A lecture on optimization
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Transcript of A lecture on optimization
BITS PilaniHyderabad Campus
Optimization Lecture 6
BITS Pilani, Hyderabad Campus
• Formulation of Linear Programming Problem (LPP) from the real world problem
• Solving LPP using Graphical method • Solving maximization LPP using Simplex method with
less than or equal to constraints• Solving minimization LPP using Simplex method with
less than or equal to constraints• Solving maximization and minimization with greater than
or equal to constraints and equal to constraints – Big-M Method
Topics we have discussed so far
BITS Pilani, Hyderabad Campus
Solving maximization and minimization with greater than or equal to constraints and equal to constraints – Two Phase Method
Today’s Topic
Second method for handling artificial variables within simplexHas two phasesPhase I : finds a basic feasible solution for the original problem, if one exists Phase II : finds an optimal solution for original problem
BITS Pilani, Hyderabad Campus
Consider a linear programming problem consisting of artificial variablesObjective: to set all artificial variables to zero at optimalityHence, solve a linear programming problem by replacing original objective function by sum of artificial variables with a minimization objectiveObserve: due to non-negativity constraint, sum of artificial variables cannot be negative, i.e., smallest possible feasible value of sum is zeroIf optimal value of modified objective function is not zero (i.e., strictly positive), conclude that problem is not feasible
Two-Phase Method
BITS Pilani, Hyderabad Campus
maximize z = -3x1 + x3
subject to
x1 + x2 + x3 + x4 = 4...........(1)
-2x1 + x2 - x3 = 1...........(2)
3x2 + x3 + 3x4 = 10..............(3)
x1 , x2 , x3 , x4 0...........(4)
Observe: not in canonical form
Two-Phase Method
BITS Pilani, Hyderabad Campus
maximize z = -3x1 + x3 subject to
x1 + x2 + x3 + x4 + x5 = 4...........(1’)
-2x1 + x2 - x3 + x6 = 1...........(2’)
3x2 + x3 + 3x4 + x7 = 10............(3’)
x1 , x2 , x3 , x4 , x5 , x6 , x7 0...(4’)
Two-Phase Method
BITS Pilani, Hyderabad Campus
minimize r = x5 + x6+ x7
i.e., maximize w = -x5 - x6 - x7
subject to
x1 + x2 + x3 + x4 + x5 = 4...........(1’)
-2x1 + x2 - x3 + x6 = 1...........(2’)
3x2 + x3 + 3x4 + x7 = 10............(3’)
x1 , x2 , x3 , x4 , x5 , x6 , x7 0...(5’)
Two-Phase Method
BITS Pilani, Hyderabad Campus
Basic Vars x1 x2 x3 x4 x5 x6 x7 RHS
w 0 0 0 0 1 1 1 0
x5 1 1 1 1 1 0 0 4
x6 -2 1 -1 0 0 1 0 1
x7 0 3 1 3 0 0 1 10
Two-Phase Method
BITS Pilani, Hyderabad Campus
Basic Vars x1 x2 x3 x4 x5 x6 x7 RHS
w 1 -5 -1 -4 0 0 0 -15
x5 1 1 1 1 1 0 0 4
x6 -2 1 -1 0 0 1 0 1
x7 0 3 1 3 0 0 1 10Perform the elementary row operation R1 = R1 - (R2 + R3 + R4 ) to make the table in canonical form
Two-Phase Method
BITS Pilani, Hyderabad Campus
Basic Vars x1 x2 x3 x4 x5 x6 x7
RHS Ratio
w 1 -5 -1 -4 0 0 0 -15x5 1 1 1 1 1 0 0 4 4/1=4
x6 -2 1 -1 0 0 1 0 1 1/1=1
x7 0 3 1 3 0 0 1 10 10/3=3.33
Two-Phase Method
BITS Pilani, Hyderabad Campus
Basic
Varsx1 x2 x3 x4 x5 x6 x7
RHS Ratio
w -9 0 -6 -4 0 5 0 -10
x5 3 0 2 1 1 -1 0 3 3/3=1
x2 -2 1 -1 0 0 1 0 1 -----
x7 6 0 4 3 0 -3 1 7 7/6=1.3
Two-Phase Method
