16.3. PROBLEM SOLUTION 87

Since K

oo

is a constant, the objective function to be minimized is:

Y = K

01

f

−1

V

−1

+ K

02

f

(1/m−1)

V

(1/n−1)

(16.8)

where

Y = Variable portion of the unit cost.

The two constraints must be developed into geometric programming format. The feed con-

straint is:

f ≤ f

max

(16.9)

where

f

max

= Maximum Feed Limit (in/rev).

The feed constraint is typically used to control the surface ﬁnish as the smaller the feed, the

better the surface ﬁnish.

The horsepower constraint is given as:

aV

b

f

c

≤ Hp (16.10)

where

a = Horsepower Constraint Constant

b = Velocity Exponent for Horsepower Constraint

c = Feed Rate Exponent for Horsepower Constraint

HP = Horsepower Limit.

These constraints can be put into geometric programming form as:

K

11

f ≤ 1 (16.11)

K

21

V

b

f

c

≤ 1 (16.12)

where

K

11

= 1/f

max

K

21

= a/Hp .

16.3 PROBLEM SOLUTION

The primal problem can be stated as:

Minimize Y = K

01

f

−1

V

−1

+ K

02

f

(1/m−1)

V

(1/n−1)

. (16.13)

Subject to the constraints:

K

11

f ≤ 1 (16.14)

K

21

V

b

f

c

≤ 1 . (16.15)

88 16. MATERIAL REMOVAL/METAL CU TTING ECONOMICS

From the coefﬁcients and signs, the signum values for the dual are:

σ

01

= 1

σ

02

= 1

σ

11

= 1

σ

21

= 1

σ

1

= 1

σ

2

= 1 .

The dual problem can be formulated as:

ω

01

+ ω

02

= 1 (16.16)

f terms − ω

01

+ (1/m − 1)ω

02

+ ω

11

+ cω

21

= 0 (16.17)

V terms − ω

01

+ (1/n − 1)ω

02

+ bω

21

= 0 . (16.18)

The degrees of difﬁculty (D) are equal to:

D = T − (N + 1) = 4 − (2 + 1) = 1 . (16.19)

From the constraint equations which have only one term it is apparent that:

ω

10

= ω

11

(16.20)

ω

20

= ω

21

. (16.21)

The dual objective function is:

Y = (K

01

/ω

01

)

ω

01

(K

02

/ω

02

)

ω

02

(K

11

)

ω

11

(K

21

)

ω

21

. (16.22)

Thus, we have 4 variables and 3 equations, or one degree of difﬁculty. To ﬁnd an additional

equation one must get the number of variables reduced to one in the dual objective function, differ-

entiate the logarithm of the equation to obtain another equation by setting the derivative to zero. It

was decided to determine the dual variables in terms of ω

02

.

From Equation (16.16) one obtains:

ω

01

= 1 − ω

02

. (16.23)

From Equation (16.18) one obtains:

bω

21

= ω

01

− (1/n − 1)ω

02

= (1 − ω

02

) − (1/n − 1)ω

02

= 1 − ω

02

/n

ω

21

= 1/b − ω

02

/bn . (16.24)

16.3. PROBLEM SOLUTION 89

From Equation (16.17) one obtains:

ω

11

= ω

01

− (1/m − 1)ω

02

− cω

21

= 1 − ω

02

− (1/m − 1)ω

02

− c(1/b − ω

02

/bn)

= (1 − c/b) + ω

02

(c/bn − 1/m)

ω

11

= (1 − c/b) + ω

02

Z (16.25)

where

Z = (c/bn − 1/m) . (16.26)

The dual can be written as:

Y = (K

01

/(1 − ω

02

))

(1−ω

02

)

(K

02

/ω

02

)

ω

02

(K

11

)

((1−c/b)+ω

02

Z)

(K

21

)

(1/b−ω

02

/bn)

. (16.27)

The log of the dual is:

Log (Y ) = (1 − ω

02

) log(K

01

/(1 − ω

02

)) + ω

02

log(K

02

/ω

02

)

+ (1 − c/b) log(K

11

) + ω

02

Z log(K

11

)

+ 1/b log(K

21

) − ω

02

/bn log(K

21

). (16.28)

Find ∂(log Y )/∂ω

02

and set it to zero to ﬁnd an additional equation.

