 0 0 0 0 00 S2 400 0 0 1 1 0 1 0 0 0 0 0 0M A1 2/5 or

40%1 0 0 0 0 0 1 0 -1 0 0 0

M A2 1/5 or 20%

0 1 0 0 0 0 0 1 0 -1 0 0

0 S5 ¼ or 25%

0 0 1 0 0 0 0 0 0 0 1 0

0 S6 3/5 or 60%

0 0 0 1 0 0 0 0 0 0 0 1

Zj 3/5 M M M 0 0 0 0 M M -M -M 0 0Cj – Zj 200- M 200-M 80 80 0 0 0 0 0 M M 0 0

TABLE II

Ci/Cj 200 200 80 80 0 0 M M 0 0 0 0SolutionVariables

Solution Values

X1 X2 X3 X4 S1 S2 A1 A2 S3 S4 S5

0 S1 999/5 1 0 0 0 1 0 0 -1 0 1 0 00 S2 400 0 0 1 1 0 1 0 0 0 0 0 0M A1 2/5 1 0 0 0 0 0 1 0 -1 0 0 0200 X2 1/5 0 1 0 0 0 0 0 1 0 -1 0 00 S5 1/4 0 0 1 0 0 0 0 0 0 0 1 00 S6 3/5 0 0 0 1 0 0 0 0 0 0 0 1

Zj 40+2/5M

M 200 0 0 0 0 M 200 -M -200 0 0

Cj – Zj 200-M 0 80 80 0 0 0 M-200 M 200 0 0

TABLE III

Ci/Cj 200 200 80 80 0 0 M M 0 0 0 0SolutionVariables

SolutionValues

X1 X2 X3 X4 S1 S2 A1 A2 S3 S4 S5 S6

0 S1 200 1 1 0 0 1 0 0 0 0 0 0 00 S2 400 0 0 1 1 0 1 0 0 0 0 0 0M A1 2/5 1 0 0 0 00 0 0 1 0 0 0M A2 1/5 0 1 0 0 0 0 0 1 0 -1 0 00 S5 ¼ 0 0 1 0 0 0 0 0 0 0 1 00 S6 3/5 0 0 0 1 0 0 0 0 0 0 0 1

Zj 3/5 M M M 0 0 0 0 0 M M -M 0 0Cj-Zj 200-M 200-M 80 80 0 0 M 0 -M M 0 0

CHAPTER 4

Exercise 4.1

Problem No. 1

Initial Feasible Solution

PROJECTSupply

Plant X 750 -1250

Plant Y -250

Plant Z 1750

Demand 1500 1750 1400 46504650

Transportation Cost:

1450 x 1000 = P 145000050 x 1500 = 750001550 x 1750 = 2712500200 x 500 = 1000001400 x 1250 = 1750000

P 6087500

Evaluation of Unused Cells:

XB = 2000 – 1750 + 1500 – 1000 = 750XC = 750 – 1250 + 500 – 1750 + 1500 – 1000 = -1250YC = 2250 – 1250 + 500 – 1750 = -250ZA = 2000 – 500 + 1750 – 1500 = 1750

Computation of Stones:

The largest negative index is -1250The smallest stone in negative position is 1400

XC = 0 + 1400 = 1400XA = 1450 – 1400 = 50YA = 50 + 1400 = 1450YB = 1550 – 1400 = 150ZB = 200 + 1400 = 1600ZC = 1400 – 1400 = 0

Optimal Solution

1000 2000 750

1500 1750 2250

2000 500 1250

50 155

200 140

PROJECTSupply

Plant X 750

Plant Y 1000

Plant Z 1750 1250

Demand 1500 1750 1400 46504650

]Transportation Cost:

50 x 1000 = P 500001450 x 1500 = 2175000150 x 1750 = 2625001600 x 500 = 8000001400 x 750 = 1050000

P 4337500

XB = 2000 – 1750 + 1500 – 1000 = 750YC = 2250 – 750 + 1000 – 1500 = 1000ZA = 2000 – 500 + 1750 – 1500 = 1750ZC = 1250 – 750 + 1000 – 1500 + 1750 – 500 = 1250

Decision: DJA Trucking Co. should follow the distribution schedule in Table II to minimize the cost to P 4,337,500

1000 2000 750

1500 1750 2250

2000 500 1250

145 150

Problem No. 2

Initial Feasible Solution

PACKING PLANTSupply

Plantation A 40

Plantation B -160

Plantation C -240 -80

Demand 130 200 190 520520

130 x 240 = P 3120040 x 160 = 6400160 x 300 = 4800090 x 200 = 18000100 x 120 = 12000

P 115600

AY = 100 – 200 + 300 – 160 = 40BW = 220 – 300 + 160 – 240 = -160CW = 40 – 240 + 160 – 300 + 200 – 100 = -240CX = 140 – 120 + 200 – 300 = -80

