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[2]. Convert the linear program to STANDARD form.
- First, look at the variable constraints: x 1 ≥ −2, x 2 , x 3 free
′ 1 =^ x^1 + 2.^ Then we have a new variable^ x
′ 1 with x
′ 1 ≥^ 0.
- x 2 free: let x 2 = x ′ 2 −^ x
′′ 2 with^ x
′ 2 ≥^ 0,^ x
′′ 2 ≥^ 0. Then we have new variables x ′ 2 , x
′′ 2 with^ x
′ 2 ≥^ 0,^ x
′′ 2 ≥^0
′ 3 −^ x
′′ 3 with^ x
′ 3 ≥^ 0,^ x
′′ 3 ≥^ 0. Then we have new variables x
′ 3 , x
′′ 3 with^ x
′ 3 ≥^ 0,^ x
′′ 3 ≥^0
- Second, change the general (first three) constraints with new variables; 2.a. 5x 1 − 2 x 2 + 6x 3 ≥ 5 : 5(x ′ 1 −^ 2)^ −^ 2(x
′ 2 −^ x
′′ 2 ) + 6(x
′ 3 −^ x
′′ 3 )^ ≥^5 ⇒ 5 x ′ 1 −^2 x
′ 2 + 2x
′′ 2 + 6x
′ 3 −^6 x
′′ 3 ≥^ 5 + 10 2.b. 3x 1 + 4x 2 − 9 x 3 = 3: 3(x
′ 1 −^ 2) + 4(x
′ 2 −^ x
′′ 2 )^ −^ 9(x
′ 3 −^ x
′′ 3 ) = 3 ⇒ 3 x
′ 1 + 4x
′ 2 −^4 x
′′ 2 −^9 x
′ 3 + 9x
′′ 3 = 3 + 6 2.c. 7x 1 + 3x 2 + 5x 3 ≤ 9: 7(x ′ 1 −^ 2) + 3(x
′ 2 −^ x
′′ 2 ) + 5(x
′ 3 −^ x
′′ 3 )^ ≤^9 ⇒ 7 x ′ 1 + 3x
′ 2 −^3 x
′′ 2 + 5x
′ 3 −^5 x
′′ 3 ≤^ 9 + 14
- We need to change the inequalities 2.a. and 2.c. to equalities by introduc- ing new variables;
3.a. 5x
′ 1 −^2 x
′ 2 + 2x
′′ 2 + 6x
′ 3 −^6 x
′′ 3 ≥^ 5 + 10: Introduce a new variable^ s^1 with s 1 ≥ 0 inducing 5x
′ 1 −^2 x
′ 2 + 2x
′′ 2 + 6x
′ 3 −^6 x
′′ 3 −^ s^1 = 5 + 10 3.b. 7x ′ 1 + 3x
′ 2 −^3 x
′′ 2 + 5x
′ 3 −^5 x
′′ 3 ≤^ 9 + 14: Introduce a new variable^ s^2 with s 2 ≥ 0 inducing 7x ′ 1 + 3x
′ 2 −^3 x
′′ 2 + 5x
′ 3 −^5 x
′′ 3 +^ s^2 = 9 + 14
- Finally, we also need to change the object function z with new variables, thus we have 4.a. z = x 1 − 5 x 2 − 7 x 3 : z = (x ′ 1 −^ 2)^ −^ 5(x
′ 2 −^ x
′′ 2 )^ −^ 7(x
′ 3 −^ x
′′ 3 ) ⇒ z = x ′ 1 −^5 x
′ 2 + 5x
′′ 2 −^7 x
′ 3 + 7x
′′ 3 −^ 2. 4.b. let ˆz = z + 2: Then we can have minimize ˆz = x
′ 1 −^5 x
′ 2 + 5x
′′ 2 −^7 x
′ 3 + 7x
′′
Thus, from (1), (2.b.), (3.a.), (3.b.), (4.b.), we finally have
minimize ˆz = x
′ 1 −^5 x
′ 2 + 5x
′′ 2 −^7 x
′ 3 + 7x
′′ 3 subject to 3 x
′ 1 + 4x
′ 2 −^4 x
′′ 2 −^9 x
′ 3 + 9x
′′ 3 = 3 + 6 5 x
′ 1 −^2 x
′ 2 + 2x
′′ 2 + 6x
′ 3 −^6 x
′′ 3 −^ s^1 = 5 + 10 7 x
′ 1 + 3x
′ 2 −^3 x
′′ 2 + 5x
′ 3 −^5 x
′′ 3 +^ s^2 = 9 + 14 x
′ 1 ≥^0 , x
′ 2 ≥^0 , x
′′ 2 ≥^0 , x
′ 3 ≥^0 , x
′′ 3 ≥^0 , s^1 ≥^0 , s^2 ≥^0
[3.a.] Convert it to STANDARD form and determine all the basic solutions
