(b) (i) This is a LP

(1)

Solutions for Operations Research HWK#1

1. (a) This is not a LP because the objective function can not be transformed to a linear combination.

(b) (i) This is a LP.

(ii) But this is not in the standard form. To convert it into the standard form, we add a slack variable associated with the first constraint, s¹ ≥ 0, and a surplus variable associated with the second constraint, e² ≥ 0. Moreover, we let

x2 = x² − bx2, x2 ≥ 0, bx2 ≥ 0.

since there is no sign restriction on x². Then we have

(log 5)x¹− (cos 4)x² = (log 5)x¹− (cos 4)(x²− bx2)

= (log 5)x¹− (cos 4)x²+ (cos 4) bx2,

e⁴x1− 7x² = e⁴x1− 7(x²− bx2)

= e⁴x1− 7x²+ 7 bx2, and

x1+ 2²x2 = x¹+ 2²(x²− bx2)

= x¹+ 2²x2− 2²xb2. Therefore in the standard form it will be

min (log 5)x¹− (cos 4)x²+ (cos 4) bx2

s.t. e⁴x1 − 7x²+ 7 bx2+ s¹ = 1

x1+ 2²x2− 2²xb2− e² = − log 0.01 x1, x2, bx2, s1, e2 ≥ 0.

(iii) The constraint matrix A is

e⁴ −7 7 1 0 1 2² −2² 0 −1

. The right hand side vector b is

1

− log 0.01

. The cost coefficient C is

log 5 − cos 4 cos 4 0 0 .

(2)

(c) (i) This is a LP.

(ii) To solve

max 7

s.t. x1− 8 ≥ 0.

is to solve

max 0 · x¹ s.t. x1− 8 ≥ 0.

since 7 is a constant.

Let

x1 = x¹− 8,

we convert the problem into the standard form, max 0 · (x¹+ 8)

s.t. x1 ≥ 0.

That is,

− min −0 · x¹ s.t. x1 ≥ 0.

0 . The right hand side vector b is

0 . The cost coefficient C is

0 . (d) (i) This is a LP.

(ii) (There may be other solutions.) To solve

max √

7 − x¹+ x2

√7 s.t. x1 − |x²| + 3|x³| ≤ 9

2x¹+ 5x²− |x³| = 1 x2 ≥ 0.

(3)

is to solve

max −x¹+ x2

√7

s.t. x1 − |x²| + 3|x³| ≤ 9 2x¹+ 5x²− |x³| = 1 x2 ≥ 0.

since √

7 is a constant.

To convert

max −x¹+ x2

√7

s.t. x1 − |x²| + 3|x³| ≤ 9 2x¹+ 5x²− |x³| = 1 x2 ≥ 0.

into the standard form, firstly we let x1 = x¹ − bx1, x1 ≥ 0, bx1 ≥ 0.

since there is no sign restriction on x¹. Secondly, we use the fact that

|x²| = x², when x² ≥ 0.

And let

x3 = |x³|, x³ ≥ 0.

Then the problem becomes

max −(x¹ − bx1) + 1

√7x2

s.t. (x¹− bx1) − x²+ 3x³ ≤ 9 2(x1 − bx1) + 5x2− x³ = 1 x1, bx1, x2, x3 ≥ 0.

After adding the slack variable associated with the first constraint, s1 ≥ 0, the problem will be converted into the standard form

max −x¹ + bx1+ 1

√7x2

s.t. x1− bx1 − x² + 3x³+ s¹ = 9 2x¹− 2 bx1+ 5x²− x³ = 1 x1, bx1, x2, x3, s1 ≥ 0.

(4)

1 −1 −1 3 1

2 −2 5 −1 0

. The right hand side vector b is

9 1

. The cost coefficient C is

h−1 1 √7¹ 0 0i . (e) (i) This is a LP.

(ii) But this is not in the standard form. To convert it into the standard form, we add slack variables associated with the constraints, s1, s², s³, s⁴ ≥ 0. And we convert the maximization problem into the minimization problem. Therefore in the standard form it will be

− min (−5x¹− 5x²− 3x³) s.t. x1 + 3x²+ x²+ s¹ = 3

−x¹ + 3x³+ x² = 2 2x¹− x²+ 2x³+ x³ = 4 2x¹+ 3x²− x³+ s⁴ = 2 x1, x2, x3, s1, s2, s3, s4 ≥ 0.







1 3 1 1 0 0 0

−1 0 3 0 1 0 0

2 −1 2 0 0 1 0

2 3 −1 0 0 0 1





.

The right hand side vector b is





 3 2 4 2





.

The cost coefficient C is

−5 −5 −3 0 0 0 0 .

(5)

2. (a) The LP problem has an optimal solution when s > 0, t > 0, and s 6= t.

(b) The LP problem is always feasible since (0, 0) is always a feasible solution.

(c) The LP problem is unbounded when t > 0, s ≤ 0, or t ≤ 0.

(d) The LP problem has multiple optimal solutions when s = t > 0.

3. (a) The problem

min x¹+ x² s.t. 2x1 − x² ≤ 1

x1 ≥ 0, x² ≥ 0

is bounded with minimum 0, but its feasible domain {(x¹, x2)|2x¹− x² ≤ 1, x¹ ≥ 0, x² ≥ 0}

is unbounded.

(b) The problem

max x¹+ x² s.t. x1+ x² ≤ 1

x1 ≥ 0, x² ≥ 0

is bounded with maximum 1, and has infinitely many optimal solutions {(x¹, x2)|x¹+ x² = 1, x¹ ≥ 0, x² ≥ 0}.

