7
SUPPLEMENT
Capacity and
Constraint
Management
PowerPoint presentation to accompany
Heizer, Render, Munson
Operations Management, Twelfth Edition
Principles of Operations Management, Tenth Edition
PowerPoint slides by Jeff Heyl
Copyright © 2017 Pearson Education, Inc.
S7 - 1
Outline
►
►
►
►
Capacity
Bottleneck Analysis and the Theory
of Constraints
Break-Even Analysis
Reducing Risk with Incremental
Changes
Copyright © 2017 Pearson Education, Inc.
S7 - 2
Outline - Continued
►
►
Applying Expected Monetary Value
(EMV) to Capacity Decisions
Applying Investment Analysis to
Strategy-Driven Investments
Copyright © 2017 Pearson Education, Inc.
S7 - 3
Learning Objectives
When you complete this supplement
you should be able to:
S7.1 Define capacity
S7.2 Determine design capacity,
effective capacity, and utilization
S7.3 Perform bottleneck analysis
S7.4 Compute break-even
Copyright © 2017 Pearson Education, Inc.
S7 - 4
Learning Objectives
When you complete this supplement
you should be able to:
S7.5 Determine the expected monetary
value of a capacity decision
S7.6 Compute net present value
Copyright © 2017 Pearson Education, Inc.
S7 - 5
Capacity
►
The throughput, or the number of units
a facility can hold, receive, store, or
produce in a period of time
►
Determines
fixed costs
►
Determines if
demand will
be satisfied
►
Three time horizons
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S7 - 6
Planning Over a Time Horizon
Figure S7.1
Options for Adjusting Capacity
Time Horizon
Long-range
planning
Intermediaterange
planning
(aggregate
planning)
Design new production processes
Add (or sell existing)
long-lead-time equipment
Acquire or sell facilities
Acquire competitors
Subcontract
Add or sell equipment
Add or reduce shifts
Short-range
planning
(scheduling)
*
Build or use inventory
More or improved training
Add or reduce personnel
*
Schedule jobs
Schedule personnel
Allocate machinery
Modify capacity
Use capacity
* Difficult to adjust capacity as limited options exist
Copyright © 2017 Pearson Education, Inc.
S7 - 7
Design and Effective Capacity
►
Design capacity is the maximum
theoretical output of a system
►
►
Normally expressed as a rate
Effective capacity is the capacity a firm
expects to achieve given current
operating constraints
►
Often lower than design capacity
Copyright © 2017 Pearson Education, Inc.
S7 - 8
Design and Effective Capacity
TABLE S7.1
Capacity Measurements
MEASURE
DEFINITION
EXAMPLE
Ideal conditions exist
during the time that
the system is
available
Machines at Frito-Lay are designed to
produce 1,000 bags of chips/hr., and the plant
operates 16 hrs./day.
Design Capacity = 1,000 bags/hr. × 16 hrs.
= 16,000 bags/day
Design capacity
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S7 - 9
Design and Effective Capacity
TABLE S7.1
Capacity Measurements
MEASURE
DEFINITION
Effective capacity
Design capacity
minus lost output
because of planned
resource
unavailability (e.g.,
preventive
maintenance,
machine
setups/changeovers,
changes in product
mix, scheduled
breaks)
Copyright © 2017 Pearson Education, Inc.
EXAMPLE
Frito-Lay loses 3 hours of output per day
(= 0.5 hrs./day on preventive maintenance,
1 hr./day on employee breaks, and 1.5
hrs./day setting up machines for different
products).
Effective Capacity = 16,000 bags/day
– (1,000 bags/hr.)
(3 hrs./day)
= 16,000 bags/day
– 3,000 bags/day
= 13,000 bags/day
S7 - 10
Design and Effective Capacity
TABLE S7.1
Capacity Measurements
MEASURE
DEFINITION
Actual output
Effective capacity
minus lost output
during unplanned
resource idleness
(e.g., absenteeism,
machine breakdowns,
unavailable parts,
quality problems)
Copyright © 2017 Pearson Education, Inc.
