Transcript CLIFF SCHEXNAYDER EMINENT SCHOLAR DEL E. WEBB …

Chapter 2
Fundamental Concepts of
Equipment Economics
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
EQUIPMENT
ECONOMICS
OBJECTIVE:
Provide the right equipment at
the right time and place, so the
work can be accomplished at
the lowest cost.
EQUIPMENT
ECONOMICS
How much does it cost
to operate the machine
on a project?
WORK AT THE
LOWEST COST
Not the same as lowest
initial cost.
Cost of Money
Interest Rate
Sources of capital funds:
 Borrow
 Earnings
 Equity
WORK AT THE
LOWEST COST
• Machine utilization
• Utilization drives purchase
or rent/lease decision.
SPECIFIC
OBJECTIVES
• Minimize ownership and
operating cost
• Increase availability
• Increase utilization
LIFE CYCLE COST
• Predicting costs of
Ownership and Usage.
• Using that information
in decision making.
TOTAL EQUIPMENT
COST
$ Depreciation
$ Operating
$ Repair
$ Overhead
25%
23%
37%
15%
EQUIPMENT COST
Question No. 1
What is the largest
single equipment
cost?
TOTAL EQUIPMENT
COST
$ Depreciation 25%
$ Operating
23%
$ Repair
$ Overhead
37%
15%
INFORMATION SYSTEM
 Machine identification
 Utilization data
 Purchase cost
 Repair cost
 Operating charges
FOG (fuel, oil, grease)
UTILIZATION DATA
 Load
 Speed
 Environment
 Underfoot
UTILIZATION
Utilization is working
time duration
not calendar
duration.
UTILIZATION DATA
Basis of costing
 Hourly
 Daily
 Weekly
 Miles
 Fuel consumption
OWNING & OPERATING
COST QUESTIONS
 Rental rates for
internal costing
OWNING & OPERATING
COST QUESTIONS
 Maintenance and PM
decisions
 Disposal/Replacement
guidance
OWNERSHIP COST
 Depreciation (Purchase
expense)
 Insurance
 Taxes
 Salvage value
 Shop expenses
OWNERSHIP COST
Purchase price*
- Salvage
+ Overhead
= Ownership expense
* A fixed cost, occurs if the
machine works 1,000 or 17,000 hrs
OWNERSHIP COST
Usage
Average Hours to
Replacement
•
•
•
•
•
•
•
Rollers
Wheel loaders
Crawler dozers
Hydraulic excavators
Graders
Scrapers
Off-highway trucks
9,500
12,300
12,500
12,500
14,300
16,100
18,300
OWNERSHIP COST
Question No. 2
Regardless of how
much a machine is
used, the owner
must pay owning
cost.
True
False
OWNERSHIP COST
Question No. 2
Regardless of how
much a machine is
used, the owner must
pay owning cost.
True
False
OWNERSHIP COST
New
$260,000
4 yr later
$120,000
Ownership expense = $140,000
OWNERSHIP COST
New
8 yr later
$260,000
$0
Ownership expense = $260,000
OWNERSHIP COST
 Include the cost of all
attachments and delivery
charges in initial machine cost
(delivered price).
 Deduct tire cost for wheeltype machines
OWNERSHIP COST
Question No. 3 Machine
owning cost includes which
of the expenses listed?
$ Repairs
$ Insurance
$ Fuel
$ Taxes
OWNERSHIP COST
Question No. 3 Machine
owning cost includes which
of the expenses listed?
$ Repairs
$ Insurance
$ Fuel
$ Taxes
OPERATING COST
 Consumables - fuel, oil,
grease & filters (FOG)
 Repair
 Maintenance
 Tires or tracks
 High wear items - cutting
edges, teeth
OPERATING COST
High wear items
 Bucket teeth
 Cutting edges
 Truck body liners
 Ripper tips
 Ripper shank protectors
OPERATING COST
Question No. 4 Machine
operating cost includes which
of the expenses listed?
$ Repairs
$ Interest
$ Tires
$ Taxes
OPERATING COST
Question No. 4 Machine
operating cost includes which
of the expenses listed?
$ Repairs
$ Interest
$ Tires
$ Taxes
OPERATING COST
Usage
Ownership Cost
Example
Determine the probable cost
per hour of owning and
operating a scraper given the
following conditions.
 Engine 350HP diesel
Ownership Cost
 Overall cost of money 10%
 Useful life 5 years
 Hours used per year 2000
 Initial cost $470,000
 Cost of tires $30,000
 Estimated salvage value $60,000
Ownership Cost
Time Value Method
Deduct tire cost from the
delivered price for large
machines.
Tires are considered a wear
item and are treated as an
operating cost.
Ownership Cost
Time Value Method
 Initial cost $470,000
 Cost of tires $30,000
$440,000
Need to calculate the uniform
series required to replace a
present value of $440,000
Uniform series capital recovery
factor [Eq. 2.6]
Uniform series capital
recovery factor [Eq. 2.6]
 i1  i  
A  P

n
 1  i   1
n
A = uniform end-of-period payments or
receipts continuing for a duration of n
periods.
Uniform series capital
recovery factor [Eq. 2.6]
 i1  i  
A  P

n
 1  i   1
n
P = a present single amount of money.
i = the rate of interest per period of
time (usually 1 year)
Uniform series capital
recovery factor [Eq. 2.6]
 i1  i  


n


1

i

1


n
 Overall cost of money 10% (i)
 Useful life 5 years (n)
Uniform series capital
recovery factor
 0.101  0.10  


