Question

Background This case study compares benefit/cost analysis and cost effectiveness analysis on the same information about highway lighting and its role in accident reduction. Poor highway lighting may be one reason that proportionately more traffic accidents occur at night. Traffic accidents are categorized into six types by severity and value. For example, an accident with a fatality is valued at approximately $4 million, while an accident in which there is property damage (to the car and contents) is valued at $6000. One method by which the impact of lighting is measured compares day and night accident rates for lighted and unlighted highway sections with similar characteristics. Observed reductions in accidents seemingly caused by too low lighting can be translated into either monetary estimates of the benefits B of lighting or used as the effectiveness measure E of lighting.

Information

Freeway accident data were collected in a 5-year study. The property damage category is commonly the largest based on the accident rate. The number of accidents recorded on a section of highway is presented here

	Number of Accident Recorded
	Unlighted		Lighted
Accident Type	Day	Night	Day	Night
Property damage	379	199	2069	836

The ratios of night to day accidents involving property damage for the unlighted and lighted freeway sections are 199/379 = 0.525 and 839/2069 = 0.406, respectively. These results indicate that the lighting was beneficial. To quantify the benefit, the accident rate ratio from the unlighted section will be applied to the lighted section. This will yield the number of accidents that were prevented. Thus, there would have been (2069)(0.525) = 1086 accidents instead of 839 if there had not been lights on the freeway. This is a difference of 247 accidents. At a cost of $6000 per accident, this results in a net annual benefit of

B = (247)($6000) = $1,482,000

For an effectiveness measure of number of accidents prevented, this results in E = 247. To determine the cost of the lighting, it will be assumed that the light poles are center poles 67 meters apart with 2 bulbs each. The bulb size is 400 watts, and the installation cost is $3500 per pole. Since these data were collected over 87.8 kilometers of lighted freeway, the installed cost of the lighting is (with number of poles rounded off):

Installation cost = $3500 (87.8 / 0.067) = 3500 (1310) = $4,585,000

There are a total of 87.8/0.067_1310 poles, and electricity costs $0.10 per kWh. Therefore, the annual power cost is

Annual power cost = 1310 poles (2 bulbs/pole)(0.4 kilowatt/bulb) x (12 hours/day)(365 days/year) x ($0.10/kilowatt-hour) = $459,024 per year

For an effectiveness measure of number of accidents prevented, this results in E = 247. To determine the cost of the lighting, it will be assumed that the light poles are center poles 67 meters apart with 2 bulbs each. The bulb size is 400 watts, and the installation cost is $3500 per pole. Since these data were collected over 87.8 kilometers of lighted freeway, the installed cost of the lighting is (with number of poles rounded off):

Installation cost = $3500 (87.8 / 0.067) = 3500 (1310) = $4,585,000

There are a total of 87.8/0.067_1310 poles, and electricity costs $0.10 per kWh. Therefore, the annual power cost is

Annual power cost = 1310 poles (2 bulbs/pole)(0.4 kilowatt/bulb) x (12 hours/day)(365 days/year) x ($0.10/kilowatt-hour) = $459,024 per year

The data were collected over a 5-year period. Therefore, the annualized cost C at i = 6% per year is

Total annual cost = $4,585,000( A/P ,6%,5) + 459,024 = $1,547,503

If a benefit/cost analysis is the basis for a decision on additional lighting, the B/C ratio is B/C = 1,482,000 / 1,547,503 = 0.96

The data were collected over a 5-year period. Therefore, the annualized cost C at i = 6% per year is

Total annual cost = $4,585,000( A/P ,6%,5) + 459,024 = $1,547,503

If a benefit/cost analysis is the basis for a decision on additional lighting, the B/C ratio is B/C = 1,482,000 / 1,547,503 = 0.96

Since B/C < 1.0, the lighting is not justified. Consideration of other categories of accidents is necessary to obtain a better basis for decisions. If a cost-effectiveness analysis (CEA) is applied, due to a judgment that the monetary estimates for lighting’s benefit is not accurate, the C/E ratio is

C/E = 1,547,503 / 247 = 6265

This can serve as a base ratio for comparison when an incremental CEA is performed for additional accident reduction proposals. These preliminary B/C and C/E analyses prompted the development of four lighting options:

W) Implement the plan as detailed above; light poles every 67 meters at a cost of $3500 per pole.

