98 17. THE OPEN CARGO SHIPPING BOX WITH SKIDS

Similarly,

H = ω

03

Y/(20L) = ω

03

Y/(20ω

05

Y/10) = (ω

03

/ω

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) = (ω

02

/ω

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

ADDITIONAL EQUATION

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

/ω

01

)

ω

01

(C

02

/ω

02

)

ω

02

(C

03

/ω

03

)

ω

03

(C

04

/ω

04

)

ω

04

(C

05

/ω

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

= (ω

01

Y/C

01

)

A

(ω

02

Y/C

02

)

B

(ω

03

Y/C

03

)

C

(ω

04

Y/C

04

)

D

(ω

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 = (ω

01

Y/C

01

)

1

(ω

02

Y/C

02

)

3

(ω

03

Y/C

03

)

3

(ω

04

Y/C

04

)

−2

(ω

05

Y/C

05

)

−5

. (17.51)

This results in

(ω

01

)(ω

02

)

3

(ω

03

)

3

(ω

01

)

−2

(ω

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

(ω

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.

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