Engineering economics

IE 451 Engineering Economy page 1

 Exam #1

Due Sunday, July 26, 2015 (5pm US Mountain Time) 

Name:_________________

Question # 1 2 3 4 5 6 7 8 Total

Possible Points 2 2 2 2 2 2 2 2 16

Your Score

Notes: a. If there is no justification on your answer, you will get ZERO credit. Thus, show

your calculation procedure and circle your final answer. b. Use the interpolation technique if the compound interest tables in Appendix C of

our text do not provide the value of interest factors. Otherwise, there will be a 0.5 point deduction.

c. You should submit a MS word file or a scanned copy or your work as an

attachment by using “Submit Assignment” button (see below). No regular email submission is accepted.

d. Only a single MS word file or a single pdf file will be accepted as your

submission. A violator for this restriction will be penalized with 10% of the total points available being taken away. If you decide to submit a scanned copy of your work with a single pdf file, be sure that your original work was printed (no cursive writing!).

e. Your exam solution should be submitted on time. Late submission will be

penalized with 5% of the total points available for each hour late. Remember that you can either type in your solution in MS word or submit a scanned copy of your hand writing with a single pdf file. For the latter case, be sure to have a scanner available to use and make the scans to a single pdf file before the submission.

IE-451-M70 Summer 2015

Final Exam

Final Exam

IE 451 Engineering Economy page 2

(1) The following table summarizes information which are associated with three new 3D Printers being considered for use in a manufacturing plant. Note that M&O stands for Maintenance & Operation Cost. A B C Useful Life (Years) 13  11  9  First Cost $2,780,000   $2,630,000   $2,300,000   Salvage Value $118,000   $97,000   $82,000   Annual Benefit $670,000   $650,000   $580,000   M&O $78,000   $71,000   $65,000   M&O Gradient $15,000   $12,500   $11,000  

The company's interest rate (MARR) is 12%. Which 3D Printer should the company choose? Use Annual Cash Flow Analysis. (2) A used car dealer in Las Cruces placed the following advertisement: $500 down now + $99 for the first 12 months + $199 for the following 48 months a. What is the price of the car if the interest rate is 12% per year compounded monthly? b. If financing is done at 12% APR, what would be the equivalent uniform monthly

payment? (3) Austin, a US Crude Company engineer recommended that US Crude purchase a special tool to reduce the cost of pumping oil out of the bayous of St. Martin Parish. As a result of Austin's recommendation, US Crude purchased the tool for $300,000 on January 1, 2010. By January 1, 2011, the tool had saved a total of $45,000 and went on line full time. After going on line full time, the tool saved US Crude $90,000 each year for the next three years and Austin was happy. However, Austin recommended the "el-cheapo" model, and it started breaking down during the early part of year five, and ended up by saving only $50,000 during year five. It was scrapped as being unusable at the end of year five, and had a zero salvage value. Austin told his boss that his recommendation had been correct. Use a MARR of 10% and evaluate the effectiveness of the tool and the correctness of Austin's recommendation. (4) You are considering two alternative plant layouts, A1 and A2, to improve its current layout. The cash flows are shown below. The first costs represent the expenses of rearranging the current layout to the alternative new layout and the annual savings represent the reduction in the production costs of the new layout compared to the current layout. Using the internal rate of return as the decision criterion, what course of action do you recommend? Use MARR = 11%.

Data Year A1 A2 First Cost 0 -$110,000 -$115,000 Annual Savings 1 to ∞ $12,500 $15,000

IE 451 Engineering Economy page 3

(5) Given the data in the table below, choose the better alternative-using present worth analysis if MARR is 9%.

Alternative A Alternative B Initial Cost $10,000 $9,000

Annual Benefits $5,200 $4,700 Annual Expenses $3,000 $2,000

Salvage Value $1,200 $1,300 Useful Life (Years) 6 4

(6) Compute the capitalized cost for the following cash flows using the minimum number of compound interest factors. Note that if you do not use the minimum number of compound interest factors, there will be a 0.5 point deduction.

P = ?

100 200

300 400

100 200

300 400

∞ i = 10%

500 500

(7) Compute the future worth for the following cash flows using the minimum number of compound interest factors. Note that if you do not use the minimum number of compound interest factors, there will be a 0.5 point deduction. 6x 5x

4x 3x

2x x i = 15 %

IE 451 Engineering Economy page 4

(8) If you invest $1,000 today, how much will it be worth in 20 years? Assume that the money will grow at the interest rates: of 8% compounded semiannually during years 1-10, and 12% compounded quarterly during the remaining years.