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Assignment 7 Due: Wed. Oct. 17, 2018 Stevens – Fall 2015

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CEE 4200 — Assignment 7 Due: Wed. Oct. 17, 2018

In Chapter 6: 6.29, 6.31, 6.41 7: 7.7, 7.15, 7.30, 7.32, 7.34

Stats 7.1 – Using the data from Stats 4.1/5.1/6.1

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Recall that you randomly called 10 different pump companies and got 10 (very!) preliminary bids, which the pump suppliers claim some annual operating cost savings due to increase power use efficiency. You determined the ROR for each pump and an ‘average’ pump last week. Now your boss(!) asks you to continue analysis of these data and (1) carry out an incremental analysis based on the ROR and procedures discussed in class to decide which pump is the best choice, and (2) for a randomly chosen step in the comparison (e.g. comparing pump 8 vs. pump 6), determine how much uncertainty in pump costs and interests rates of 10% would influence your

decision about which pump of those two is a better choice. Explain this to your boss (who knows even less than you do about statistics). You may use any information from Stats 4.1 – 6.1 to help and consider using the first order error analysis discussed in class. Here’s the section from the exam Stats Review.

I. Uncertainty propagation

A. First Order Error Analysis

1. linear combinations – , if the x are uncorrelated,

B. non-linear combinations – use 1st order Taylor series approximations where


Pump Pump 1 Pump 2 Pump 3 Pump 4 Pump 5 Pump 6 Pump 7 Pump 8 Pump 9 Pump 10

Cost 7.55 6.97 6.79 6.75 6.82 8.23 7.73 8.49 6.65 7.6

Life 10 10 8 12 11 9 9 10 10 10

Annual 1.283 1.027 1.487 1.487 1.38 1.34 1.187 1.397 1.347 1.22

Savings 0.883 0.627 1.087 1.087 0.98 0.94 0.787 0.997 0.947 0.82

Salvage 0.683 1.074 0.686 0.527 0.884 0.935 0.532 0.722 0.815 0.867

if f aixi i 1=


= f2 ai2 i2 i 1=



f k f k0 xi f kxi

i 1=


+= f2 xi f 2

i 2

i 1=



  • CEE 4200 — Assignment 7 Due: Wed. Oct. 17, 2018


Sheet 1

Defender Challenger
pi 10 9 Initial cost
a_c 1 0.9 annual cost
a_b 1.1 0.99 annual benefit
a_net 0.1 0.09 benefit-cost
f -3 -2.4 Future maintenance cost at year 10
nf 10 10 Maintenance year
s 0.00 0.00 Salvage
n 20 20
ROR -12% By changing this value
AW $0.007 $0.006
Diff, % 0.0006744324 Set this value to zero

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