12-04-2023

Equipment Cost Management

Investment Cost Interest rates, taxes, insurances, storage cost

Construction Equipment Owning Cost Depreciation Cost Physical, Economic, Technical Factors

Management
Major Repair Cost 80-200% of depreciation cost

Fuel Cost Engine BHP, Load factor, Engine Condition

Equipment Cost

Lubricant Cost Capacity of crank, pistons, oil change freq.

CMT 522 Servicing & Maintenance Cost Tyre, tube, battery, electricals, cleaning etc.
Session 18 – Equipment Economics – Present Worth Operating Cost
Labor Cost Salaries, wages etc. of operators and staff
Analysis Method using interest factors
Cost of Field Repairs Minor repairs, spare parts replacements

Overheads Light, water, security charges etc.

Aritra Halder
| Aritra Halder | Assistant Professor
Assistant Professor | School of Construction

Fuel Cost Lubricant Cost

• Usually diesel engines or electric motors are used in construction. The • Lubricants include-
consumption depends on – • Engine Oil
• Engine BHP • Air Filter Oil
• Load factor i.e. the extent at which the engine operates at full power. • Transmission Oil
• The conditions of the engine. • Hydraulic Oil
• Optimum fuel consumption per HP per hour = 0.27XLoad factor • Greases etc.
Excellent working Average working Severe working H.P. ×f×0.006×4.5 C
Sl. No. Type of Equipment • Quantity of oil required per hour= + lit/hour
condition condition condition 7.4 t
1 Wheel type (highway) 0.25 0.30 0.40 • Where, Q = Qty of oil consumed in lit/hr
2 Wheel type (off highway) 0.5 0.55 0.60 • HP = Rated horse power of the engine
3 Track type tractors 0.5 0.63 0.75 • f = Operating Factor
4 Excavators 0.5 0.55 0.60 • C = Capacity of crank case
Construction Plant and Machinery Committee-1972, Recommendations for load factor • t = number of hours between oil changes
| Aritra Halder | Assistant Professor | Aritra Halder | Assistant Professor

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Comparison of Alternatives Comparison of Alternatives – Other Methods

• Investment Alternatives – Select the highest positive option • Rate of return or Internal Rate of Return (IRR) – The rate of
–E.g. – Purchasing an equipment and putting it into use return technique is one of the methods used in selecting an
alternative for a project.
• Cost Alternatives – Select the option with minimum –In this method, the interest rate per interest period is
negative cash outflow determined, which equates the equivalent worth (either present
–E.g. – Highway projects by Government worth, future worth or annual worth) of cash outflows (i.e. costs
or expenditures) to that of cash inflows (i.e. incomes or revenues)
of an alternative.
[Problems to be worked out – Ref Notes (PW,FW,AW Methods)]

| Aritra Halder | Assistant Professor | Aritra Halder | Assistant Professor

IRR & MARR Interest Factors

• After determination of the rate of return for a given
Interest Factors
alternative, it is compared with minimum attractive rate of Single Payment Category Uniform Series Category Unequal Payment Series Category
return (MARR) to find out the acceptability of this
alternative for the project. If the rate of return i.e. ir is greater Single Payment Compound
Amount Factor
Uniform Series Compound
Amount Factor
Arithmetic Gradient Factor
than or equal to MARR, then the alternative will be selected
Single Payment Present Uniform Series Present
or else it will not be selected. The MARR is the minimum Worth Factor Worth Factor
Geometric Gradient Factor

rate of return from the investment, which is acceptable. Uniform Series Sinking Fund
Factor

Uniform Series Capital

Recovery Factor

| Aritra Halder | Assistant Professor | Aritra Halder | Assistant Professor

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Formulations for Interest Factor Calculation Formulations for Interest Factor Calculation
P F=? F=?
• Single Payment Compound Amount Factor: (SPCAF): It • Uniform Series Compound Amount Factor: (USCAF): A A A A A
is the factor by which a single payment (P) is multiplied This factor converts a uniform series of payments (A)
to get an amount (F) in future for a given interest rate 0 1 2 3 n-1 n to its compound amount (F) in future for a given
and time period. 0 1 2 3 n-1 n
interest rate and time period.
• 𝐹 𝑃, 𝑖, 𝑛 = (1 + 𝑖)
( )
• 𝐹 𝐴, 𝑖, 𝑛 =
• Single Payment Present Worth Factor: (SPPWF): P=?
It is the factor by which a single payment in future (F) is • Uniform Series Present Worth Factor: (USPWF):
multiplied to get its present worth (P) for a given P=? F This factor converts a uniform series of payments (A) A A A A A
interest rate and time period. to its present worth (P) for a given interest rate and
• 𝑃 𝐹, 𝑖, 𝑛 = 0 1 2 3 n-1 n
time period.
0 1 2 3 n-1 n
( )
• 𝑃 𝐴, 𝑖, 𝑛 =
( )

| Aritra Halder | Assistant Professor | Aritra Halder | Assistant Professor

Formulations for Interest Factor Calculation Formulations for Interest Factor Calculation
F = given
• Uniform Series Sinking Fund Deposit Factor: (USSFDF): • Arithmetic Gradient Factor: (AGF):
This is the factor by which a future sum (F) is multiplied to get A A A=? A A When the decrease or increase in installments follow (n-1)G
the uniform sum (A) to be set aside regularly in order to get a an arithmetic pattern, this factor will convert the (n-2)G
final deposit equal to (F) for a given interest rate and time decrease/increase (G) into a uniform series of G 2G
period. 0 1 2 3 n-1 n payments (A) for a given interest rate and time
period. A A A=? A A
• 𝐴 𝐹, 𝑖, 𝑛 = ( ) • 𝐴 𝐺, 𝑖, 𝑛 = − 0 1 2 3 n-1 n

