MSM3M02a: Autumn 2010

Solution to Integer Programming: Problem Set 5

Dr. Y.B. Zhao (Mathematics, University of Birmingham)

Q1. Proof.

Let x ∈ S. Then

a0x0 +n∑

j=1

ajxj ≤ b, lj ≤ xj ≤ uj, j = 0, 1, ..., n.

(i) If a0 > 0, then

x0 ≤ 1

a0

b−

n∑

j=1

ajxj

=1

a0

b− ∑

j:aj>0

ajxj −∑

j:aj<0

ajxj

≤ 1

a0

b− ∑

j:aj>0

ajlj −∑

j:aj<0

ajuj

Similarly, if a0 < 0, then

x0 ≥ 1

a0

b−

n∑

j=1

ajxj

=

1

a0

b− ∑

j:aj>0

ajxj −∑

j:aj<0

ajxj

≥ 1

a0

b− ∑

j:aj>0

ajuj −∑

j:aj<0

ajlj

(ii) If ∑

j:aj>0

ajuj +∑

j:aj<0

ajlj ≤ b

Then

a0x0 +n∑

j=1

ajxj =∑

j:aj>0

ajxj +∑

j:aj<0

ajxj ≤∑

j:aj>0

ajuj +∑

j:aj<0

ajlj ≤ b

thus, a0x0 +∑n

j=1 ajxj ≤ b is redundant.

(iii) If ∑

j:aj>0

ajlj +∑

j:aj<0

ajuj > b

then

a0x0 +n∑

j=1

ajxj =∑

j:aj>0

ajxj +∑

j:aj<0

ajxj ≥∑

j:aj>0

ajlj +∑

j:aj<0

ajuj > b

so there is no feasible point to S, i.e., S = ∅.

(iv) When aj ≥ 0 for all i = 1, ..., m and cj < 0,we may decrease the variable xj (sincedecreasing xj can improve the objective value and keep the constraint Ax ≤ b beingsatisfied). Thus we may decrease xj until it reach its lower bound. Thus, xj = lj canbe fixed. Similarly, if aij ≤ 0 for all i = 1, ..., m and cj > 0, we may increase thevariable to enlarge the function cT x and still keep feasible Ax ≤ b, until it reach theupper bound. Thus, xj = uj.

Q2. We start with the third constraint

4x2 − 4x3 − 2x4 + 2x5 ≤ −6

From this constraint

(a)x2 = 1 =⇒ x3 = 1, thus x2 ≤ x3 (1)

(b)x2 = 1 =⇒ x4 = 1, thus x2 ≤ x4 (2)

(c)x5 = 1 =⇒ x3 = 1, thus x5 ≤ x3 (3)

(d)x5 = 1 =⇒ x4 = 1, thus x5 ≤ x4 (4)

(e)No solution satisfying x2 = x3 = 0, thus x2 + x3 ≥ 1 (5)

(f)No solution satisfying x4 = x5 = 0, thus x4 + x5 ≥ 1 (6)

Combining (1) and (5) implies that

2x3 ≥ x2 + x3 ≥ 1 =⇒ x3 ≥ 1/2, so x3 = 1.

Similarly, combining (4) and (6) leads to

2x4 ≥ x4 + x5 ≥ 1 =⇒ x4 ≥ 1/2, so x4 = 1.

Substituting the indentities x3 = 1 and x4 = 1 into the original problem, we get thefollowing simplified problem.

max −1 + 8x1 − 6x2 − 7x5

s.t. 2x1 + x2 + 2x5 ≤ 5−3x1 − 5x2 − 3x5 ≤ −74x2 + 2x5 ≤ 0x1, x2, x5 ∈ {0, 1}

The third constraint 4x2 + 2x5 ≤ 0 implies that x2 = x5 = 0 which contradicts withthe second constraint. Thus the original problem is infeasible (no feasible solution).

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Q3. Divide both sides of the inequality by 10,

12

10y1 +

18

10y2 + y3 +

6

10y4 ≥ 55

10.

Rounding up the left coefficients to the nearest integer yields

2y1 + 2y2 + y3 + y4 ≥ 55

10

which is a valid inequality for P . Since the right 5510

= 5.5. by rounding up it to thenearest integer. We have

2y1 + 2y2 + y3 + y4 ≥ 6

which is valid for X.

Q4. Consider the inequality4y1 + 5y2 + 9y3 + 12y4 ≤ 34.

Dividing the inequality by 5 gives the valid inequality for X:

4

5y1 + y2 +

9

5y3 +

12

5y4 ≤ 34

5.

As y ≥ 0, rounding down the coefficients on the left to the nearest integer yields

y2 + y3 + 2y4 ≤ 4

5y1 + y2 +

9

5y3 +

12

5y4 ≤ 34

5

so, we get the valid inequlity:

y2 + y3 + 2y4 ≤ 34

5= 6.8.

Since y ∈ Z4+, the left-hand-side must be integer, and thus we have

y2 + y3 + 2y4 ≤ b6.8c = 6,

which is a valid inequality for X.

Q5. (i) When b < 0, the set X is empty.

(ii) When∑n

j=1 aj = b, then∑n

j=1 ajxj ≤ b is redundant.

(iii) When aj > b, any feasible point must satisfy that xj = 0, and thus xj = 0 is valid.

(iv) When ai + aj > b, there is no feasible point satisfying xi = xj = 1, and thusxi + xj ≤ 1 which is a valid inequality for X.

Q6. (i)* Let α ∈ (12, 1], i.e.,1

2< α ≤ 1. Consider the inequality

αx1 + (1− α)x2 ≤ y.

We now prove that this inequality is valid for X and cut off the point (x1, x2, y) =(1, 0, 1

2). In fact, (1, 0, 1

2) does not satisfy the above inequality. So we only need to

prove that it is a valid inequality. In fact, X can be represented as

X = {(0, 0, 0), (x1, x2, 1) with 0 ≤ x1 ≤ 1, 0 ≤ x2 ≤ 1}.

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Since if 0 ≤ x1, x2 ≤ 1,

αx1 + (1− α)x2 ≤ α + (1− α) = 1.

This means all points in X satisfy the inequality

αx1 + (1− α)x2 ≤ y,

and hence it is valid for X.

(ii) This is 2-dimensional case. If we represent X graphically. It is very easy to find avalid inequality cutting off the point (9, 9/4). Notice that

X = {(0, 0); (x, 1) with 0 ≤ x ≤ 4; (x, 2) with 0 ≤ x ≤ 8;

(x, y ≥ 3) with 0 ≤ x ≤ 9}

The line crossing (9, 3) and (8, 2) will cut off the point (9, 9/4) and be valid for X.

(iii) Dividing the inequality

9x1 + 12x2 + 8x3 + 17x4 + 13x5 ≥ 50

by 12 leads to9

12x1 + x2 +

8

12x3 +

17

12x4 +

13

12x5 ≥ 50

12= 4.016.

Rouding up the left coefficients to the nearest integer yields

x1 + x2 + x3 + 2x4 + 2x5 ≥ 50

12,

when x ∈ X, the left is integer, thus the valid inequality is given by

x1 + x2 + x3 + 2x4 + 2x5 ≥ 5.

However, the point x = (0, 256, 0, 0, 0) is cut off by this inequality.

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