This document discusses the integration of engineering and economic principles for effective decision-making in construction projects, emphasizing the importance of evaluating alternatives based on economic efficiency and life cycle costs. It presents a systematic five-step tool for decision-making, which includes identifying problems, alternatives, criteria, evaluation, and final decision-making, alongside methods for economic evaluation such as the payback period and net present worth. The role of cash flow diagrams in representing cash transactions over time is also highlighted, underscoring the need for both economic and non-economic considerations in selecting the best project alternatives.

1. PREPARED BY:-
Karm Balar
ASST. Prof.
S.S.A.S.I.T.
Construction Economics
Construction management

2. introduction
Economic
decision-
making
Time value of
money
Cash-flow
diagrams
Effect of
taxation
Evaluating
alternatives by
equivalence
Benefit-cost
ratio

4. • The Relationship Between Engineering And Economic is Close And in
The Present Scenario, Engineer Are Expected To Not Only Create
Technical Alternative But Also Evaluate Them By Economic Efficiency.
• With The Growing Maintenance Cost, in The infrastructure Sector, Its
Realized That Not Only The initial Cost, But Overall Life Cycle Cost of A
Project Should Be Taken into Account When We Evaluating Options.
• in Construction Projects, Economy influence Dynamically From
Multiple Points of View Like The Quantity of Materials, Buy And Rejecting
of Equipments, Reward Or Punishment Conditions, Etc.

5. • Its, Extremely Important That Engineers And Construction Manager Have
A Working Knowledge of Economic Principle, Terminology And The
Different Method That Can Be Used For Decision Making Process.
• in This Chapter Some Important Aspects of Decision Making Are Discuss.

7. Decision making is the process of identifying alternative
courses of action and selecting.
◦ This definition presents two Important Parts:
1. Identifying alternative courses of action means that an ideal solution
may not exist or might not be identifiable.
2. Selecting an appropriate alternative implies that there may be a
number of alternatives and that alternatives are to be evaluated and
rejected.

9. WHEN I GO FOR ECONOMIC DECISION-MAKING ?
There are various situations such as…

10. Comparison of Designs Or Elimination of Over-design
Designing For Economy of Production/Maintenance/Transportation
Economy of Selection
Economy of Perfection
Economy of Relative Size
Economy And Location
Economy And Standardization And Simplification

11. SIR, HAVE ANY COMMMAN MODEL FOR THIS?
YES

12.  A very useful 5 step tool to help managers to reach
project economic decisions.
 The five steps are:
◦ Problem
◦ Alternatives
◦ Criteria
◦ Evaluate
◦ Decision

13. P. What is the PROBLEM?
 What decision are your trying to make?
 What is the issue at hand?
A. What are the ALTERNATIVES?
 What actions are you considering?
 What options are available to you in this decision?
C. What are the CRITERIA important to the decision?
 What characteristics are you looking for in your result?
 Which criteria are more important than others? How do you
rank them?

14. E. EVALUATE each alternative.
Evaluate each alternative on the basis of each criterion.
Give each alternative a plus (+) or a minus (-) according to how
well it meets each criterion.
D. Make a DECISION.
Calculate the net value of each alternative; which alternative
best meets your highest-ranking criteria?
What do you gain with each alternative?
What do you give up with each alternative?
 An engineer has to take a decision among the competing alternatives.
 in such situations, The decision maker may not consider Time Value for
money.

15. out-of-pocket method
payback period
Avg. annual rate of
return
03
02
01
three most common methods of evaluation are…

16. • The out of the pocket commitment is the total expanse required for
an alternative.
• For example. Let a pre cast concrete factory has to produce 1,00,000
railway sleeper per year an economic choice to be made using. Steel
formwork and wooden formwork.
• For steel formwork
• (1,00,000*10) + (4,00,000*1)
• 14,00,000
• For wooden formwork
• (50,000*12) + (1,00,000*9)
• 15,00,000
Details Steel formwork Wooden
formwork
Cost for preparing one set
of formwork
Rs. 4,00,000/- Rs. 50,000/-
Labour charge (fixing,
removing of FW.)
Rs. 10/- Rs. 9/-
Life of Formwork 1 year. 1 month.

17. • Payback period for an investment may be taken as the
number of the years it takes to repay the original invested
capital.
• This method is very simple for evaluating project and
investments.
• It is understood that the shorter the payback period, the
higher the likelihood of the project being profitable.