• Uniform Series Capital Recovery Factor: (USCRF): P = given • Geometric Gradient Factor: (GGF):
This is the factor by which a present sum (P) is multiplied to When the decrease or increase in installments follow A(1+g)n-1
A(1+g)n-2
get the uniform sum (A) to be invested regularly in order to A A A=? A A a geometric pattern, this factor will convert the A(1+g)2
recover the present capital sum (P) for a given interest rate A(1+g)
decrease/increase (g) into equivalent present sum
and time period. (P) for a given interest rate and time period. A
0 1 2 3 n-1 n ( )
• 𝐴 𝑃, 𝑖, 𝑛 = ( ) • 𝑃 𝑔, 𝑖, 𝑛 =
( )
0 1 2 3 n-1 n
( )

| Aritra Halder | Assistant Professor | Aritra Halder | Assistant Professor

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 12-04-2023

Effective Interest Rate for multiple compounding per year Evaluation by Equivalence
• For interest rate ‘i’ compounded in more than one • Present Worth Comparison : In this method, the present worth
interval per year, effective annual interest rate can be (at time zero) of the cash flow in terms of equivalent single sum
obtained by the following formula – is determined using an interest rate (also called discount rate).
The method is based on the following assumptions-
• Number of intervals = m, rate of interest per interval = a) Cash flows are known
inom b) Cash flows are not affected by inflation
c) Interest rate is known
• Hence, effective interest rate 𝑖 = ((1 + × ) −1) × 100 d) Comparisons are made with before tax cash flows
e) No intangible considerations are accounted for
f) No considerations for availability of funds to implement the
alternatives

| Aritra Halder | Assistant Professor | Aritra Halder | Assistant Professor

Illustrative Example – Present Worth Comparison Illustrative Example – Present Worth Comparison
• An alternative A requires initial investment of Rs. 500,000 and annual • Present cost of alternative A : 5,00,000 + 2,50,000 * (P|A, 10%, 10)
expense of Rs. 250,000 for the next 10 years. Alternative B requires an ( . )
initial investment of Rs. 750,000 and an annual expense of Rs. 200,000 = 5,00,000 + 2,50,000*( . . )

for the next 10 years. Which alternative would you prefer if interest rate
were 10 per cent? = 5,00,000 + 2,50,000*6.1446
CA S H FLO W DI A GRA M FO R A LTE RN ATI VE A CA S H FLO W DI A GRA M FO R A LTE RN ATI VE B

0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 = 20,36,150
0 0
• Present cost of alternative B : 750,000 + 200,000 * (P|A, 10%, 10)
-1
-1
-2
= 750,000 + 200,000 * 6.1446
Cash Flow in Lakhs

-2 -2 -2 -2 -2 -2 -2 -2 -2 -2
Cash Flow in Lakhs

-2
-3
-2.5 -2.5 -2.5 -2.5 -2.5 -2.5 -2.5 -2.5 -2.5 -2.5
-3 -4 = 19,78,920
-5
-4
-6
-5
• For present cost analysis in case of only cash outflow, alternative with lower cost will be
-5 -7
preferred. Hence opt for alternative B.
-6 -8 -7.5
| Aritra Halder | Assistant Professor | Aritra Halder | Assistant Professor

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Interest Table Interest Table

| Aritra Halder | Assistant Professor | Aritra Halder | Assistant Professor

Illustrative Example Illustrative Example

• Solve the problem using (a) present worth method, (b) annual worth method, • Present Worth Method
(c) incremental rate of return method for the given cash flow data for purchase • NPW for equipment X = -50000 + 5000X(P|F,10%,1) + 17500X(P|F,10%,2) + 30000X(P|F,10%,3)
of a machine. Minimum attractive rate of return is 10%. + 42500X(P|F,10%,4)

End of year  0 1 2 3 4 = -50000 + 5000X0.9091 + 17500X0.82645 +30000X0.75131 + 42500X0.6830

= 20,575
Equipment X -50,000 5000 17,500 30,000 42,500

Equipment Y -50,000 40,000 15,000 15,000 15,000 • NPW for equipment Y = -50000 + 40000X(P|F,10%,1) + 15000X(P|F,10%,2) +
15000X(P|F,10%,3) + 15000X(P|F,10%,4)
Cash Flow for Equipment X Cash Flow for Equipment Y
= -50000 +40000X0.9091 +15000X0.82645 +15000X0.75131 + 15000X0.6830
60000 50000 40000
42500 40000
40000 30000
= 20,274
30000
17500 20000 15000 15000 15000
20000 10000
5000
0
0 0 1 2 3 4
0 1 2 3 4
-10000
-20000 -20000
-30000
• Hence choose X since the net present worth for the alternative X is more.
-40000 -40000
-50000
-60000 -50000 -60000 -50000

| Aritra Halder | Assistant Professor | Aritra Halder | Assistant Professor

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 12-04-2023

Thank You!
Contact Me
at
ahalder@nicmar.ac.in
+91-8504017164

| Aritra Halder | Assistant Professor