ENGINEERING ECONOMICS

(BNP 30402/BPK 30902)

CHAPTER 3:
COST ESTIMATION
TECHNIQUES (CET)

TS. DR. ROSLINDA BINTI ALI

FAKULTI TEKNOLOGI KEJURUTERAAN
UNIT TECHNIQUE
⦿ Involves using a ‘per unit factor’ that can be
estimated effectively.

⦿ Examples:
• Capital cost of plant per kilowatt of capacity

• Revenue per mile

• Capital cost per installed telephone

• Revenue per customer served

2
UNIT TECHNIQUE (CONT.)
⦿ An often used example is the cost of a particular
house. Using a per unit factor of, say, RM120 per
square foot, and applying that to a house with 3,000
square feet, results in an estimated cost of RM120 x
3,000 = RM360,000.

⦿ This techniques is useful in preliminary estimates,

but using average costs can be very misleading. In
general, more detailed methods will result in greater
estimation accuracy.

3
FACTOR TECHNIQUE
⦿ An extension of the unit technique where the products of several
quantities are summed and then added to components estimated
directly.

C = cost being estimated

Cd = cost of the selected component d estimated directly
fm = cost per unit of component m
Um = number of units of component m

⦿ Useful when the complexity of the estimating situation does not

require a WBS, but several different parts are involved.

4
FACTOR TECHNIQUE (CONT.)
⦿ Example 1:

▪ Suppose that we need a slight refined estimate of the

cost of a house consisting of 2,000 sq. meter, 2 porches,
and a garage. Using unit factor of RM85 per sq. meter,
RM10,000 per porch, and RM8,000 per garage for 2
directly components, calculate the total estimate of
building the house.

5
FACTOR TECHNIQUE (CONT.)
⦿ Solution 1:

▪ Given: unit factor: RM85/sq.meter; RM10,000/porch;

RM8,000/garage.
▪ Calculate the estimation for a house of 2,000 sq.meter;
2 porches; and 1 garage. Therefore:

(RM85 x 2,000sq.meter) + (2 x RM10,000) + RM8,000

= RM198,000

6
FACTOR TECHNIQUE (CONT.)
⦿ Example 2:

⦿ A commercial office building has 15,000 gross square feet of retail space in
the first floor, and the second floor has the same amount planned for office
use. Also, the size and location of the parking lot and the prime road
frontage available along the property may offer some additional revenue
resources. Analyse the potential revenue impacts based on the following
details:
1. The retail space should be designed for two different uses:
a. 60% for a restaurant operation (utilisation: 79%, yearly rent:
RM23/sq.feet)
b. 40% for a retail clothing store (utilisation: 83%, yearly rent:
RM18/sq.feet)
2. There is a high probability that all the office space in second floor will
be leased to one client (utilisation: 89%, yearly rent: RM14/sq.feet)
3. An estimated 20 parking spaces can be rented on a long term basis to
two existing businesses that adjoin the property (rate per month per
parking space: RM22)
4. One spot along the road frontage can be leased to a sign company, for
erection of a billboard (rate per month per billboard: RM65).

7
FACTOR TECHNIQUE (CONT.)
⦿ Solution 2:

1. First floor: 15,000sq.feet

a. Restaurant: [60%(15,000)](79%)(RM23) = RM163,530
b. Clothing store: [40%(15,000)](83%)(RM18) = RM89,640

2. Second floor: 15,000sq.feet

Rental: 15,000(89%)(RM14) = RM186,900

3. Parking space: 20(RM22)(12months) = RM5,280

4. Billboard: 1(RM65)(12months) = RM780

Total revenue: RM163,530 + RM89,640 + RM186,900 + RM5,280 + RM780

= RM446,130

8
PARAMETRIC COST ESTIMATING
⦿ Parametric cost estimating is the use of historical cost data and
statistical techniques to predict future costs.

⦿ Statistical technique are used to develop Cost Estimating

Relationships (CERs) that tie the cost/price of an item(e.g. a
product/ service/activity) to one or more independent variables
(i.e. drivers)

⦿ Parametric models are used in the early design stages to get an

idea of how much the product (project) will cost, on the basis of
few attributes (e.g. weight, volume and power).

⦿ The output of parametric models is used to gauge the impact

design decisions on the total cost.

