A problem is provided which is solved by using graphical and analytical method of linear programming method and then it is solved by using geometrical concept and algebraic concept of simplex method.
A problem is provided which is solved by using graphical and analytical method of linear programming method and then it is solved by using geometrical concept and algebraic concept of simplex method.
This slide set is a work in progress and is embedded in my Principles of Finance course site (under construction) that I teach to computer scientists and engineers
http://awesomefinance.weebly.com/
Solve given LP problem using simplex method and find maximum value o.pdfaminbijal86
Solve given LP problem using simplex method and find maximum value of objective function.
Write all
steps.
max : = 4.71 + 6x2 s.t. -x1 + 2 11 12s 27 21 52 90 i 1.2
Solution
the LPP problem according to Simplex method is given as follows:
Let, S1, S2, and S3 are slack variables three constraints. The standard form of given LPP is given
as follows:
Max Z = 4X1 + 5X2 + 0S1 + 0S2 + 0S3
Subject To:
-X1 + X2 + S1 + 0S2 + 0S3 = 11
X1 + X2 + 0S1 + S2 + 0S3 = 27
2X1 + 5X2 + 0S1 + 0S2 + S3 = 90
X1, X2, S1, S2, S3 >= 0
2)
To form an initial basic solution, consider the structural variables as zero, in other words X1 =
X2 = 0. Thus initial basic variables remains are S1, S2, and S3.
Z = 0, S1 = 11, S2 = 27, and S2 = 90
3)
The Initial Tableau is constructed as follows:
Cj
4
5
0
0
0
B (Quantity)
Cb
Xb
x1
x2
S1
S2
S3
0
S1
-1
1
1
0
0
11
0
S2
1
1
0
1
0
27
0
S3
2
5
0
0
1
90
Iteration 1:
Cj
4
5
0
0
0
B (Quantity)
RR
Cb
Xb
x1
x2
S1
S2
S3
0
S1
-1
1
1
0
0
11
11/1 = 11
0
S2
1
1
0
1
0
27
27/1 = 27
0
S3
2
5
0
0
1
90
90/5 = 18
Zj
0
0
0
0
0
0
j = Cj – Zj
4
5
0
0
0
0
zj=cBj x xj
j= (cj - zj)
As the j is positive the optimal solution is not obtained.
X2 is having highest j, thus incoming variable is X2. It realizes highest profit for next iteration.
Pivot Column = X2, pivot elements are 1, 1, and 5.
Determine replacement ratio by dividing B value by respective pivot column elements.
The replacement ratio for basis variable S1 is lowest positive, this variable is outgoing variable.
S1 is completely utilized if 11 units of X2 are to be produced.
Pivot row = S1
Key element (KE) = 1
The elements of next variables are obtained as follows:
Incoming Variable value for X1 is obtained as:
Other basis variable is obtained as:
1st row/KE
Old element – (Key column element x I.V. row element)
Cj
4
5
0
0
0
B (Quantity)
RR
Cb
Xb
x1
x2
S1
S2
S3
5
X2
-1
1
1
0
0
11
11/-1
= -11
0
S2
1-(1*-1)
=2
1-(1*1)
= 0
0-(1*1)
= - 1
1-(1*0)
= 1
0-(1*0)
= 0
27-(1*11)
16
16/2
= 8
0
S3
2-(5*-1)
= 7
5-(5*1)
= 0
0-(5*1)
= -5
0-(5*0)
= 0
1-(5*0)
= 1
90-(5*11)
= 35
35/7
= 5
Zj
-5
5
5
0
0
55
j = Cj – Zj
9
0
-5
0
0
As the j is positive the optimal solution is not obtained.
X1 is having highest j, thus incoming variable is X1. It realizes highest profit for next iteration.
Pivot Column = X1, pivot elements are -1, 2, and 7.
The replacement ratio for basis variable S3 is lowest positive, this variable is outgoing variable.
S3 is completely utilized if 5 units of X1 are to be produced.
