 98 17. THE OPEN CARGO SHIPPING BOX WITH SKIDS
Similarly,
H = ω
03
Y/(20L) = ω
03
Y/(20ω
05
Y/10) =
03
05
)(C
05
/C
03
)
= (0.133/0.111)(10/20) = 0.599 (versus 0.500) (17.28)
and
W = ω
02
Y/(10L) = ω
02
Y/(10ω
05
Y/10) =
02
05
)(C
05
/C
02
)
= (0.133/0.111)(10/10) = 1.198 (versus 1.000) (17.29)
Therefore, the primal becomes
C = C
01
/(LW H ) + C
02
LW + C
03
LH + C
04
HW + C
05
L (17.30)
= C
01
/(1.284 0.599 1.198) + C
02
(1.284 1.198) + C
03
(1.284 0.599)
+ C
04
(0.599 1.198) + C
05
(1.284)
= 43.41 + 15.38 + 15.38 + 28.70 + 12.84
= 115.71 versus 115.72 . (17.31)
The volume of the box is 1.284 × 0.599 × 1.198 = 0.921 cubic yards versus the 1.0 cubic yard
volume in the original problem. The number of trips will increase from 400 in the original problem
to 434.3 or 435 trips. The shipping cost increases from \$40 to \$43.41, but the box cost (side, end
and bottom) totals \$59.46 versus \$60 and the primary increase is the cost of the skids, that is \$12.84.
17.4 DIMENSIONAL ANALYSIS APPROACH FOR
The dimensional analysis approach starts with the primal and dual equations which are:
C = C
01
/(LW H ) + C
02
LW + C
03
LH + C
04
HW + C
05
L (17.2)
and
Dual (Y ) ={(C
01
01
)
ω
01
(C
02
02
)
ω
02
(C
03
03
)
ω
03
(C
04
04
)
ω
04
(C
05
05
)
ω
05
} (17.12)
The primal dual relationships are:
L
1
W
1
H
1
= ω
01
Y/C
01
(17.32)
LW = ω
02
Y/C
02
(17.33)
LH = ω
03
Y/C
03
(17.34)
HW = ω
04
Y/C
04
(17.35)
L = ω
03
Y/C
03
(17.36)
These relationships can be combined to give:
(L
1
W
1
H
1
)
A
(LW )
B
(LH )
C
(H W )
D
(L)
E
= 1
=
01
Y/C
01
)
A
02
Y/C
02
)
B
03
Y/C
03
)
C
04
Y/C
04
)
D
05
Y/C
05
)
E
(17.37) 17.4. DIMENSIONAL ANALYSIS APPROACH FOR ADDITIONAL EQUATION 99
The equations from dimensional analysis are:
L terms A + B + C + E = 0 (17.38)
W terms A + B + D = 0 (17.39)
H terms A + C + D = 0 (17.40)
Y + A + B + C + D + E = 0 . (17.41)
From Equations (17.39) and (17.40), one obtains that:
B = C (17.42)
Subtracting Equation (17.38) from Equation (17.41), one obtains:
D =−2A (17.43)
Subtracting Equation (17.38) from Equation (17.39) results in
D = C + E (17.44)
From Equation (17.38) one observes that:
C + E = A B (17.45)
Thus,
D =−2A = A B,
(17.46)
so
B = 3A (17.47)
C = 3A (17.48)
D =−2A (17.49)
and via Equation (17.38)
E =−5A. (17.50)
Thus, if A = 1, then B = 3,C = 3,D =−2, and E =−5.
Thus, from Equation (17.37)
1 =
01
Y/C
01
)
1
02
Y/C
02
)
3
03
Y/C
03
)
3
04
Y/C
04
)
2
05
Y/C
05
)
5
. (17.51)
This results in
01
)(ω
02
)
3
03
)
3
01
)
2
01
)
5
= (C
01
)(C
02
)
3
(C
03
)
3
(C
04
)
2
(C
05
)
5
= (40)(2)
3
(10)
3
(40)
1
(10)
5
(17.52)
or in terms of ω
01
01
)(3ω
01
1)
3
(3ω
01
1)
3
(1 2ω
01
)
2
(2 5ω
01
)
5
= 2 . (17.53)
Equation (17.53) is the same as Equation (17.19) and the solution procedure from that point on
would be the same as that used for the constrained derivative approach.

Get Geometric Programming for Design and Cost Optimization, 2nd Edition now with O’Reilly online learning.

O’Reilly members experience live online training, plus books, videos, and digital content from 200+ publishers.