Replacement
Models
11.1
Introduction
The
replacement
problems
are concerned with the situations that
arise
when
some
items
such
as
men,
machines.
electric
light
bulbs
ctc
need
replacement
due
to
their
decreased
efficiency,
failure
or
brcakdown.
Srch decreased efficiency
or
complete breakdown may either be graduai
or
all
of
a
sudden.
The
replacement
problem
arises
because
of
the
following
factors
i)
The
old
item
has
become
worse
or
requires
expensive
maintenance.
ii)
The
old
item
has
failed
due
to
accident.
ii)
A
more
efficient
design
of
equipment
has
become
available
in
market.
Thus
the
provlem
of
replacement
is
to
decide
the
best
policy
to
dctermine
the
age
at
which
the
replacement
is
most
economical
instcad
of
continuing
at
increased
cost
due
to
factor
like
maintenance.
The
objective
is
to
find
the
optimum
period
of
replacement.
We
shall
discuss
the
following
main
type
of
replacement
situations:
i)
Replacement
of
items
that
deterioráte
with
time.
Replacement
of
items
which
do
not
deteriorate
but
fail
after
ii)
certain
amount
of
use.
For
items
which
do
not
deteriorate
but
fail
all
of
a
sudden,
following
are
the
two
types
of
replacement
policies
i)
Individual
replacement
policy
:
Under
this
policy,
an
item
is
replaced
immediately
afier
its
failure.
iiGroup
replacement
policy:
Under
this
policy,
we
take
decisions
as
to
when
all
the
items
must
be
replaced,
irrespective
of
the
fact
that
items
have
failed
or
have
not
failed,
with
the
provision
that,
if
any
item
fails
before
the
replacement
time
it
may
be
individually
replaced.
11.4
KEsource
tvlaliagol
Remarks
In
numerical
problems
we
consider
the
minimum
valte
of
determine
the
Plc
the
average
annual
cost
(i.e.,)
minimum
of
gptimum
replacement
period.
Example
1:
A
machine
owner
finds
fron
s
past
records
that
the
costs
per
year
of
maintaining
a
machine
whae
purchase
price
is
Rs.
6000
are
as
given
below:
Year
1
2
3
4
5
Maintenance
Cost (Rs)
Resale
1000
1200.
1400
2300
2800
1500
750
375
200
200
S
3000
Value
(Rs)
Determine
at
what
age
is
replacement
due
?
IMU.
MBA
Apr
97,
MU.
BE.
80
S
scrap
value.
Solution :
See
Table below. Here C
Rs.
6000,
Year Main. cost
Totat
(R)
C-S
Total
Cost
Ave.
cost
(n)
Main.
cost
(Rs)
=
(3)
+(4)
Pn)
(Rs)
2R,
(Rs)
(1)
=
P(n)
n
(1)
(2)
(4)
5)
(6)
1000
1000
3000
4000
4000
2
1200
2200
4500
6700
3350
1400
3600
5250
8850
2950
5400
5625
2756
2700
4
1800
11,025
5 2300
7700
5800
13,500
6
2800
10,500
5800
2717
Minimum
cost is in
5th
year
optimum
replacement
plan
:
16,300
replace
the
machine
at
the
end
of
5th
year
Example
2:
The
cost
of
a machine
is
Rs
6100
and its serap value
Rs.100.
The
maintenance
costs
found
from
experience
are
as
follows:
Year
2
3
5
6
Main.
Cost (Rs)
100
250
400
600
900
1200
1600
2000
When
should
the
machine
be
replaced?
[MU.
BE.
Mech
Oct
96
Solution:
Since
the
scrap
value
of
the
machine
is
Rs.100
resale
value
of
the
machine
alter
a
year
remains
constant
throughout.
The
costs
required
can
now be calculated as follows
Ave. cost
Year Main.
costumulative
C-S
Total
Cost
Mai.
cost
)
T-(3)+(4)
TA
(5).
(1)
(2)
(3)
(4)
(5)
say
(6)
100
100
6000
6100
6100
2
250
350
6000
6350
3175
3
400
08750
6000
6750
2250
4
600
1350
6000
7350
1837
5
900
2250
G000
8250
1650
1200
3450
6000
9450
1575*
6
7
1600
5050
6000
11,050
1579
8
2000
7050
6000
13,050
1631
Here
Ta
is
minimum
during
6th
year.