EXAMPLE
On average, machines at Frito-Lay are not
running 1 hr./day due to late parts and
machine breakdowns.
Actual Output = 13,000 bags/day
– (1,000 bags/hr.)
(1 hr./day)
= 13,000 bags/day
– 1,000 bags/day
= 12,000 bags/day
S7 - 11
Utilization and Efficiency
Utilization is the percent of design
capacity actually achieved
Utilization = Actual output/Design capacity
Efficiency is the percent of effective
capacity actually achieved
Efficiency = Actual output/Effective capacity
Copyright © 2017 Pearson Education, Inc.
S7 - 12
Bakery Example
Design
Capacity
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 1,200 rolls per hour
Bakery operates 7 days/week, 3 - 8 hour shifts
Design capacity = (7 x 3 x 8) x (1,200) = 201,600 rolls
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S7 - 13
Bakery Example
Utilization
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 1,200 rolls per hour
Bakery operates 7 days/week, 3 - 8 hour shifts
Design capacity = (7 x 3 x 8) x (1,200) = 201,600 rolls
Utilization = 148,000/201,600 = 73.4%
Copyright © 2017 Pearson Education, Inc.
S7 - 14
Bakery Example
Efficiency
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 1,200 rolls per hour
Bakery operates 7 days/week, 3 - 8 hour shifts
Design capacity = (7 x 3 x 8) x (1,200) = 201,600 rolls
Utilization = 148,000/201,600 = 73.4%
Efficiency = 148,000/175,000 = 84.6%
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S7 - 15
Bakery Example
Design
Capacity
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 201,600 rolls per line
Efficiency = 84.6%
Expected output of new line = 130,000 rolls
Design capacity = 201,600 x 2 = 403,200 rolls
Copyright © 2017 Pearson Education, Inc.
S7 - 16
Bakery Example
Effective
Capacity
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 201,600 rolls per line
Efficiency = 84.6%
Expected output of new line = 130,000 rolls
Design capacity = 201,600 x 2 = 403,200 rolls
Effective capacity = 175,000 x 2 = 350,000 rolls
Copyright © 2017 Pearson Education, Inc.
S7 - 17
Bakery Example
Actual
Output
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 201,600 rolls per line
Efficiency = 84.6%
Expected output of new line = 130,000 rolls
Design capacity = 201,600 x 2 = 403,200 rolls
Effective capacity = 175,000 x 2 = 350,000 rolls
Actual output = 148,000 + 130,000 = 278,000 rolls
Copyright © 2017 Pearson Education, Inc.
S7 - 18
Bakery Example
Utilization
Efficiency
Actual production last week = 148,000 rolls
Effective capacity = 175,000 rolls
Design capacity = 201,600 rolls per line
Efficiency = 84.6%
Expected output of new line = 130,000 rolls
Design capacity = 201,600 x 2 = 403,200 rolls
Effective capacity = 175,000 x 2 = 350,000 rolls
Actual output = 148,000 + 130,000 = 278,000 rolls
Utilization = 278,000/403,200 = 68.95%
Efficiency = 278,000/350,000 = 79.43%
Copyright © 2017 Pearson Education, Inc.
S7 - 19
Capacity and Strategy
►
Capacity decisions impact all 10
decisions of operations management
as well as other functional areas of
the organization
►
Capacity decisions must be integrated
into the organization’s mission and
strategy
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S7 - 20
Capacity Considerations
1. Forecast demand accurately
2. Match technology increments and
sales volume
3. Find the optimum operating size
(volume)
4. Build for change
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S7 - 21
Economies and Diseconomies of
Scale
Average unit cost
(sales per square foot)
Figure S7.2
1,300 sq ft
store
Economies
of scale
1,300
2,600 sq ft
store
Diseconomies
of scale
2,600
Number of square feet in store
Copyright © 2017 Pearson Education, Inc.