CY = 100 – 100 = 0BY = 90 + 100 = 190BX = 160 – 100 = 60AW = 130 – 100 = 30AX = 40 + 100 = 140CW = 0 + 100 = 100

240 160 100

220 300 200

40 140 120

Table II

PACKING PLANTSupply

Plantation A 40

Plantation B -160

Plantation C 180 260

Demand 130 200 190 520520

30 x 240 = P 7200140 x 160 = 2240060 x 300 = 18000190 x 200 = 38000100 x 40 = 4000

P 89600

AY = 100 – 200 + 300 – 160 = 40BW = 220 – 300 + 160 – 240 = - 160CX = 140 – 160 + 240 – 40 = 180CY = 120 – 200 + 300 – 160 + 240 – 40 = 260

BW = 0 + 30 = 30BY = 60 – 30 = 30AW = 30 – 30 = 0AX = 140 + 30 = 170

240 160 100

220 300 200

40 140 120

Optimal Solution

PACKING PLANTSupply

Plantation A 160 40

Plantation B 250

Plantation C 20 100

Demand 130 200 190 520520

170 x 160 = P 2720030 x 220 = 660030 x 300 = 9000190 x 200 = 41800100 x 40 = 4000

P 88600

CX = 140 – 300 + 220 – 40 = 20CY = 120 – 200 + 220 – 40 = 100AW = 240 – 160 + 300 – 220 = 160AY = 100 – 200 + 300 – 160 = 40

Decision: The banana plantation should follow the distribution schedule in Table III to minimize the transportation cost to P 88,600.

240 160 100

220 300 200

40 140 120

Problem No. 3

Initial Feasible Solution – Optimal Solution

DEPARTMENT STORESSupply(D)

Makati(E)

Cubao(F)

Sta. Cruz(G)

Ermita(H)

Plantation A 11.50

16.50 7.507250

Plantation B 1.50

Plantation C 1.00

Demand 8700 5800 2175 2900 2175 2175021750

7250 x 5.00 = P 36250.001450 x 11.00 = 15950.005800 x 6.50 = 37700.002175 x 5.00 = 10875.00725 x 4.00 = 2900.002175 x 5.50 = 11962.00

P 115637.50

AE = 12.00 – 6.50 + 11.00 – 5.00 = 11.50AF = 11.00 – 5.00 + 11.00 – 5.00 = 12.00AG = 14.50 – 4.00 + 11.00 – 5.00 = 16.50AH = 0 – 0 + 5.50 – 4.00 + 11.00 – 5.00 = 7.50BH = 0 – 0 + 5.50 – 4.00 = 1.50CE = 9.00 – 6.50 + 4.00 – 5.50 = 1.00

Decision: Rachel Mae should follow the distribution schedule in Table I to minimize the cost to P115637.50

5.00 12.00 11.00

11.00 6.50 5.00

-- 9.00 --

145 580 217 725

217 217

Problem No. 4

Table 1

To From

PA PB Supply

Plant M 1600 1000

4300 2540 1000

Plant N 2000 1300

2160 - 1300

Plant P 2040 840

1360 1200 1200

Plant Dummy 0 160

0 200 200

Demand 2300 1400 3700

Total transportation cost:

1000 (1600) + 1300 (2000) + 2160 + 1200 (1360) + 200 = P 5, 832, 000

MB = 4300 – 2160 + 2000 - 1600 = 2540

PA = 2040 - 2000 + 2160 - 1360 = 840

PB = 0 – 0 + 2160 - 2000 = 160

Exercise 4.2

Problem No. 1

From Supply To DemandA 380 Q 300B 260 R 260C 140 S 220

From To Q R SA P140 P180 P60B P80 P160 P140C P40 P100 P180

Table 1.

K1 = 140 K2 = 180 K3 = 160From To Q R S Supply

R1 = 0 A 300 80 -100 380R2 = -20 B -40 180 80 260R3 = 20 C -120 -100 140 140

Demand 300 260 220 780

Solution:

Q, A = R1 + K1 = 140 S, B = R2 + K3 = 140

Q, A = 0 + K1 = 140 S, B = -20 + K3 = 140

Q, A = K1 = 140 S, B = K3 = 140 + 20

R, A = R1 + K2 = 180 S, B = K3 = 160

R, A = 0 + K2 = 180 S, C = R3 + K3 = 180

R, A = K2 = 180 S, C = R3 + 160 = 180

R, B = R2 + K2 = 160 S, C = R3 = 180 - 160

R, B = R2 + 180 = 160 S, C = R3 = 20

R, B = R2 = 160 -180

R, B = -20

Q, B = 80 – (-20) – 140 = -40

Q, C = 40 – 20 – 140 = -120

R, C = 100 – 20 – 180 = -100

S, A = 60 – 0 – 160 = -100

Q, C = 0 + 140 = 140

Q, A = 300 – 140 = 160

R, A = 80 + 140 = 220

R, B = 180 – 140 = 40

S, B = 80 + 140 = 220

S, C = 140 – 140 = 0

Table 2

K1 = 140 K2 = 180 K3 = 160From To Q R S Supply

R1 = 0 A 160 220 -100 380R2 = -20 B -40 40 220 260R3 = 20 C 140 -100 0 140

Demand 300 260 220 780

Problem No. 2

Projects Demand Plant AvailableA 1500 X 1450B 1750 Y 1600C 1400 Z 1600

From To A B CX P1000 P2000 P750Y P1500 P1750 P2250Z P2000 P500 P1250

Table 1

K1 = 1000 K2 = 1250 K3 = 2000From To A B C Supply

R1 = 0 X 1450 750 -1250 1450R2 = 500 Y 50 1550 -250 1600R3 = -750 Z 1750 200 1400 1600

Demand 1500 1750 1400 4650Transportation cost: P6, 087, 500

A, X = 1, 450, 000

A, Y = 75, 000

B, Y = 2, 712, 500

B, Z = 100, 000

C, Z = 1, 750, 000

TC 6, 087, 500

Solution

A, X = R1 + K1 = 1000 B, Z = R3 + K2 = 500

A, X = 0 + K1 = 1000 B, Z = R3 + 1250 = 500

A, X = K1 = 1000 B, Z = R3 = 500 – 1250

A, Y = R2 + K1 = 1500 B, Z = R3 = -750

A, Y = R2 + 1000 = 1500 C, Z = R3 + K3 = 1250

A, Y = R2 = 1500 – 1000 C, Z = -750 + K3 = 1250

A, Y = R2 = 500 C, Z = K3 = 1250 + 750

B, Y = R2 + K2 = 1750 C, Z = K3 = 2000

B, Y = 500 + K2 = 1750

B, Y = K2 = 1750 – 500

B, Y = K2 =1250

A, Z = 2000 – (-750) – 1000 =1750

X, B = 2000 – 0 -1250 = 750

C, X = 750 – 0 – 2000 = -1250

C, Y = 2250 – 500 – 2000 = -250

C, X = 0 + 1400 = 1400

C, Z = 1400 – 1400 = 0

B, Z = 200 + 1400 = 1600

B, Y = 1550 – 1400 = 150

A, Y =50 + 1400 = 1450

A, X = 1450 – 1400 = 50

Table 2.

R1 = 0 X 50 750 1400 1450R2 = 500 Y 1450 150 -250 1600R3 = -750 Z 1750 1600 0 1600

Demand 1500 1750 1400 4650

C, Y = 0 + 1400 = 1400

A, Y = 1450 – 1400 = 50

A, X = 50 + 1400 + 1450

C, X = 1400 – 1400 = 0

Table 3.

R1 = 0 X 1450 750 0 1450R2 = 500 Y 50 150 1400 1600R3 = -750 Z 1750 1600 0 1600

Demand 1500 1750 1400 4650Transportation cost: P 5737, 500

A, X = 1, 450, 000

A, Y = 75, 000

B, Y = 262, 500

B, Z = 800, 000

C, X = 3, 150, 000

TC 5, 737, 500

Problem No. 3

Projects Demand From SupplyA 750 Cavite 100B 200 Batangas 800C 500 Laguna 150

1450 1050

From To Project A Project B Project CCavite P500 P100 P700Batangas P600 P400 P600Laguna P300 P200 P500

Table 1.

K1 = 500 K2 = 300 K3 = 600From To Project A Project B Project C Supply

R1 = 0 Cavite 100 -200 100 100R2 = 100 Batangas 650 150 -100 800R3 = -100 Laguna -100 50 100 150R4 = -600 Dummy (X) 1100 900 400 400

Demand 750 200 500 1450

Transportation Cost: P 560, 000

A, Cavite = 50, 000

A, Batangas = 390, 000

B, Batangas = 60, 000

B, Laguna = 10, 000

C, Laguna = 50, 000

TC 560, 000

Solution:

A, Cavite = R1 + K1 = 500 B, Laguna = R3 + K2 = 200

A, Cavite = 0 + K1 = 500 B, Laguna = R3 + 300 = 200

A, Cavite = K1 = 500 B, Laguna = R3 = 200 - 300

A, Batangas = R2 + K1 = 600 B, Laguna = -100

A, Batangas = R2 + 500 = 600 C, Laguna = R3 + K3 = 500

A, Batangas = R2 = 600 – 500 C, Laguna = -100 + K3 = 500

A, Batangas = R2 = 100 C, Laguna = K3 = 500 + 100

B, Batangas = R2 + K2 = 400 C, Laguna = K3 = 600

B, Batangas = 100 + K2 = 400 C, X = R4 + K3 = 0

B, Batangas = K2 = 400 – 100 C, X = R4 + 600 = 0

B, Batangas = K2 = 300 C, X = R4 = - 600

A, Laguna = 300 – (-100) – 500 = -100

A, X = 0 – (-600) -500 = 1100

B, Cavite = 100 – 0 – 300 = -200

B, X = 0 – (-600) – 300 = 900

C, Cavite = 700 – 0 – 600 = 100

C, Batangas = 600 – 100 – 600 = -100

B, Cavite = 0 + 100 = 100

B, Batangas = 150 – 100 = 50

A, Batangas = 650 +100 = 750

A, Cavite = 100 -1 00 = 0

Table 2.