Pivot row = S3
Key element (KE) = 7
Cj
4
5
0
0
0
B (Quantity)
RR
Cb
Xb
x1
x2
S1
S2
S3
5
X2
-1-(-1*1)
= 0
1-(-1*0)
= 1
1-(-1*-5/7)
= 2/7
0
0-(-1*1/7)
= 1/7
11-(-1*5)
= 16
16/(2/7)
= 56
0
S2
2-(2*1)
= 0
0-(2*0)
= 0
- 1-(2*-5/7)
= 3/7
1
0-(2*1/7)
= -2/7
16-(2*5)
= 6
16/(3/7)
= 37.33
4
X1
7/7
= 1
0/7
= 0
-5/7
0
1/7
35/7
= 5
5/(-5/7)
= -7
Zj
4
5
-10/7
0
9/7
(5*16+4*5)
= 100
j = Cj – Zj
0
0
10/7
0
-9/7
As the j is positive the optimal solution is not obtained.
S1 is having highest j, thus incoming variable is S1. It realizes highest profi.
Fundamentals of AlgebraChu v. NguyenIntegral ExponentsDustiBuckner14
Fundamentals of Algebra
Chu v. Nguyen
Integral Exponents
Exponents
If n is a positive integer (a whole number, i.e., a number without decimal part) and x is a number, then
The number x is called the base and n is called the exponent.
The most common ways of referring to are “ x to the nth power,”
“ x to the nth,” or “the nth power of x.”
Integral Exponents (cont.)
For any non-zero number x and a positive integer n
and
Note: is not defined
and
Rules Concerning Integral Exponents
Following are five rules in which m and n are positive integers:
Rule 1: ; for example,
Rule 2: ; for example
or
Rules Concerning Integral Exponents (Cont.)
Rule 3: ; for example
or
Rule 4: ; for example
or
Rule 5: ; for example
or
Basic Rules for Operating with Fractions
Since dividing by zero is not defined, we assume that the denominator
is not zero.
Following are the eight basic rules for operating with fractions.
Rule 1: ; for example
Rule 2: ; for example
Rule 3: ; for example
Basic Rules for Operating with Fractions (cont.)
Rule 4: ; for example
Rule 5: ; for example
Rule 6: ; for example
Basic Rules for Operating with Fractions (cont.)
Rule 7: ; for example
Rule 8: ; for example
Notes: a*b +a*x may be expressed as a(b + x)
a*b + 1 may be written as a(b + ), and
m*x – y may be expressed as m(x - )
Square Root
Generally, for a>0 , there is exactly one positive number x such that
, we say that x is the root of a, written as
for
When n = 2, we say that x is the square root of “a” and is denoted by
or or
For example:
or
Practices
Carrying out the following operations:
24 ; 2-2 ; 2322, ; 252-5 ; and (2x3)5
; ; ; and
3.
4.
5.
n
m
n
m
x
x
x
+
=
n
x
1
0
)
2
(
2
2
2
=
=
=
-
+
-
x
x
x
x
1
0
=
x
0
0
n
n
n
x
x
x
x
=
=
+
0
0
n
n
x
x
1
=
-
3
2 ...
Students, digital devices and success - Andreas Schleicher - 27 May 2024..pptxEduSkills OECD
Andreas Schleicher presents at the OECD webinar ‘Digital devices in schools: detrimental distraction or secret to success?’ on 27 May 2024. The presentation was based on findings from PISA 2022 results and the webinar helped launch the PISA in Focus ‘Managing screen time: How to protect and equip students against distraction’ https://www.oecd-ilibrary.org/education/managing-screen-time_7c225af4-en and the OECD Education Policy Perspective ‘Students, digital devices and success’ can be found here - https://oe.cd/il/5yV
How to Create Map Views in the Odoo 17 ERPCeline George
The map views are useful for providing a geographical representation of data. They allow users to visualize and analyze the data in a more intuitive manner.
Palestine last event orientationfvgnh .pptxRaedMohamed3
An EFL lesson about the current events in Palestine. It is intended to be for intermediate students who wish to increase their listening skills through a short lesson in power point.
The Art Pastor's Guide to Sabbath | Steve ThomasonSteve Thomason
What is the purpose of the Sabbath Law in the Torah. It is interesting to compare how the context of the law shifts from Exodus to Deuteronomy. Who gets to rest, and why?
Ethnobotany and Ethnopharmacology:
Ethnobotany in herbal drug evaluation,
Impact of Ethnobotany in traditional medicine,
New development in herbals,
Bio-prospecting tools for drug discovery,
Role of Ethnopharmacology in drug evaluation,
Reverse Pharmacology.