8,000 sq ft
store
8,000
S7 - 22
Managing Demand
►
►
►
Demand exceeds capacity
►
Curtail demand by raising prices, scheduling
longer lead times
►
Long-term solution is to increase capacity
Capacity exceeds demand
►
Stimulate market
►
Product changes
Adjusting to seasonal demands
►
Produce products with complementary
demand patterns
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S7 - 23
Complementary Demand
Patterns
Figure S7.3
Sales in units
4,000 –
Combining the
two demand
patterns reduces
the variation
3,000 –
Snowmobile
motor sales
2,000 –
1,000 –
Jet ski
engine
sales
JFMAMJJASONDJFMAMJJASONDJ
Time (months)
Copyright © 2017 Pearson Education, Inc.
S7 - 24
Tactics for Matching Capacity
to Demand
1. Making staffing changes
2. Adjusting equipment
►
Purchasing additional machinery
►
Selling or leasing out existing equipment
3. Improving processes to increase throughput
4. Redesigning products to facilitate more throughput
5. Adding process flexibility to meet changing product
preferences
6. Closing facilities
Copyright © 2017 Pearson Education, Inc.
S7 - 25
Service-Sector Demand
and Capacity Management
►
Demand management
►
►
Appointment, reservations, FCFS rule
Capacity
management
►
Full time,
temporary,
part-time
staff
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S7 - 26
Bottleneck Analysis and the
Theory of Constraints
►
Each work area can have its own unique
capacity
►
Capacity analysis determines the
throughput capacity of workstations in a
system
►
A bottleneck is a limiting factor or constraint
►
►
A bottleneck has the lowest effective capacity in a
system
The time to produce a unit or a specified
batch size is the process time
Copyright © 2017 Pearson Education, Inc.
S7 - 27
Bottleneck Analysis and the
Theory of Constraints
►
The bottleneck time is the time of the
slowest workstation (the one that takes
the longest) in a production system
►
The throughput time is the time it takes
a unit to go through production from start
to end, with no waiting
Figure S7.4
A
B
C
2 min/unit
4 min/unit
3 min/unit
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S7 - 28
Capacity Analysis
►
Two identical sandwich lines
►
Lines have two workers and three operations
►
All completed sandwiches are wrapped
First assembly line
Bread
Fill
Toaster
15 sec/sandwich
20 sec/sandwich
40 sec/sandwich
Wrap/
Deliver
Bread
Fill
Toaster
37.5 sec/sandwich
15 sec/sandwich
20 sec/sandwich
40 sec/sandwich
Order
30 sec/sandwich
Second assembly line
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S7 - 29
Capacity
Analysis
Order
30 sec
Bread
Fill
Toaster
15 sec
20 sec
40 sec
Wrap
Bread
Fill
Toaster
37.5 sec
15 sec
20 sec
40 sec
►
The two lines are identical, so parallel
processing can occur
►
At 40 seconds, the toaster has the longest
processing time and is the bottleneck for
each line
►
At 40 seconds for two sandwiches, the
bottleneck time of the combined lines = 20
seconds
►
At 37.5 seconds, wrapping and delivery is
the bottleneck for the entire operation
Copyright © 2017 Pearson Education, Inc.
S7 - 30
Capacity
Analysis
Order
30 sec
Bread
Fill
Toaster
15 sec
20 sec
40 sec
Wrap
Bread
Fill
Toaster
37.5 sec
15 sec
20 sec
40 sec
►
Capacity per hour is 3,600 seconds/37.5
seconds/sandwich = 96 sandwiches per
hour
►
Throughput time is 30 + 15 + 20 + 40 + 37.5
= 142.5 seconds
Copyright © 2017 Pearson Education, Inc.
S7 - 31
Capacity Analysis
►
Standard process for cleaning teeth
►
Cleaning and examining X-rays can happen
simultaneously
Hygienist
cleaning
Check in
Takes
X-ray
Develops
X-ray
24 min/unit
2 min/unit
2 min/unit
4 min/unit
X-ray
exam
Dentist
Check
out
8 min/unit
6 min/unit
5 min/unit
Copyright © 2017 Pearson Education, Inc.