18. For example let a contractor have two brands of excavators..
• A & B.
The brand is available for down payment of Rs. 4 lakhs. Both
can be Useful for a period of life 5 Years.
BRAND A (YEAR) GIVE RETURN in Rs.
1 50,000
2 1,00,000
3 1,00,000
4 1,50,000
5 1,50,000
TOTAL 5,50,000
BRAND B (YEAR) GIVE RETURN in Rs.
1 1,00,000
2 1,50,000
3 1,50,000
4 1,00,000
5 1,00,000
TOTAL 6,00,000
From above table we see that brand A return invested Rs.
Within 4 years…
Where brand B return invested Rs. Within 3 years…
so its good to go with brand B…

19.  The time value of money is important when one is interested either in
investing or borrowing the money.
 If a person invests his money today in bank savings, by next year he will
definitely accumulate more money than his investment.
 This accumulation of money over a specified time period is called as time
value of money.
 Similarly if a person borrows some money today, by tomorrow he has to
pay more money than the original loan. This is also explained by time
value of money.

20.  The time value of money is generally expressed by interest amount.
 The original investment or the borrowed amount (i.e. loan) is known as
the principal.
 The amount of interest indicates the increase between principal amount
invested or borrowed and the final amount received or owed.

21.  Quantifying alternatives for any item is the most important aspect of
decision making for selecting the best option.
 For example, a construction company is planning to purchase a new
concrete mixer for preparing concrete at a construction site.
 Let’s say there are two alternatives available for purchasing the mixer;
 an automatic concrete mixer
 a semi-automatic concrete mixer.
 Then the task is to find out best alternative that the company will
purchase that will yield more profit.

22.  For this purpose one has to quantify both the alternatives by the following
parameters…
 The initial cost that includes purchase price, sales tax, cost of delivery
and cost of assembly and installation.
 Annual operating cost.
 Annual profit which will depend on the productivity i.e. quantity of
concrete prepared.
 The expected useful life.
 The expected salvage value.
 Other expenditure or income (if any) associated with the equipment.
 Income tax benefit

23.  on the basis of this criteria, the best alternative is selected by calculating
the present worth or future worth or the equivalent uniform annual worth
of both alternatives by the appropriate interest rate per year and the
number of years (i.e. the comparison must be made over same number of
years for both alternatives).
 Then the concrete mixer with least cost or higher net income is
considered for purchase.
 In addition to economic parameters as mentioned above, the non-
economic parameters namely environmental, social, & legal and the
related regulatory body also be considered for the evaluation and
selection of the best alternative.

24.  These non-economic parameters are required for the selection of the
alternative for the infrastructure and heavy construction projects like
 dams
 Bridges
 roadways etc.
 also other publicly and privately funded projects namely
 office buildings
 Hospitals
 apartment building
 shopping malls etc.
 When the available alternatives exhibit the same equivalent cost or same
net income, then the non-economic parameters may play a vital role in
the selection of the best alternative.

25. How can we represent the
value??

26.  The graphical representation of the cash i.e. both cash outflows and cash
inflows with respect to a time scale is generally referred as cash flow
diagram.
 Number of interest periods is shown on the time scale.
 The interest period may be a quarter, a month or a year.
 The cash flows generally occur at different time intervals within an
interest period (i), for easy calculation, all the cash flows are assumed to
occur at the end of an interest period.
 Thus in Fig., the numbers on the time scale represent the END OF YEAR
(EOY).

27. EOY 10
0 1 2 3 4 5 6 7 8 9 10
Year 1 Year 7
Time
Cash inflow
Cash outflow
End of year 1 35000 80000 45000
1,50,000 15000 25000

28. • To ensure that sufficient cash is available to meet the demand.
• To ensure the cash resources are fully utilize for the benefit of the
owner and for company.
• To know project funding pattern.

29. Cash flow consist two part…
1) cash out flow
2) cash in flow
1 cash outflow
Payment made by contractor over a period of time for the resources
he uses in a project is known as outflow of cash and its mention as –
ve value over graph.
And its include…….
Material purchase
Labour wages
Equipment purchase/hire etc.

30. 2 Cash inflow
• Money receive from the owner, company, builder, etc. over the
time is represent as inflow of cash, and mention as +ve value over
graph.
• in the cash diagram time is drawn on the x axis with appropriate
scale, in terms of week, month, year, etc.
• When on y axis the amount involved in the transaction.
• Cash inflow drawn on the positive side of y axis, while outflow
drawn negative side of y axis.
Time
Cash inflow
Cash outflow

32. MONTH RECEIPTS (Rs.) EXPEND
APRIL 1,50,000 1,00,000
MAY 1,80,000 1,25,000
JUNE 2,40,000 2,60,000
EX:- The details of the financial transactions during the following
month for delta construction pt. Ltd. Are given below. Draw the CFD in
a single unit showing all transactions.

