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