### 興建A工程 - 國立交通大學

```工程專案生命週期之成本估價方法

2



（National Nano Device Laboratories；NDL），若無法適用於



( CIC + NDL )
7F~10F
CIC
(CUB:廠務+C10K)
(FAB:C100)
2F~6F
NDL
B2F~1F 停車空間

【資料來源：國家奈米元件實驗室新建工程，2004】
3

（國內成本估價教科書之回顧）

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1979
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1980
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1983
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1987
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1989
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1990
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1990
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1995
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1996
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1997
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1998
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2001
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2002
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4


（國外成本估價教科書之回顧）

1
Construction
estimate and costs
Pulver
1960
－
●
●

2
Cost Estimating for Engineering
and Management
Ostwacd
1974
●
●
●

3
Estimating in Building
Construction
Dagostino
1989
－
●
●

4
Preconstruction Estimating
Budget Through Bid
O’Brien
1994
●
●
●

5
Construction Estimates From
Take-off To Bid
Trauner,et al.
1995
－
●
●

6
Project Cost Estimating
Smith
1995
●
●
●

7
Estimating Construction Cost
Peurifoy,
berlender
2002
●
●
●

5

3.1 成本估價作業流程
1. 工程專案之成本估價作業流程（通案情況）
2. NDL之成本估價作業流程（案例情況）
3.2 成本估價方法之案例說明
1. 單位基準法（單位面積法、單位設備法、單位柱間法、單位

2. 成本指標法
3. 成本對應產量估價法
4. 曲線估價法 (Curve estimate)
5. Factored estimate
3.3 工程會作法-建築工程案例
6
Three key questions in choosing an estimating
method:
 What is the purpose of the estimate?
 How precise of an estimate do I need?
 How much time do I have?
Order-of-Magnitude Estimates (粗估)
Intermediate Estimates (概估)
Definitive Estimates
(+ - 50%)
(+ - 30%)
(+ - 20%)
7
3種估算方式(PMBOK)：

Top Down (Analogous) ：由上往下 (類比估算)

Parametric cost estimate：參數估算

Bottom Up：由下往上
8
Top Down (Analogous)
• To compare the project with one from your past.
• Is it half as big or 5 times bigger? The focus is on what is
different from the previous project.
• This technique gives the widest margin of uncertainty.
Estimating steps:
1. Identify similar projects.
2. Obtain historical data. How much did the similar project cost?
3. Re-evaluate and document the similarities and differences.
4. Apply managers’ expert judgment and decide on estimate.
5. Document the estimate and estimating assumptions.
need not be individually identified)
9
Parametric cost estimate:




the one that uses Cost Estimating Relationships and
associated mathematical algorithms (or logic) to establish
cost estimates.
Normally associated with construction.
Based on historical data.
For example:
 If historically it takes 40 hours to design, create, build
and implement a report, 10 reports is going to take 400
hours.
Note:
Detailed cost estimating process is laborious and time consuming.
The need to reengineer business processes and reduce cost in the Department of
Defense (DoD) has led to a parametric cost estimating initiative. (around 1994)
10
Bottom Up





Requires a more detailed analysis of the project.
The Work Breakdown Structure (WBS) is defined to identify
the tasks associated with the work products.
Each task is estimated and the cost for all of the individual
pieces is combined to create the overall total.
 more accurate in the detailed tasks.