33. April
Month
1,50,000
1,25,000 2,60,000
may June
1,00,000
1,80,000 2,40,000
+ve in flow
-ve out flow
Cash flow diagram alternative - 1
April
Month
50,000
20,,000
may June
55,,000
+ve in flow
-ve out flow
Cash flow diagram alternative - 2

34.  The project cash flow is basically a graph of receipts versus time. The project
cash flow can be prepared form different perspectives of contractor, owner
etc.,
 Some projects involve initial capital investment i.e. cash outflow at the
beginning and show increased income or revenue i.e. cash inflow in the
subsequent years. The alternatives having this type of cash flow are known as
investment alternatives.
 The cost or expenses are generally known as cash outflows whereas revenue
or incomes are generally considered as cash inflows. Thus in the economic
comparison of alternatives, cost or expenses are considered as negative cash
flows. On the other hand the income or revenues are considered as positive
cash flows.

35. mobilization advance
The margin in a project
Retention
Extra claims
Distribution of margin
Certification type
Certification period,

36. MOBILIZATION ADVANCE it is an advance taken by contractor (2 to 5%) from client for
various resources at a site for smooth work and maintain cash flow.
MARGIN the margin (profit margin or contributions) is the excess over cost. Higher the margin
in a project the better it is for cash flow.
RETENTION whenever any claims for payment of a sum of money arises out of the contract
against the contractor, the site engineer handle such claimed amount. This amount known as.....
MEASUREMENT PERIOD – It is usual for the contractors to be paid on a monthly basis. The
payment can be made fortnightly or sometimes bimonthly as well. These conditions can be found
under ‘terms of payment’ given in the tender document.
CERTIFICATION TIME TAKEN BY THE OWNER – in normal conditions owner takes
about 3-4 weeks' time to process the bill and release the payment to the constructor or contractor.

37. • Equivalence indicates that different amount of money at different
time periods are equivalent by considering the time value of
money.
• The following simple example will explain the meaning of equivalence.
• EX:- What are the equivalent amounts of Rs.10000 (today) at an
interest rate of 10% per year for the following cases?
a)1 year from now (future)
b)1 year before

38. Solution:
a) At interest rate of 10% per year, Rs.10000 (now) will be equivalent
to Rs.11000 one year from now as shown below….
Amount accumulated at the end of one year:-
Rs.(10000*1.10) = Rs.11000/-
b) Similarly Rs.10000 now was equivalent to Rs.9090.90 one year ago
at interest rate of 10% per year.
Amount one year before:-
(Rs.10000/1.10) = Rs.9090.90

39. • Once the concept of a cash flow diagram is understood the next step is to
proceed to the extension of this diagram to compare different engineering
alternative from an economic point of view.
• CM is a world of cost comparison between alternatives of different
engineering efficiency namely one with a high initial cost and low
operational and maintenance cost compared to another with a low initial
cost but high operation and maintenance cost.
• Using the time value of money and cash flow diagram for illustrative
purpose, equivalence is studied to identify the better alternative.
• For most of the engineering projects, there are more than one alternative. It
is the duty of the project management team (comprising of engineers,
designers, project managers etc.) to select the best alternative that involves
less cost and results more revenue.

40. The economic comparison of mutually exclusive alternatives can be carried out
by different equivalent worth methods namely
1
•Present worth method
2
•Rate of Return Method
3
•Annual cost & worth method.
4
•Future worth method

41.  In this method, the present worth (at time zero) of the cash flow in terms of
equivalent single sum is deter-mined using an interest rate, sometimes
also called the discounting rate.
 The method is based on the following assumptions…
(a) Cash flows are known.
(b) Cash flows do not include effect of inflation. (The discussion on inflation
follows later in the text.)
(c) The interest rate (discounting rate) is known.
(d) Comparisons are made with before-tax cash flows. (The concept of tax has
been discussed in later sections.)
(e) Comparisons do not include intangible considerations.
(f) Comparisons do not include consideration of the availability of funds
to implement alternatives.

