Title: ENGR 112
1ENGR 112
2Engineering Economic Analysis
- Time Value of Money
- 1 today is more valuable than 1 a year later
- Engineering economy adjusts for the time value of
money to balance current and future revenues and
cost
0
5
10
Years
3Recap
- Capital vs. Interest
- Types of Interest
- Simple vs. Compound
- Cash Flow Diagrams
- Calculate a Future given a Principal and vice
versa - P/F and F/P ? F P(1 i)N
- Frequency of Compounding
- Quarterly, monthly, daily
- n m x N ? F P(1 i/m)mN
4Annuity
- A series of uniform cash flows
- Begins at end of period one
- Continues through the end of period N
- Annuity given a Principal (A/P)
- Annuity given a Future (A/F)
5IDDI Company Case
- The I Dint Do It (IDDI) Company plans to borrow
150,000 from the Trust Me Bank (TMB) at an
interest rate of 9 per year, compounded
quarterly, in order to finance the installation
of pollution reduction equipment. The TMB will
agree to the loan only if IDDI makes quarterly
payments during the next 4 years. Use the A/P
formula to determine the quarterly payment.
P 150,000
i 9 per year, compounded quarterly
m 4
N 4
6IDDI Company Case Solution
Quarterly compounding ? i/4 0.09/4 0.0225
7IDDI Company Cash Flow Diagram
8In Class Problem
- What is the total amount of interest paid over
the 4 years of the loan period?
9Analyzing a Project
- Present Worth
- Internal Rate of Return (IRR)
10Analyzing a Project
- Present Worth
- Value at time zero (e.g., today) that is
equivalent to the cash flow series of a proposed
project or alternative - Easy to understand and use
- Matches our intuitive understanding of money
- PW gt 0 ? Desired Value!!
- PW 0 ? Economic indifference
- PW lt 0 ? Should be avoided, if possible
11Present Worth
- Useful when
- Setting a price to buy or sell a project
- Evaluating an investment or project when the
price to invest or the first cost is given - Calculating and equivalent value for an irregular
series of cash flows
12Present Worth
- Assumptions
- 1. Cash flows occur at the end of the period,
except for first costs and pre-payments - i.e., insurance, leases
- 2. Cash flows are known, certain values
- 3. The interest rate (i) is given
- 4. The problems horizon (N) is given
13Voice Recognition for Construction Inspection
- A large construction job will take 4 years to
complete. The costs of the many required
inspections can be reduced by purchasing a voice
recognition system to act as the front end for
a word processor. Then inspectors can be trained
to record their comments and revise the written
output of an automated form. - The firm interest rate is 12. What is the PW
of the cash flows for the voice recognition
system? The first cost for purchase and training
is 100K. Savings are estimated at 30K, 40K,
65K, and 35K for the 4 years of construction.
14VR Project Cash Flow Diagram
15VR Project Solution
PW -100K 30K(P/F,0.12,1) 40K(P/F,0.12,2)
65K(P/F,0.12,3) 35K(P/F,0.12,4)
PW 27,182
16In Class Problem
- Find the present worth of the 4 different years
using formulas in Excel. - Find the present worth of the 4 different years
using Excel financial functions.
17Analyzing a Project
- Internal Rate of Return (IRR)
- Interest rate that makes the PW equal zero
- Firms, government agencies and individuals must
compute the cost of financial projects - Solution
- Calculate the IRR for a project or for a
financial source - Then, compare the resulting IRR with an interest
rate for the time value of money
18Golden Valley Manufacturing
- By replacing its assembly line conveyor with an
asynchronous conveyor, Golden Valley
Manufacturing will save 50,000 per year in
rework, inspection, and labor costs. The
asynchronous conveyor will cost 275,000, and its
life is 15 years (when its salvage value equals
0). If Golden Valley uses a 10 interest rate,
should the asynchronous conveyor be installed?
19GVM Cash Flow Diagram
20GVM Solution
- PW 0 -275K 50K(P/A, IRR, 15)
- (P/A, IRR, 15) 275K / 50K 5.500
- From tables
- (P/A, 0.16, 15) 5.575
- (P/A, IRR, 15) 5.500
- (P/A, 0.17, 15) 5.324
21In Class Problem 4
- Use Excel to find the value of IRR of return
using interpolation
22GVM Solution
16.30 gt 10 GOOD INVESTMENT!!
23In Class Problem 5
24Internal Rate of Return
PW
Present Worth PW
0
-PW
i
Interest rate i