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Economic Analysis in Transportation Systems. Tapan K. Datta, Ph.D., P.E. CE 7640: Fall 2002. Chapter 5 Interest and Vestcharge. Economic cost of owning or using one product as compared to other alternatives can be assessed correctly if a factor for cost of money or lost opportunity is used. - PowerPoint PPT Presentation
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Economic Analysis in Transportation Systems
Tapan K. Datta, Ph.D., P.E.
CE 7640: Fall 2002
Chapter 5Interest and Vestcharge
Economic cost of owning or using one product as compared to other alternatives can be assessed correctly if a factor for cost of money or lost opportunity is used.
Concept of Interest Interest: Money paid by borrower for
the money loaned to him or herRate of Interest: Rate based on
amount of loan per unit time
Risk and how lenders often vary rates based on credit risk
Return: Monetary return received from operating an enterprise Net profit received
Rate of Return: Quotient of the net return for one year divided by invested capital.
Dividends: Stock type ventures pay Dividends to stockholders
Vestcharge: Charge for the use of money invested in physical assets it is in lieu of cash returns Vestcharge is generally less than the
interest ratePrime Rate: The best rate the banks
give to their best customers.
Vestcharge
Interest: may be considered as monetary rent on borrowed money.
Return: Is expected monetary compensation for investing money.
Vestcharge: means an economic charge against the investment.
Concept of Vestcharge
In economic analysis the word Vestcharge is used exactly as the terms:
1. Discount Factor2. Minimum Attractive Rate of Return
(MARR)3. Minimum Attractive Discount Rate
Interest and Vestcharge
Why use Vestcharge?
1. Money invested in durable and long lasting goods should pay the penalty for failure to be consumed immediately.
2. The investor, the public has alternative uses of the highway tax money for dividend or profit returning investment
3. The risk of all uses of money varies, so there is a need of a factor in the analysis, which provides a measure of alternative risks.
4. The disbursement dates vary for a public work project, so a discount factor is needed to bring them all on similar timelines.
Why use Vestcharge?
Method of Calculating MARR-Winfrey Budget = $ 28 M
Budget allows only up to project No. 40 to be built. The lowest rate of return is 11%.
Project No.
Estimated Cost
Cumulative Costs
Probable Rate of Return
1 540 K 540 K 22% 2 1,610 K 2,150 K 22% 3 954K 3,104 K 21% : : : :
38 3,610 K 26,936 K 12% 39 200 K 27,136 K 12% 40 805 K 27,941 K 11% 41 512 K 28,453 K 11% : : : :
92 726 K 65,762 K 7%
Winfrey’s Process (MARR)1. Estimate cost of project2. Estimate project benefits3. Estimate Rate of Return for each project4. Rank order projects from highest to lowest5. Cut projects as soon as it reaches the
budget amount.6. The lowest Rate of Return is called the
MARR and can be used as the vestcharge rate.
Alternative Method
1. Calculate Rate of Return observed in similar projects.
2. Rank order them highest to lowest3. Use lowest observed Rate of Return
(ROR) as vestcharge.
Example ProblemBasic Data Project A Project B Project C
Construction Cost
$1,000,000 $1,250,000 $1,100,000
Annual O&M Costs
$150,000 $145,000 $140,000
Annual VMT 20,800,000 19,600,000 21,400,000
Total User Cost per Vehicle Mile
$0.335 $0.330 $0.340
Existing Total User Cost per Vehicle Mile = $0.35. Analysis Period = 20 years, Terminal Value = 0.What is the MARR?
Costs and BenefitsCosts:• Initial Investment (construction costs)• Annual Operation and Maintenance (O&M) costs
Benefit:• Reduction in Road User cost For example, for Project A, the benefit is (0.350-0.335)*(20,800,000)
= $312,000 per year.
Steps to Follow1. Set up the EUAC, PW or FW equation for each
of the alternatives equal to zero.
Project A:EUAC = - 1,000,000*(CR)i
n=20 - 150,000 - 312,000 = 0Project B:EUAC = - 1,250,000*(CR)i
n=20 - 145,000 + 392,000 = 0Project C:EUAC = - 1,100,000*(CR)i
n=20 - 140,000 + 214,000 = 0
Initial Investment Annual Costs
Analysis Period = n
Annual Benefits
Steps to Follow2. Try with an interest rate, say, i = 15%.
