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Quantitative Techniques for Business ManagementTRANSCRIPT
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)
X= -3
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
(0,3)
3 3 Y = 3
(9,0)
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
(8,0)
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
Y = 2
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
y = 4
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
3. Minimize: 6x1+24x2
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
4. Minimize: 32x1+36x2
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
6. Minimize: 15x1+12x2
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
IV. Graphical method
Equation No. 1 Equation no. 2
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
IV. Graphical method
Equation No. 1 Equation no. 2
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
0
X 0 10.67
Y 16
0
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
Outgoing variable: S2
Entering variable: X2
Optimum column
Ci
Cj
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
Ci
Cj
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
2
1
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
Ci
Cj
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
Outgoing variable: S1
Entering variable: X1
Ci
Cj20,80
015,00
00 0
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
0 0
Ci
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
4
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
Ci
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
Ci
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
2
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
₱ 6 ₱ 7 0 0Solution variables
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)
₱ 6 ₱ 7 0 0Solution variables
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
₱ 50 ₱ 30 0 0Solution variables
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
outgoing row Optimum column
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
₱ 6 ₱ 7 0 0Solution variables
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
outgoing row Optimum column
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)
₱ 50 ₱ 30 0 0Solution variables
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
Ci \ Cj ₱ 30.50 ₱ 29.00 0 0 0 0Solution Variable
Solution Values X1 X2 S1 S2 S3 S4
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
1.)
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
4.)
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)
Ci \ Cj ₱ 30.50 ₱ 29.00 0 0 0 0Solution Variable
Solution Values X1 X2 S1 S2 S3 S4
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
1.)
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
Ci \ Cj ₱ 25 ₱ 22.50 ₱ 30 0 0 0Solution Variable
Solution Values X1 X2 X3 S1 S2 S3
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
Outgoing row optimum column
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
4.)
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
Ci \ Cj ₱ 25 ₱ 22.50 ₱ 30 0 0 0Solution Variable
Solution Values X1 X2 X3 S1 S2 S3
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
Outgoing row optimum column
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
4.)
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
Ci \ Cj ₱ 25 ₱ 22.50 ₱ 30 0 0 0Solution Variable
Solution Values X1 X2 X3 S1 S2 S3
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
2.)
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
Outgoing row optimum column
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
5
Table 2
CjCi ₱50 ₱55 0 0
Solution Variables
Solution Values X1 X2 S1 S2
0 S1 66 0 1 -⅕
0 S2 14 ⅕ 1 0 ⅕
Zj 770 11 55 11
Cj - Zj 39 0 0 -11
Outgoing row optimum column
(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
9/5
CjCi ₱50 ₱55 0 0
Solution Variables
Solution Values X1 X2 S1 S2
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
Cj
Ci
₱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
2
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
Solution Values X1 X2 S1 S2
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
3/2
Table 3
CjCi ₱3 ₱2 0 0
Solution Variables
Solution Values X1 X2 S1 S2
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
S6
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
To
From
PROJECTSupply
A B C
Plant X 750 -1250
1450
Plant Y -250
1600
Plant Z 1750
1600
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
145
50 155
200 140
To
From
PROJECTSupply
A B C
Plant X 750
1450
Plant Y 1000
1600
Plant Z 1750 1250
1600
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
Evaluation of Unused Cells:
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
50
145 150
160
140
Problem No. 2
Initial Feasible Solution
To
From
PACKING PLANTSupply
W X Y
Plantation A 40
170
Plantation B -160
250
Plantation C -240 -80
100
Demand 130 200 190 520520
Transportation Cost:
130 x 240 = P 3120040 x 160 = 6400160 x 300 = 4800090 x 200 = 18000100 x 120 = 12000
P 115600
Evaluation of Unused Cells:
AY = 100 – 200 + 300 – 160 = 40BW = 220 – 300 + 160 – 240 = -160CW = 40 – 240 + 160 – 300 + 200 – 100 = -240CX = 140 – 120 + 200 – 300 = -80
Computation of Stones:
The largest negative index is -240The smallest stone in negative position is 100
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
130
160
40
90
100
Table II
To
From
PACKING PLANTSupply
W X Y
Plantation A 40
170
Plantation B -160
250
Plantation C 180 260
100
Demand 130 200 190 520520
Transportation Cost:
30 x 240 = P 7200140 x 160 = 2240060 x 300 = 18000190 x 200 = 38000100 x 40 = 4000
P 89600
Evaluation of Unused Cells:
AY = 100 – 200 + 300 – 160 = 40BW = 220 – 300 + 160 – 240 = - 160CX = 140 – 160 + 240 – 40 = 180CY = 120 – 200 + 300 – 160 + 240 – 40 = 260
Computation of Stones:
The largest negative index is -160The smallest stone in negative position is 30
BW = 0 + 30 = 30BY = 60 – 30 = 30AW = 30 – 30 = 0AX = 140 + 30 = 170
240 160 100
220 300 200
40 140 120
30
60
140
190
100
Optimal Solution
To
From
PACKING PLANTSupply
W X Y
Plantation A 160 40
170
Plantation B 250
Plantation C 20 100
100
Demand 130 200 190 520520
Transportation Cost:
170 x 160 = P 2720030 x 220 = 660030 x 300 = 9000190 x 200 = 41800100 x 40 = 4000
P 88600
Evaluation of Unused Cells:
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
30
170
190
100
30
Problem No. 3
Initial Feasible Solution – Optimal Solution
To
From
DEPARTMENT STORESSupply(D)
Makati(E)
Cubao(F)
Sta. Cruz(G)
Ermita(H)
Dummy
Plantation A 11.50
12.00
16.50 7.507250
Plantation B 1.50
10150
Plantation C 1.00
4350
Demand 8700 5800 2175 2900 2175 2175021750
Transportation Cost:
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
Evaluation of Unused Cells:
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 --
14.50
4.00
5.50
0
0
0
725
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
3700
Total transportation cost:
1000 (1600) + 1300 (2000) + 2160 + 1200 (1360) + 200 = P 5, 832, 000
Evaluation of Unused Cells:
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.
K1 = 1000 K2 = 1250 K3 = 2000From To A B C Supply
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.
K1 = 1000 K2 = 1250 K3 = 2000From To A B C Supply
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.
K1 = 500 K2 = 300 K3 = 600From To Project A Project B Project C Supply
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.
K1 = 500 K2 = 300 K3 = 600From To Project A Project B Project C Supply
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.
K1 = 500 K2 = 300 K3 = 600From To Project A Project B Project C Supply
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
P10
P6P5
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
P3
1
P5P5
1P
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
Job Opportunity cost table
Financing Corporation
Personnel Pampanga Bataan Bulacan
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
Revised Opportunity Cost
Financing Corporation
Personnel Pampanga Bataan Bulacan
P1,000
1,000
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
Job Opportunity cost table
6
4 5 6
2 5
Job Machine 1 2 3 4 5 6
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
Revised Opportunity Cost
Job Machine 1 2 3 4 5 6
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
1
1
1
Job Machine 1 2 3 4 5 6
A
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
Job Opportunity cost table
Job Machine 1 2 3 4 5 6
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
Revised Opportunity Cost
1
11
9
Job Machine 1 2 3 4 5 6
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:
Las Piñas Quezon City Caloocan San Juan Mandaluyong Pasig
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
Las Piñas Quezon City Caloocan San Juan Mandaluyong Pasig
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
Las Piñas Quezon City Caloocan San Juan Mandaluyong Pasig
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
Job Opportunity cost table
Machines
Jobs 1 2 3
1 P0 P0 P0
2 90 100 120
3 100 130 70
Machines1 Machines1 Machines3
P50
P70
P70
P30
P600
P30
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
Revised Opportunity Cost
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
14
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
Job Opportunity cost table
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
2
2
6
2
2
2
6
2
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
Revised Opportunity Cost
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
2
2
2
2
2
2
2
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
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
5 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
Job Opportunity cost table
Personnel Job#1 Job#2 Job#3
David 0 0 0
Hazel 9 hours 10 hours 12 hours
Rachel 10 hours 13 hours 7 hours
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
Revised Opportunity Cost
Personnel Job#1 Job#2 Job#3
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:
COST INFORMATION 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
JOB OPPORTUNITY COST TABLE
Jobs Machine 1 Machine 2 Machine 3 Machine 4
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
TOTAL OPPORTUNITY COST TABLE
Jobs Machine 1 Machine 2 Machine 3 Machine 4
A P270 P 0 P 0 P 80
B 0 0 0 0
C 0 0 0 60
SOLUTION:
JOB
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
Department Store 3 600 790 790
Use the assignment method to determine the least-cost purchasing plan.
ANSWER:
COST INFORMATION TABLE
Item # 1 Item # 2 Item # 3
Department Store 1 P 650 P 800 P 780
Department Store 2 660 780 790
Department Store 3 600 790 790
TOTAL OPPORTUNITY COST TABLE
Item # 1 Item # 2 Item # 3
Department Store 1 P50 P 20 P 0
Department Store 2 60 0 10
Department Store 3 0 10 10
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
Find:
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
cost
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.
Find:
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
cost
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,
000
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.