Hence
the
machine
should
be
replaced at the end
of
6th
year.
Example
3:
A
taxi
owner
estimates
from
his
past
records
that
the
costs
per
year
for
operating
a
taxi
whose
purchase
price
when
new
is
Rs.
60,000
are
as
given
below:
3
5
Age
1
2
Operting
Cost
(Rs)
15,000
18,000
20,000
10,000
12,000
After
5
years,
the
operating
cost
is
Rs.
6000
k
where
k6,
7,
8,
9,
10
(k
denoting
age
in
years).
If
the
resale
value
decreases
by
10%
of
purchase
price
each
year,
what
is
the
best
IMU.
BE.
Apr
91/
replacement
policy
?
Cost
of
the
money
is
zero.
10
Rs.
60,000
x
100
Solution:
10%
of
purchase
price
Rs.
6000
Thus
the
resale
value
decreases
by
Rs.
6000
every
year,
which
nneans
(C-S)
increases
by
Rs.
6000
every
year.
Average
annual
cost
of
the
taxi
is
computed
as
below.
(i1)
In
a
similar
fashion
we
prepare
a
table
for
B.
(3)
(4)
(5)
2)
Running
cost R
(6)
()
()
(1)
Commutative
year
running
costC-S
(5)
(3)+(4)
(
400
400
10,000
10,400
10,400
2
1200
1600
10,000
11,600
5800
2000
3600
10,000
13,600
4533
4
2800
6400
10,000
16,400
4100
3600
10,000
10,000
20,000
4000*
b
4400
14,400
10,000
24,400
4066
The above Table indicates that machine 'B° should be replaced at
the
end
of
5"
year.
Since the lowest average cost
of
Rs.4000 for machine B is less
than
the lowest average cost
of
Rs.5200 for
machine
A, machine A can
be
replaced
by
machine
B.
Now
we
have to determine as to when A
should
be replaced. Machine
A should
be
replaced when the cost for next
year
of
running
this
machine
becomes
more
than
the
average
yearly
cost
for
machine
B
Now
total
cost
for
machine
A
in
the
first
year
=
Rs.
9200
Total cost for machine A
in
the
II
year
Rs. 11,400 Rs.
9200
Rs.2200
III
year
Rs.
4200
150
4
IV
year
=
Rs.
6200(
1-
Sb)
As
the
cost
of
running
machine
A
in
lIl
year
(Rs.4200)
is
mo
thar
e
the
average
yearly
cost for machine B
(Rs
4000)
;
machine
A should
replaced
at the end of two years (i.e.,)
one
year
after
it
is
one year oid
Replacement
Models
11.9
Example
nle
6:
A
machine
shop
has
a
press
which
is
to
be
replaced
wears
out.
A
new
press
is
to
be
installed
now.
Further
an
optimum
a
m
replacement
is
to
be
found
for
next
7
years
after
which
the
press
is
no
longer
required.
The following
data
is
given
after
Installation
cost
at
Salvage
Value
Operating
cost
Year
beginning
of
year
(Rs)
at
end
of
year
during
the
year
(Rs)
(Rs)
1
200
100
60
2
210
50
80
3
220
30
100
240
20
120
5
260
15
150
290
10
180
320
0
230
Find
the
optimum
replacement
plan
and
the
corresponding
minimum
cost.
Solution:
Using
the
given
information,
the
minimum
average
annual
cost
of
the
press
is
computed
in
the
following
table.
C-S
T
TA
Year
(n)
60
200-
100
100
160
160
60
140
210-50
160
300
150
2
80
190
430
143*
3
100
240
120
360
220
580
145
4
245
755
151
5
150
510
280
970
162
6
180
690
1240
177
920
320
7
230
Since
T^
is
minimum
for
n
=
3,
the
machine
should
be
replaced
every
third
year.
Ans
ne
Value,
Present
worth
factor
(pwf)
and
Repla
cement
Models
113
1
Discount
Rate
11.3
honCy
Value
:
Since
money
has
a
value
over
time,
we
often
speak:
)In
one
way,
spending
Rs.100
today
would
be
equivalent
to
h10%
per
year.
This
can
be
explained
in
the
following
2
money
1s
worth
Ways.
equivalent
to
spending
Rs.110
in
a
year's
time
lconsequently
one
rupee
after
a
year
from
now
is
equivalent
to
now
is
equivalent
to
(1:1rupee
today.