S7 - 32
Capacity
Analysis
►
►
►
►
►
►
Hygienist
cleaning
Check
in
Takes
X-ray
Develops
X-ray
24 min/unit
2 min/unit
2 min/unit
4 min/unit
X-ray
exam
Dentist
Check
out
8 min/unit
6 min/unit
5 min/unit
All possible paths must be compared
Bottleneck is the hygienist at 24 minutes
Hourly capacity is 60/24 = 2.5 patients
X-ray exam path is 2 + 2 + 4 + 5 + 8 + 6 = 27
minutes
Cleaning path is 2 + 2 + 4 + 24 + 8 + 6 = 46
minutes
Longest path involves the hygienist cleaning
the teeth, patient should complete in 46
minutes
Copyright © 2017 Pearson Education, Inc.
S7 - 33
Theory of Constraints
►
Five-step process for recognizing and
managing limitations
Step 1: Identify the constraints
Step 2: Develop a plan for overcoming the constraints
Step 3: Focus resources on accomplishing Step 2
Step 4: Reduce the effects of constraints by offloading
work or expanding capability
Step 5: Once overcome, go back to Step 1 and find
new constraints
Copyright © 2017 Pearson Education, Inc.
S7 - 34
Bottleneck Management
1. Release work orders to the system at the
pace of set by the bottleneck’s capacity
►
Drum, Buffer, Rope
2. Lost time at the bottleneck represents lost
capacity for the whole system
3. Increasing the capacity of a nonbottleneck
station is a mirage
4. Increasing the capacity of a bottleneck
increases the capacity of the whole system
Copyright © 2017 Pearson Education, Inc.
S7 - 35
Break-Even Analysis
►
►
►
Technique for evaluating process and
equipment alternatives
Objective is to find the point in dollars
and units at which cost equals
revenue
Requires estimation of fixed costs,
variable costs, and revenue
Copyright © 2017 Pearson Education, Inc.
S7 - 36
Break-Even Analysis
►
Fixed costs are costs that continue even
if no units are produced
►
►
Depreciation, taxes, debt, mortgage
payments
Variable costs are costs that vary with
the volume of units produced
►
Labor, materials, portion of utilities
►
Contribution is the difference between
selling price and variable cost
Copyright © 2017 Pearson Education, Inc.
S7 - 37
Break-Even Analysis
►
Revenue function begins at the origin
and proceeds upward to the right,
increasing by the selling price of each
unit
►
Where the revenue function crosses
the total cost line is the break-even
point
Copyright © 2017 Pearson Education, Inc.
S7 - 38
Break-Even Analysis
–
Total revenue line
900 –
800 –
700 –
Cost in dollars
Total cost line
Break-even point
Total cost = Total revenue
600 –
500 –
Variable cost
400 –
300 –
200 –
100 –
|
Figure S7.5
0
Fixed cost
|
|
|
|
|
|
|
|
|
|
|
100 200 300 400 500 600 700 800 900 1000 1100
Volume (units per period)
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S7 - 39
Break-Even Analysis
Assumptions
► Costs and revenue are linear
functions
►
►
We actually know these costs
►
►
Generally not the case in the real
world
Very difficult to verify
Time value of money is often
ignored
Copyright © 2017 Pearson Education, Inc.
S7 - 40
Break-Even Analysis
BEPx = break-even point
in units
BEP$ = break-even point
in dollars
P = price per unit
(after all
discounts)
x = number of units
produced
TR = total revenue = Px
F = fixed costs
V = variable cost per unit
TC = total costs = F + Vx
Break-even point occurs when
TR = TC
or
Px = F + Vx
Copyright © 2017 Pearson Education, Inc.
BEPx =
F
P–V
S7 - 41
Break-Even Analysis
BEPx = break-even point
in units
BEP$ = break-even point
in dollars
P = price per unit
(after all
discounts)
F
BEP$ = BEPx P =
P
P–V
F
= (P – V)/P
=
x = number of units
produced
TR = total revenue = Px
F = fixed costs
V = variable cost per unit
TC = total costs = F + Vx
Profit
= TR - TC
= Px – (F + Vx)
= Px – F – Vx
= (P - V)x – F
F
1 – V/P
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S7 - 42
Break-Even Example
Fixed costs = $10,000
Direct labor = $1.50/unit
BEP$ =
Material = $.75/unit
Selling price = $4.00 per unit
$10,000
F
=
1 – [(1.50 + .75)/(4.00)]
1 – (V/P)
$10,000
=
= $22,857.14
.4375
Copyright © 2017 Pearson Education, Inc.