R1 = 0 Caviite 0 100 100 100R2 = 100 Batangas 750 50 -100 800R3 = - 100 Laguna -100 50 100 150R4 = -600 Dummy (x) 1100 900 400 400

Demand 750 200 500 1450

A, Laguna = 0 +50 =50

A, Batangas = 750 – 50 = 700

B, Batangas = 50 +50 = 100

B, Laguna = 50 -50 = 0

Table 3.

R1 = 0 Cavite 0 100 100 100R2 = 100 Batangas 700 100 -100 800R3 = -100 Laguna 50 0 100 150R4 = -600 Dummy (x) 1100 900 400 400

Demand 750 200 500 1450

C, Batangas = 0 + 100 = 100

C, Laguna = 100 – 100 = 0

A, Laguna = 50 + 100 = 150

A, Batangas = 700 – 100 = 600

Table 4.

R1 = 0 Cavite 0 100 100 100R2 = 100 Batangas 600 100 100 800R3 = -100 Laguna 150 0 0 150R4 = -600 Dummy (x) 1100 900 400 400

Demand 750 200 500 1450Transportation cost: P515, 000

A, Batangas = 360, 000 C, Batangas = 60, 000

A, Laguna = 45, 000

B, Cavite = 10, 000

B, Batangas = 40, 000

Exercise 4.3

Problem No. 1

Cost Information

Machines

Jobs A B C

J-19 P11 P14 P90

K-20 100 11P8P5

L-21 9 12 7

Jobs MachineA Machine B Machine C

J-19 P11-8=3 P14-10=4 P6-6=0K-20 8-8=0 10-10=0 11-6=5L-21 9-8=1 12-10=2 7-6=1

Job Opportunity cost table

Machines

Jobs A B C

J-19 P3 P4 P0

K-20 0 0 120

L-21 1 2 1

Jobs Machine A Machine B Machine C

J-19 P3-3=0 P4-3=1 0K-20 0 0 0L-21 0 1 0

Revised Opportunity Cost

Machines

Jobs A B C

J-19 P0 P1 P0

K-20 0 0 0

L-21 0 1 0

Assigned Cost

Job#1 to Machine 1 P 11Job#2 to Machine 3 10Job#3 to Machine 2 7Total Assignment cost P 28

Problem No. 2

Cost Information

Financing Corporation

Personnel Pampanga Bataan Bulacan

Mel P21,000 P18,000

Ben 25,000 23,000

Fred 33,000

Jobs Pampanga Bataan Bulacan

Mel P21,000-20,000=1,000 P22,000-22,000=0 P18,000-15,000=3,000Ben 25,000-20,000=5,000 23,000-22,000=1,000 15,000-15,000=0Fred 20,000-20,000=0 33,000-22,000=11,000 15,000-15,000=0

20,000

P22,000

15,000P50

15,000

Mel P0 P3,000

Ben 5,000 0

Fred 0 0

Jobs Pampanga Bataan Bulacan

Mel P1,000-1,000=0 0 P3,000-1,000=2,000Ben 5,000-1,000=4,000 1,000-1,000=0 0Fred 0 11,000-11,000=0 0

P1,000

11,000

Mel P0 P0 P2,000

Ben 4,000 0 0

Fred 0 0 0

Assigned Cost

Mel to Pampanga P21,000Ben to Bataan 23,000Fred to Bulacan 15,000Total Assignment cost P 59,000

Problem No. 3

3-A.Cost Information

Job Machine 1 2 3 4 5 6

A7 8 5

B8 4 6

C9 9 8 12 10 6

Jobs 1 2 3 4 5 6

A 7-6=1 6-6=0 2-2=0 8-5=3 5-4=1 5-5=0B 6-6=0 8-6=2 4-2=2 5-5=0 4-4=0 6-5=1C 9-6=3 9-6=3 8-2=6 12-5=7 10-4=6 6-5=1

A0 0 3 0

B0 2 2 0 0

C3 3 6 7 6

Jobs 1 2 3 4 5 6

A 1-1=0 0 0 3-1=2 1-1=0 0B 0 2-1=1 2-1=1 0 0 1-1=0C 3-1=2 3-1=2 6-1=5 7-1=6 6-5=1 1-1=0

A0 0 0 2 0 0

B0 1 1 0 0 0

C2 2 5 6 1 0

Assigned Cost

A to 1 P7 hoursB to 4 5 hoursC to 6 6 hoursTotal Assignment cost 18 hours

3-B.Cost Information

B16 18 10 14 19 12

C12 14 12 18 20 24

Jobs 1 2 3 4 5 6

A 1-1=0 3-3=0 2-1=1 2-2=0 1-1=0 1-1=0B 16-1=15 18-3=15 10-1=9 14-2=12 19-1=18 12-1=11C 12-1=11 14-3=11 12-1=11 18-2=16 20-1=19 24-1=23