∂(log Y )/∂ω

02

= (1 − ω

02

)[1/{K

01

/(1 − ω

02

)}][K

01

(−1)(1 − ω

02

)

−2

(−1)]

+[log(K

01

/(1 − ω

02

))](−1)

+ ω

02

[1/{K

02

/ω

02

}][K

02

(−1)ω

−2

02

(1)]+[log(K

02

/ω

02

)](1)

+ 0 + Z log(K

11

) + 0 − (1/bn) log(K

21

)

∂(log Y )/∂ω

02

= (−1) −[log(K

01

/(1 − ω

02

))]+1 +[log(K

02

/ω

02

)]

+ 0 + Z log(K

11

) + 0 − (1/bn) log(K

21

)

∂(log Y )/∂ω

02

= log[(K

02

/K

01

)((1 − ω

02

)/(ω

02

))]+log(K

11

)

Z

− log(K

21

)

1/bn

∂(log Y )/∂ω

02

= log[(K

02

/K

01

)((1 − ω

02

)/(ω

02

))]+log[(K

11

)

Z

/ log(K

21

)

1/bn

]

∂(log Y )/∂ω

02

= log[(K

02

/K

01

)((1 − ω

02

)/(ω

02

))(K

11

)

Z

/(K

21

)

1/bn

] (16.29)

Setting the expression equal to zero and then use the anti-log, one goes through various steps

to obtain an expression for ω

02

:

1 = (K

02

/K

01

)((1 − ω

02

)/(ω

02

))(K

11

)

Z

/(K

21

)

1/bn

(1 − ω

02

)/(ω

02

) = (K

01

/K

02

)((K

21

)

1/bn

/(K

11

)

Z

)

1/ω

02

− 1 = (K

01

/K

02

)((K

21

)

1/bn

/(K

11

)

Z

)

1/ω

02

= (K

01

/K

02

)((K

21

)

1/bn

/(K

11

)

Z

) + 1

ω

02

= (K

02

K

Z

11

)/(K

01

K

1/bn

21

+ K

02

K

Z

11

). (16.30)

Now the values for other three dual variables can be obtained in terms of ω

02

as:

ω

01

=−ω

02

(16.31)

ω

21

= 1/b − ω

02

/bn (16.32)

ω

11

= (1 − c/b) + ω

02

Z. (16.33)

90 16. MATERIAL REMOVAL/METAL CU TTING ECONOMICS

Substituting the values of the dual variables into the dual objective function, the dual objective

function can be expressed in terms of ω

02

and is:

Y = (K

01

/(1 − ω

02

))

(1−ω

02

)

(K

02

/ω

02

)

ω

02

(K

11

)

((1−c/b)+ω

02

Z)

(K

21

)

(1/b−ω

02

/bn)

(16.34)

and the minimum cost would be:

C

u

= Y + K

00

(16.35)

or C

u

= K

00

+ (K

01

/(1 − ω

02

))

(1−ω

02

)

(K

02

/ω

02

)

ω

02

(K

11

)

((1−c/b)+ω

02

Z)

(K

21

)

(1/b−ω

02

/bn)

. (16.36)

By using the primal-dual relationships of the objective function, one can obtain the following

equation for the primal variables. Starting with

K

01

f

−1

V

−1

= ω

01

Y (16.37)

and

K

02

f

1/m−1

V

1/n−1

= ω

02

Y. (16.38)

One can obtain equations for the primal variables in terms of the dual variables, dual objective

function and constants. These equations are for the case when both constraints are binding.

f = (ω

02

K

01

/(ω

01

K

02

))

m

(K

01

/ω

01

)

(m(m−1)/(n−m))

(ω

02

K

02

)

((mm)/(n−m)

Y

(m/(n−m))

(16.39)

V = (K

01

/ω

01

)

(n(m−1)/(m−n))

(ω

02

/K

02

)

((mn)/(m−n))

Y

(n/(m−n))

. (16.40)

Upon examination of the equations for the dual variables, it is possible for either ω

11

or ω

21

to be negative, which means that the constraints are not binding. Thus, the dual variable must be set

to zero if the constraint is not binding. If ω

11

is zero, then Equations (16.16), (16.17), and (16.18)

become:

ω

01

+ ω

02

= 1 (16.41)

f terms − ω

01

+ (1/m − 1)ω

02

+ cω

21

= 0 (16.42)

V terms − ω

01

+ (1/n − 1)ω

02

+ bω

21

= 0 . (16.43)

The degree of difﬁculty becomes zero with the variable ω

11

removed, and the new values for

the dual variables become:

ω

02

= (b − c)/(b/m − c/n) (16.44)

ω

01

= 1 − ω

02

(16.45)

ω

11

= 0 (loose constraint) (16.46)

ω

21

= (1/m − 1/n)/(b/m − c/n) . (16.47)

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