42. Following figure are show all three typical types of problem that are encounter
in the present worth analysis and that three set of problem are given as...
prresent worth
problems
type:-1
alternatives with
equal lives
type:-2
alternatives with
unequal lives
common multiple
method
study period
method
type:-3
alternatives with
infinite lives

43. TYPE:-1 alternatives with equal lives
 As the name suggest, in such problem the compacting alternatives have
equal life span. [The useful lives (life span) of alternatives are equal]
 for evaluating the alternatives, the present worth of both the compacting
alternatives are found out.
 the alternatives with the maximum present worth is the most economical
alternative.
 for cost dominated cash flow diagrams, the alternative with the lowest
present cost is chosen.
 in case of cash flow diagram involving both cost and revenue, the net or
difference of present worth of revenue and cost are found.
 this is referred to as net present worth or net present value (NPV).
 the method of comparison of (NPV) is quite popular for evolution of
alternative.

44. EX.:- There are two alternatives for purchasing a concrete mixer. Both the
alternatives have same useful life. The 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.

45. 0 1 2 3 4 5
Solution: Both alternatives have the same life span of 5 years, so the present
worth of the alternatives will be compared over a period of 5 years.
The cash flow diagram of Alternative-1…
Rs.3,00,000
Income
Expenditure
(Year)
Rs.20,000
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.
Rs.1,25,000

46. The equivalent present worth of Alternative-1 i.e. PW1 is calculated as follows…
(cash outflow)
The initial cost, P = Rs.3,00,000
Annual Op. & Mant. cost, A = Rs.20,000
(cash inflow).
Salvage value (Also refer future value), F = Rs.1,25,000
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)
* When you have the value of A & need to find the Value of P, it will be use.
( when you have Uniform series of present worth, it's also called A.)
** When you have the value of F & need to find the Value of P it will be use.
FOLLOW NEXT SLIDE FOR EQ.

47. Name of the
factor
(1)
Abbreviation
(2)
Functional
representatio
n (3)
Mathematical
expression
(4)
Given
(5)
To find out
(6)
= (4) x (5)
Single payment
compound
amount factor SPCAF
(F / P,i, n)
P F
Single payment
present worth
factor SPPWF
(P / F,i, n)
F P
Uniform series
present worth
factor USPWF
(P / A,i, n)
A P
Capital recovery
factor CRF
( A / P,i, n)
P A
Uniform series
compound
amount factor USCAF
(F / A,i, n)
A F
Sinking fund
factor
SFF ( A / F,i, n) F A

48. (P/A, i, n)* =
I = 10% = 0.1
N = 5 Yr.
= 3.79
(P/F, i, n)** = = 0.621
I = 10% = 0.1
N = 5 Yr.
1.61

49. Now putting the mathematical expressions of different compound interest
factors (as mentioned in table) in the above expression for PW1 (in Rs.) results
in the following;
PW1 = -3,00,000 - 20,000*3.7908 +1,25,000*0.6209
PW1 = - 3,00,000 -75,816 + 77,613
PW1 = - Rs.2,98,203

50. 0 1 2 3 4 5
The cash flow diagram of Alternative-2…
Rs.2,00,000
Income
Expenditure
(Year)
Rs.35,000
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.
Rs.70,000

51. The equivalent present worth of Alternative-2 i.e. PW2 is calculated as follows…
(cash outflow)
The initial cost, P = Rs.2,00,000
Annual Op. & Mant. cost, A = Rs.35,000
(cash inflow).
Salvage value (Also refer future value), F = Rs.70,000
PW1 = -2,00,000 – 35,000(P/A, i, n)* + 70,000(P/F, i, n)**
PW1 = - 2,00,000 – 35,000(P/A, 10%, 5) + 70,000(P/F, 10%, 5)
PW1 = -2,00,000 - 35,000*3.7908 +70,000*0.6209
PW2 = -Rs.2,89,215 PW1 = - Rs.2,98,203

52. Comparing the equivalent present worth of both the
alternatives..
it is observed that Alternative-2 will be selected as it
shows lower negative equivalent present worth
compared to Alternative-1 at the interest rate of 10%
per year.

53. EX.:- There are two alternatives for purchasing a concrete mixer. Both the
alternatives have same useful life. The 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
annual revenue generated from production of concrete = Rs.50,000
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
annual revenue generated from production of concrete = Rs.45,000
Useful life = 5 years.
Compute the equivalent present worth of the alternatives at i=10% per year and
find out the economical alternative.

54. The cash flow diagram of Alternative-1…
PW1 = - Rs.1,08,663

55. The cash flow diagram of Alternative-2…
PW2 = - Rs.1,18,629 PW1 = - Rs.1,08,663

56. Comparing the equivalent present worth of the both
the alternatives, it is observed that
Alternative-1 will be selected as it shows lower cost
compared to Alternative-2. at the interest rate of
10% per year.