For Project A:EUAC = - 1,000,000*(CR)i=15
n=20 - 150,000 + 312,000
= - 1,000,000*0.15976 - 150,000 + 312,000
= + $2,240For i = 16%, EUAC = -$6,670 The value of i lies somewhere between 15% and 16%.
Steps to Follow3. Use linear interpolation to find the exact
value of the interest rate. + $2,240
- $6,670
A B15% 16%
X
(0.16 – X)/6670 = (X – 0.15)/2240By solving the equation, we find X = 0.1525 Rate of Return = 15.25%.
15.25%
Steps to Follow4. Similarly, find the Rate of Return for other
projects.In this problem, RORA = 15.25%
RORB = 19.16%
RORC = 5.68%
5. Rank the projects based on highest to lowest Rate of Return.
6. Conclusion:The Minimum Attractive Rate of Return (MARR) is 5.68%.
Chapter 6: Compound Interest Equations
P = a present sum of money; a single payment; the present worth of an annuity or a single sum.
F = a sum at a future date, ‘n’ interest payment periods from the present. This is equivalent to P with compound interest rate “i” over ‘n’ periods
Chapter 6: Compound Interest Equations
P
Fn-1 n1 2 3
F = P(1 + i)n
Compound the amount at the end of ‘n’ periods
CA = (1 + i)n
Compound Interest Equations
Fn-1 n1 2 3
Present worth of single
sum to be withdrawn at
the end of “n” periods in the
future. 1
1
(1+i)n
P
(1+i)nP = F
PW =
Compound amount at the end of “n” periods to which a series of paymentsof ‘n’ uniform depositswill accumulate.
(1+i)n –1
i
(1+i)n –1
i
Compound Interest EquationsF
n-1 n0 1 2 3
A
F = A
SCA =
Sinking Fund i
(1+i)n - 1
SF = i (1+i)n - 1
A = End of Period annuity amortization paymenti = interest raten = number of interest periods
Compound Interest Equations
Fn-1 n 0 1 2 3
A
A = F
Compound Interest Equations
n-1 n 0 1 2 3
A
P
Present worth of a series of ‘n’ uniform period end withdrawals (1 + i)n - 1
SPW = (1 + i)n - 1 i (1 + i)n
P = A i (1 + i)n
Capital Recovery with interestamenity which will return ‘n’ period end uniform receipts
i(1+i)n
(1+i)n-1
i(1+i)n (1+i)n-1
n-1 n 0 1 2 3
A
P
Compound Interest Equations
A = P
CR =
Examples $ 100 kept at 6% compound interest for 7 years,
what will be the amount at end of 7 years?F = P(1+i)n =
F = 100(CAF) in
= 100(CAF)i =6
= 100 (1.5036)F = $ 150.36
What will be the amount after 12 years, using 6% interest
F = 100 (2.0121) = $ 201.21
n =7(CAF)i =6 = 1.5036
n =7
From Compound Interest Table
(CAF)i =6 = 2.0121n =12
Examples• $ 100 due in 7 years. What is its present worth?
P = 100
P = 100 (.6651)
P = $ 66.51
•What is the present worth of a $100 bond after 12 years?
P = 100
= 100 (0.4970) = $ 49.70
*(PWF)i n
(PWF)i =6 = 0.6651n =7
*(PWF)i =6n =12
Examples Annual savings of $ 100 for 7 years at 6%?
What will be annual deposit that will accumulate to $ 100 at 6% after 7 years?A = 100 (SF)6
7 = 100 (0.11914) A = $ 11.91
F = 100
= 100 (8.394) = $ 839.38
(SCA)i =6n =7
ExamplesWhat is the present worth at 6% interest
of $ 100 receivable each year for 7 years?P = 100 (SPW)6
7 = 100 (5.582)
P = $ 558.20
Examples
A $ 100 initial fund that earns interest at 6% on its balance could be exactly paid out in 7 yearsA = 100 (CR)6%
7
= 100 (0.17913)A = $ 17.91
Chapter 7- Methods of Economic Analysis
Economic analysis is performed for Project SelectionProject EvaluationCost Effectiveness Analysis is
performed for $ Cost/Fatality Savings or$ Cost/Injury Savings
Traditional Analysis Methods1. Equivalent Uniform Annual Cost Method
(EUAC)2. Present Worth of Costs Method (PWOC)3. Equivalent Uniform Annual Net Return
Method (EUANR)4. Net Present Value Method (NPV)5. Benefit to Cost Ratio (B/C)6. Rate of Return Method (ROR)* Incremental Benefit-Cost Ratio* Cost/Effectiveness Method
EUAC = -I(CR)in + T(SF)i
n - K - UI = Initial costT = Terminal valueK = Total uniform annual costsU = Uniform annual road user costs
PWOC = -I + T(PW)in - K(SPW)i
n - U(SPW)in
EUANR = -I(CR)in + T(SF)i
n - K + RR = Uniform annual gross benefit
Economic Analysis
Economic Analysis
B/C = PWOB/PWOCor
B/C = EUAB/EUAC
Economic Analysis ProblemHighwayProject Industrial Plant
A1Base
A2Prop
B3Base
B4Prop
I Initial Investment 140K 160K 120K 200KT Terminal Value 40K 50K 10K 18KK Total Administ.