11
3.1 概念與可行性研究階段之成本估價作業流程（通案情況）

－
－
－

－
－

－
－

－
－

－

A/E

、

【資料來源：整理自楊世清（1998）、鄭正光（2002）】
12
3.1 概念與可行性研究階段之成本估價作業流程（公共工程）


0
30％
100％

【資料來源：各機關辦理公有建築物作業手冊（2001）、公共建設工程經費估算編列手冊（2001）】
13


（可行性研究）

(地方政府)

（每年元月）

(地方民意代表會或議會)

1.專業審議，核定實需總工程費
2.審查第一年度工程費

【資料來源：公共建設工程經費估算編列手冊（2001）】
14
3.1 概念與可行性研究階段之成本估價作業流程（案例情況）
NDL之成本估價流程

NDl之作業流程

(可行性

（基準估價）

1.主辦機關進行可行性研究及

2.主管機關審查同意後，報請

－
－

－
－

－
－

－
NDL 概念與可行性研究階段之成本估價作業流程
15
3.2 概念與可行性研究階段成本估價方法之案例說明（方法簡介）
1.單位基準法

（1）單位面積法： 總工程費＝建築物之樓地板面積總量×

（2）單位設備法： 總工程費＝建築物設備之總數 ×

（3）單位柱間法： 總工程費＝建築物等柱間之總數 ×

（4）單位體積法： 總工程費＝建築物體積之總數 ×

16
2.成本指標法

3.成本對應產量估價法

X
17
1.單位基準法－(1)單位面積法（定義整理說明）

C＝ X × Y

Y：興建工程之樓地板面積總量（M2）
C：興建工程之總工程費（元）
1.越相似之工程類型，其估算準確度相對提高。
2.單價由大量歷史之成本資料中取得。
3.單價易受建築物之形狀及性質之不同，為主要誤差來源。
（1）計算興建工程之樓地板面積總量（Y ）

（2）計算單位樓地板面積之單價（ X ）
（3）計算興建工程之總工程費（C）
【資料來源：整理自洪億萬（1993）、Clough,et al.（2000）】
18



19

（1）計算興建A工程之樓地板面積總量，主辦機關於此階段

1

1,650
2

1,650
3

1,650
4

1,650
5

1,650
6

1,650
7

1,650
8

2,310
9

2,310

16,170M2（4,900坪）
20

（2）計算過去B工程之單位樓地板面積之單價

1.

1
1,702,600
1,702,600
2.

1
4,025,330
4,025,330
3.

1
17,502,572
17,502,572
4.

1
6,577,970
6,577,970
5.

1
2,801,234
2,801,234
6.

1
1,361,760
1,361,760
7.

1
750,265
750,265

34,721,731

1
2,777,738
2,777,738

1
1,874,973
1,874,973

39,374,442

21

（3）計算興建A工程之總工程費


A之樓地板

B之單位樓地

A之



（275,339,631元）為基準，利用單位面積法所推估之總工程費之



22
1.單位基準法－(2)單位設備法（定義整理說明）

C＝ X × Y

Y：興建工程設備之總量（如式, 人）
C：興建工程之總工程費（元）
1.越相似之工程類型，其估算準確度相對提高。
2.需對興建工程進行詳細之調查。

（1）計算興建工程設備之總數（Y）

（2）計算單位設備之單價（X）
（3）計算興建工程之總工程費（C）
【資料來源：整理自洪億萬 (1993)、Clough,et al. (2000)、林文天(1985)】
23





（稱為興建A工程）之實際案例之總工程費，說明單位







24

（1）興建A工程規劃之教室間數計算如下：

＝4（層）× 12（間/層）
＝48（間）
（2）利用過去B工程之總工程費推估單位教室間數之單價

B之教室間數總數＝20 (間)
B之總工程費＝29,599,899 (元)
B之單位教室間數之單價
＝29,599,899 / 20
＝1,479,995 (元/間)
25

（3）計算興建A工程之總工程費


A之總教


B之單位教

A之總

A之單位設

A之

A之

26
1.單位基準法－(3)單位柱間法（定義整理說明）

C ＝ (X × Y)

Y：興建工程等柱間之總數（M2）
C：興建工程之總工程費（元）
1.適用規則性較高之建築工程類型(如工廠、教室、百貨商店)。
2.由於建築物立柱間距離相等，故估價較為準確。
1.計算興建工程等柱間之總數（Y）

2.計算單位等柱間樓面積之單價（X）
3.計算興建工程之總工程費（C）
【資料來源：整理自洪億萬(1970)、林文天(1985)】
27





28

（1）首先，計算興建A工程等柱間樓面積之數量，以下為其平面圖。

29

（1）計算興建A工程等柱間之總數。

(1)
4（1～4層）× 14（個/層）＝56 個
(2)
4（1～4層）× 2（個/層）＋1（地下層）× 2（個/層）＝10 個
(3)
4（1～4層）× 2（個/層）＋1（地下層）× 2（個/層）＝10 個
(4)
4（1～4層）× 1（個/層）＝4 個
（2）計算興建A工程單位等柱間樓面積之單價。

(1)
9.00×12.92＝116.28
116.28 × 7,660＝890,705
(2)