X) Install poles at twice the distance apart (134 meters). This is estimated to cause the accident prevention benefit to decrease by 40%.

Y) Install cheaper poles and surrounding safety guards, plus slightly lowered lumen bulbs (350 watts) at a cost of $2500 per pole; place the poles 67 meters apart. This is estimated to reduce the benefit by 25%.

Z) Install cheaper equipment for $2500 per pole with 350-watt lightbulbs and place them 134 meters apart. This plan is estimated to reduce the accident prevention measure by 50% from 247 to 124.  

Case Study Exercises Determine if a definitive decision on lighting can be determined by doing the following:

1. Use a benefit/cost analysis to compare the four alternatives to determine if any are economically justified.

2. Use a cost-effectiveness analysis to compare the four alternatives. From an understanding viewpoint, consider the following:

3. How many property-damage accidents could be prevented on the unlighted portion if it were lighted?

4. What would the lighted, night-to-day accident ratio have to be to make alternative Z economically justified by the B/C ratio?

5. Discuss the analysis approaches of B/C and C/E. Does one seem more appropriate in this type of situation than the other? Why? Can you think of other bases that might be better for decisions for public projects such as this one

please answer just 1,2 and 5... thank you very much

Answer 1

A)

For alternative W

from the data given in the paragraph it is clear that

the ratio of night to day accident involving property damage for the unlighted freeway section is

199/379= 0.525

and for lighted section is 839/2069 = 0.406

so if there would have been no light in the lighted section then no of accidents would have been 2069*0.525= 1086

but there are actuaally 839 accidents so

Effective measure of number of accidents= 1086-839= 247

now

benefit from saving one accident is 6000

so benefit from saving 247 accident is

247*6000

= 1,482,000

Now

two types of costs are invlved in the process of lighting

i) installation cost and ii). annual power cost

to calculate installation cost you need to know rate of installing per pole and total no of poles installed

since total distance is 87.8 Km and each pole is at distance of 67 m

so total number of pole is 87.8*1000/67

= 1310

and installing each pole cost 3500

so total installation cost

=3500*1310

= 4,585,000

now Annual power cost calaculation

we have to calculate total unit consumed by all the bulbs in all the poles

so total no of bulbs= no of poles* no of bulbes per pole

= 1310*2= 2620 bulbs

power of each bulb is 400W= 0.4KW

so enrgy consumed power*time

here time is no of hours in 1 year

= 2620*0.4 *12*365 kW yearly

so rate = enrgy consumed* rate per unit

= 0.1*2620*0.4*12*365

=459024

Now it is to be kept in mind that this datas are for five year but installation cost is only at the beggining of the year

so

annualised cost C at 6% interest rate is given by

so EAC(Effective annual cosT)

we have to apply the method for this

Calculate v:

v  =	1
	1 + Cost of Capital

v  =	1
	1 + 0.06

v  =	1
	1.06

v = 0.9434

Calculate Discount Factor for installation cost

a5|0.06  =	(1 - vAsset Lifetime)
	Cost of Capital

a5|0.06  =	(1 - 0.94345)
	0.06

a5|0.06  =	(1 - 0.747273118149)
	0.06

a5|0.06  =	0.252726881851
	0.06

a_5|0.06 = 4.2121
In Microsoft Excel, this function can be written in a cell as =PV(0.06,5,-1)