Oper. & Maint. 7K 8K 500K 485.5KU Uniform Annual
Road user Cost 74K 70KUT
Uniform RoadUser Taxes 15K 14.5K
I Vestcharge 8% 8% 8% 8%n Analysis Period 10 10 10 10
These are two mutually exclusive pairs with equal levels of service
EUAC MethodEUACA1= -140,000 * CR i=8
n=10 + 40,000 * SF i=8
n=10 - 7,000 - 74,000= - 140,000 *(0.149029) + 40,000* (0.0629) - 81,000
EUACA1= - 99,103
EUACA2= -160,000 * CR i=8 n=10 + 50,000 *
SF i=8 n=10 -8,000 - 70,000
= -160,000*(0.149029) + 50,000* (0.069029) -78,000
EUACA2= - 98,394
EUACA1 - EUACA2 = +99,103 - 98,394
= 709
A2 Alternative has $709 less annual costs
A2 is better than A1.
EUAC Method
PWOC Method
PWOCA1 = - 140,000 + 40,000 PW810 -
7000 * SPW810 - 74,000 *
SPW810
= -140,000 + 40,000*(0.46319) - 7,000 * (6.710081) -
74,000* (6.710081)PWOCA1 = -664,989
PWOC MethodPWOCA2 = - 160,000 + 50,000 (PW8
10) - 8000 * (SPW810) -
70,000 * (SPW810)
PWOCA2 = -660,227
PWOCA2 - PWOCA1 = -660,227 - (-664,989)
= $4,762The cost of A1 Alternative is $4762
more than A2
A2 is better.
Reduction in Road user costs is equivalent to cash income.
EUANRA2 = -(Ip - IB) (CR810) + (Tp - TB) *
(SF810) - (Up - UB) -(Kp - KB)
EUANRA2 = - (160K - 140K) (0.149029) +
(50K- 40K) (0.069029) – (70 K - 74 K) - (8K- 7K) = $709A2 is better.
Equivalent Uniform Annual Net Return Method
Net Present Value Method
NPVA2 = -(Ip - IB) + (Tp - TB) * (PW810) –
[(Up - UB) + (Kp - KB)] * (SPW810)
= -20 K + 10K (0.463193)- (-K+1K)*(6.71008)NPVA2= 4,762
A2 is better.
B/C Ratio
B/C = EUAB/EUACor
B/C = PWOB/PWOCBenefit- Reduction in road user costs and reduction in
annual expenses together is the benefit- Initial cost and Terminal value should be
considered as costs
B/C Ratio (Continued)EUAB = - (UP - UB) - (KP - KB) = - (70,000 - 74,000) - (8,000 - 7,000)EUAB = 3000EUAC = -(Ip - IB)*(CR8
10) + (Tp - TB)*(SF810)
= -(160K-140K)(0.149029) + (50K- 40K)(.069029)
EUAC = - 2291B/C = + 3000/-2291 = 1.311.31 is considered positive since the numerator is
positive
Rate of Return MethodAssume, i = 10% on the basis of present worth
0 = B – C 0 = -(160,000 – 140,000) + (50,000 –
40,000)*(PW1010) – (70,000 – 74,000)
(SPW1010) - (8,000 – 7,000) (SPW10
10)
= -20,000 + 10,000*(0.385543) + 4,000(6.144567) – 1,000 (6.144567) = $2,288
Rate of Return Method (Continued)
Second Iteration: Assume i = 12%0 = -20,000 + 10,000(0.321973) + 4,000(5.65023) – 1,000 (5.65023) = 171Rate of Return A2= 10 + 2*[(2288 -171)/2288]Rate of Return = 12.15%
Small Improvement Project
All/ Year
Before After Reduction
PDO Injury PDO Injury PDO Injury
7 Mile & Ryan 49 18.67 29 6.5 20 12.17
7 Mile & John R 35 16.67 24.57 4.57 10.43 12.1
Hubbell &
Puritan21.67 13.33 12.83 7.01 8.84 6.32
Total 39.27 30.59
National Safety Council (NSC)
For 1996Injury = $ 34,100PDO = $ 6,400
Reduction in road user costs related to traffic crashes
= 39.27(6,400) + 30.59 (34,100) = 251,328 + 1043119
= $ 1,294,447
Project Cost Location 1 = $ 34,100
Location 2 = $ 35,200 Location 3 = $ 29,400
Total = $ 98,700Annual Operation and Maintenance cost
= $ 1000 per Intersection per yearAssume i = 7% and n = 15 years
PWOC and PWOB
PWOC = -98,700 – (1000)*(SPW715)
= -98,700 – 1000 (9.1079) = -107,807.9
PWOB = 1,294,447*(SPW715)
= 11,789,693B/C = 11 ,789,693/107,807.9
B/C = 109:1
Incremental B/C RatioAssume that you want to pick 1 alternative amongst
5 solutions.Say, the alternatives were ranked in increasing cost.