- Standard form (Ax = b, x ≥ 0):
2 x 1 + x 2 + s 1 = 100 x 1 + x 2 + s 2 = 80 x 1 + s 3 = 40 x 1 , x 2 , s 1 , s 2 , s 3 ≥ 0
- Determine the basic solutions (feasible or infeasible)
- Separate x = (x 1 , x 2 , s 1 , s 2 , s 3 )T^ into m basic variables xB (basis) and n − m non-basic variables xN (values are zero); m= row rank of the constraint matrix A, n=number of components of x.(In this case, m = 3, n = 5)
- We have (n m) cases to check (here, there are 10 cases):
- (1). xN = {x 1 = x 2 = 0}, xB = {s 1 , s 2 , s 3 } ⇒ x = (0, 0 , s 1 , s 2 , s 3 )T must satisfy Ax = b
s 1 = 100 s 2 = 80 s 3 = 40
Thus, x = (0, 0 , 100 , 80 , 40)T^ is a basic feasible solution.
- (2). xN = {x 1 = s 1 = 0}, xB = {x 2 , s 2 , s 3 } ⇒ x = (0, x 2 , 0 , s 2 , s 3 )T must satisfy Ax = b
x 2 = 100 x 2 + s 2 = 80 s 3 = 40
Thus, x = (0, 100 , 0 , − 20 , 40)T^ is a basic infeasible solution.
- (3). xN = {x 1 = s 2 = 0}, xB = {x 2 , s 1 , s 3 } ⇒ x = (0, x 2 , s 1 , 0 , s 3 )T must satisfy Ax = b
x 2 + s 1 = 100 x 2 = 80 s 3 = 40
Thus, x = (0, 80 , 20 , 0 , 40)T^ is a basic feasible solution.
- Choose two directions such as d = (1, 0)T^ , (2, 3)T^ , and prove that they are really directions of unboundedness;
- Prove that d = (1, 0)T^ is a direction of unboundedness; let x be any feasible point and γ ≥ 0. Then we need to prove that x + γd satisfies all the constraints. With x + γd = (x 1 + γd 1 , x 1 + γd 2 ) = (x 1 + γ, x 2 ),
−3(x 1 + γ) + 2(x 2 ) = (− 3 x 1 + 2x 2 ) − 3 γ ≤ 30 − 3 γ ≤ 30 −2(x 1 + γ) + x 2 = (− 2 x 1 + x 2 ) − 2 γ ≤ 12 − 3 γ ≤ 12
Thus, d = (1, 0)T^ is a direction of unboundedness.
- We can prove that d = (2, 3)T^ is a direction of unboundedness by the same way. (You should do this)
- And since one is not (non-zero) constant multiple of the other, these two vectors are linearly independent.
[5.c.] Using the standard form
- Standard form:
− 3 x 1 + 2x 2 + s 1 = 30 − 2 x 1 + x 2 + s 2 = 12 x 1 , x 2 , s 1 , s 2 ≥ 0
- From the standard form, we need to find a direction of unboundedness d = (d 1 , d 2 , d 3 , d 4 )T^ ; - Since the first two components of d are given in [5.b], we have d = (1, 0 , d 3 , d 4 )T^ , d = (2, 3 , d 3 , d 4 )T^. - Determine d 3 , d 4 using the standard form (Ad = 0). Then, we have d = (1, 0 , 3 , 2)T^ , d = (2, 3 , 0 , 1)T^. - Check if Ad = 0 and d ≥ 0.