S7 - 43
Break-Even Example
Fixed costs = $10,000
Direct labor = $1.50/unit
BEP$ =
Material = $.75/unit
Selling price = $4.00 per unit
$10,000
F
=
1 – [(1.50 + .75)/(4.00)]
1 – (V/P)
$10,000
=
= $22,857.14
.4375
$10,000
F
BEPx =
=
= 5,714
4.00 – (1.50 + .75)
P–V
Copyright © 2017 Pearson Education, Inc.
S7 - 44
Break-Even Example
50,000 –
Revenue
Dollars
40,000 –
Break-even
point
30,000 –
Total
costs
20,000 –
Fixed costs
10,000 –
|
|
|
|
|
|
0
2,000
4,000
6,000
8,000
10,000
Units
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S7 - 45
Break-Even Example
Multiproduct Case
Break-even
F
point in dollars =
éæ V ö
ù
(BEP$)
åêêç1- Pi ÷ ´ Wi úú
ëè
û
i ø
( )
where
= variable cost per unit
= price per unit
= fixed costs
= percent each product is of total dollar sales
expressed as a decimal
i = each product
V
P
F
W
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S7 - 46
Multiproduct Example
Fixed costs = $3,000 per month
ANNUAL FORECASTED
SALES UNITS
PRICE
COST
Sandwich
9,000
$5.00
$3.00
Drink
9,000
1.50
.50
Baked potato
7,000
2.00
1.00
ITEM
1
2
3
4
ITEM (i)
ANNUAL
FORECASTED
SALES UNITS
SELLING
PRICE (Pi)
VARIABLE
COST (Vi)
5
6
7
8
9
(Vi/Pi)
1 - (Vi/Pi)
ANNUAL
FORECASTED
SALES $
% OF SALES
(Wi)
WEIGHTED
CONTRIBUTION
(COL 6 X COL 8)
Sandwich
9,000
$5.00
$3.00
.60
.40
$45,000
.621
.248
Drinks
9,000
1.50
0.50
.33
.67
13,500
.186
.125
2.00
1.00
.50
.50
14,000
.193
.097
$72,500
1.000
.470
Baked
potato
7,000
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S7 - 47
F
MultiproductBEP
Example
=
éæ
ö
$
Fixed costs = $3,000 per month
ANNUAL FORECASTED
SALES UNITS
ITEM
Sandwich
9,000
Drink
9,000
Baked potato
7,000
1
2
3
4
ITEM (i)
ANNUAL
FORECASTED
SALES UNITS
SELLING
PRICE (P)
VARIABLE
COST (V)
ù
Vi
åêêç1- P ÷ ´ Wi úú
ëè
û
i ø
( )
PRICE x 12
COST
$3,000
= $5.00
= $76,596
$3.00
.47
1.50
.50
$76,596
Daily
2.00
1.00
sales = 312 days = $245.50
5
6
7
(V/P)
1 - (V/P)
ANNUAL
FORECASTED
SALES $
8
9
% OF SALES
WEIGHTED
CONTRIBUTION
(COL 5 X COL 7)
Sandwich
9,000
$5.00
$3.00
.60
.40
$45,000
.621
.248
Drinks
9,000
1.50
0.50
.33
.67
13,500
.186
.125
2.00
1.00
.50
.50
14,000
.193
.097
$72,500
1.000
.470
Baked
potato
7,000
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S7 - 48
Figure S7.6
Reducing Risk with
Incremental Changes
(b) Leading demand with a
one-step expansion
Expected
demand
Demand
(c) Lagging demand with
incremental expansion
New
capacity
Copyright © 2017 Pearson Education, Inc.