B C B-CA1 10 2 8A2 20 14 6A3 50 25 25A4 80 30 50A5 90 45 45
The method involves examining the ratio Bi+1 - Bi
Ci+1 - Ci
If it is > 1, then alt. Ai+1 is accepted and compared with Bi+2
If it is < 1, then alt. Ai is accepted and compared with Ai+2
Comparison Incremental Benefits
Incremental Costs
Inc. Benefits/ Inc. Costs
Decision in Favor of
A1 to A0 10 2 5 A1 A2 to A1 10 12 5/6 A1
A3 to A1 40 23 40/23 A3
A4 to A3 30 5 6 A4
A5 to A4 10 15 2/3 A4
You select A4
If benefits and costs are in different units, this method cannot be applied.
Cost/Effectiveness MethodThis is a comparison of cost to achievement of
a given unit of effect. This method gives the cost of alternatives per
unit benefitSometimes it is difficult to assign dollar values
to benefit items, such as accident costs driver comfort travel time
You need data for initial cost annual cost terminal value service life interest rate annual benefits
You can not consider more than one benefit items in the comparison.
Steps to Follow1. Estimate the following:
initial costs (I) annual costs (K) terminal value (T) service life (n)
2. Select interest rate3. Calculate EUAC of each alternative4. Select units of effectiveness5. Estimate average annual benefit in the desired
units of effectiveness6. Calculate C/E ratios
Example ProblemThree similar traffic safety projects involve the following data:
Compare the projects using Cost-Effectiveness Method.
Project Initial Cost
Annual O&M
Terminal Value
Annual Average Crash Reductions
X 1,500K 100K 40K Total=30, Injury=6, Fatal=1
Y 1,400K 120K 30K Total=25, Injury=9,Fatal=0.5
Z 1,300K 130K 25K Total=20, Injury=5,Fatal=1.2
Example ProblemAssume a service life of 5 years and a vestcharge rate of 8%.
Costs as per NSC, 1999 data: $6,400/PDO crash $35,300/Injury crash $970,000/Fatal crash
Calculations for Project X We have:I = $1,500,000K = $100,000T = $40,000N = 5 yearsVestcharge rate, i = 8%EUAC = -1,500,000(CR)i=8
n=5 – 100,000 + 40,000(SF) i=8n=5
= -1,500,000*(0.25046) – 100,000 +40,000*(0.17046) = - $468,872
Calculations for Project XLet us select the Units of Effectiveness as
follows: Number of total crashes prevented Number of injury crashes prevented Number of fatal crashes prevented
Calculations for Project XBenefit, BT = 30 total crashes prevented
Benefit, BI = 9 injury crashes prevented
Benefit, BF = 1 fatal crash prevented
Calculations for Project XC/E for total crashes = EUAC/BT = 468,872/30 = $15,629 per total crash prevented, OrC/E for injury crashes = EUAC/BI = 468,872/6 = $78,145 per injury crash prevented, OrC/E for fatal crashes = EUAC/BF = 468,872/1 = $468,872 per fatal crash prevented, OrSimilarly, the C/E Calculations are done for other two projects and summarized
as follows.
Summary of C/E Values
Project Per Total Crash Prevented
Per Injury Crash Prevented
Per Fatal Crash Prevented
X $15,629 $78,145 $468,872
Y $18,621 $51,726 $931,060
Z $22,567 $90,267 $376,114