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Integer Type Unknown Variables

2.8 Solution Procedure for General IP Problem with Inequality Constraints

2.8.1 Integer Type Unknown Variables

The critical point of f (x, s) based on the variable x1 is obtained as follows:

f (x1) = x1(q11x1+ q12x2+ q21x2+ q13s1+ q31s1+ q14s2+ q41s2+ q15s3+ q51s3+ r1) + c.

(2.72) The partial derivative with respect to the unknown variable x1 is:

∂f

∂x1 = 2q11x1+ q12x2+ q21x2+ q13s1+ q31s1+ q14s2+ q41s2+ q15s3+ q51s3+ r1. (2.73)

Since q12= q21, q13= q31, q14= q41, and q15= q51:

∂f

∂x1 = 2q11x1+ 2q12x2+ 2q13s1+ 2q14s2+ 2q15s3+ r1. (2.74)

The critical point x?1 for the variable x1 is obtained by equating the partial derivative to zero:

∂f

∂x1

= 0 =⇒

2q11x1+ 2q12x2+ 2q13s1+ 2q14s2+ 2q15s3+ r1 = 0, 2q11x1 = −(2q12x2+ 2q13s1+ 2q14s2+ 2q15s3+ r1), x1 = 2q12x2+ 2q13s1+ 2q14s2+ 2q15s3+ r1

−2q11 ,

or, x?1 = 2q12x2+ 2q13s1+ 2q14s2+ 2q15s3+ r1

−2q11 .

(2.75)

Substituting the original values from (2.70) in (2.75), we obtain:

x?1 = 2P (a>1a2)x2+ 2P a11s1+ 2P a21s2+ 2P a31s3+ (d1− 2P (b>a1))

−2P (a>1a1) , (2.76)

which is a critical point at which the function f (x, s) is at relative extremum (relative minimum for our minimization problem).

The change in f (x, s) due to change in x1 is calculated as follows:

f (x1) = x1(q11x1+ q12x2+ q13s1+ q14s2+ q15s3 + q21x2+ q31s1+ q41s2+ q51s3+ r1) + c.

(2.77) If x1 changes to x01, then f (x01) is given by:

f (x01) = x01(q11x01+ q12x2+ q13s1+ q14s2+ q15s3 + q21x2+ q31s1+ q41s2+ q51s3+ r1) + c.

(2.78) The difference between (2.78) and (2.77) gives the change in value of function when x1 changes to x01. Therefore,

∆fx0

1 = f (x01) − f (x1),

= (x01− x1)(q11(x01+ x1) + q12x2+ q13s1+ q14s2+ q15s3+ q21x2+ q31s1+ q41s2+ q51s3+ r1),

= q11x012− q11x21+ (q12x2+ q13s1+ q14s2+ q15s3+ q21x2+ q31s1+ q41s2+ q51s3+ r1)x01 (q12x2+ q13s1+ q14s2+ q15s3+ q21x2+ q31s1+ q41s2+ q51s3+ r1)x1.

(2.79)

Since q12= q21, q13= q31, q14= q41, and q15= q51:

∆fx0

1 = q11x012− q11x21+

(2q12x2+ 2q13s1+ 2q14s2+ 2q15s3+ r1)x01− (2q12x2+ 2q13s1+ 2q14s2+ 2q15s3+ r1)x1.

(2.80)

Note that the partial derivative of (2.80) with respect to x01 is the same as (2.74):

∂∆fx0

1

∂x01 = 2q11x01 + 2q12x2+ 2q13s1+ 2q14s2+ 2q15s3+ r1. (2.81)

If the value of x1 is updated to x?1, we obtain ∆f due to change in x1 to x?1 as follows:

Substituting the original values from (2.70) in (2.82), we obtain:

∆fx?

1 for a problem with n integer variables and m inequality constraints shown in Figure 2.2 is:

x?i = 2Pi−1

Substituting the original values from (2.70) in (2.85) we obtain:

In the previous section, we calculated the value of x?i which is the critical value of xi at which the function f (x, s) equals zero, we then calculated ∆fx0

i when the variable changes from xi to x0i, and finally we calculated ∆fx?

i which shows the change in f (x, s) if the variable changes from xi to x?i. Based on this information we will show that updating variable xi to a new value is based up on xi and x?i, and using graphical representations we will show the feasible and infeasible regions for the new value of xi.

Recall from the previous section that if x1 changes to x01 then ∆fx0

i is given by:

1 is equal to zero when one of the following condition is met:

∆fx0

1 = 0 =⇒ (2.88a)

x01− x1 = 0, (2.88b)

q11(x01+ x1) + q12x2+ q13s1+ q14s2+ q15s3+ q21x2+ q31s1+ q41s2+ q51s3+ r1 = 0.

(2.88c)

From (2.88b), if x01− x1 = 0 then the variable is not changing, since x01 = x1, and ∆fx0

1

is always 0 in this case. But, from (2.88c), we obtain the following:

q11(x01+ x1) + q12x2+ q13s1+ q14s2+ q15s3 + q21x2+ q31s1+ q41s2+ q51s3+ r1 = 0, q11(x01) = −(q11(x1) + q12x2+ q13s1+ q14s2+ q15s3+ q21x2+ q31s1+ q41s2 + q51s3+ r1), x01 = −(q11(x1) + q12x2+ q13s1+ q14s2+ q15s3+ q21x2+ q31s1+ q41s2+ q51s3+ r1)

q11 ,

x01 = −(q12x2+ q13s1+ q14s2+ q15s3+ q21x2+ q31s1+ q41s2+ q51s3+ r1) q11

− x1, x01 = −(2q12x2 + 2q13s1 + 2q14s2 + 2q15s3+ r1)

q11 − x1,

x01 = 2x?1− x1.

