 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/m1)
V
(1/n1)
(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/m1)
V
(1/n1)
. (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
02
+ ω
11
+
21
= 0 (16.17)
V terms ω
01
+ (1/n 1
02
+
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
01
)
ω
01
(K
02
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:
21
= ω
01
(1/n 1
02
= (1 ω
02
) (1/n 1
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
02
21
= 1 ω
02
(1/m 1
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
))
(1ω
02
)
(K
02
02
)
ω
02
(K
11
)
((1c/b)+ω
02
Z)
(K
21
)
(1/bω
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
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
02
}][K
02
(1
2
02
(1)]+[log(K
02
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
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
02
1 = (K
01
/K
02
)((K
21
)
1/bn
/(K
11
)
Z
)
1
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
))
(1ω
02
)
(K
02
02
)
ω
02
(K
11
)
((1c/b)+ω
02
Z)
(K
21
)
(1/bω
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
))
(1ω
02
)
(K
02
02
)
ω
02
(K
11
)
((1c/b)+ω
02
Z)
(K
21
)
(1/bω
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/m1
V
1/n1
= 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 =
02
K
01
/(ω
01
K
02
))
m
(K
01
01
)
(m(m1)/(nm))
02
K
02
)
((mm)/(nm)
Y
(m/(nm))
(16.39)
V = (K
01
01
)
(n(m1)/(mn))
02
/K
02
)
((mn)/(mn))
Y
(n/(mn))
. (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
02
+
21
= 0 (16.42)
V terms ω
01
+ (1/n 1
02
+
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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