10.75×12.92＝138.89
138.89 × 7,660＝1,063,898
(3)
10.75×11.10＝119.33
119.33 × 7,660＝914,030
(4)
6.00×12.92＝77.52

77.52 × 7,660＝593,803

與單位面積法相同
30

（3）計算興建A工程之總工程費
A等柱間

(1)
56（個）× 890,705（元/個） ＝ 49,879,480
(2)
10（個）× 1,063,898（元/個）＝ 10,638,980
(3)
10（個）× 914,068（元/個） ＝ 9,140,680
(4)
4（個）× 593,803（元/個） ＝ 2,375,212


A等柱間

72,034,352元

A之單位柱

A之

A之

31










32

(圖1)
33
(圖2)
(圖3)
• 單位柱間法 --- 如圖2所示，考量了左右兩邊牆柱數目的不同，將分

• 單位面積法 --- 如圖3所示，其柱牆將視為均勻擺設，僅考慮樓地板

• 故兩種估算法在圖1右邊將會在此產生差異。
34
1.單位基準法－(4)單位體積法（定義整理說明）

C＝X×Y

Y＝興建工程體積之總量（M3）
C＝興建工程之工程費（元）
1.越相似之工程類型，其估算準確度相對提高。
2.體積計算較為複雜。
3.單位體積法較單位面積法準確?
1.計算興建工程體積之總量（Y）

2.計算單位體積之單價（X）
3.計算興建工程之總工程費（C）
【資料來源：整理自洪億萬（1993）、Clough,et al.（2000）、林文天(1985)】
35



B工程）之實際案例，推估桃園縣某國小新

36

（1）計算興建A工程之立體示意圖之體積總量

＝36,544.1 (M3)
（2）計算過去B工程之立體示意圖之單位體積單價

B之體積總量＝14,836.6 (M3)
B之總工程費＝29,599,899 (元)
B之單位體積單價
＝ 29,599,899 / 14,836.6
＝ 1,995 (元/M3)
37

（3）計算興建Ａ工程之總工程費


A之


B之單位體

A之

A之單位體

A之

A之

38

Ct  (U avg 
B )  E
v
c
Ct ：總建造費
U avg：每立方公尺空間平均單價
 B：建築物總體積
v
Ec：特殊工程費(挖土、基礎及附屬設備等之特殊工程費)


 無閣樓時 Bv  A  ( H  h / 2)




A=面積、H=樓高、h=樓頂高(尖屋頂或斜屋頂之樓高)

39


(http://www.cpacltd.com/list/Immovable-Property-technical-specification_5.html)

40
2.成本指標法（定義整理說明）

I 
Cc   c   C r
 Ir 

Cr：已知設備容量之成本
Ic：預估新設備之建造年指標
Ir：已知設備之基準年指標
1.計算快速。
2.需注意成本指標之適用性。
（1）計算已知、預估新設備容量之成本指標（Ir）、（ Ic ）

（2）計算預估新設備容量之成本（Cc ）
【Ostwald (1992), Clough,et al. (2000), Barrie and Paulson (1992)】
41



1月
2月
3月
4月
5月
6月
7月
8月
9月
10月
11月
12月

2001
6281
6272
6279
6286
6288
6318
6404
6389
6391
6397
6410
6390
6343
2002
6462
6462
6502
6480
6512
6532
6605
6592
6589
6579
6578
6563
6538
2003
6581
6640
6627
6635
6642
6694
6695
6733
6741
6771
6794
6782
6694
2004
6825
6862
6957*
7017
7065
7109

1月
2月
3月
4月
5月
6月
7月
8月
9月
10月
11月
12月

2001
3545
3536
3541
3541
3547
3572
3625
3605
3597
3602
3596
3577
3574
2002
3581
3581
3597
3583
3612
3624
3652
3648
3655
3651
3654
3640
3623
2003
3648
3655
3649
3652
3660
3677
3683
3712
3717
3745
3765
3757
3693
2004
3767
3802
3859*
3908
3956
3996
【資料來源：Engineering News Record，2004】
42



（稱為興建A工程）之總工程費，說明成本

43

（1）計算過去B工程之基準年指標、成本


Ir＝ 100.26
B之單位樓地板面積之單價＝ 14,681（元/M2）
（2）計算興建A工程之建造年指標、成本


Ic＝100.20
A之單位樓地板面積之單價＝(100.20/100.26)×14,681(元/M2)＝14,672(元/M2)
88年12月
89年1月
89年2月
89年3月
89年4月
89年5月
89年6月
89年7月

104.15
103.47
103.59
103.99
104.08
104.10
104.06
103.88

107.35
106.25
106.42
107.11