Present
worih
factor
(pw)
IMU.
MBA
Nov.96
to
(1
1)
have
just
seen
abOve,
one
rupee
a
year
from
now
is
equivalent
nee
today
at
the
interest
rate
10%
per
year.
one
rupee
spent
Pars
from
now
1s
equivaient
to
(l:1)
today.
Similarly
we
can
sav
Ine
spent
'n
years
from
noW
Is
equivalent
to(1
1)-n
today
The
on
quantity
(1
1)
is
called
present
worth
factor
(pwf)
or
present
value
of
one
rupee
spent
n
years
from
now.
Discount
rate
(Depreciation
Value)
The
present
worth
factor
of
unit
amount
to
be
spent
after
one
year
is
oiven
by
V
=
(1
+r)
were
r
is
the
interest
rate.
Then
V
is
called
discount
IMU.
MBA
Nov.96]
rate
(technically
known
as
depreciation
value)
Example
7:
Let
the
value
of
the
money
be
10%
per
year
and
Suppose
that
machine
A
is
replaced
after
every
3
years
whereasS
machine B is
replaced
after
every
six
years.
1ne
yearly
cost
of
both
machines
are
given
as
under
5
6
1
2
3
4
Age
Machine
A:
1000
200
400
1000
200
400
Machine
B:
1700
100
200
300
400
500
Determine
which
machine
should
be
purchased
?
IMU.
BE
Nov
93
100
10
Solution:
Present
worth
factor
V =
100+
10IT
Total discount cost (present worth)
of
A
for
3 years
is
10
Rs.
1000+
200
x+400
xT
=
Rs.
1512
[nearly]
Again
the
total
discount
cost
of
B
for
six
years
is
Rs.
1700
100
x
(+200x
300-
+400xi
Rs. 2765
Average
yearly
cost
of
A = = Rs. 504
1512
3
-22
Rs.
461
and
average
yearly
cost
of
B
=
6
Although
this
shows
that
the
apparent
advantage
with
B,
but
the
comparison
is
unfair
because
the
periods
of
consideration
are
different.
So
if
we
consider 6
year
period
of
machine
A
also,
then
the
total
discount
of
A
will
be
1000+
200x..
+
400
x
=
Rs.
2647
which
is
118
Rs
less
costlier
than
machine
B
over
the
same
period.
Machine
A
should
be
purchased
[Ans]
Example 8 A
pipeline
is
due
for
repairs.
It
will
cost
Rs.10,000
and
lasts
for 3
years.
Alternatively
a
new
pipeline
can
be
laid
at
a
cost
of
Rs.
30,000
and
lasts
for
10
years.
Assuming
cost of
capital
to
De
10%
and
iguoring
salvage
value,
which
alternative
should
be
chosen.
IMU.
BE.
May
96
Solution
:
Consider
two
types
of
pipeline
for
infinite
replacenct
cycles
of
10
years
for
the
new
pipeline, and 3 years
for
the
existmg
pipeline
100
The
present
worth
factor is V =
T00+
10
=
0.9091.
Replacement
Models
11.17
1et
D,
denote
the
discounted
value
of
all
future
costs
associated
with
a
nolicy
of
replacing
they
equipment
after
'n'
years.
Then
if
we
designate
olicy
of
replac
the
initial
outlay
by
C,
D
D
C+V
C+
y2n
C
+.
C[l+Vn
+V2
+
...
...]
C
1-V
Now
substituting
the
values ofC,
V,
n
for
two
types
of pipelines ;
the
discounted
value
for
the
existing
pipeline
is given by
10,000
D3
1-(0.9091)3
Rs.
40,258
and for the new pipeline
30,000
1-
(0.909110
Rs,
48820
Since
D
D10.
the existing pipeline
should
be
continued.
[Ans]
D1o
Alternatively,
the
comparison
may
be
made
over
3 x
10
30
years.
Example
9
The
cost
patterns
of
2
machines
A
and
B,
when
money
value
is
not
considered
is
given
beloww
Year
Machine
A
Machine
B
1
900
1400
2
600
100
700
700
3
Find
the
cost
patterns
for
each
machines
when
money
is
worth
10%
per
year,
and
hence
find
which
machine
is
less
costly.
