Maximize :P = 4x, + 3x2 + 2x3 3x, + 2x, + 5x, + s, = 23 2x, + x2 + X3 +Sz = 8 X, + X2 +2x, + S3 = 7 |- 4x, – 3x, - 2x3 + P = 0 %3D 3x, + 2x, +5x, 5 23 2х, + х, + x, s8 System: ST : X, + X2 +2x3 <7 х, X2, х, 2 0
Maximize :P = 4x, + 3x2 + 2x3 3x, + 2x, + 5x, + s, = 23 2x, + x2 + X3 +Sz = 8 X, + X2 +2x, + S3 = 7 |- 4x, – 3x, - 2x3 + P = 0 %3D 3x, + 2x, +5x, 5 23 2х, + х, + x, s8 System: ST : X, + X2 +2x3 <7 х, X2, х, 2 0
Practical Management Science
6th Edition
ISBN:9781337406659
Author:WINSTON, Wayne L.
Publisher:WINSTON, Wayne L.
Chapter2: Introduction To Spreadsheet Modeling
Section: Chapter Questions
Problem 20P: Julie James is opening a lemonade stand. She believes the fixed cost per week of running the stand...
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![Simplex Method Example
Maximize :P = 4x, + 3x, +2x3
|3x, + 2х, + 5х, < 23
2x, + x, + x, <8
X, + x, + 2x, s7
3x, +2x, +5x, +s, = 23
2x, + x, + X3 +S2 = 8
x, + X, +2x, +s, =7
(- 4x, – 3x, -2x, +P = 0
System:
ST :
X1, X2, X3 2 0
X2 X3 s, S2 s, P
1 0 0 0| 23
0 1 0 0
23 +3 = 7.6
8+2=(4)
3
2
5
s2 [2]
1
1
8
1
S3
1
2
0 0 1 0
7
7÷1= 7
Р(-4) -3 -2 0 о 01
- 3R, + R, → R,
1
Row Operations:R, → R,
- R, + R, → R,
and
4R, + R4 → R4
X2
X3
S2
S3 P
[o 1/2 7/2 1 -3/2 0 0|11]11+(1/2) = 22
1/2 1/2 0
o 0| 4 4+ (1/2) = 8
X|1
s3 0 {1/2} 3/2 0 -1/2 1 0 3 3-(1/2) = 6
P0 (-1) o 0
X1 enters to become a basic variable, s2 exits to become a nonbasic variable
(pivot column identifies the entering variable and pivot row identifies the exiting variable)
1/2
2
0 1 16
R, + R¸ → R,
Row Operations:
2R, → R, and
R3 + R2 → R,
R, + R, → R4
S3 P
1 -1 -1 0
-1 0
X, X2 X3
S, S2
S0 0
Final Tableau x, 1 0 -1 0
2
8
1
1
2 0 6
2 1 22
X2 0 1
3 0 -1
PO 0 3 0 1
X2 enters to become a basic variable, s3 exits to become a nonbasic variable
(pivot column identifies the entering variable and pivot row identifies the exiting variable)
s2,53 and x3 are nonbasic {=0}; So x1,X2, S1 and P are basic.
Optimal Solution: Maximum P=22 at x1=1, x2=6, x3=0 (s1=8,s2=0,s3=0)
Teacher's Note: Notice that x3 never entered and thus is 0. This is a viable solution.
Suppose x1,x2,x3 represented 3-speed, 5-speed, 10-speed bicycles. In this problem it is
more profitable not to manufacture 10-speed bikes.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4819039e-d0fd-4b39-aecc-3598f549ab59%2Fb6718cae-6311-4d0f-a515-9c73a4a4ed5a%2F7h0qn19_processed.png&w=3840&q=75)
Transcribed Image Text:Simplex Method Example
Maximize :P = 4x, + 3x, +2x3
|3x, + 2х, + 5х, < 23
2x, + x, + x, <8
X, + x, + 2x, s7
3x, +2x, +5x, +s, = 23
2x, + x, + X3 +S2 = 8
x, + X, +2x, +s, =7
(- 4x, – 3x, -2x, +P = 0
System:
ST :
X1, X2, X3 2 0
X2 X3 s, S2 s, P
1 0 0 0| 23
0 1 0 0
23 +3 = 7.6
8+2=(4)
3
2
5
s2 [2]
1
1
8
1
S3
1
2
0 0 1 0
7
7÷1= 7
Р(-4) -3 -2 0 о 01
- 3R, + R, → R,
1
Row Operations:R, → R,
- R, + R, → R,
and
4R, + R4 → R4
X2
X3
S2
S3 P
[o 1/2 7/2 1 -3/2 0 0|11]11+(1/2) = 22
1/2 1/2 0
o 0| 4 4+ (1/2) = 8
X|1
s3 0 {1/2} 3/2 0 -1/2 1 0 3 3-(1/2) = 6
P0 (-1) o 0
X1 enters to become a basic variable, s2 exits to become a nonbasic variable
(pivot column identifies the entering variable and pivot row identifies the exiting variable)
1/2
2
0 1 16
R, + R¸ → R,
Row Operations:
2R, → R, and
R3 + R2 → R,
R, + R, → R4
S3 P
1 -1 -1 0
-1 0
X, X2 X3
S, S2
S0 0
Final Tableau x, 1 0 -1 0
2
8
1
1
2 0 6
2 1 22
X2 0 1
3 0 -1
PO 0 3 0 1
X2 enters to become a basic variable, s3 exits to become a nonbasic variable
(pivot column identifies the entering variable and pivot row identifies the exiting variable)
s2,53 and x3 are nonbasic {=0}; So x1,X2, S1 and P are basic.
Optimal Solution: Maximum P=22 at x1=1, x2=6, x3=0 (s1=8,s2=0,s3=0)
Teacher's Note: Notice that x3 never entered and thus is 0. This is a viable solution.
Suppose x1,x2,x3 represented 3-speed, 5-speed, 10-speed bicycles. In this problem it is
more profitable not to manufacture 10-speed bikes.
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