5


1

0
.
10

1


5
 Overall cost of money 10%
 Useful life 5 years
Uniform series capital
recovery factor
 0.161051 
 0.610510 
 0.263797
Ownership Cost
Time Value Method
A = $440,000  [0.263797]
A = $116,071 per year
P = a present amount of money.
Ownership Cost
Time Value Method
• Estimated salvage value $60,000
Need to calculate the uniform
series required to replace a end
of period amount of $60,000
Uniform series sinking fund
factor [Eq. 2.4]
Uniform series sinking
fund factor


i
A  F

n


1

i

1


F = a future single amount after
n periods of time.
Uniform series sinking
fund factor


0.10
A  F

5
 1  0.10   1
F = $60,000
Uniform series sinking
fund factor
A = $60,000  [0.163797]
A = $9,828 per year
Ownership Cost
Time Value Method
$116,071- $9,828
=
2,000 hr / yr
$53.12 / hour
Ownership Cost
Average Annual
Investment Method
P(n  1)  S(n  1)
AAI 
2n
P = purchase price
S = salvage value
n = life in years
Ownership Cost
Average Annual
Investment Method
$440,000(5  1)  $60,000(5  1)

25
P = $440,000
S = $60,000
n = 5 years
Ownership Cost
Average Annual
Investment Method
$440,000( 6)  $60,000( 4 )

10
AAI = $288,000/yr
Investment Cost
$288,000 / yr  10%
=
2,000 hr / yr
$14.40 / hour
Depreciation
$470,000
- $30,000 tires
- $60,000 salvage
= $380,000
Hourly Depreciation
$380,000

5 yr  2,000 hr / yr
$38.00 / hr
Ownership Cost
Average Annual
Investment Method
$14.40 + $38.00 =
$52.40/hr
Ownership Cost
Time Value Method $53.12/hr
Average Annual
Investment Method $52.40/hr
Difference between the
methods $0.72
Operating Cost
Example
 Annual cost of repairs equals
70% of str. Line depreciation
 Operating factor, 0.5
 Cost of fuel $1.02 per gal
Operating Cost
 Crankcase capacity, 14 gal
 Time between oil changes,
200 hr
 Cost of lube oil $2.50 per gal
 Cost of other oils and grease
$0.45 per hour
Operating Cost
 Repairs to tires 14% of
tire depreciation
 Life of tires 4,000 hours
OPERATING
COST
Repair
Hourly depreciation $38.00
$38.00 X 70% =
$26.60 per hour
OPERATING COST
Fuel
 Operating factor, 0.5
 Cost of fuel $1.02 per gal
See discussion top of page 36.
0.04 X 350hp X 0.5 = 7 gal/hr
7 gal X $1.02/gal = $7.14/hr
OPERATING COST Fuel
Manufacturer’s consumption data
can be used to estimate fuel.
OPERATING COST
Oil & Grease
Equation 2.12, page 37
What the engine burns
Oil changes
350hp  0.5  0.006 14 gal
q

7.4
200 hr
q = 0.1418919 + 0.070 = 0.212 gal/hr
OPERATING COST
Oil & Grease
 Cost of lube oil $2.50 per gal
 Cost other oils and grease $0.45/hr
0.212 gal/hr X $2.50/gal = $0.53/hr
Other oils and grease
= $0.45/hr
Total cost O&G = $0.98/hr
OPERATING COST
Tire Repair
 Tire repairs 14% of tire depreciation
 Life of tires 4,000 hours
$30,000
 $7.50 per hour
4,000 hr
$7.50 X 14% = $1.05 per hour
OPERATING COST
Tire Depreciation
How many tire replacements
Life of tires 4,000 hours
5 yr  2,000 hr / yr
 2.5 sets
4,000 hr
Therefore 3 sets
OPERATING COST
Tire Depreciation
First set: (purchased at time 0)
Uniform series capital recovery factor
[Eq. 2.6]
$30,000  (0.263797)
 $3.96/hr
2,000 hr
OPERATING COST
Tire Depreciation
Second set: (purchased at
time 2 years in future) must
first calculate value at time
zero.
OPERATING COST
Tire Depreciation
Second set: 2 years in future
If a future amount F is given, the
present amount P equals [Eq. 2.2]:
F
P
n
1  i 
Present Worth Compound
Amount Factor
1
n
1  i 
 Overall cost of money 10% (i)
 n = 2 years
Present Worth Compound
Amount Factor
1

0
.
826446
2
1  0.10 
 Overall cost of money 10%
 n = 2 years
OPERATING COST
Tire Depreciation
Second set: value at time zero.
P = $30,000 X (0.826446)
= $24,793
$24,793  (0.263797)
 $3.27/hr
2,000 hr
OPERATING COST
Tire Depreciation
Third set: (purchased at time 4 yr)
must first calculate value at time
zero.
P = $30,000 X (0.683013)
= $20,490
OPERATING COST
Tire Depreciation
Third set: purchased 4 years in
the future $20,490
$20,490  (0.263797)
 $2.70/hr
2,000 hr
OPERATING COST
Tire Depreciation
First set:
$3.96/hr
Second set: $3.27/hr
Third set:
$2.70/hr
Total Tire Dep. $9.93/hr
Operating Cost
 Repair
 Fuel
 Lube Oils
 Tire repair
 Tire dep.
$26.60/hr
7.14
0.98
1.05
9.93
 Total Oper $45.70/hr
OPERATING COST
OPERATOR WAGES