* Create Basic-Simplex-Solution Basics on Solving Linear Problems using the Primal Simplex Method and additional information on Alternative Simplex Algorithms. * Rename guide/english/mathematics/Basic-Simplex-Solution to guide/english/mathematics/linear-equations/basic-simplex-solution/index.md
		
			
				
	
	
		
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			92 lines
		
	
	
		
			3.1 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
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| title: Basics of Simplex Solutions
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| ---
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| 
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| ## Reason for Sthe Simplex Method
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| In linear algebra, some problems have multiple solutions that can be accepted as feasible, but in order to get the Optimal solution it becomes necessary to use a method such as the Simplex method.
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| Note: Alternative methods such as a Graphical LP can also be used but this is not always possible.
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| 
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| ## Different Simplex Methods:
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| Here are the 3 most used methods for solving a LP using Simplex:
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| 1. Primal Simplex - Used to solve MAXIMUM Problems.
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| 2. Dual Simplex - Used when a derived problem exists within a Primal Simplex Solution (Duality).
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| 3. 2-Phase Simplex - Used to solve MIN Problems.
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| 
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| ## Solving Linear Programming Problems using Simplex
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| 
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| In order to use the Simplex method, first the Linear model needs to be converted into canonical form.
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| The canonical for is when all inequality expressions are changed into equal expressions.
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| 
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| Example:
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| Basic LP
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| MAX Z = 5X + 4Y
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| Subject to Constraints:
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|                       1. 8X + 6Y <= 120
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|                       2. 2X + 1Y <= 50
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|                       3. X >= 10
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|                       4. X, y >= 0
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|                       
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| Canonical Form
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| 
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| MAX Z = -5X - 4Y                                      Note: Z Row becomes negative.
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| ST:
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|     1. 8X + 6Y + S1 = 120                             Note: In order to set a Smaller-than equation to equal a Slack variable is introduced.
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|     2. 2X - 1y + S2 = 50
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|     3. X - E1 + A1 = 10                               Note: In order to set a Larger-than equation to equal a Excess variable is added and an Artificial variable is subtracted.
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|     
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| Now the Initial Tablau can be created:
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| 
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| Note: X & Y will be the Non-Basic-Variables as they are Negative and Slack/Axcess/Artificial variables do not count in this case.
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| 
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|       *
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| T0  | X  | Y  | S1  | S2  | E1  | A1  ||  RHS | Ratio
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| Z   | -5 | -4 | 0   | 0   | 0   | 0   ||  0   | ---
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| 1   | 8  | 6  | 1   | 0   | 0   | 0   ||  120 | 15
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| 2   | 2  | 1  | 0   | 1   | 0   | 0   ||  50  | 25
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| 3   | 1  | 0  | 0   | 0   | -1  | 1   ||  10  | 10    *
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| 
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| Note: * Represents the Column with the Smallest Negative and the Column with the Smallest Positive.
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| 
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| Steps:
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| 1. Find the Column with the Smallest Negative.
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| 2. Devide the RHS with the selected column to calculate the Ratio.
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| 3. Find the Row with the Smallest Positive Ratio
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| 4. Pivot on the selected Row & Column.
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| 5. Continue doing this until no more negative NBVs remain.
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| 
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| Optimal Table:
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| T2  | X  | Y  | S1  | S2  | E1  | A1  ||  RHS 
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| Z   | 0  | 0  | 0.67| 0   | 0.3 | -0.3||  76.67   
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| 1   | 0  | 1  | 1.67| 0   | 1.3 | -1.3||  6.67 
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| 2   | 0  | 0  | -0.1| 1   | 0.6 | -0.6||  23.33  
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| 3   | 1  | 0  | 0   | 0   | -1  | 1   ||  10  
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| 
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| Steps:
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| 1. Identify all Basic Variables:
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|   1.1. Only Columns with a single '1' and rest '0' can be a Basic Variable.
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|   1.2. Order of identification is determined by Row number.
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| 2. The RHS value corresponding to the '1' value for each BV is the Value of that variable.
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| 
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| Eg:
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| cBV:  Y = 6.67 ;  S1 = 23.33  ; X = 10
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| 
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| Thus the Optimal Solution to the LP is Z = 5(10) + 4(6.67)
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