Expected
demand
Demand
New
capacity
New
capacity
Expected
demand
(d) Attempts to have an average
capacity with incremental
expansion
Demand
Demand
(a) Leading demand with
incremental expansion
New
capacity
Expected
demand
S7 - 49
Reducing Risk with
Incremental Changes
(a) Leading demand with incremental
expansion
Figure S7.6
Demand
New
capacity
Expected
demand
1
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2
3
Time (years)
S7 - 50
Reducing Risk with
Incremental Changes
(b) Leading demand with a one-step
expansion
Figure S7.6
Demand
New
capacity
Expected
demand
1
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2
3
Time (years)
S7 - 51
Reducing Risk with
Incremental Changes
(c) Lagging demand with incremental
expansion
Figure S7.6
New
capacity
Demand
Expected
demand
1
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2
3
Time (years)
S7 - 52
Reducing Risk with
Incremental Changes
(d) Attempts to have an average capacity with
incremental expansion
Figure S7.6
New
capacity
Demand
Expected
demand
1
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2
Time (years)
3
S7 - 53
Applying Expected Monetary
Value (EMV) and Capacity
Decisions
►
►
Determine states of nature
►
Future demand
►
Market favorability
Assign probability values to states
of nature to determine expected
value
Copyright © 2017 Pearson Education, Inc.
S7 - 54
EMV Applied to Capacity
Decision
▶Southern Hospital Supplies capacity
expansion
EMV (large plant) = (.4)($100,000) + (.6)(–$90,000)
= –$14,000
EMV (medium plant) = (.4)($60,000) + (.6)(–$10,000)
= +$18,000
EMV (small plant) = (.4)($40,000) + (.6)(–$5,000)
= +$13,000
EMV (do nothing) = $0
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S7 - 55
Strategy-Driven Investments
►
Operations managers may have to
decide among various financial
options
►
Analyzing capacity alternatives
should include capital investment,
variable cost, cash flows, and net
present value
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S7 - 56
Net Present Value (NPV)
In general:
F = P(1 + i)N
where
F
P
i
N
= future value
= present value
= interest rate
= number of years
Solving for P:
F
P=
(1 + i)N
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S7 - 57
Net Present Value (NPV)
In general:
F = P(1 + i)N
where
F
P
i
N
= future value
While
= present
value this works fine,
is cumbersome for
= interest rate
= number oflarger
years values of N
it
Solving for P:
F
P=
(1 + i)N
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S7 - 58
NPV Using Factors
F
P=
= FX
N
(1 + i)
where
X = a factor from Table S7.2 defined
as = 1/(1 + i)N and F = future
value
TABLE S7.2
Present Value of $1
YEAR
6%
8%
10%
12%
14%
1
.943
.926
.909
.893
.877
2
.890
.857
.826
.797
.769
3
.840
.794
.751
.712
.675
4
.792
.735
.683
.636
.592
5
.747
.681
.621
.567
.519
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Portion of
Table S7.2
S7 - 59
Present Value of an Annuity
An annuity is an investment that
generates uniform equal payments
S = RX
where
X = factor from Table S7.3
S = present value of a series of uniform
annual receipts
R = receipts that are received every year
of the life of the investment
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S7 - 60
Present Value of an Annuity
TABLE S7.3
Present Value of and Annuity of $1
YEAR
6%
8%
10%
12%
14%
1
.943
.926
.909
.893
.877
2
1.833
1.783
1.736
1.690
1.647
3
2.673
2.577
2.487
2.402
2.322
4
3.465
3.312
3.170
3.037
2.914
5
4.212
3.993
3.791
3.605
3.433
Portion of
Table S7.3
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Present Value of an Annuity
▶River Road Medical Clinic equipment investment
$7,000 in receipts per year for 5 years
Interest rate = 6%
From Table S7.3
X = 4.212
S = RX
S = $7,000(4.212) = $29,484
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Limitations
1. Investments with the same NPV may have
different projected lives and salvage
values
2. Investments with the same NPV may have
different cash flows
3. Assumes we know future interest rates
4. Payments are not always made at the end
of a period
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