(2.89) From (2.89), if the value of variable is changed from x1 to x01 where x01 = 2x?1− x1, then

∆fx0

1 = 0.

To generalize our findings, x?i represents the critical value at which the function f (x, s) is at relative extremum (relative minimum for our minimization problem). The value of

∆fx0

i = 0 when:

• x0i = xi (new value equals old value).

• x0i = 2x?i − xi (new value = 2x?i - old value).

The variable update procedure is simply based on xi and x?i, and the four possible combina-tions of these two parameters give the following cases. Note that the figures shown in these cases are parabolic since ∆fxi is quadratic in nature.

• Case 1: xi = 0 and x?i ≤ 0

∆fx

x xi

x < 0

x0i =2x?i

∆fx?

i

Figure 2.3. Change in ∆fx when: xi= 0 and x?i ≤ 0.

Figure 2.3 represents the change in ∆fx when xi = 0 and x?i ≤ 0. Based on this figure, the best new value for xi is x?i where ∆fx?i is obtained, but since x?i < 0 the new value of xi is kept at 0.

• Case 2: xi = 0 and x?i > 0

∆fx

x xi

∆fx < 0 x0i = 2x?i

∆fx?i

Figure 2.4. Change in ∆fx when: xi= 0 and x?i > 0.

Figure 2.4 represents the change in ∆fx when xi = 0 and x?i > 0. Based on this, the best new value for xi is x?i, since maximum absolute value ∆fx?

i is obtained at x?i. If x?i is not integer then the best integer value based on ∆fx that is closer to x?i is chosen.

Also, in an attempt to escape local optimality of the objective function, xi can be updated to an integer value of x, where xi < x ≤ 2x?i.

• Case 3: xi > 0 and x?i ≤ 0

∆fx

x xi

x < 0

∆fx < 0 x0i = 2x?i − xi

∆fx?i

x?i = 0 x?i < 0

Figure 2.5. Change in ∆fx when: xi> 0 and x?i ≤ 0.

Figure 2.5 represents two different curves based on the value of x?i.

If x?i = 0, represented by the red curve, the best new value of xi is x?i since the maximum absolute value ∆fx?

i is obtained at x?i. A new value for xi can also be updated to an integer value of x, where 0 ≤ x < xi to escape local optimality of the objective function value.

If x?i < 0, represented by the blue curve, the best new value of xi is x?i since the maximum absolute value ∆fx?

i is obtained at x?i. Since x?i < 0, the new value of xi is updated to 0, or to an integer value of x where 0 ≤ x < xi.

• Case 4: xi > 0 and x?i > 0

∆fx

xi x

x < 0

∆fx?

i

x0i = 2x?i − xi

∆fx< 0

(2x?i − xi) < 0 0 ≤ (2x?i − xi) < xi (2x?i − xi) > xi

Figure 2.6. Change in ∆fx when: xi> 0 and x?i > 0.

Figure 2.6 represents three different curves based on the value of 2x?i − xi.

If (2x?i − xi) > xi, represented by blue curve, the best new value for xi is x?i since maximum absolute value ∆fx?i is obtained at x?i. If x?i is not integer then the best integer value based on ∆fx that is closer to x?i is chosen. Also, in an attempt to escape local optimality of the objective function, xi can be updated to an integer value of x, where xi < x ≤ 2x?i − xi.

If 0 ≤ (2x?i − xi) < xi, represented by red curve, the best new value for xi is x?i since maximum absolute value ∆fx?

i is obtained at x?i. If x?i is not integer then the best integer value based on ∆fx that is closer to x?i is chosen. Also, in an attempt to escape

local optimality of the objective function, xi can be updated to an integer value of x, where 2x?i − xi ≤ x < xi.

If (2x?i − xi) < 0, represented by green curve, the best new value for xi is x?i since maximum absolute value ∆fx?i is obtained at x?i. If x?i is not integer then the best integer value based on ∆fx that is closer to x?i is chosen. Also, in an attempt to escape local optimality of the objective function, xi can be updated to an integer value of x, where 0 ≤ x < xi.

In Table 2.3, we summarize the update rules from the four cases. Here xi represents the current value of the variable and x0i represents the new value of the same variable. Note that an integer value is always selected for the new value of xi.

Case xi x?i (2x?i − xi) Best x0i Acceptable x0i

1 0 ≤ 0 ≤ 0 0 0

2 0 > 0 > 0 x?i xi < x ≤ 2x?i 3 > 0 0 < 0 0 0 ≤ x < xi

> 0 < 0 < 0 0 0 ≤ x < xi 4 > 0 > 0 > xi x?i xi < x ≤ 2x?i − xi

> 0 > 0 ≥ 0 & < xi x?i 2x?i − xi ≤ x < xi

> 0 > 0 < 0 x?i 0 ≤ x < xi

Table 2.3: Updating integer type unknown variable based on critical value.

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