Calculate Discounted Investment for installation cost

Discounted Investment for Item 1  =	Investment Cost
	a5|0.06

Discounted Investment for Item 1  =	$4,585,000.00
	4.2121

Discounted Investment for Item 1 = $1,088,530.66

Calculate EAC for installation cost
EAC1 = Discounted Investment + Maintenance Cost
EAC1 = $1,088,479
EAC1 = $1,088,479

thus total annual cost

= $1,088,479+459024

= 1,547,503

so B/C= 1,482,000/1,547,503 =0.96

Since B/C<1 the lightning is not justified as cost is greater than the benefit

For ALTERNATIVE X

since benefit decrease by 40%

original benefit was 1482,000

so now benefit= 0.6*1,482,000

=889,200

Now

two types of costs are invlved in the process of lighting

i) installation cost and ii). annual power cost

to calculate installation cost you need to know rate of installing per pole and total no of poles installed

since total distance is 87.8 Km and each pole is at distance of 134 m

so total number of pole is 87.8*1000/134

= 655

and installing each pole cost 3500

so total installation cost

=3500*655

= 2,292,500

now Annual power cost calaculation

we have to calculate total unit consumed by all the bulbs in all the poles

so total no of bulbs= no of poles* no of bulbes per pole

= 655*2= 1310 bulbs

power of each bulb is 400W= 0.4KW

so enrgy consumed power*time

here time is no of hours in 1 year

= 1310*0.4 *12*365 kW yearly

so rate = energy consumed* rate per unit

= 0.1*13100*0.4*12*365

=229,512

Now it is to be kept in mind that this datas are for five year but installation cost is only at the beggining of the year

so

annualised cost C at 6% interest rate is given by

so EAC(Effective annual cosT)

we have to apply the method for this

Calculate v:

v  =	1
	1 + Cost of Capital

v  =	1
	1 + 0.06

v  =	1
	1.06

v = 0.9434

Calculate Discount Factor for Item 1:

a5|0.06  =	(1 - vAsset Lifetime)
	Cost of Capital

a5|0.06  =	(1 - 0.94345)
	0.06

a5|0.06  =	(1 - 0.747273118149)
	0.06

a5|0.06  =	0.252726881851
	0.06

a_5|0.06 = 4.2121
In Microsoft Excel, this function can be written in a cell as =PV(0.06,5,-1)

Calculate Discounted Investment for Item 1:

Discounted Investment for Item 1  =	Investment Cost
	a5|0.06

Discounted Investment for Item 1  =	$2,292,500.00
	4.2121

Discounted Investment for Item 1 = $544,265.33

Calculate EAC for Item 1
EAC1 = Discounted Investment + Maintenance Cost
EAC1 = $544,265.33 + $229,512.00
EAC1 = $773,777.33

consider item 1 as lighning

so B/C= 889,200/773,777 =1.149

Since B/C>1 the lightning is economically justified as cost is less than the benefit.

For ALTERNATIVE Y

since benefit decrease by 25%

original benefit was 1482,000

so now benefit= 0.75*1,482,000

= 1,111,500

Now

two types of costs are invlved in the process of lighting

i) installation cost and ii). annual power cost

to calculate installation cost you need to know rate of installing per pole and total no of poles installed

since total distance is 87.8 Km and each pole is at distance of 67 m

so total number of pole is 87.8*1000/67

= 1310

and installing each pole cost 2500

so total installation cost

=2500*1310

= 3,275,000

now Annual power cost calaculation

we have to calculate total unit consumed by all the bulbs in all the poles

so total no of bulbs= no of poles* no of bulbes per pole

= 1310*2= 2620 bulbs

power of each bulb is 350W= 0.35KW

so enrgy consumed power*time

here time is no of hours in 1 year

= 2620*0.35 *12*365 kW yearly

so rate = energy consumed* rate per unit

= 0.1*2620*0.35 *12*365

=401,646 peryear

Now it is to be kept in mind that this datas are for five year but installation cost is only at the beggining of the year

so

annualised cost C at 6% interest rate is given by

so EAC(Effective annual cosT)

we have to apply the method for this

Calculate v:

v  =	1
	1 + Cost of Capital

v  =	1
	1 + 0.06

v  =	1
	1.06

v = 0.9434

Calculate Discount Factor for Item 1:

a5|0.06  =	(1 - vAsset Lifetime)
	Cost of Capital

a5|0.06  =	(1 - 0.94345)
	0.06

a5|0.06  =	(1 - 0.747273118149)
	0.06

a5|0.06  =	0.252726881851
	0.06

a_5|0.06 = 4.2121
In Microsoft Excel, this function can be written in a cell as =PV(0.06,5,-1)

Calculate Discounted Investment for Item 1:

Discounted Investment for Item 1  =	Investment Cost
	a5|0.06

Discounted Investment for Item 1  =	$3,275,000.00
	4.2121

Discounted Investment for Item 1 = $777,521.90

Calculate EAC for Item 1
EAC1 = Discounted Investment + Maintenance Cost
EAC1 = $777,521.90 + $401.00
EAC1 = $777,923

thus B/C for this is

1,11,500/777,523 = 1.429

so it is economically feasible

as cost is less than the benfit...

