92 lines
3.1 KiB
Markdown
92 lines
3.1 KiB
Markdown
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---
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title: Basics of Simplex Solutions
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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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## 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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## Solving Linear Programming Problems using Simplex
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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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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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Canonical Form
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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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Now the Initial Tablau can be created:
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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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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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Note: * Represents the Column with the Smallest Negative and the Column with the Smallest Positive.
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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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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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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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Eg:
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cBV: Y = 6.67 ; S1 = 23.33 ; X = 10
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Thus the Optimal Solution to the LP is Z = 5(10) + 4(6.67)
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----------------------------------------------------------------------------------------------------------------------
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