107.26
107.36
107.30
106.96

99.36
99.27
99.29
99.31
99.31
99.23
99.21
99.22

100.26
99.68
99.88
100.12
100.21
100.20
100.12
99.90

107.22
106.47
106.49
107.05
107.14
107.19
107.19
107.03

【資料來源：營建物價，2003】
44

（3）計算興建A工程之總工程費


A之

A之單位樓

A之





45



I 
Cc   c   C r
 Ir 
 I ci
Y   Cri 
 Ir
 i





推估現有工程項目之成本

Y :總建造費
i :工程項目i
Cri :某基準年工程項目成本
I r :某基準年成本指數
I c :推估年成本指數
i
i
46





Assume that we have an estimate on file for a similar
structure that we completed in 1978 including design and
owners expense for a cost of \$4,200,000. We were
planning to build the new warehouse in 1991.
The Engineering News Records (ENR) index for 1978,
relative to a base date of 1967, was 1674/676 = 2.48 (or
248%).
The projected index for 1991 was 2730/676 = 4.04.
Therefore, the estimated cost for a project of similar size
and quality will be (4.04/2.48)* \$4,200,000= \$6,840,000
47
3.成本對應產量估價法（定義整理說明）

capacity對成本之影響，預估新設備之總工程費。
Q
C 2  C1  2
 Q1




X

C1：已知設備容量之成本
Q2：預估新設備之容量
Q1：已知設備之容量
X：專案類型之成本容量因素
Has been most widely used in the petrochemical construction industry.
X=0.6 is typical to some types of plants. 【Barrie and Paulson 1992】
（1）計算專案類型之成本容量因素（X）

（2）計算預估新設備容量之成本（C2）
【Oberlender 2000; Barrie and Paulson 1992; Clark and Lorenzoni 1985 】
48



49

（1）計算專案類型之成本容量因素（利用過去B、C、D工程）


(元)

B

60,000,000
5,000
12,000
C

83,080,000
6,200
13,400
D

112,500,000
7,500
15,000
X計算說明如下：
1.C-B辦公大樓：13,400＝12,000×（6,200/5,000）XC-B, XC-B＝0.51 。
2.D-B辦公大樓：1,5000＝12,000×（7,500/5,000）XD-B , XD-B≒0.55 。

50

（2）計算興建A工程之總工程費


＝12,000×（6,000 / 5,000）0.53 ＝13,218 (元/M2)。

＝6,000 (M2) ×13,218 (元/M2)＝79,308,000元。
51
Cost indices vs. Cost capacity factors





Cost indices: focuses on cost changes over time
Cost-capacity factors: focuses on changes in size,
scope, or capacity of similar projects
X: based on historical records
Q: such as
 Maximum barrels per day  refinery
 Tons of steel per day  steel mill
 Gross floor area  warehouse
To consider changes in both time and capacity:
 I  Q
C  C  
 I  Q
2
2
2
1
1
1



X
52
An example
Department is conducting a preliminary estimate of the cost
of building a 600-MW fossil-fuel plant. It is known that a
200-MW plant cost \$100 million 20 years ago when the
approximate cost index was 400, and that cost index is now
1,200. The cost capacity factor for a fossil-fuel power plant
is 0.79.