IMU.
BE.
Nov 91)
Solution:
The
total
outlay
for
three
years
for
Machine
A
=
900
+600
+
700
Rs.
2200
and
also
for
=
Rs.
2200
Machine
B
=
1400+100+
700
Here
we
observe
that
the
total
outlay
for
either
maçhine
is
same
for
Unree
years
when
money
value
is
not
taken
into
account,
Hence
both
the
machines appear equally good.
11.18
Resource
Managoll
ear
and
from
Now
consider
the
money
value
at
the
rate
of
10%
per
the
following
table
we
get
the
discounted
costs
for
A,
B.
2
3
Total
cost
Ycar
Machine
A
|900
600x
110
700x
S
2023.93
Machine
B1400
600x
10
100
02
Rs.
2023.93
700
11
(100)
2
2069.43
100
Machine
A
is
preferred
Ans
be
Example
10:
Assume
that
the
present
value
of
one
runee
t
spent
in
a
year's
time
is
Re
0.9
and
C =
Rs.
3000,
capital
enet
of
cquipment
and
the
running
costs
are
given in
the
table
below. Wh
should the machine be
repaced?
[BRU. BE.
Apr
97, MSU. BE.
4pr
97
4
en
Year
1
2
3
5
6
ung.
cost
(Rs):
500 600 800 1000 1300 1600 2000
Solution:
Consider
the
following
table:
Year
n R
V-R,V=|ER,V-C+R,V-y-
Wn
(1)
(2)
(3)
(4)
(5)
(6)
(7)
1
500
500
500
3500
3500
600
0.90
1040
4040
1.90
2126.32
2
540
800 0.81
648
1688
4688
2.71
1729.88
3
1000
0.73
730
2418
5418
3.44
1575
4
1300
0.66
858
3276
6276
4.10
1530.73
6
1600
0.59
944.784
4220.78
7220.78
4.69
1539.61
Since
W(n)
is
minimum
at
6th
year
optimum
replacement
plan
is
end
of
sixth
year
Example
11:
The
cost
of
a
new
machine
is
Rs.5000.
The
maintenance
cost
of
n"
year
is
given
by
C,
=
500
(n-1);
n
=
1,2
.
suppose
that
money
is
worth
5%
per
year,
after
how
many
years
WI
it be
economical
to
replace
the
machine
by
a
New
one?
IMU.
BE.
Apr
961
Solution:
The
present
worth
of
the
money
to
be
spent
a
year
ro
V
=
(1+0.05)l
=
0.9523
now
is
Replacement
Models
11.19
The
optimum
replacement
time
is
determined
in
the
following
table.
Year
n R
Vn-1R.Vn-1|
C+ER,V-1
5
Zv-
Wn
2
3
6 7
1.0000
0
5000
1.000
5000
1.9523
2805
500
0.9523
1000
0.9073
1500
0.8638
1296
2
476
6476
3
907
6383
2.8593 2232
4
7679
3.7231 2063
4.5458 2061
5.3293
2117
Since
W(n)
is
mininum
for
n = 5
and
R =
1500
< w
(5)
as
well
as
20000
0.8227
1645
-
9324
6
2500
0.7835
1959
11,283
w5)>
R6
2500,
it
is
economical to replace
the
machine
by
a new one at
the
end
of
5 years.
[Ans)
Example
12:
A production machine installed has initial
investment
of
Rs.
30,000
and
its
salvage
value
at
the
end
of
i
years
of
30,000
its
use
is
estimated
as Rs.
The
annual
operating and
maintenance
cost
in
the
first
year
is Rs. 15,000
and
increases
by
Rs.
1000
in
each
subsequent
years
for
first
five
years
and
increases
by Rs.
5000 in
each
year
thereafter.
Replacement
policy is
to
be
planned
over
a
period
of
seven
years.
During
this
period
cost
of
capital
may
be
taken
as
10%
per year.
Solve
the
problem
for
optimal
replacement.
Ans:
Here
C
=
30,000,
V=
O
IC-S,V
Year
R
V-lR,
V2R,V"S
S,"+R,
y-
S
(n)
2V
1.000
15000
1.0000
15000
15000
I5000
13635
31365
8260
26864
1.909
16000
0.909
14544
29544
10000
2
2.735
3
17000
|
0.826
14042
43586
7500
5633
24846
3.486
4
18000
0.751
13518
57104
6000
4098
23811
5000
3105
23261*
4.169
19000
0.683
12977
70081
23500
4.790
24000
0.621
14904
84985
4286
2422
6
1926
24173
5.355
29000
0.565
16385
101370
3750
Optimum
replacement
plan
is
after
5
years.