⦿ Two commonly used estimating relationships:

1. The power-sizing technique
2. The learning curve

9
POWER-SIZING TECHNIQUE
⦿ Sometimes referred to as an exponential model – used for
developing capital investment estimates for industrial
plants and equipment.

(both in RM as of the point in time for which

the estimate is desired)

(both in the same physical units)

10
POWER-SIZING TECHNIQUE (CONT.)
⦿ Example 3:

▪ An aircraft manufacture desires to make a

preliminary estimate of the cost of building a
600-MW fossil-fuel plant for the assembly of its
new long-distance aircraft . A 200-MW cost
RM100 million 20 years ago with cost index 400.
The cost index now 1,200 and the cost-capacity
factor is 0.79.

11
POWER-SIZING TECHNIQUE (CONT.)
⦿ Solution 3:

Before using the power-sizing model to estimate the

cost of the 600-MW plant (CA), update the known cost of
200-MW plant 20 years ago to a current cost (CB).
▪ CA-200 = CB-200 (SA-200/SB-200)
RM100mil = CB-200 (400/1,200)
CB-200 = RM100mil (1,200/400)
= RM300 mil

Current cost CB-200 is RM300mil

12
POWER-SIZING TECHNIQUE (CONT.)
⦿ Solution 3 (cont.):

Therefore, the estimation for CA-600:

CA = RM300mil (600-MW/200-MW)0.79
= RM300mil (2.38)
= RM715mil

13
LEARNING CURVE
⦿ A learning curve is a mathematical model that explains the
phenomenon of increased worker efficiency and improved
organisational performance with repetitive production of a
good or service.

⦿ Is also called an experience curve or a manufacturing

progress function.

⦿ Basic concept: some input resources (e.g. energy costs,

labour hours, material costs, engineering hours) decrease,
on a per-output-unit basis, as the number of units
produced increases.

14
LEARNING CURVE
⦿ Most learning curves are based on the assumption that a constant
percentage reduction occurs in, say, labour hours, as the number
of units produced is doubled.

⦿ E.g. if 100 labour hours required to produce the first output unit
and a 90% learning curve is assumed, then 100(0.9) = 90 labour
hours would be required to produce the second unit.

⦿ Similarly, 100(0.9)2 = 81 labour hours would be needed to

produce the fourth unit, 100(0.9)3 = 72.9 labour hours would be
needed to produce the eight unit, and so on.

⦿ Therefore, a 90% learning curve results in a 10% reduction in

labour hours each time the production quantity is doubled.

15
LEARNING CURVE (CONT.)
⦿ Most learning curves assume a constant percentage
reduction occurs as the number of units produced is
doubled.

16
LEARNING CURVE (CONT.)
⦿ The total time to produce x units, Tx is given by:

17
LEARNING CURVE (CONT.)
⦿ Example 4:

⦿ Assume the first unit of production required 3

hours time for assembly. The learning rate is
75%. Find:
a. the time to assemble the 8th unit
b. the time needed to assemble the first 6 units.

18
LEARNING CURVE (CONT.)
⦿ Solution 4:

a. Z8 = 3(8)log 0.75/log 2
= 3(8)-0.415
= 1.27 hours

b. Time needed to produce 6 units

x x
Tx = ∑ Z u ; Tx = ∑ K(un)
u=1 u=1
6
log 0.75/log 2
T6 = ∑ 3(u )
u=1

= 3[ 1-0.415 + 2-0.415 + … 6-0.415]

= 11.8 hours

19
LEARNING CURVE (CONT.)
⦿ Example 5:

⦿ A construction company has ordered 10 specialised test

units capable of field checking 15 separate elements in
potable water in emergency situations. The company took
200 hours to build the first unit. If direct and indirect
labour costs average RM50 per hour, and an 80% learning
rate is assumed, estimate:
1. The time needed to complete units 5 and 25.
2. The total labour cost for the 10 units.

20
LEARNING CURVE (CONT.)
⦿ Solution 5:
1. Z5 = 200 (5) log0.8/log2
= 200 (5) -0.322
= 119 hours

Z25 = 200 (25) log0.8/log2

= 71 hours

10
2. T10 = ∑ 200 (u) -0.322
= 1263 hours
u=1

Total labour cost = (RM50) (1263)

= RM63,150

21
THANK YOU…

22