BITS Pilani, Hyderabad Campus
Basic
Varsx1 x2 x3 x4 x5 x6 x7
RHS Ratio
w 0 0 0 -1 3 2 0 -1
x1 1 0 2/3 1/3 1/3 -1/3 0 1 1/1/3=3
x2 0 1 1/3 2/3 2/3 1/3 0 3 9/2=4.5
x7 0 0 0 1 -2 -1 1 1 1/1=1
Two-Phase Method
BITS Pilani, Hyderabad Campus
Basic Vars x1 x2 x3 x4 x5 x6 x7
RHS
w 0 0 0 0 1 1 1 0x1 1 0 2/3 0 1 0 -1/3 2/3x2 0 1 1/3 0 2 1 -2/3 7/3x4 0 0 0 1 -2 -1 1 1
•Observe: all artificial variables are now non-basic, i.e., problem is feasible
•Hence x5= x6= x7 = 0
•Initial basic feasible solution: x1=2/3, x2=7/3, x4=1
•End of Phase I
Two-Phase Method
BITS Pilani, Hyderabad Campus
In Phase 2 we solve the original problem Maximize z = -3x1 + x3 Subject to x1 + (⅔)x3 = ⅔ x2 + (⅓) x3 = ⅓ x4 = 1x1 , x2 , x3 ≥ 0with initial basic variables as x1 , x2 , x4
Two-Phase Method
BITS Pilani, Hyderabad Campus
Basic
Varsx1 x2 x3 x4
RHS
z 3 0 -1 0 0x1 1 0 2/3 0 2/3x2 0 1 1/3 0 7/3x4 0 0 0 1 1
Drop –w rowDrop all artificial variable columns
Two-Phase Method
Not in canonical form Hence perform ERO : R1 = R1 – 3R2
BITS Pilani, Hyderabad Campus
Basic
Varsx1 x2 x3 x4
RHS Ratio
z 0 0 -3 0 -2x1 1 0 2/3 0 2/3 1x2 0 1 1/3 0 7/3 7x4 0 0 0 1 1 ---
Two-Phase Method
BITS Pilani, Hyderabad Campus
Basic Vars x1 x2 x3 x4
RHS
z 9/2 0 0 0 1
x3 3/2 0 1 0 1
x2-1/2 0 0 0 2
x4 0 0 0 1 1
Coefficients of all non-basic variables in the first row ≥ 0
Hence we have optimal solution
z=1, x2=2, x3=1, x4=1, x1= 0
Two-Phase Method
BITS Pilani, Hyderabad Campus
Infeasible Solution
Infeasibility is detected in the simplex method when an artificial variable remains positive in the final tableau.
For Example consider the LPPMAX z = 2x1 + 6x2 s. t. 4x1 + 3x2 < 12
2x1 + x2 > 8
x1, x2 > 0
BITS Pilani, Hyderabad Campus
Infeasible Solution
MAX z = 2x1 + 6x2 s. t. 4x1 + 3x2 + S1 = 12
2x1 + x2 - S2 + A1 = 8
x1, x2 , S1, S2 , A1 > 0
BITS Pilani, Hyderabad Campus
Basic Vars x1 x2 S1 S2 A1
RHS
z -2 1 0 0 M 0S1 4 3 1 0 0 12A1 2 1 0 -1 1 8
Basic Vars x1 x2 S1 S2 A1
RHS Ratio
z -2-2M 1-M 0 M 0 -8MS1 4 3 1 0 0 12 3A1 2 1 0 -1 1 8 4
Infeasible Solution
BITS Pilani, Hyderabad Campus
Infeasible Solution
Basic Vars x1 x2 S1 S2 A1
RHS
z 0 (5+M)/2 (1+M)/2 M 0 6-2M
x1 1 3/4 1/4 0 0 3
A1 0 -3/4 -1/4 0 0 3/2
BITS Pilani, Hyderabad Campus
Infeasible Solution
Basic Vars x1 x2 S1 S2 A1
RHS
r 0 0 0 0 1 0
S1 4 3 1 0 0 12
A1 2 1 0 -1 1 8
Min r = A1
s. t. 4x1 + 3x2 + S1 = 12
2x1 + x2 - S2 + A1 = 8x1, x2 , S1, S2 , A1 > 0
BITS Pilani, Hyderabad Campus
Infeasible Solution
Basic Vars x1 x2 S1 S2 A1
RHS
r -2 -1 0 1 0 8
S1 4 3 1 0 0 12
A1 2 1 0 -1 1 8
BITS Pilani, Hyderabad Campus
Solve the following linear programming problem by using the simplex method:Min Z =2 X1 + 3 X2
S.t.½ X1 + ¼ X2 ≤ 4X1 + 3X2 20X1 + X2 = 10X1, X2 0
BITS Pilani, Hyderabad Campus
SolutionStep 1: standard formMin Z, s.t.Z – 2 X1 – 3 X2 - M A1 -M A2 = 0 ½ X1 + ¼ X2 + S1 = 4 X1 + 3X2 - S2 + A1 = 20 X1 + X2 + A2 = 10X1, X2 ,S1, S2, A1, A2 0 Where: M is a very large number
BITS Pilani, Hyderabad Campus
M, a very large number, is used to ensure that the values of A1 and A2, …, and An will be zero in the final (optimal) tableau as follows:
1. If the objective function is Minimization, then A1, A2, …, and An must be added to the RHS of the objective function multiplied by a very large number (M).
