Linear Programming, 1 Max c 1 *X 1 +…+ c n *X n = z s.t. a 11 *X 1 +…+ a 1n *X n b 1 … a m1 *X 1 +…+ a mn *X n b m X 1 , X n 0 Standard form
Dec 14, 2015
Linear Programming, 1
Max c1 *X1 +…+ cn *Xn = zs.t. a11*X1 +…+ a1n*Xn b1
… am1*X1 +…+ amn*Xn bm
X1 , Xn 0Standard form
Linear Programming, 2
Max c1 *X1 +…+ cn *Xn = zs.t. a11*X1 +…+ a1n*Xn b1
… am1*X1 +…+ amn*Xn bm
X1 , Xn 0n +1 Variables
Linear Programming, 3
Max c1 *X1 +…+ cn *Xn = zs.t. a11*X1 +…+ a1n*Xn b1
… am1*X1 +…+ amn*Xn bm
X1 , Xn 0Objective function
Linear Programming, 4
Max c1 *X1 +…+ cn *Xn = zs.t. a11*X1 +…+ a1n*Xn b1
… am1*X1 +…+ amn*Xn bm
X1 , Xn 0
n Objective function coefficients (data)
Linear Programming, 5
Max c1 *X1 +…+ cn *Xn = zs.t. a11*X1 +…+ a1n*Xn b1
… am1*X1 +…+ amn*Xn bm
X1 , Xn 0m Constraints
Linear Programming, 6
Max c1 *X1 +…+ cn *Xn = zs.t. a11*X1 +…+ a1n*Xn b1
… am1*X1 +…+ amn*Xn bm
X1 , Xn 0
m*n technical coefficients (data)
Linear Programming, 7
Max c1 *X1 +…+ cn *Xn = zs.t. a11*X1 +…+ a1n*Xn b1
… am1*X1 +…+ amn*Xn bm
X1 , Xn 0m resource limits (data)
Objective Function Isoclines - 1
Max 2*X1 + 3*X2 = zs.t. X1 + 2*X2 10 X1 , X2 0
X2
X1
10
5
X2 = z/3 – 2/3*X1
z=6X2=2–2/3*X1
Objective Function Isoclines - 2
Max 2*X1 + 3*X2 = zs.t. X1 + 2*X2 10 X1 , X2 0
X2
X1
10
5
X2 = z/3 – 2/3 * X1
z=15z=6z=0
Graphical Solution
Max 2*X1 + 3*X2 = zs.t. X1 + 2*X2 10 X1 , X2 0
X2
X1
10
5
X2 = z/3 – 2/3 * X1
z=20
Solving Arithmetically
Max c1 *X1 +…+ cn *Xn = zs.t. a11*X1 +…+ a1n*Xn b1
… am1*X1 +…+ amn*Xn bm
X1 , Xn 0
Convert Inequalities
Max c1 *X1 +…+cn *Xn +0*S1 +…+0*Sm = zs.t. a11*X1 +…+a1n*Xn +1*S1 +…+0*Sm = b1
… am1*X1 +…+amn*Xn +0*S1 +…+1*Sm = bm
X1 , X2 , S1 , Sm 0
Max c1 *X1 +…+cn *Xn = zs.t. a11*X1 +…+a1n*Xn b1
… am1*X1 +…+amn*Xn bm
X1 , X2 0
Arithmetic Problem, 1
• Number of unknown variables: n+m
• Number of equations (constraints): m
Can only solve m equations for m unknown variables
1 Constraint 1 Non-Zero Variable
Max Objective s.t. X1 + 2*X2 + S = 10 X1 , X2 , S 0
X2
X1
10
5
X1 10
X2 0
S 0
10S
02X
01X
0S
52X
01X
2 ConstraintsX2
X1
10
5
5
10 Max Objective s.t. X1 + 2*X2 + S1 = 10 2*X1 + X2 + S2 = 10 X1 , X2 , S1, S2 0
02S
51S
02X
51X
Max Objective s.t. X1 + 2*X2 + S1 = 10 2*X1 + X2 + S2 = 10 X1 , X2 , S1, S2 0
X2
X1
10
5
102S
101S
02X
01X
52S
01S
52X
01X
02S
01S3
102X3
101X
10
5
2 Constraints 2 Non-Zero Variables
Arithmetic Problem, 3
Extreme points have only m non-zero variables and n-m zero variables
Thus, if we knew which n-m variables are zero, we could solve for the remaining m variables using our m equation system
New task: Which variables should remain zero?
Arithmetic Solution
1. Pick n-m variables and set to zero
2. Solve mxm equation system
3. Compute optimality indicators
4. If optimal, done; otherwise
Optimal solution is extreme point
• Number of non-zero variables = number of equations
• Non-zero (positive) variables are called basic variables
• All remaining variables are called non-basic variables
• Non-basic variables are zero.
Solver tasks
• Determine which variables form basis
• Invert coefficient matrix to calculate
optimal variable values
If optimal solution is an extreme point, why not simply calculate
all extreme points and choose the one with highest objective
function value?