57. Type 2: Alternatives with Unequal Lives
In such problems, the alternatives do not have an equal life period of
service-in other words, they have not same life span.
A relevant example would be a decision to choose between two batching
plants that may have different service lives-say, 5 years and 10 years.
Needless to say, a simple comparison of the two alternatives would not be
accurate, as in one case there would be a need to replace the plant at the end
of five years, and any cost likely to be incurred at that point in time should be
appropriately accounted for in the budgeting at the outset.
The common multiple method and the study period method are two
approaches to discuss this.

58. • The benefit-cost analysis method is mainly used for economic evaluation
of public projects which are mostly funded by government organizations.
• As the name indicates, this method involves the calculation of ratio of
benefits to the costs involved in a project.
• Costs are the expenditures namely initial capital investment, annual
operating cost, annual maintenance cost etc. to be incurred by the owner
of the project and salvage value if any is subtracted from the costs.
• However, this fund is generally taxpayers ‟money i.e. tax collected by
government from general public.
• thereby the actual owners of public projects are the general public.
• Thus in case of public projects, the cost is incurred by the government
whereas the benefits and dis-benefits are mostly experienced by the
general public.

59. • The objective of private project is to maximize the project profit, but
objective of public project it to provide services /goods to the public at the
minimum costs.
• Public projects here mean those, which are funded by Government (State or
Centre)
• The Government has a responsibility towards public welfare. Some
examples of Public projects are:
• Dams
• Defense projects
• Highways
• Education
• Health
• Unlike private projects, many Government projects cannot be evaluated
strictly in commercial terms. Profit, taxes, payoff periods take a back
seat in public projects.

60.  Consider a public sector organization is planning to set up a thermal
power plant at a particular location.
 The costs to be incurred by the public sector organization are…
 the cost of purchasing the land required for the thermal power plant
 cost of construction of various facilities
 cost of purchase and installation of various Equipment
 annual operating and maintenance cost
 and other recurring costs, etc.
 The benefits associated with the project are…
 a generation of electric power that will cater to the need of the public
 generation of revenue by supplying the electricity to the customers
 job opportunities for local residents
 development of other infrastructure in the nearby areas, etc.

61.  The time value of money is taken into account for calculating the
equivalent worth of the costs and benefits associated with a project.
 The benefit-cost ratio of a project is calculated by taking the ratio of the
equivalent worth of benefits to that of the costs associated with that
project Either of present worth, annual worth or future worth methods
can be used to find out the equivalent worth of costs and benefits
associated with the project.
 The benefit-cost ratio of projects is determined in different forms namely
conventional benefit-cost ratio and modified benefit-cost ratio.
 The benefit-cost ratio is generally designated as B/C ratio.

62. Conventional B/C ratio
The conventional benefit-cost ratio of a project is mentioned as follows…
Conventional B/C ratio =
Equivalent worth of Benefits - Equivalent worth of Dis-benefits
Equivalent worth of total cost - Equivalent worth of salvage value
so.,
Conventional B/C ratio =
PW of Benefits - PW of Dis-benefits - PW of operating and maintenance cost
Initial cost - PW of salvage value
OR
Conventional B/C ratio =
AW of Benefits - AW of Dis-benefits - AW of operating and maintenance cost
AW of initial cost - AW of salvage value
AW = Annual Worth
PW = Present Worth
If You have the value of FW = Future Worth Then
change AW by FW in above equations, & remain same.

63.  if B/C ratio ≥ 1 means benefits outweighs cost thus, the investment
is justified.
 if B/C ratio ≤ 1 means benefits decreased from the project & more
cost required to invest thus investment is not justifiable.

64. The cash flow details of a public project is as follows…
initial cost = Rs.2,10,00,000
Annual operating cost = Rs.16,00,000
Worth of annual benefits = Rs.50,00,000
Worth of annual dis-benefits = Rs.11,00,000
Salvage value = Rs.40,00,000
Interest rate per year = 8% and useful lie = 30 Years
Using benefit-cost ratio method, find out the economical acceptability of the
public project. Find out the equivalent worth of costs, benefits and dis-benefits.

65. Conventional B/C ratio =
PW of Benefits - PW of Dis-benefits - PW of operating and maintenance cost
Initial cost - PW of salvage value
= 50,00,000 (P/A, i, n) – 11,00,000 (P/A, i, n) -16,00,000 (P/A, i, n)
2,10,00,000 - 40,00,000 (P/F, i, n)
I = 8% , n=30
(P/A, i, n)* = (P/F, i, n)** =
Modified B/C ratio = 1.257

66. 66