Example: if the objective function is Min Z = X1+2X2, then the obj. function should be Min Z = X1 + X2+ MA1 + MA2+ …+ MAn
OR Z – X1 - X2- MA1 - MA2- …- MAn = 0
2. If the objective function is Maximization, then A1, A2, …, and An must be subtracted from the RHS of the objective function multiplied by a very large number (M).
Example: if the objective function is Max Z = X1+2X2, then the obj. function should be Max Z = X1 + X2- MA1 - MA2- …- MAn
OR Z - X1 - X2+ MA1 + MA2+ …+ MAn = 0
N.B.: When the Z is transformed to a zero equation, the signs are chang
BITS Pilani, Hyderabad Campus
Basic variables
X1
2
X2
3
S1
0
S2
0
A1
M
A2
M
RHS
S1 ½ ¼ 1 0 0 0 4
A1 1 3 0 -1 1 0 20A2 1 1 0 0 0 1 10
Z -2 -3 0 0 -M -M 0
BITS Pilani, Hyderabad Campus
To correct this violation before starting the simplex algorithm, the elementary row operations are used as follows:New (Z row) = old (z row) ± M (A1 row) ± M (A2 row)In our case, it will be positive since M is negative in the Z row, as following:Old (Z row): -2 -3 0 0 -M -M 0M (A1 row): M 3M 0 -M M o 20MM (A2 row): M M 0 0 0 M 10MNew (Z row):2M-2 4M-3 0 -M 0 0 30M
BITS Pilani, Hyderabad Campus
Basic variables
X1
2
X2
3
S1
0
S2
0
A1
M
A2
M
RHS
S1 1/2 1/4 1 0 0 0 4
A1 1 3 0 -1 1 0 20A2 1 1 0 0 0 1 10
Z 2M-2 4M-3 0 -M 0 0 30M
BITS Pilani, Hyderabad Campus
Basic variables
X1
2
X2
3
S1
0
S2
0
A1
M
A2
M
RHS
S15/12 0 1 1/12 -1/12 0 7/3
X21/3 1 0 -1/3 1/3 0 20/3
A22/3 0 0 1/3 -1/3 1 10/3
Z 2/3M-1 0 0 1/3M-1 1-4/3M 0 20+10/3M
BITS Pilani, Hyderabad Campus
Basic variables
X1 X2 S1 S2 A1 A2 RHS
S10 0 1 -1/8 1/8 -5/8 1/4
X20 1 0 -1/2 1/2 -1/2 5
X11 0 0 1/2 -1/2 3/2 5
Z 0 0 0 -1/2 ½-M 3/2-M 25
BITS Pilani, Hyderabad Campus
•In the final tableau, if one or more artificial variables (A1, A2, …) still basic and has a nonzero value, then the problem has an infeasible solution•If there is a zero under one or more nonbasic variables in the last tableau (optimal solution tableau), then there is a multiple optimal solution.•When determining the leaving variable of any tableau, if there is no positive ratio (all the entries in the pivot column are negative and zeroes), then the solution is unbounded.