(c) The problem

min x¹− x² s.t. x¹− x² ≥ 0

x1 ≥ 0, x² ≥ 0

is bounded with minimum 0, but its optimal soluiton set {(x¹, x2)|x¹ = x², x1 ≥ 0, x² ≥ 0}

is unbounded.

(d) Each optimal soluiton belongs to the feasible domain. The optimal solution set can not be unbounded when the feasible domain is bounded.

(6)

4. (a)

0 0.2 0.4 0.6 0.8 1 1.2

0 0.2 0.4 0.6 0.8 1

optimal solution x+2y=1

x+2y=2 x+y=1

feasible domain

(b)

|x − 1| + y =

(x − 1) + y, if x ≥ 1;

−(x − 1) + y, if x < 1.

0 0.5 1 1.5 2 2.5 3

x+2y=2 x+2y=3

−(x−1)+y=1 (x−1)+y=1

optimal solution

feasible domain

(7)

(c)

||x − 1| − 2| + |y − 3| =

|x − 1| − 2 + |y − 3|, if |x − 1| ≥ 2;

−|x − 1| + 2 + |y − 3|, if |x − 1| < 2.

=









(x − 1) − 2 + |y − 3|, if |x − 1| ≥ 2, x ≥ 1;

−(x − 1) − 2 + |y − 3|, if |x − 1| ≥ 2, x < 1;

−(x − 1) + 2 + |y − 3|, if |x − 1| < 2, x ≥ 1;

(x − 1) + 2 + |y − 3|, if |x − 1| < 2, x < 1.

=











(x − 1) − 2 + (y − 3), if |x − 1| ≥ 2, x ≥ 1, y ≥ 3;

(x − 1) − 2 − (y − 3), if |x − 1| ≥ 2, x ≥ 1, y < 3;

−(x − 1) − 2 + (y − 3), if |x − 1| ≥ 2, x < 1, y ≥ 3;

−(x − 1) − 2 − (y − 3), if |x − 1| ≥ 2, x < 1, y < 3;

−(x − 1) + 2 + (y − 3), if |x − 1| < 2, x ≥ 1, y ≥ 3;

−(x − 1) + 2 − (y − 3), if |x − 1| < 2, x ≥ 1, y < 3;

(x − 1) + 2 + (y − 3), if |x − 1| < 2, x < 1, y ≥ 3;

(x − 1) + 2 − (y − 3), if |x − 1| < 2, x < 1, y < 3.

=











x+ y − 6, if x ≥ 3, y ≥ 3;

x− y, if x ≥ 3, y < 3;

−x + y − 4, if x ≤ −1, y ≥ 3;

−x − y + 2, if x ≤ −1, y < 3;

−x + y, if 1 ≤ x < 3, y ≥ 3;

−x − y + 6, if 1 ≤ x < 3, y < 3;

x+ y − 2, if − 1 < x < 1, y ≥ 3;

x− y + 4, if − 1 < x < 1, y < 3.

0 1 2 3 4 5 6 7 8 9 10

x+y-6=3

x+y-2=3

-x-y+6=3

-x+y=3

x-y=3 x-y+4=3

feasible domain

y-1/4x=1

y-1/4x=0 optimal

solution

(8)

5. (a) Following is the three-dimensional graph for the three systems.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0 0.2 0.4 0.6 0.8 1

−2

−1.5

−1

−0.5 0 0.5 1

X1+X 2=1

feasible domian

(b) To show that the three systems are equivalent, we show the equivalence between (f¹, s1) and (f², s2), the equivalence between (f², s2) and (f³, s3), and the equivalence between (f³, s3) and (f¹, s1).

the equivalence between (f¹, s1) and (f², s2):

Consider x³ in the problem (f², s2) as a slack variable associated with the first constraint in (f1, s1), then

(f¹, s1) : min −2x¹+ x² s.t. x1 + x² ≤ 1

x1, x2 ≥ 0.

is equivalent to

(f², s2) : min −2x¹+ x² s.t. x1+ x²+ x³ = 1

x1, x2, x3 ≥ 0.

the equivalence between (f², s2) and (f³, s3) : By the constraints of (f², s2), we have

x1 = 1 − x²− x³ ≥ 0.

Therefore the objective function of (f², s2),

−2x¹+ x2, becomes

−2(1 − x²− x³) + x² = 3x²+ 2x³− 2

(9)

and the constraints are

1 − x²− x³ ≥ 0 (or x²+ x³ ≤ 1) x2, x3 ≥ 0.

This shows that

(f², s2) : min −2x¹+ x² s.t. x1+ x²+ x³ = 1

x1, x2, x3 ≥ 0.

is equivalent to

(f³, s3) : min 3x²+ 2x³− 2 s.t. x² + x³ ≤ 1

x2, x3 ≥ 0.

the equivalence between (f3, s3) and (f1, s1):

To show the equivalence between (f³, s3) and (f¹, s1), we let 1 − x²− x³ = x¹.

Then we have

3x²+ 2x³− 2 = −2(1 − x²− x³) + x²

= −2x¹+ x² for the objective function of (f³, s3) and

1 − x²− x³ = x¹ ≥ 0 (which is equivalent to 1 − x²− x³ = x¹ with x²+ x³ ≤ 1.) x2, x3 ≥ 0.

for the constraints. That is,

(f³, s3) : min 3x²+ 2x³− 2 s.t. x² + x³ ≤ 1

x2, x3 ≥ 0.

is equivalent to

(f1, s1) : min −2x¹+ x2

s.t. x1 + x² ≤ 1 x1, x2 ≥ 0.