Transcript MIE 754 Manufacturing & Engineering Economics

MIE 754 - Class #5
Manufacturing & Engineering
Economics
• Concerns and Questions
•Quick Recap of Previous Class
• Today’s Focus:
– Chap 3 Comparing Alternatives with
Different Useful Lives
– Chap 4 Rate of Return Methods
• Hmwk #3 Due in 1 Week:
– Chap 2 - 3, 4, 5, 6, 11, 12, 15,
16, 24, 30, 37
Concerns and Questions?
Quick Recap of Previous Class
 Effective
Interest
 Comparison
of Alternatives
 Procedure
 Defining
 Useful
Investment Alternatives
Life versus Study Period
Useful Life versus Study Period
 Comparison
must be over the same
study period for ALL alternatives!
 Useful
Lives = Study Period
 Useful Lives are Different Among
Alters.
UL < SP
UL > SP
Same Study Period Required!!
 Use
either:
 Repeatability Assumption
 Cotermination Assumption
Example
A
Capital Investment -$3,500
Annual Revenues
1,900
Annual Expenses
-645
Useful Life
4
Market Value
0
B
-$5,000
2,500
-1,020
6
0
Several In-class Examples
The Effect of Compounding

Benjamin Franklin, according to the American
Bankers Association, left $5,000 to the
residents of Boston in 1791, with the
understanding that it should be allowed to
accumulate for a hundred years. By 1891 the
$5,000 had grown to $322,000. A school was
built, and $92,000 was set aside for a second
hundred years of growth. In 1960, this second
century fund had reached $1,400,000. As
Franklin put it, in anticipation: "Money makes
money and the money that money makes
makes more money."
The “Ben Franklin” Problem Solution
Given: P=$5,000 N=100
F = $322,000
Find: i'%
F = P(F|P, i'%, 100)
$322,000 = $5000(F|P, i'%, 100)
therefore, (F|P, i'%, 100) = 64.4
From tables, (F|P, 4%, 100) = 50.5049
(F|P, 5%, 100) = 131.501
F = P(1+i')N therefore, 322 = 5(1+i')100
64.4 = (1+i')100 or (1+i') = 1.0425 so that i' =
Chapter 4 - Rate of Return Methods
 Compare
against minimum standard
of desirability - minimum attractive
rate of return (MARR)
 Internal
Rate of Return (IRR) Method
Solves for the interest rate that
equates the equivalent worth of a
project's cash outflows (expenditures) to
the equivalent worth of cash inflows
(receipts or savings).
IRR is Like a “Break-Even” Problem
 Find
 FW
 Can
i' such that
FW(neg, i') = FW(pos, i')
= 0 = FW(pos, i') - FW(neg, i')
use any one of the EW methods for
IRR:
 PW(i' %) = 0
 AW(i' %) = 0
 FW(i' %) = 0
 Why?
Can Solve for i' by:




Trial and error
Linear interpolation
An equation solver
Computer program
 You
will need to know how to interpolate
in this course!
Evaluating Project with IRR
 Compare
IRR to MARR to determine
whether or not the project is acceptable
with respect to profitability.
 IRR
= i'  MARR
acceptable
 IRR
= i' < MARR
unacceptable
(reject)
 Difficulties
with the IRR Method
 The IRR Method assumes that recovered
funds are reinvested at the IRR rather than
the MARR
 Possible multiple IRRs
 Why
should you learn the RR
Methods?
 The majority of companies favor the RR
methods for evaluating investment projects
Example Problem Using IRR
Cost/Revenue Estimates
Initial Investment:
Annual Revenues:
Annual Operating Costs:
Salvage Value @ EOY 5:
Study Period:
MARR
$50,000
$20,000
$2,500
$10,000
5 years
20%
Solution using IRR method
Find i'% such that the PW(i'%) = 0
0=-50,000+17,500(P|A, i'%,5)+10,000(P|F,
i'%,5)
PW(20%) = 6354.50 tells us that i' > 20%
PW(25%) = 339.75 > 0, tells us that i'% > 25%
PW(30%) = -4,684.24 < 0, tells us that i'% <
30%
25% < i' < 30%
Use linear interpolation to estimate i'% or
Linear Interpolation example
i%
25
i'
30
PW
339.75
0
-4684.24
i' =  25.3% > MARR, accept
Comparing Mutually Exclusive
Alternatives (MEAs) with RR Methods
 Fundamental
Purpose of Capital
Investment:
• Obtain at least the MARR for every dollar
invested.
 Basic
Rule:
• Spend the least amount of capital possible
unless the extra capital can be justified by
the extra savings or benefits.
(i.e., any increment of capital spent above
the minimum must be able to pay its own
Why not select the investment opportunity
that maximizes IRR? See example below
A
B
B-A()
-$100 -$10,000 -
Investment
$9,900
Lump-Sum $1,000 $15,000 $14,000
Receipt
Next Year
IRR
900%
50% 41.4%
If MARR = 20%, would you rather have A
or B if comparable risk is involved?
Comparing MEAs - using the IRR
method cont'd.
If MARR = 20%, PWA = $733 and PWB =
$2,500
* Never simply maximize the IRR.
* Never compare the IRR to anything except
the MARR.
IRRA->B:
PWA->B = 0 = -9,900 + 14,000(P|F, i'%, 1)
9,900/14,000 = (P|F, i'%, 1)
i' = 41.4% > MARR
Rate of Return Methods for Comparing
Alternatives MUST use an Incremental
Approach!
Step 1. Rank order alternatives from
least to greatest initial investment.
Step 2. Compare current feasible
alternative with next challenger in the
list
Step 3. Compute RR (IRR or ERR) and
compare with MARR. If RR < Marr
choose the least initial investment
alternative. If RR  MARR choose the
greater initial investment alternative
Step 4. Remove rejected alternative
Example Problem: Given three MEAs
and MARR = 15%
1
2
3
Investment (FC) -28,000 -16,000 23,500
Net Cash Flow/yr 5,500
3,300
4,800
Salvage Value
1,500
0
500
Useful Life
10 yrs 10 yrs 10 yrs
Study Period
10 yrs 10 yrs 10 yrs
Example Problem Cont.
Step 1. DN -> 2 -> 3 -> 1
Step 2. Compare DN -> 2
 cash flows
 Investment
-16,000 - 0 = -16,000
 Annual Receipts
3,300 - 0 = 3,300
 Salvage Value
0-0=
0
Compute  IRRDN->2
Step 3. Since i' > MARR, keep alt. 2
(higher FC) as current best alternative.
Drop DN from further consideration.
Step 4. Next comparison: 2 -> 3
 Investment -23,500 - (-16,000) =
7,500
 Annual Receipts 4,800 - 3,300 =
1,500
 Salvage Value
500 - 0 =
500
Computing  IRR2->3
PW(i') = 0
0= -7,500 + 1,500(P|A, i'%, 10) +
500(P|F, i'%, 10)
i'2->3  15.5%
Since i' > MARR, keep Alt. 3 (higher
FC) as current best alternative. Drop
Alt. 2 from further consideration.
Next comparison: 3 -> 1  cash flows
 Investment -28,000 - (-23,500) =
4,500
 Annual Receipts 5,500 - 4,800 =
700
 Salvage Value
1,500 - 500 =
1,000
Compute  IRR3->1
PW(i') = 0
0= -4,500 + 700(P|A, i'%, 10) +
Since i' < MARR, keep alt. 3 (lower FC)
as current best alternative. Drop alt. 1
from further consideration.
Step 5. All alternatives have been
considered.
Recommend alternative 3 for investment.
Graphical Interpretation of Example