MATH C (Business Statistics) Simplex Method in Excel

1. OUTPUT IN MATHC
LINEAR PROGRAMMING

2. GRAPHICAL METHOD
 Pretty Ladies Company produces two types of
beauty soaps: papaya and carrot soap, which must
be processed through mixing ,shaping, and finishing
department. Mixing department is available for
eighteen (18) hours in every production period,
Shaping department is available for forty-two (42)
hours in every production and Finishing department
is available for hours twenty-four (24) hours of
work. Producing papaya soap requires two (2)
hours in mixing, two (2) hours in shaping and three
(3) hours in finishing. Carrot soap requires one (1)
hour in mixing, three (3) hours in shaping and one
(1) hour in finishing. Papaya soap contributes
₱80.00 profit while carrot soap contributes ₱60.00
profit. The problem is to determine the no. of each
soaps to make per production period in order to
maximize profit.

3. REPRESENTATION
 Let x = no.of papaya soaps
 Let y = no. of carrot soaps
Mixing Shaping Finishin
g
Profit
papaya 2x 2x 3x 80.00
carro
t
1y 3y 1y 60.00
18(max) 42(max) 24 (max)
GRAPHICAL
METHOD

4. GRAPHICAL METHODMaximize: 80x + 60y
2x + y ≤ 18
2x + 3y ≤ 42
3x + y ≤ 24
x ≥ 0 , y ≥ 0
EQ.1 EQ.2 EQ.3
2x+y≤18 2x+3y≤42 3x+y≤24
2x+y=18 2x+3y=42 3x+y=24
let x=0 let x=0 let x=0
2(0)+y=18 2(0)+3y=42 3(0)+y=24
Y=18 3y=42
3 3
Y=24
(0,18)
Y=14
(0,14) (0,24)
Let y=0 Let y=o Let y= o
2x+(0)=18 2x+3(0)=42 3x+ (0)=24
2x=18
2 2
2x=42
2 2
3x=24
3 3
X=9 X=21 X=8
(9,0) (21,0) (8,0)

5. GRAPHICAL METHOD
 2x+y≤18 EQ.1 (0,18)
2(0)+(0)≤18 (9,0)
0≤18
TRUE
 2x+3y≤42 EQ.3 (0,24)
2(0)+3(0) ≤ 42 (8,0)
0≤42
TRUE
 3x+y≤2
3(0)+(0) ≤24 EQ.2 (O,14)
0≤24 (21,0)
TRUE

6. EQ

 80(0)+60(0)=0
 80(0)+60(14)=840
 80(3)+60(12)=960
 80(6)+60(6)=840
 80(8)+60(0)=640
EQ.1 AND EQ. 2 EQ.1 AND EQ. 3
2x+y=18(-1)
2x+3y=42(1)
-2x-y=-18
2x+3y=42
2y=24
2 2
Y=12
2x+y=18(3)
3x+y=24(-2)
6x+3y= 54
-6x-2y=-48
Y=6
2x+3(12)=42
2x+36=42
2x=42-36
2x=6
2 2
X=3
(3,12)
3x+(6)=24
3x=24-6
3x=18
3 3
X=6
(6,6)
Intersection
point
Coordinates
(x,y)
Profit
O (0,0) 0
C (0,14) 840
G (3,12) 960
H (6,6) 840
F (8,0) 640

7.  Therefore, Pretty Ladies Co. can produce 3
papaya soaps and 12 carrot soaps per
production period and has a maximum profit
of 960.00

8. SIMPLEX METHOD
 My Lady Manufacturing Co. makes two types of
bag; ordinary and special edition. The bags are
sold to the independent dealers at a profit of
₱200.00 per ordinary edition and ₱300.00 per
special edition. An ordinary edition bag requires
30 hours for assembly, 20 hours for designing and
10 hours for finishing. A special edition bag
requires 75 hours for assembly ,25 hours for
designing and 5 hours for finishing. A production
run generally has 15000 hours available for
assembly, 6500 hours available for designing and
2500 for finishing. Determine the maximum profit .