11.4
Group
Replacement
Policy
Lrample
15:
The
following
failure
rates
have
been
observed
for
certain items.
2
3
End
of
month
0.85
1.00
Probability
of
failure
todate
0.10
0.30
0.55
The
cost
of
replacing
an
individual
item
is
Rs.1.25.
The
decision
is
a0e
to
replace
all
items
simultaneously
at
fixed
intervals
and
also
place
individual
items
as
they
fail.
If
the
cost
of
group
replacement
is
50
paise,
what
is
the
best
interval
for
group
replacement.
A
group
replacement
per
item,
would
a
policy
of
strictly
indiast
a
replacement
become
preferable
to
the
adopted
policy.
11.26
Resource
Management
T
echniqucs
t
what
ndividual
IMU.
BE.
Nov
941
at
Solution:
Assume
that
items
failing
during
a
month
are
renlan
.
the end
of
the month.
that
Suppose
that
there
are
1000
items
in
use.
Let
pj
be
the
probability
that
rin
h
an
item,
which
was
new
when
placed
in
position for use, fails durino
dh
month
of
its
life. Thus,
we
have
P1
0.10
P2
0.30
-
0.10
0.20
P30.55-0.30
=
0.25
P40.85
-0.55
=
0.30
Ps1.00-0.85
=
0.15
Since
the
sum
of
probabilities
is
one,
all
the
probabilities
beyond
p
will
be
taken
as
zero.
Let
N,
denote
the
number
of
replacements
at
the
end
of
the
month.
Then
we
have
No=
Number
of
items
in
the
beginning
=
1000
N
NP=
100
N
NoP2
+
NiP-
200+
10
210
Ng
NoP3
+
NP2
+Nop
=
250
+20
+21
=
291
N4
NoP4
+
N\P3
+Nop
+NaP
396
Ns
NPs
+
NIP4
t+
NoP
+NP2
tNaP
=
331
The
expected
life
of
each
item
=2ip,
1xp+2
x
p2t
3
xpt
4
x
pat
5
x
Ps
=
Ix0.10
+2
x0.20
+3
x
0.25+4x0.30
+5
x
0.15
=
3.20
Replacement
odels,e11.27
Average number
of
failures per month
1000
3.2
313
(app)
replacement
of
all
the
1000
items
simultaneously
costs
O
ner
item
and
the
replacement
of
individual
item
on
failure
costs
2
Since
the
Rs
0
s.0.50
per
item
and
the
average
cost
for
different
group
replacement
policies
is
s.1.25,
shown
below:
End
of
1tL
Individual
Total
cost
(Rs)
individual+
group
replacement
Average
month
cost (Rs)
1000x
0.50+1.25
x
100
100
625
625
1000
x
0.50
+
1.25
x
310
310
=
887.7
443.8
210
601
1000x
0.50
+
1.25
x
601
417.1*
=
1251.3
1000 x
0.50+1.25
x 997
4
997
436.6
=
1746.3
Since
the
average
cost
is
lowest
in the
3rd
month,
it
is
optimal
to
have
a
group
replacement
after
every
3rd
month.
Further,
since
the
average
cost
is
more
than
Rs.
391.25
for
individual
replacement,
the
policy
of
individual
replacement
is
preferable.
Example
16:
Let
p()
be
the
probability
that
a
machine
in a
group
30
machines
would
break
down
in
period
t.
The
cost
of
repairing
broken
machine
is
Rs.200.00
Preventive
maintenance
is
performed
by
servicing
all
the
30
machines
at
the
end
T
units
of
time.
Preventive
maintenance
cost
is
Rs.15
per
machine.
Find
optimal
T
which
will
minimize
the
expected
total
cost
per
period
of
servicing,
given
that
0.03
for
t=1
P1-1)+
0.01
for
t=2,
3,.,
10
0.13
for
t=
l1,
12,
13,
...
p)
IMU.
BE.
Apr
98
Solution:
Here
2
3
4
56
7
8
9
10
11
12
P)
03
.04
.05
.06
.0708
.09
0.