This is a presentation by Dada Robert in a Your Skill Boost masterclass organised by the Excellence Foundation for South Sudan (EFSS) on Saturday, the 25th and Sunday, the 26th of May 2024.
He discussed the concept of quality improvement, emphasizing its applicability to various aspects of life, including personal, project, and program improvements. He defined quality as doing the right thing at the right time in the right way to achieve the best possible results and discussed the concept of the "gap" between what we know and what we do, and how this gap represents the areas we need to improve. He explained the scientific approach to quality improvement, which involves systematic performance analysis, testing and learning, and implementing change ideas. He also highlighted the importance of client focus and a team approach to quality improvement.
We all have good and bad thoughts from time to time and situation to situation. We are bombarded daily with spiraling thoughts(both negative and positive) creating all-consuming feel , making us difficult to manage with associated suffering. Good thoughts are like our Mob Signal (Positive thought) amidst noise(negative thought) in the atmosphere. Negative thoughts like noise outweigh positive thoughts. These thoughts often create unwanted confusion, trouble, stress and frustration in our mind as well as chaos in our physical world. Negative thoughts are also known as “distorted thinking”.
The Indian economy is classified into different sectors to simplify the analysis and understanding of economic activities. For Class 10, it's essential to grasp the sectors of the Indian economy, understand their characteristics, and recognize their importance. This guide will provide detailed notes on the Sectors of the Indian Economy Class 10, using specific long-tail keywords to enhance comprehension.
For more information, visit-www.vavaclasses.com
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
1. Simplex Method By Dr. Eng. / Tamer Atteya Tanta Univ. – EGYPT Facebook link
2. Example Max. Z = 13x 1 +11x 2 Subject to constraints: 4x 1 +5x 2 < 1500 5x 1 +3x 2 < 1575 x 1 +2x 2 < 420 x 1 , x 2 > 0 دالة الهدف القيود أو الشروط
3. Rewrite objective function so it is equal to zero We then need to rewrite Z = 13x 1 +11x 2 as: Z -13x 1 -11x 2 =0
4. Convert all the inequality constraints into equalities by the use of slack variables S 1 , S 2 , S 3 . 4x 1 +5x 2 + S 1 = 1500 5x 1 +3x 2 +S 2 = 1575 x 1 +2x 2 +S 3 = 420 x 1 , x 2 , S 1 , S 2 , S 3 > 0
5. Z - 13x 1 -11x 2 = 0 Subject to constraints: 4x 1 +5x 2 + S 1 = 1500 5x 1 +3x 2 +S 2 = 1575 x 1 +2x 2 +S 3 = 420 Next, these equations are placed in a tableau Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z S 1 S 2 S 3 Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 -13 -11 0 0 0 0 S 1 0 4 5 1 0 0 1500 S 2 0 5 3 0 1 0 1575 S 3 0 1 2 0 0 1 420
