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[1.a.] Choose a direction of unboundedness, let’s say d = (1, 0)T^. Then x − γd = yb for some γ > 0 and a boundary point yb. Since γ = 6, yb = (6, 12)T^ − 6(1, 0)T^ = (0, 12)T^. Thus, x = yb + 6(1, 0)T^ where yb is a convex combination of extreme points and 6(1, 0)T^ is a direction of unboundedness.
[1.b.] Since (1, 0)T^ is a direction of unboundedness, x = α(1, 0)T^ is feasible for all α > 0. And at x = α(1, 0)T^ , the objective function z = − 5 x 1 − 7 x 2 = − 5 α, and z = − 5 α → −∞ as α → ∞. Thus, there’s no finite optimal solution.
[2.a.] Let S = {x : Ax = b, x ≥ 0 }. Suppose that −d is a direction of unbound- edness of S. Then, (a) x + γ(−d) ≥ 0 for all x ∈ S and γ ≥ 0. And since d is a direction of unboundedness, (b) (x + γd) ≥ 0 for all x ∈ S and all γ ≥ 0. From (a), d ≤ xr , and from (b), d ≥ − xr ⇒ (d) − xr ≤ d ≤ xr. By taking γ → ∞ in (d), d → 0, which is impossible. Thus, −d is not a direction of unboundedness.
[2.b.] Let x be any feasible point, and γ > 0. Show that x + γd satisfy all the constraints;
A(x + γd) = A(x + γ
αidi)
= A(
αi(x + γdi))
=
αiA(x + γdi)
=
αib = b
αi = b
In addition,
x + γd = x + γ
αidi
=
αi(x + γdi)
≥
αi · 0 = 0
Thus, d is a direction of unboundedness. Note: I used the Definition of ”a di- rection of unboundedness”, but it’s ok to use the theorem ”Ad = 0, d ≥ 0 iff d is a direction of unboundedness of the set {x : Ax = b, x ≥ 0 } ”.
[4] Simplex method
Standard form of the constraints
minimize zˆ = − 7 x 1 − 8 x 2 4 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
- Start with a basic feasible solution x with xB = {s 1 , s 2 , s 3 }, xN = {x 1 , x 2 } ⇒ x = (0, 0 , 100 , 80 , 40)T^ , which corresponds to the extreme point xa = (0, 0)T^ in the graph. Then, we can rearrange the constraints such as zˆ = − 7 x 1 − 8 x 2 s 1 = 100 − 4 x 1 − x 2 s 2 = 80 − x 1 − x 2 s 3 = 40 − x 1
- If either x 1 and x 2 is increased from zero, then ˆz = − 7 x 1 − 8 x 2 decreases.
⇒ x = (0, 0 , 100 , 80 , 40)T^ is not optimal.
- Since z decreases faster in the x 2 direction, x 2 is increased with x 1 = 0 fixed. Then s 1 = 100 − x 2 , s 2 = 80 − x 2 , decreases and s 3 = 40 is unaffected, and to maintain the non-negativity of the variables, x 1 can be increased until x 1 = 80.
- When x 2 = 80; since s 2 = 0, we have xB = {x 2 , s 1 , s 3 }, xN = {x 1 , s 2 }, and the basic feasible solution x = (0, 80 , 20 , 0 , 40)T^. And since x 2 is replacing s 2 in the basis we use the 2nd constraint to express x 2 in terms of the non-basic variables x 1 , s 2 , i.e.,
x 2 = 80 − x 1 − s 2
- Using x 2 = 80 − x 1 − s 2 , we have
ˆz = − 7 x 1 − 8 x 2 = − 7 x 1 − 8(80 − x 1 − s 2 ) = −640 + x 1 + 8s 2.
- ˆz has a minimum when x 1 = s 2 = 0. Thus, ˆz has a minimum value − 640 at x = (0, 80 , 20 , 0 , 40)T^.
⇒ x = (0, 80 , 20 , 0 , 40)T^ is an optimal solution.
Conclusion: Back to the original problem, z = −zˆ has maximum value 640 at x = (0, 80 , 20 , 0 , 40)T^.
[5.a] Standard form of the linear program
−x 1 + x 2 + s 1 = 1 x 1 − 2 x 2 + s 2 = 2 x 1 , x 2 , s 1 , s 2 ≥ 0
- Start with a basic feasible solution x with xB = {s 1 , s 2 }, xN = {x 1 , x 2 } ⇒ x = (0, 0 , 1 , 2)T^ , which corresponds to the extreme point xa = (0, 0)T^ in the graph. The constraints is written with the ”basic” variables expressed in terms of the ”non-basic” variables: s 1 = 1 + x 1 − x 2 s 2 = 2 − x 1 + 2x 2