Alternative Z

Since accident prevention measure become 124 so total benefit

124*6000

= 744,000

two types of costs are invlved in the process of lighting

i) installation cost and ii). annual power cost

to calculate installation cost you need to know rate of installing per pole and total no of poles installed

since total distance is 87.8 Km and each pole is at distance of 134 m

so total number of pole is 87.8*1000/134

= 655

and installing each pole cost 2500

so total installation cost

=2500*655

= 1,637,500

now Annual power cost calaculation

we have to calculate total unit consumed by all the bulbs in all the poles

so total no of bulbs= no of poles* no of bulbes per pole

= 655*2= 1310 bulbs

power of each bulb is 350W= 0.35KW

so enrgy consumed power*time

here time is no of hours in 1 year

= 1310*0.35 *12*365 kW yearly

so rate = energy consumed* rate per unit

= 0.1*13100*0.35*12*365

=200,823

Now it is to be kept in mind that this datas are for five year but installation cost is only at the beggining of the year

so

annualised cost C at 6% interest rate is given by

so EAC(Effective annual cosT)

we have to apply the method for this

Calculate v:

v  =	1
	1 + Cost of Capital

v  =	1
	1 + 0.06

v  =	1
	1.06

v = 0.9434

Calculate Discount Factor for Item 1:

a5|0.06  =	(1 - vAsset Lifetime)
	Cost of Capital

a5|0.06  =	(1 - 0.94345)
	0.06

a5|0.06  =	(1 - 0.747273118149)
	0.06

a5|0.06  =	0.252726881851
	0.06

a_5|0.06 = 4.2121
In Microsoft Excel, this function can be written in a cell as =PV(0.06,5,-1)

Calculate Discounted Investment for Item 1:

Discounted Investment for Item 1  =	Investment Cost
	a5|0.06

Discounted Investment for Item 1  =	$1,637,500.00
	4.2121

Discounted Investment for Item 1 = $388,760.95

Calculate EAC for Item 1
EAC1 = Discounted Investment + Maintenance Cost
EAC1 = $388,760.95 + $200,823.00
EAC1 = $589,584

so

B/C ratio is 744000/589584 = 1.2619

thus B/C>1

so economically it is justified

Q.2

we have

alternative	Cost	E	C/E
W	1547503	247	6265
X	773771	148	5228
Y	777523	185	4203
Z	589584	124	4755

we have taken cost from part A as we have calculated for each part...

E for W is given

for X it is 60% of this =0.6*247=148

for Y it is 75% of this = 0.75*247 =185

for X it is 124 given

so Y has minimum C/E ratio... thus it is best of these alternatives

also all those alternatives whose C/E ratio is less than the base given in question that one is better as compared to W

Q.3).

from the data given in the paragraph it is clear that

the ratio of night to day accident involving property damage for the unlighted freeway section is

199/379= 0.525

and for lighted section is 839/2069 = 0.406

so if there would have been light in the nolight section then no of accidents would have been 379*0.406= 154

but there are actuaally 199 accidents so

Effective measure of number of accidents= 199-154= 45

so prevented damage

= 45*6000

= 270,000

Q.5.

Benefit/Cost (B/C) Analysis is defined as a systematic process for calculating and comparing benefits and costs of a project for two purposes:

1. To determine if it is a sound investment (justification/feasibility); and
2. To see how it compares with alternate projects (ranking/priority assignment). (National Academies Transportation Research Board (TRB) Economics Committee.)

Benefit/Cost analysis is also commonly referred to as Cost-Benefit Analysis, CBA, Benefit/Cost Analysis, and BCA. The analysis is identical despite the naming differences. Benefit/costs analysis is one type of economic valuation