6. a) Choose the most negative number from Z-row. That variable ( x 1 ) is the entering variable . b) Calculate Ratio = (value-col.) / (entering -col.) c) Choose minimum +ve Ratio. That variable (S 2 ) is the departing variable . Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 -13 -11 0 0 0 0 S 1 0 4 5 1 0 0 1500 S 2 0 5 3 0 1 0 1575 S 3 0 1 2 0 0 1 420
7. a) Choose the most negative number from Z-row. That variable ( x 1 ) is the entering variable . b) Calculate Ratio = (value-col.) / (entering -col.) c) Choose minimum +ve Ratio. That variable (S 2 ) is the departing variable . Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 -13 -11 0 0 0 0 S 1 0 4 5 1 0 0 1500 1500/4=375 S 2 0 5 3 0 1 0 1575 1575/5=315 S 3 0 1 2 0 0 1 420 420/1=420
8. Make the pivot element equal to 1 Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 -13 -11 0 0 0 0 S 1 0 4 5 1 0 0 1500 x 1 0 5 3 0 1 0 1575 S 3 0 1 2 0 0 1 420
9. Make the pivot element equal to 1 Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 -13 -11 0 0 0 0 S 1 0 4 5 1 0 0 1500 x 1 0 5/5 3/5 0 1/5 0 1575 / 5 S 3 0 1 2 0 0 1 420
10. Make the pivot element equal to 1 Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 -13 -11 0 0 0 0 S 1 0 4 5 1 0 0 1500 x 1 0 1 3/5 0 1/5 0 315 S 3 0 1 2 0 0 1 420
11. Make the pivot column values into zeros Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 -13 -11 0 0 0 0 S 1 0 4 5 1 0 0 1500 x 1 0 1 3/5 0 1/5 0 315 S 3 0 1 2 0 0 1 420
12. Make the pivot column values into zeros Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 -13 -11 0 0 0 0 S 1 0 4 5 1 0 0 1500 x 1 0 1 3/5 0 1/5 0 315 S 3 0 -0 1 -1 2 -3/5 0 -0 0 -1/5 1 -0 420 -315
13. Make the pivot column values into zeros Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 -13 -11 0 0 0 0 S 1 0 -4(0) 4 -4(1) 5 -4(3/5) 1 -4(0) 0 -4(1/5) 0 -4(0) 1500 -4(315) x 1 0 1 3/5 0 1/5 0 315 S 3 0 0 7/5 0 -1/5 1 105
14. Make the pivot column values into zeros Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 -13 -11 0 0 0 0 S 1 0 0 13/5 1 -4/5 0 240 x 1 0 1 3/5 0 1/5 0 315 S 3 0 0 7/5 0 -1/5 1 105
15. Make the pivot column values into zeros Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 0 -16/5 0 13/5 0 4095 S 1 0 0 13/5 1 -4/5 0 240 x 1 0 1 3/5 0 1/5 0 315 S 3 0 0 7/5 0 -1/5 1 105
16. Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 0 -16/5 0 13/5 0 4095 S 1 0 0 13/5 1 -4/5 0 240 x 1 0 1 3/5 0 1/5 0 315 S 3 0 0 7/5 0 -1/5 1 105
17. Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 0 -16/5 0 13/5 0 4095 S 1 0 0 13/5 1 -4/5 0 240 92.3 x 1 0 1 3/5 0 1/5 0 315 525 S 3 0 0 7/5 0 -1/5 1 105 75
18. Optimal Solution is : x 1 = 270, x 2 = 75, Z= 4335 Z x 1 x 2 S 1 S 2 S 3 Value Ratio Z 1 0 0 0 15/7 16/7 4335 S 1 0 0 0 1 -3/7 -13/7 45 x 1 0 1 0 0 2/7 -3/7 270 x 2 0 0 1 0 -1/7 5/7 75
19. Example Max. Z = 3x 1 +5x 2 +4x 3 Subject to constraints: 2x 1 +3x 2 < 8 2x 2 +5x 3 < 10 3x 1 +2x 2 +4x 3 < 15 x 1 , x 2 , x 3 > 0
20. Cont… Let S 1 , S 2 , S 3 be the three slack variables. Modified form is: Z - 3x 1 -5x 2 -4x 3 =0 2x 1 +3x 2 +S 1 = 8 2x 2 +5x 3 +S 2 = 10 3x 1 +2x 2 +4x 3 +S 3 = 15 x 1 , x 2 , x 3 , S 1 , S 2 , S 3 > 0 Initial BFS is : x 1 = 0, x 2 = 0, x 3 =0, S 1 = 8, S 2 = 10, S 3 = 15 and Z=0.