A0 0 0 0 0 0

B15 15 12 18 11

C16 19 23

Jobs 1 2 3 4 5 6

A 0 0 0 0 0 0B 15-9=6 15-9=6 9-9=0 12-9=3 18-9=9 11-9=2C 11-11=0 11-11=0 11-11=0 16-11=5 19-11=8 23-11=12

A0 0 0 0 0 0

B6 6 0 3 9 2

C0 0 0 5 8 12

Assigned Cost

A to 1 1 hoursB to 3 10 hoursC to 2 14hoursTotal Assignment cost 25 hours

Problem No. 4

COST INFORMATION TABLE

SALESADVISERS

O F F I C E S

Las Piñas Quezon City Caloocan San Juan Mandaluyong Pasig

Ramirez 8 14 12 10 13 9

Arambulo 7 13 14 11 12 8

Cordero 6 12 12 9 10 9

Cereza 9 11 10 9 12 11

Diaz 8 10 9 8 9 7

Tellez 10 14 11 10 14 18

SOLUTION:

8 - 6 = 2 14 - 10 - 4 12 - 9 = 3 10 - 8 = 2 13 - 9 = 4 9 - 7 = 2

7 - 6 = 1 13 - 10 = 3 14 - 9 = 5 11 - 8 = 3 12 - 9 = 3 8 - 7 = 1

6 - 6 = 0 12 - 10 = 2 12 - 9 = 3 9 - 8 = 1 10 - 9 = 1 9 - 7 = 2

9 - 6 = 3 11 - 10 = 1 10 - 9 = 1 9 - 8 = 1 12 - 9 = 3 11 - 7 = 4

8 - 6 = 2 10 - 10 = 0 9 - 9 = 0 8 - 8 = 0 9 - 9 = 0 7 - 7 = 0

10 - 6 = 4 14 - 10 = 4 11 - 9 = 2 10 - 8 = 2 14 - 9 = 5 18 - 7 = 11

SALE'S ADVISERS OPPORTUNITY COST TABLE

SALESADVISERS

O F F I C E S

Ramirez 2 4 3 2 4 2

Arambulo 1 3 5 3 3 1

Cordero 0 2 3 1 1 2

Cereza 3 1 1 1 3 4

Diaz 2 0 0 0 0 0

Tellez 4 4 2 2 5 11

SOLUTION :

Ramirez 2 - 2 = 0 4 - 2 = 2 3 - 2 = 1 2 - 2 = 0 4 - 2 = 2 2 - 2 = 0

Arambulo 1 - 1 = 0 3 - 1 = 2 5 - 1 = 4 3 - 1 = 2 3 - 1 = 2 1 - 1 = 0

Cordero 0 - 0 = 0 2 - 0 = 2 3 - 0 = 3 1 - 0 = 1 1 - 0 = 1 2 - 0 = 2

Cereza 3 - 1 = 2 1 - 1 = 0 1 - 1 = 0 1 - 1 = 0 3 - 1 = 2 4 - 1 = 3

Diaz 2 - 0 = 0 0 - 0 = 0 0 - 0 = 0 0 - 0 = 0 0 - 0 = 0 0 - 0 = 0

Tellez 4 - 2 = 2 4 - 2 = 2 2 - 2 = 0 2 - 2 = 0 5 - 2 = 3 11 - 2 = 9

TOTAL OPPORTUNITY COST TABLE

SALESADVISERS

O F F I C E S

Ramirez 0 2 1 0 2 0

Arambulo 0 2 4 2 2 0

Cordero 0 2 3 1 1 2

Cereza 2 0 0 0 2 3

Diaz 2 0 0 0 0 0

Tellez 2 2 0 0 3 9

ASSIGNED: COST

Ramirez to San Juan 10

Arambulo to Pasig 8

Cordero to Las Piñas 6

Cereza to Quezon City 11

Diaz to Mandaluyong 9

Tellez to Caloocan 11

TOTAL ASSIGNMENT COST 55

Exercise 4.4

Solve the following problems

1.Three jobs has to be done on three machines. Each job can be assigned to one and only one machines. The cost of each job on each machine is given in the following table.

Machines

Jobs 1 2 3

1 P50 P70 P90

2 140 100 120

3 150 130 160

Find the job assignment that will minimize cost?