 I 2  Q2 

C2  C1  
 I1  Q1 
X
 1200  600MW 
 100


 400  200MW 
0.79
 30030.79  714.6
53
In the earlier example

Considering only cost indices


Considering only cost-capacity factor





Warehouse cost = \$6,840,000 = (4.04/2.48)*4,200,000
assume that cost varies closely with floor area; assume that X = 0.8
assume that floor area = 120,000 square feet for the past project
assume that floor area = 150,000 square feet for the current project
Warehouse cost = \$42,000,000*(150,000/120,000)^0.8 =
\$5,020,000
Considering both cost indices and cost-capacity factor

Warehouse cost = \$42,000,000 * (4.04/2.48) *
(150,000/120,000)^0.8 = \$8,176,000
54



Example: Estimate cost of 100,000 bbl/day pipestill at

A pipestill is a first stage, key process plant in a refinery that
distills the crude oil into heating oil, gasoline, etc.

The company has built a similar plant near the city of Rotterdam in
1979.
55

A good example:

40*150%=60 million





New plant does not have some
extra cost items
Escalation
Location factor
Proration factor (0.55) to
reflect the size difference
Pollution requirement
Higher competitiveness 
5% deduction in total cost
56

estimate
57
4.曲線估價法 (Curve estimate)


Same example: Estimate cost of 100,000 bbl/day
The curve is plotted from historical data (e.g., final cost)
for similar pipestills. only for on-site facilities




【Clark and Lorenzoni 1985】
Y axis: total erected cost (TEC)
X axis: variable that impact the most on the cost (i.e., the feed
(進料器) to the unit and the amount of overhead product)
Off-site facilities (auxiliary or support equipment) =
percentage of on-sites.
In the example,


The curve reflects a base year (1960) and a standard location =
25 million.
An index (354%) is used to reflect a specific time (year 1986)
58
(百萬桶/天)
59
On-site and off-site curves can be done in more details
 a more accurate estimate.
60




Need significant historical data
Really is a hardware predictor
More on-site oriented
Accuracy range = ±20% (?)
61
5. Factored estimate (Hendrickson and Au 1989)



Commonly used in process industries.
An industrial process requires several major equipment
components (such as towers, drums, and pumps in a
chemical processing plant) plus ancillary items (such as
piping, valves, and electrical elements).
Total cost (CTotal) = costs of major equipment components
(Ci) + costs of their ancillary items (fi × Ci)

CTotal = (Ci) + (fi × Ci)
where fi is a fraction or multiple of Ci
62
Example 5-14: Conceptual estimate for a chemical processing plant
In making a preliminary estimate of a chemical processing plant, several major
types of equipment are the most significant parameters in affecting the
installation cost. The cost of piping and other ancillary items for each type of
equipment can often be expressed as a percentage of that type of equipment for
a given capacity. The standard costs for the major equipment types for two
plants with different daily production capacities are as shown in Table 1.
 It has been established that the installation cost of all equipment for a plant with
daily production capacity between 100,000 bbl and 400,000 bbl can best be
estimated by using linear interpolation of the standard data.

TABLE 1: Cost Data for Equipment and Ancillary Items
Equipment
type
Furnace (火爐)
Tower
Drum
Pump, etc.
