Engineering Economy

Programming 101

ES323

Introduction:

       Engineering economics, previously known as engineering economy, is a subset of economics for application to engineering projects. Engineers seek solutions to problems, and the economic viability of each potential solution is normally considered along with the technical aspects. Fundamentally, engineering economics involves formulating, estimating, and evaluating the economic outcomes when alternatives to accomplish a defined purpose are available.

1.   Comparison by present worth method:- 

Now some examples showing the use of present worth method for comparison of mutually exclusive alternatives are presented. First the comparison of equal life span mutually exclusive alternatives by present worth method will be illustrated followed by comparison of different life span alternatives. The following examples are formulated only to demonstrate the use of different methods for comparison of alternatives. The values of different cost and incomes mentioned in the examples are not the actual ones pertaining to a particular item. In addition it may also be noted here that the cash flow diagrams have been drawn not to the scale. These are merely graphical representations.

Example -1   

There are two alternatives for purchasing a concrete mixer. Both the alternatives have same useful life. The cash flow details of alternatives are as follows;

Alternative-1: Initial purchase cost = Rs.3,00,000, Annual operating and maintenance cost = Rs.20,000, Expected salvage value = Rs.1,25,000, Useful life = 5 years.

Alternative-2: Initial purchase cost = Rs.2,00,000, Annual operating and maintenance cost = Rs.35,000, Expected salvage value = Rs.70,000, Useful life = 5 years.

Using present worth method, find out which alternative should be selected, if the rate of interest is 10% per year.

Solution: 
Since both alternatives have the same life span i.e. 5years, the present worth of the alternatives will be compared over a period of 5 years. The cash flow diagram of Alternative-1 is shown in Fig. 2.1.
As already mentioned Module-1, the cash outflows i.e. costs or expenditures are represented by vertically downward arrows whereas the cash inflows i.e. revenue or income are represented by vertically upward arrows. The same convention is adopted here.

Fig. 2.1 Cash flow diagram of Alternative-1

The equivalent present worth of Alternative-1 i.e. PW₁ is calculated as follows;
The initial cost, P = Rs.3,00,000 (cash outflow),
Annual operating and maintenance cost, A = Rs.20,000 (cash outflow),
Salvage value, F = Rs.1,25,000 (cash inflow).
PW₁ = - 3,00,000 - 20,000(P/A, i, n) + 1,25,000(P/F, i, n)
PW₁ = - 3,00,000 - 20,000(P/A, 10%, 5) + 1,25,000(P/F, 10%, 5)
Now putting the mathematical expressions of different compound interest factors (as mentioned in Module-1) in the above expression for PW₁ (in Rs.) results in the following;

PW₁ = - 3,00,000 -75,816 + 77,613
PW₁= - Rs.2,98,203

2.    Comparison of alternatives by future worth method:

In the future worth method for comparison of mutually exclusive alternatives, the equivalent future worth (i.e. value at the end of the useful lives of alternatives) of all the expenditures and incomes occurring at different periods of time are determined at the given interest rate per interest period. As already mentioned, the cash flow of the mutually exclusive alternatives may consist of expenditures and incomes in different forms. Therefore the equivalent future worth of these expenditures and incomes will be determined using different compound interest factors namely single payment compound amount factor, uniform series compound amount factor and future worth factors for arithmetic and geometric gradient series etc.

The use of future worth method for comparison of mutually exclusive alternatives will be illustrated in the following examples. Similar to present worth method, first the comparison of equal life span alternatives by future worth method will be illustrated followed by comparison of different life span alternatives. Some of the examples already worked out by the present worth method will be illustrated using the future worth method in addition to some other examples.

Example -6 (Using data of Example-1)

There are two alternatives for purchasing a concrete mixer. Both the alternatives have same useful life. The cash flow details of alternatives are as follows;

Alternative-1: Initial purchase cost = Rs.300000, Annual operating and maintenance cost = Rs.20000, Expected salvage value = Rs.125000, Useful life = 5 years.

Alternative-2: Initial purchase cost = Rs.200000, Annual operating and maintenance cost = Rs.35000, Expected salvage value = Rs.70000, Useful life = 5 years.

Using future worth method, find out which alternative should be selected, if the rate of interest is 10% per year.

Solution:

The future worth of the mutually exclusive alternatives will be compared over a period of 5 years. The equivalent future worth of the alternatives can be obtained either by multiplying the equivalent present worth of each alternative already obtained by present worth method with the single payment compound amount factor or determining the future worth of expenditures and incomes individually and adding them to get the equivalent future worth of each alternative.

The equivalent future worth of Alternative-1 is obtained as follows;

PW₁ is the equivalent present worth of Alternative-1 which is equal to - Rs.298203 (referring to Example-1). (F/P, i, n) is the single payment compound amount factor.

Now putting the value of single payment compound amount factor in the above expression;

FW₁ = -Rs.480256

3.    Comparison of alternatives by annual worth method:

In this method, the mutually exclusive alternatives are compared on the basis of equivalent uniform annual worth. The equivalent uniform annual worth represents the annual equivalent value of all the cash inflows and cash outflows of the alternatives at the given rate of interest per interest period. In this method of comparison, the equivalent uniform annual worth of all expenditures and incomes of the alternatives are determined using different compound interest factors namely capital recovery factor, sinking fund factor and annual worth factors for arithmetic and geometric gradient series etc. Since equivalent uniform annual worth of the alternatives over the useful life are determined, same procedure is followed irrespective of the life spans of the alternatives i.e. whether it is the comparison of equal life span alternatives or that of different life span alternatives. In other words, in case of comparison of different life span alternatives by annual worth method, the comparison is not made over the least common multiple of the life spans as is done in case of present worth and future worth method. The reason is that even if the comparison is made over the least common multiple of years, the equivalent uniform annual worth of the alternative for more than one cycle of cash flow will be exactly same as that of the first cycle provided the cash flow i.e. the costs and incomes of the alternative in the successive cycles is exactly same as that in the first cycle. Thus the comparison is made only for one cycle of cash flow of the alternatives. This serves as one of greater advantages of using this method over other methods of comparison of alternatives. However if the cash flows of the alternatives in the successive cycles are not the same as that in the first cycle, then a study period is selected and then the equivalent uniform annual worth of the cash flows of the alternatives are computed over the study period.