Cost Information

Machines

Jobs 1 2 3

1 P50 P70 P90

2 140 100 120

3 150 130 160

Machines1 Machines2 Machines3Job #1 P50-50= P0 P70-70=P0 P90-90=P0Job #2 140-50=90 100-70=30 120-90=30Job #3 150-150=100 130-70=60 160-90=70

Machines

Jobs 1 2 3

1 P0 P0 P0

2 90 100 120

3 100 130 70

Machines1 Machines1 Machines3

Job #1 0 0 0Job #2 90-30=60 30-30=0 30-30=0Job #3 100-60=40 60-60=0 70-60=10

Machines

Jobs 1 2 3

1 P0 P0 P0

2 60 0 0

3 40 0 10

Assigned Cost

Job#1 to Machine 1 P 50Job#2 to Machine 3 120Job#3 to Machine 2 130Total Assignment cost P300

2. The Ramirez Reality Company wishes to reshuffle and asign six of its top-notch real-estate sales advisors to one of six areas where their low-cost housing projects are located. Using the demography of their six areas and the past sales performance of their six sales advisor, the company estimates that the sales of property, in homes per year would be as follows.

SalesAdvisor

Areas 1 2 3 4 5 6

Carol26 18 16 20 24 28

Ben24 16 14 22 28 26

Oscar20 18 12 18 24 24

Mel24 22 18 18 20 22

Minnie18 14 16 16 18 20

Art24 24 20 20 22 28

Determine to which area each of the six should be assigned in order to maximize the total annual sales of houses.

Cost Information

SalesAdvisor

Areas 1 2 3 4 5 6

Carol26 18 16 20 24 28

Ben24 16 14 22 28 26

Oscar20 18 12 18 24 24

Mel24 22 18 18 20 22

Minnie18 14 16 16 18 20

Art24 24 20 20 22 28

Carol Ben Oscar Mel Minnie ArtAreas 1 26-16=10 24-14=10 20-12=8 24-18=6 18-14=4 28-20=8Areas 2 18-16=2 16-14=2 18-12=6 22-18=4 14-14=0 24-20=4Areas 3 16-16=0 14-14=0 12-12=0 18-18=0 16-14=2 20-20=0Areas 4 20-16=4 22-14=8 18-12=6 18-18=0 16-14=2 20-20=0Areas 5 24-16=8 28-14=14 24-12=12 20-18=2 18-14-4 22-20=2Areas 6 28-16=12 26-14=12 24-12=12 22-18=4 20-14=6 28-20=8

SalesAdvisor

Areas 1 2 3 4 5 6

Carol10 18 0 4 8 12

Ben10 16 0 8 14 12

Oscar8 18 0 18 12 12

Mel6 4 0 0 20 2

Minnie4 0 16 1 4 6

Art8 4 0 0 22 8

Carol Ben Oscar Mel Minnie ArtAreas 1 10-2=8 10-2=8 8-6=2 6-2=4 4-2=2 8-2=6Areas 2 2-2=0 2-2=0 6-6=0 4-2=2 0 4-2=2Areas 3 0 0 0 0 2-2=0 0Areas 4 4-2=2 8-2=6 6-6=0 0 2-2=0 0Areas 5 8-2=6 14-2=12 12-6=6 2-2=0 4-2-2 2-2=0Areas 6 12-2=10 12-2=10 12-6=6 4-2=2 6-2=4 8-2=6

Areas- Opportunity cost table

SalesAdvisor

Areas 1 2 3 4 5 6

Carol8 0 0 4 6 10

Ben8 0 0 8 12 10

Oscar2 0 0 0 6 6

Mel4 2 0 0 0

Minnie2 0 0 0 4 4

Art6 2 0 0 0 6

Carol Ben Oscar Mel Minnie ArtAreas 1 8-2=6 8-6=2 2-2=0 4-2=2 2-2=0 6-2=4Areas 2 0 0 0 2-2=0 0 2-2=0Areas 3 0 0 0 0 0 00Areas 4 2-2=0 6-6=0 0 0 0 0Areas 5 6-2=4 12-6=6 6-2=4 0 2-2-0 0Areas 6 10-2=8 10-6=4 6-2=4 2-2=0 4-2=0 6-2=4

SalesAdvisor

Areas 1 2 3 4 5 6

Carol6 0 0 0 4 8

Ben2 0 0 0 6 4

Oscar0 0 0 0 4 4

Mel2 0 0 0 0 0

Minnie0 0 0 0 0 2

Art4 0 0 0 0 4

Assigned CostCarol to Areas 1 18Ben to Areas 2 14Oscar to Areas 3 20Mel to Areas 4 22Minnie to Areas 5 16Art to Areas 6 22Total Assignment cost 112

3. Given the following assignment table:

Personnel Job#1 Job#2 Job#3

David 5 hours 7 hours 9 hours

Hazel 14 hours 10 hours 12 hours

Rachel 15 hours 13 hours 16 hours

Find the assignment program that will minimize the number of hours spent in doing the jobs?