10
0.11
0.
12
0.15
0
.28
Resource
Managemen
one
this
Since
the
sum
of
all
probabilities
can
never
be
greater
than
means
P120,
P130
etc.
A
machine
which
has
lasted
upto
Iperiod
is
sure
to
fail
in
10th
Let
N,
be
the
number
of
machines
at
the
end
of
fth
period
30
period.
No
N
NoP
30
x
0.03
=
0.9
1
N
N2
NP2
+
NP1
30
x
0.041x
0.03
1.23 1
N
NP+NIP2+
NP2
=
30
x
0.05+
1 x
0.04
+ 1 x
0.03
1.57
2
N
NoP4+
NIP3+
NP2+
NPI
N4
=
1.95
2
Ns=
2,
Ng
=3,
N7
=3,
Ng4,
Ns
N9
4,
N10
5,
N1
=6.
Similarly
11
Since
the
expected
life
of
each
machine,
2
ip
=
6.41
time
units
we
30
have
average
number
of machines
failed
per
period
is
641
5
(app)
Cost
of
individual replacement
Rs.
5 x
200
Rs.
1000
Group maintenance cost
is
computed below:
Average cost
of
Maintenance
per
period
End of
Cost
of
Maintenance
Period
in
groupP
Rs
(30x
15)
+
1
x
200
650
Rs.
650
2
Rs
(30x
15)
+ 2 x
200
850
Rs.
425
3
Rs
(30x
15)
+
4x
200
1250
Rs.
417
4
Rs
(30x
15)
+
6
x
200
1650
Rs.
412
5
Rs
(30x
15)
+8
x
200
2050
Rs.
410*
6
Rs
(30x
15)
+
I1
x
200
2650
Rs. 442
Replacement
Models
1.29
Since
the
minimum
cost
occurs
in
the
5th
period
it
is
optimal
to
maintain
all the machines upto
5th
period.
Example
17:
There
is
a
large
number
of
light
bulbs,
all
of
which
LAns
ct
be
kept
in
working
order.
If
a
bulb
fails
in
service,
it
costs
Re.1
te replace it,
but
if
all
the
bulbs
are
replaced
in
the
same
operation,
it
cOsts
only 35 paise a bulb.
If
the proportion of bulbs failing
in
successive
time
intervals
is
known,
decide
on
the
best
replacement
policy
and give reason. The following mortality rates for light bulbs
have
been
observed.
Proportion
failing
during
first
week
=0.09
Proportion
failing
during
second
week
=
0.165
Proportion
failing
during
third
week
=0.24
Proportion
failing
during
fourth
week
=
0.36
Proportion
failing
during
fifth
week=
0.12
Proportion
failing
during
sixth
week
=
0.03
Solution:
Let
number
of
bulbs
initially
be
No=
10000
(say)
Ifp, denote the probability
of
failure during i
week
then
p = 0.09,
P20.16,
p3
=
0.24,
P4=0.36,
ps=0.12,
P =
0.03,
Now
N,
denote
the
number
of
replacement
at
the
end
of
fh
week.
Then
No
10000
N
Nop
=
1000
x
0.09
900
Na
NP2+NIP
=
1681
Ng
NoP3
+
NP2
t
Nap=2695
Similarly
N4=4324,
Ns
=2747,
N4
N
2599
The
expected
life
of
each
bulbs
2
ip,
=
3.35
10,000
3.35
2985
app.
Average
number
of
failures
per
week
he
cost
of
individual
replacement
2985
x
1=
Rs.
2985
1.30
Resource
Managenment
Techniques
Now
the
average
cost
of
different
group
replacement
is
as
follows:
End
of
Individual
Average
Total cost (Rs)
individual
t
group
week
replacement
cost
(Rs)
10,000 x
0.35
+900x
900 4400
-4400
10,000
x
0.35+
(2581
x
)-
6081
2581
3041
10,000
x
0.35+
(5276
x
1)-
8776
5276
923
10,000x
0,35
+
(9550x
)
13,050
9550
10,000
(),35
(2,207)
12,297
9s
optimul
to
bave
Hroup
eplavement every
Jd
week
A
Is
avernge
00sl
Jean
than
N.
2983
o
lnulivldual
replavenment,
the
adiuy
ul
Hp
veplauemet
in
preleed
4