21. Cont… Therefore, x 2 is the entering variable and S 1 is the departing variable. Basic Variable Coefficients of: Sol. Ratio Z x 1 x 2 x 3 S 1 S 2 S 3 Z 1 -3 -5 -4 0 0 0 0 S 1 0 2 3 0 1 0 0 8 8/3 S 2 0 0 2 5 0 1 0 10 5 S 3 0 3 2 4 0 0 1 15 15/2
22. Cont… Therefore, x 3 is the entering variable and S 2 is the departing variable. Basic Variable Coefficients of: Sol. Ratio Z x 1 x 2 x 3 S 1 S 2 S 3 Z 1 1/3 0 -4 5/3 0 0 40/3 x 2 0 2/3 1 0 1/3 0 0 8/3 - S 2 0 -4/3 0 5 -2/3 1 0 14/3 14/15 S 3 0 5/3 0 4 -2/3 0 1 29/3 29/12
23. Cont… Therefore, x 1 is the entering variable and S 3 is the departing variable. Basic Variable Coefficients of: Sol. Ratio Z x 1 x 2 x 3 S 1 S 2 S 3 Z 1 -11/15 0 0 17/15 4/5 0 256/15 x 2 0 2/3 1 0 1/3 0 0 8/3 4 x 3 0 -4/15 0 1 -2/15 1/5 0 14/15 - S 3 0 41/15 0 0 2/15 -4/5 1 89/15 89/41
24. Cont… Optimal Solution is : x 1 = 89/41, x 2 = 50/41, x 3 =62/41, Z= 765/41 Basic Variable Coefficients of: Sol. Z x 1 x 2 x 3 S 1 S 2 S 3 Z 1 0 0 0 45/41 24/41 11/41 765/41 x 2 0 0 1 0 15/41 8/41 -10/41 50/41 x 3 0 0 0 1 -6/41 5/41 4/41 62/41 x 1 0 1 0 0 -2/41 -12/41 15/41 89/41
25. Example Min.. Z = x 1 - 3x 2 + 2x 3 Subject to constraints: 3x 1 - x 2 + 3x 3 < 7 -2x 1 + 4x 2 < 12 -4x 1 + 3x 2 + 8x 3 < 10 x 1 , x 2 , x 3 > 0
26. Cont… Convert the problem into maximization problem Max.. Z’ = -x 1 + 3x 2 - 2x 3 where Z’= -Z Subject to constraints: 3x 1 - x 2 + 3x 3 < 7 -2x 1 + 4x 2 < 12 -4x 1 + 3x 2 + 8x 3 < 10 x 1 , x 2 , x 3 > 0
27. Cont… Let S 1 , S 2 and S 3 be three slack variables. Modified form is: Z’ + x 1 - 3x 2 + 2x 3 = 0 3x 1 - x 2 + 3x 3 +S 1 = 7 -2x 1 + 4x 2 + S 2 = 12 -4x 1 + 3x 2 + 8x 3 +S 3 = 10 x 1 , x 2 , x 3 > 0 Initial BFS is : x 1 = 0, x 2 = 0, x 3 =0, S 1 = 7, S 2 = 12, S 3 = 10 and Z=0.
28. Cont… Therefore, x 2 is the entering variable and S 2 is the departing variable. Basic Variable Coefficients of: Sol. Ratio Z’ x 1 x 2 x 3 S 1 S 2 S 3 Z’ 1 1 -3 2 0 0 0 0 S 1 0 3 -1 3 1 0 0 7 - S 2 0 -2 4 0 0 1 0 12 3 S 3 0 -4 3 8 0 0 1 10 10/3
29. Cont… Therefore, x 1 is the entering variable and S 1 is the departing variable. Basic Variable Coefficients of: Sol. Ratio Z’ x 1 x 2 x 3 S 1 S 2 S 3 Z’ 1 -1/2 0 2 0 3/4 0 9 S 1 0 5/2 0 3 1 1/4 0 10 4 x 2 0 -1/2 1 0 0 1/4 0 3 - S 3 0 -5/2 0 8 0 -3/4 1 1 -
30.
31. Example Max.. Z = 3x 1 + 4x 2 Subject to constraints: x 1 - x 2 < 1 -x 1 + x 2 < 2 x 1 , x 2 > 0
32. Cont… Let S 1 and S 2 be two slack variables . Modified form is: Z -3x 1 - 4x 2 = 0 x 1 - x 2 +S 1 = 1 -x 1 + x 2 +S 2 = 2 x 1 , x 2 , S 1 , S 2 > 0 Initial BFS is : x 1 = 0, x 2 = 0, S 1 = 1, S 2 = 2 and Z=0.
33. Cont… Therefore, x 2 is the entering variable and S 2 is the departing variable. Basic Variable Coefficients of: Sol. Ratio Z x 1 x 2 S 1 S 2 Z 1 -3 -4 0 0 0 S 1 0 1 -1 1 0 1 - S 2 0 -1 1 0 1 2 2
34. Cont… x 1 is the entering variable, but as in x 1 column every no. is less than equal to zero, ratio cannot be calculated. Therefore given problem is having a unbounded solution . Basic Variable Coefficients of: Sol. Ratio Z x 1 x 2 S 1 S 2 Z 1 -7 0 0 4 8 S 1 0 0 0 1 1 3 - x 2 0 -1 1 0 1 2 -