Cost Information

David 5 hours 7 hours 9 hours

5 hours

7 hours

Personnel Job#1 Job#1 Job#1David 5hours- 5hours = 0 7hours- 7hours = 0 9hours- 9hours = 0Hazel 14 hours -5hours =9 hours 10hours- 7hours =3 hours 12hours- 9hours =3 hoursRachel 15 hours -5hours =10 hours 13hours- 7hours = 6 hours 16hours- 9hours = 7 hours

David 0 0 0

Personnel Job#1 Job#1 Job#1David 0 0 0Hazel 9 hours -3hours =6 hours 3hours- 3hours =0 3hours- 3hours =0Rachel 10 hours -6hours =4 hours 6hours- 6hours =0 7hours- 6hours = 1 hours

David 0 0 0

Hazel 6 hours 0 0

3 hours

6 hours

3 hours

Rachel 4 hours 0 1 hours

Assigned Cost

David to Job#1 5 hoursHazel to Job#2 12 hoursRachel toJob#3 13 hoursTotal Assignment cost 30 hours

4. The manager of a company wishes to assign three jobs to four machines on one-on-one basis. The cost of each job on each machine is given on the following table:

Jobs Machine 1 Machine 2 Machine 3 Machine 4

A P 650 P 480 P 560 P 640

B 160 260 340 340

C 200 300 380 440

Find the assignment that will minimize the assignment cost.

ANSWER:

A P 650 P 480 P 560 P 640

B 160 260 340 340

C 200 300 380 440

JOB OPPORTUNITY COST TABLE

A P 490 P 220 P 220 P 300

B 0 0 0 0

C 40 40 40 100

SOLUTION:

Machine 1 Machine 2 Machine 3 Machine 4650 – 160 = 490480 – 260 = 220560 – 340 = 220640 – 340 = 200160 – 160 = 0 260 – 260 = 0 340 – 340 = 0 340 – 340 = 0200 – 160 = 40 300 – 260 = 40 380 – 340 = 40 440 – 340 = 100

A P270 P 0 P 0 P 80

B 0 0 0 0

C 0 0 0 60

SOLUTION:

A 490 – 220 220 – 220 = 0 220 – 220 = 0 300 – 220 = 80

B 0 – 0 = 0 0 – 0 = 0 0 – 0 = 0 0 – 0 = 0

C 40 – 40 =0 40 – 40 = 0 40 – 40 = 0 100 – 40 = 60

ASSIGNED COST

Job A to Machine 2 480Job B to Machine 1 160Job C to Machine 3 380TOTAL 1020

5. Given the following table:

Item # 1 Item # 2 Item # 3

Department Store 1 P 650 P 800 P 780

Department Store 2 660 780 790

Use the assignment method to determine the least-cost purchasing plan.

ANSWER:

Department Store 1 P 650 P 800 P 780

Department Store 1 P50 P 20 P 0

SOLUTION:

Item #1 Item #2 Item #3650 – 600 = 50 800 – 780 = 20 780 – 780 = 0660 – 600 = 0 780 – 780 = 0 790 – 780 = 10600 – 600 = 0 790 – 780 = 10 790 – 790 = 10

ASSIGNED COST

Dept. Store 1 to Item #3 780Dept. Store 2 to Item #2 780Dept. Store 3 to Item #1 600

P 2160

CHAPTER 5

Exercise 5.1

Solve the following problems:

1. The cost of placing an order is P60 and the carrying cost is P30 per unit per year. The estimated annual demand is 10, 000 units. Find the economic order quantity.

P= P60 C= P30/ units D= 10, 000 units

Qu=√ 2PDCQu=√ 2(60)(10,000)30

Qu = 200

2. The DJA Company uses 100, 000 units per year of a product. The carrying cost per unit is P3 per year. The cost of ordering a batch is P60.

a. What is optimum order size ( EOQ )?b. If the ordering cost is P60 per order, how many units should be ordered at one time?

D= 100, 000 units C= P3/units P= P60

a. Qu=√ 2PDC b. Qn=√ DC2 PQu=√ 2(60)(100,000)3

Qn=√ 100,000 (3)2(60)

Qu= 2, 000 ` Qn= 50

3. Hazel Rae, the owner of Hazel’s Boutique is planning to order maternity dresses for her store. The dress manufacturer informed her that if the maternity dress order received this month are worth P30, 000 or more, the manufacturer will sponsor the one-day visit by Dr. Alene del Rosario, an OB-Gyne who will come to the store and talk with prospective mothers about child care. Hazel orders P350, 000 worth of maternity dresses per year. Her carrying costs are 20% of average inventory and her ordering costs is P300 per order.

a. If Hazel places her order using the EOQ model, will the value of her order be sufficient to qualify for a visit by Dr. del Rosario?

b. How much is the worth of maternity dresses per order?c. How many days interval per order?