Now the comparison of mutually exclusive alternatives by annual worth method will be illustrated in the following examples. First the data presented in Example-2 will be used for comparison of the alternatives by the annual worth method.

Example -10 (Using data of Example-2)

There are two alternatives for purchasing a concrete mixer and following are the cash flow details;

Alternative-1: Initial purchase cost = Rs.300000, Annual operating and maintenance cost = Rs.20000, Expected salvage value = Rs.125000, Useful life = 5 years.

Alternative-2: Initial purchase cost = Rs.200000, Annual operating and maintenance cost = Rs.35000, Expected salvage value = Rs.70000, Useful life = 5 years.

The annual revenue to be generated from production of concrete (by concrete mixer) from Alternative-1 and Alternative-2 are Rs.50000 and Rs.45000 respectively. Compute the equivalent uniform annual worth of the alternatives at the interest rate of 10% per year and find out the economical alternative.

Solution:

The cash flow diagram of Alternative-1 i.e. Fig. 2.3 is shown here again for ready reference.

Fig. 2.3 Cash flow diagram of Alternative -1

The equivalent uniform annual worth of Alternative-1 i.e. AW₁ is computed as follows;

Here Rs.20000 and Rs.50000 are annual amounts.

Now putting the values of different compound interest factors;

AW₁= - Rs.28665

CONCEPTUAL FRAMEWORK

(FLOW CHART)

Proposed Solution:

Pseudo code:

start

          econ1=int(input("Investment(PW):"))

          econ2=int(input("Market Value(FW):"))

          econ3=int(input("Annual Receipt(AW):"))

          econ4=int(input("Annual Disburstment(AW):"))

          i=float(input("Interest Rate:"))

          n=int(input("Number of Period:"))

          present_econ1=econ1

          present_econ2=econ2/((1+i)**(n))

          present_econ3=econ3*((((1+i)**(n))-1)/((i)*((1+i)**(n))))

          present_econ4=econ4*((((1+i)**(n))-1)/((i)*((1+i)**(n))))

          present_worth=(-present_econ1+present_econ2+present_econ3-

          present_econ4)

          future_econ1=econ1*((1+i)**(n))

          future_econ2=econ2

          future_econ3=econ3*((((1+i)**(n))-1)/(i))

          future_econ4=econ4*((((1+i)**(n))-1)/(i))

          future_worth=(-future_econ1+future_econ2+future_econ3-future_econ4)

          annual_econ1=econ1*(((i)*((1+i)**(n)))/(((1+i)**(n))-1))

          annual_econ2=econ2*((i)/(((1+i)**(n))-1))

          annual_econ3=econ3

          annual_econ3=econ4

          print 'Present Worth is:' ,present_worth

          print 'Future Worth is:' ,future_worth

          print 'Annual Worth is:' ,annual_worth 

End

Python Code:

class engineering_economy:

    def engineering_economy(self):

        print 'Welcome to Engineering Economy'

e=engineering_economy()

e.engineering_economy()

econ1=int(input("Investment(PW):"))

econ2=int(input("Market Value(FW):"))

econ3=int(input("Annual Receipt(AW):"))

econ4=int(input("Annual Disburstment(AW):"))

i=float(input("Interest Rate:"))

n=int(input("Number of Period:"))

class present_worth(engineering_economy):

    def present_worth(self):

        print ' '

p=present_worth()

p.present_worth()

present_econ1=econ1

present_econ2=econ2/((1+i)**(n))

present_econ3=econ3*((((1+i)**(n))-1)/((i)*((1+i)**(n))))

present_econ4=econ4*((((1+i)**(n))-1)/((i)*((1+i)**(n))))

present_worth=(-present_econ1+present_econ2+present_econ3-present_econ4)

print 'Present Worth is:' ,present_worth

if present_worth >0:

        print('economically justifiable')

elif present_worth <0:

        print('not economically justifiable')

class future_worth(engineering_economy):

    def future_worth(self):

        print ' '

p=future_worth()

p.future_worth()

future_econ1=econ1*((1+i)**(n))

future_econ2=econ2

future_econ3=econ3*((((1+i)**(n))-1)/(i))

future_econ4=econ4*((((1+i)**(n))-1)/(i))

future_worth=(-future_econ1+future_econ2+future_econ3-future_econ4)

print 'Future Worth is:' ,future_worth

if future_worth >0:

        print('economically justifiable')

elif future_worth <0:

        print('not economically justifiable')

class annual_worth(engineering_economy):

    def annual_worth(self):

        print ' '

p=annual_worth()

p.annual_worth()

annual_econ1=econ1*(((i)*((1+i)**(n)))/(((1+i)**(n))-1))

annual_econ2=econ2*((i)/(((1+i)**(n))-1))

annual_econ3=econ3

annual_econ4=econ4

annual_worth=(-annual_econ1+annual_econ2+annual_econ3-annual_econ4)

print 'Annual Worth is:' ,annual_worth

if annual_worth >0:

        print('economically justifiable')

elif annual_worth <0:

        print('not economically justifiable')

Conclusion

    This programming project about Engineering concludes that by using my program it help me and the other students to easily answers the engineering economy problems and this formulas are the most useful formula that you can use in getting the present worth, future worth and annual worth.