9. REPRESENTATION
 Let x=no. of ordinary edition bags
Y=no.of special edition bags
Assembl
y
Designin
g
Finishing Profit
ordinary 30x 20x 10x 200.00
speci
al
75y 25y 5y 300.00
15000
(MAX)
6500
(MAX)
2500
(MAX)

10. Linear program New Program
MAXIMIZE=
200x+300y 200x+300y+0S1 +0S2
+0S3
30x+75y≤15000 30x+75y+S1 +0S2
+0S3=15000
20x+25y≤6500 20x+25y+0S1 +S2
+0S3=6500
10x+5y≤2500 10x+5y+0S1 +0S2
+S3=2500
X,y≥0 x,y≥0

11. FIRST TABLE OC
Cj 200 300 0 0 0
Pm Qty X Y S1 S2 S3
PR 0 S1 15000 30 75 1 0 0
0 S2 6500 20 25 0 1 0
0 S3 2500 10 5 0 0 1
Zj 0 0 0 0 0 0
Cj-Zj 200 300 0 0 0
PR= Qty divided by OC
15000/75=200
6500/25=260
2500/5=500

12. SECOND TABLE OC
Cj 200 300 0 0 0
Pm Qty X Y S1 S2 S3
300 Y 200 ⅖ 1 1/75 0 0
PR 0 S1 1500 10 0 -1/3 1 0
0 S3 1500 8 0 -1/15 0 1
Zj 60000 120 300 4 0 0
Cj-Zj 80 0 -4 0 0
PR= Qty divided by OC Zj= 300(200)+0(1500)+0(1500)=60000
200/⅖=500 300(2/5)+0(10)+0(8) =120 300(0)+0(1)+0(0)=0
1500/10=150 300(1)+0(0)+0(0) =300 300(0)+0(0)+0(1)=0
1500/8=187 1/2 300(1/75)+0(-1/3)+0(-1/15) = 4
S1 OLD(element)-(PR(old) x NEW) S3 OLD(element)-
(PR(old) x NEW)
6500 -
(25x200)=1500
20- (25X2/5)=10
25- (25x1)=0
0-(25x1/75)=-1/3
1-(25x0)=1
0-(25x0)=0
2500-(5x200)=1500
10-(5x2/5)=8
5- (5x1)=0
0- (5x1/75)=-1/15
0-(5x0)=0
1-(5x0)=1

13. THIRD TABLE
Cj 200 300 0 0 0
Pm Qty X Y S1 S2 S3
200 X 150 1 0 -1/30 1/10 0
300 Y 140 0 1 2/75 -1/25 0
0 S3 300 0 0 1/5 -4/5 1
Zj 72000 200 300 1 1/3 8 0
Cj-Zj 0 0 -1 1/3 -8 0
200-(2/5x150)=140
2/5-(2/5x1)=0
1-(2/5x0)=1
1/75-(2/5x-
1/30)=2/75
0-(2/5x1/10)=-1/25
0-(2/5x0)=0
1500-(8x150)=300
8-(8x1)=0
0-(8x0)=0
-1/15(8x-1/30)=1/5
0-(8x1/10)=-4/5
1-(8x0)=1
Zj= 200(150)+300(140)+0(300)=72000 200(-1/30)+300(2/75)+0(1/5)= 1 1/3
200(1)+300(0)+0(0)= 200 200(1/10)+300(-1/25)+0(4/5)=8
200(0)+300(1)+0(0)= 300 200(0)+300(0)+0(1)=0
OLD(element)-(PR(old) x NEW) OLD(element)-(PR(old) x
NEW)

14.  Therefore, My Lady Manufacturing Co. can
produce 150 ordinary edition bags and140
special edition bags with the maximum profit
of ₱72000.00