Equipment Cost (\$1000)
100,000 bbl
3,000
2,000
1,500
1,000
400,000 bbl
10,000
6,000
5,000
4,000
Cost of ancillary items
as % of equipment cost (\$1000)
100,000 bbl
400,000 bbl
40%
30%
45%
35%
50%
40%
60%
50%
63
A new chemical processing plant with a daily production capacity of 200,000
bbl is to be constructed (熔爐)
in Memphis, TN in four years. Determine the total
preliminary cost estimate of the plant including the building and the
equipment on the following basis:
1.The installation cost for equipment was based on linear interpolation from
Table 1, and adjusted for inflation for the intervening four years. We expect
inflation in the four years to be similar to.…
the period 1990-1994.
2.The location index for equipment installation is 0.95 for Memphis, TN, in
field labor
comparison with theFLOH:
standard
cost.
3.An additional cost of \$500,000 was required for the local conditions in
Memphis, TN.
64
The solution of this problem can be carried out according to
the steps as outlined in the problem statement:
• The costs of the equipment and ancillary items for a plant with a
capacity of 200,000 bbl can be estimated by linear interpolation
of the data in Table 1 and the results are shown in Table 2.
Table 2: Results of Linear Interpolation for an Estimation Example
Equipment
type
Furnace
Tower
Drum
Pumps, etc.
Equipment Cost
Percentage for
(in \$1,000)
ancillary items
\$3,000 + (1/3)(\$10,000-\$3,000) = \$5,333 40% - (1/3)(40%-30%) = 37%
\$2,000 + (1/3)(\$6,000-\$2,000) = \$3,333 45% - (1/3)(45%-35%) = 42%
\$1,500 + (1/3)(\$5,000-\$1,500) = \$2,667 50% - (1/3)(50%-40%) = 47%
\$1,000 + (1/3)(\$4,000-\$1,000) = \$2,000 60% - (1/3)(60%-50%) = 57%
65



Hence, the total project cost in thousands of current dollars is
given as: CTotal = (Ci) + (fi × Ci) = [Ci (1+ fi )]
= (\$5,333)(1.37) + (\$3,333)(1.42) +(\$2,667)(1.47) + (\$2,000)(1.57)
= \$2,307 + \$4,733 + \$3,920 + \$3,140
= \$ 19,100
The corresponding cost in thousands of four year in the future
dollars is (considering inflation) :
(\$19,100)(105/94) = \$21,335
The total cost of the project after adjustment for location is
(0.95)(\$21,335,000) + \$500,000 = \$20,768,250
66


Advantage: this estimate reflects specific design
A factored estimate has a tendency to be biased low.




The factored estimate is a series of building blocks. If any of
these blocks are omitted, the estimate will be low.
The interface between building blocks is not always clearly
defined.
“Should” account for all items
Accuracy range = ±15% (?)
67
(Clark and Lorenzoni 1985)
68
3.3 工程會作法-建築工程案例

69

70

 單位面積直接成本概估法
 依據各地區相關工程近期發包資訊，對於相同樓層數、同級建

 預估直接工程成本(A)＝單位面積造價×總樓地板面積
 地區修正係數(B)＝因施工現場所在位置，考慮勞工、材料之供

 特別修正係數(C)＝高難度施工或特殊指定建材或特定承包商或

 修正後金額＝A×B×C


 將主要工程項目予以量化，再依個別情況及條件求得單價，以

71

72
73
74
75



 基地面積： 2,100M2 （約635坪）
 總樓地板面積： 8,264M2 （約2,500坪）
 地下室面積： 980M2
 屋頂面積： 607M2
 外牆面積： 5,300M2
76









77

=137,636*0.035
=137,636*0.1
78

79

80


 b = (1+a)^(n-1)
 a: 物價調整年增率


 物價調整費＝(137,636K+13,764K+13,764K)×(1.018－
1.00)＝(165,164K×0.018)=2,973K元
81

82

=138,125千元
83

84
```