C= 20% P= P300/ units A= P350, 000

a. Qu=√ 2PAC b. Qp=√ 2 APC c. Qd=365√ 2 PAC Qu=√ 2(P300)(P350,000)20%

Qp=√ 2(P350,000)(P300)20%

Qd=365√ 2(P300)P350,000(20%)

Qu=P 32, 404/ Yes Qp= P 32, 404 Qd= 34 days

4. This year Arielle Rachel Mae, the owner of Rachel’s Novelty Store in Makati has ordered a particular novelty item according to EOQ calculations. She places 10 orders per year. Her ordering cost is P46 per order, while her carrying cost is 25%. Next year, annual demand for the same item is expected to increase by 20%. How many orders will Rachell Mae should be placing next year?

EOQ = 10 orders per year P = P46/ order C = 25%

Qn=√ AC2P [10]2 = √ .25 A92 .25 = 92(100) A = 36, 800 annual

demand now

10=√ A(.25)2(46)

100 = .25 A92

.25 A.25

=9 ,200.25

A1= 36, 800 + 20 %( 36, 800 )

A1= 44 160

5. The following data are for an inventory item in which the EOQ model applies.

Given:

D= 10, 000 units ( annual demand )

R = P10 per unit

P = P500 per order

C = 25% of average inventory

a. The number of order per yearb. The economic order quantityc. The amount in pesos per orderd. The number of days interval between orderse. The total inventory costs.

a. Qn=√ DC2 P b.Qu=√ 2PDC c. Qp=√ 2PDC /R

Qn=√ P10,000(25% )2(P500)

Qu=√ 2(P500)(P10,000)25% Qp=√ 2(P500)(P10,000)25% /10

Qn= 2 Qu= P 6, 325 Qp= P 20, 000

d. Qd=365√ 2PDC e. Ordering Cost + Carrying Cost = Total inventory

Qd=365√ 2(P500)P10,000 (25%)

Qn x P DQnx 12xc

Qd= 231 days 2 x 500 = 1, 000 10,0002

x 12x25% = 625 1, 625

6. The Arambulo’s Garment Factory purchases their raw materials from one of the textile mills in Metro Manila. One of these raw materials is being used in the manufacture of children’s dresses. The factory uses 5, 000 yards of this raw material per year. The cost of ordering this raw material from the Company is P500 per order. Carrying cost is 20% of average inventory.

a. The no. of order/yearb. The EOQc. Amount in pesos/orderd. No. of days interval between ordere. Total inventory cost

D= 5, 000 yards P= P500/ order C= 20%

a. Qn=√ DC2 P b. Qu=√ 2PDC c. Qp=√ 2PDCQn=√ P5,000(20% )

2(P500) Qu=√ 2(P500)(P5,000)20%

Qp=√ 2(P500)(P5,000)20%

Qn= 1 Qu= P 5, 000 Qp= P 5, 000

d. Qd=365√ 2PDC e. Ordering Cost + Carrying Cost = Total inventory

Qd=365√ 2(P500)P5,000 (20%)

Qn x P DQnx 12xc

Qd= 365 days 1 x 500 = 500 5,0001

x 12x20% = 500 1,

Exercise 5.2

1. EOQ before discount

Qu=√2 (160 )(9000).12

Qu = 4, 899 units/ order

Considering discount

1. Incremental price benefit

P15(9000) = P 135, 000

2. Incremental carrying cost

C= 190205 x (.28 x 205 ) = P 22.8 with discount

22.8 (30024.6

−24.6 x 489924.6

=278−4899=−¿ P 4621

3. Incremental benefit of reduced ordering cost

160(90004899

¿−160( 9000300 )=294−4800=−P 4506

Therefore net incremental benefit = P 135, 000 – (-P4621) + (-P4506) = P 135, 115/ Yes the owner should take the offer.

2. A. Nr=√21900 (.28)2(3000)

=1 runyear

approximately

B. TC = (Qn x P) + (A/Qn) x ½ x c

(√21900 ( .28 )2 (3000 )

x P3000)+( 21,9001 ) x 12 x .28Total inventory cost:

P3, 000 + P3,066 = P 6, 066 / year

3. A. Nr=√300 ,000 ( .28 )2 (1350 )

=5.5777∨6 runs / year

B. 365¿ √2(1350)300,000(.28)

=34.6days

4. Nr=√1 ,440 ,000(.28)2 (1000)

=14.1986∨14 runs / year

5. No, because from the given formula of EOQ, R, which is the cost per unit can only be found at the formula of finding number of units per order (Qu). But because Qu is also not given, it’s impossible to derived a formula having two entities of the equation missing so therefore, it’s not.

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