Benefit / Cost Ratios

A method for analyzing the desirability of public works projects, or any other project where benefits and costs can be quantified.

Benefits (B) - Advantages to the owner (The owner may be the public)

Disbenefits (D) - Disadvantages to the owner when the project under consideration is       
                             implemented.

Costs (C) - Anticipated expenditures for construction, Operation &Maintenance, etc.

B/C = Benefits-Disbenefits
                      Costs
If  B/C > 1.0 then the project is advantageous.
B,C and D must all be in same dollar units --- PW, FAC or FW

Costs are not negative.

Modified B/C =   Benefits-Disbenefits-M&O Costs
                                    Initial investments - Salvage

                        =    Net Benefits
                                 Net Cost

Examples of B/C components:

      Benefits                               Costs                                      Disbenefits                          
Reduced Accidents            Rebuilt highway          Less tow truck and hospital work
Electricity, irrigation            Building dam              Paying for flooded land. Lost use of land
Water,tourism.


Example: Suppose a tunnel is proposed to be built to save travel time and road construction cost around a mountain. In order to save on first cost, a twin tunnel system is proposed with 1/2 capacity in each shaft.
Construction = $3,000,000  --- Shaft 1 -- Now
                       $4,000,000  --- Shaft 2 --Built in 20 years ( N = 20 )

Maintenance (Shaft lining) = $160,000/Half  -- every 10 years

Benefits   = $260,000/yr    N =    1-20
                  $300,000/yr    N = 21-50

If  i = 5% and   N=50yrs, is the project justified?

Solution :  1st cost       PW = $3,000,000 + $4,000,000(P/F,5,20) = $4,507,600

       Relining cost       PW = $160,000(P/F,5,10+20) + $320,000(P/F,5,30+40)
                                          =  $160,000(0.9908) + $320,000(0.3734)
                                          =  $278,016
Note: (P/F,5,10 + 20) = (P/F,5,10) + (P/F,5,20)

Benefits  PW = $260,000(P/A,5,20) + $300,000(P/A,5,30)(P/F,5,20)
                    = $260,000(12.4622) + $300,000(15.3725)(0.3769)
                     =  $4,978,340

B/C =        4,978,340         = 1.040
          4,507,600 + 278,016

Modified B/C =   4,978,340 - 278,016 = 1.042
                                4,507,600

The project is justified in either case, since B/C >1.0
Note that if the B/C ratio is greater than 1.0, the modified B/C will also be greater than 1.0

Example:  What if $2,000,000 additional funds are found to construct a full capacity tunnel now for 
$5,000,000?

Relining = $200,000 --- Every 10 years
Benefits = $300,000/year

However, to determine if this full capacity tunnel is a viable option, we now must use an incremental B/C ratio to see if the additional investment is justified.
       
1st cost (Full)  PW = $5,000,000
     
Relining (Full)  PW = $200,000(P/F,5,10+20+30+40) = $200,000(1.3642) 
                             = $272,840

Benefits(Full)  PW = $300,000(P/A,5,50) = $300,000(18.2559)
                             = $5,476,770

Incremental values :

Cost (Full- 1/2) = $5,000,000 - $4,507,600 = $492,400

Maintenance(Full - 1/2) = $272,840 - $278,016 = -$5176 (A reduction in cost is negative)

Benefits(Full - 1/2) = $5,476,770 - $4.978,390 = $498,430

B/C(Full - 1/2) =    498,430   = 1.023>1.0
                        492,400 - 5176

Thus the incremental cost is justified .Choose the full tunnel option.

What about B/C(Full)?

B/C(Full) =  5,476,770  = 1.038 < 1.040 = B/C(1/2)
                   5,272,840

This would appear to require the selection of the 1/2 tunnel option. However,B/C is not a ranking value!

Why is this?

The B/C analysis is an incremental analysis, even with only one alternative to consider. The increment is against the Do-nothing alternative. An  incremental analysis must always be used to compare alternates when using a B/C ratio approach.

If only costs are given, a reduction in costs can be used as benefits for the higher-cost option.

Example : A current process costs $50,000/yr.   N = 5    i = 10%  

System X         1st cost = $10,000
                  Annual cost = $45,000/yr

System Y        1st cost = $15,000
                  Annual cost = $42,000/yr

First, analyze the lowest initial cost option.
    
PW Cost (X-0) = $10,000  (0 = Do-Nothing)
PW Benefits (X-0) = (-45,000-(-50,000)) = 5000/yr * (P/A,10,5) = $18,954 

(Benefits are a reduction of annual costs) B/C(X-0) = 18,954 = 1.89>1   OK
                                                                                  10,000

Next, analyze the incremental costs and benefits for the next higher initial cost option.

Cost(Y-X) = $15,000-$10,000 = $5000

Benefits(Y-X) = $8000-$5000 = $3000/year * (P/A,10,5) = $11,372  

The current process  B/C(Y-X) = 11,372 = 2.27>1  is OK
                                                     5000

The incremental expense is justified and thus choose alternative Y.

The question begs to be asked, if we are using the PW or EAW of costs and benefits, why do we need to bother with calculating the B/C ratio of an option?

The reasons are simple:

1)The public is not educated in the ways of engineering economics.
2)PW or EAW have no clear cut meaning to the public in terms of gauging the relative merits of different options.
3)The B/C ratio gives a mental barrier(1.0) that must be cleared to be acceptable.

Public projects lend themselves to the B/C approach.

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Added to the Web: May 20, 2002.
Last updated: April 26, 2002.
Web page design by Dan Solarek.
http://cset.sp.utoledo.edu/engt3600/