Replacement Problem
Introduction
• The problem of replacement is felt when the job
performing units such as men, machines,
equipments, parts etc., become less effective or
useless due to either sudden or gradual deterioration
in their efficiency, failure or breakdown.
• By replacing them with new ones at frequent
intervals, maintenance and other overhead costs can
be reduced.
• However, such replacements would increase the
need of capital cost for new ones.
Introduction…
Example:
• A machine becomes more and more expensive to
maintain after a no. of years, a railway time table
gradually becomes more and more out of date, an
electric light bulb fails all of a sudden, pipeline is
blocked, or an employee loses his job etc.
• In all such situations, there is a need to formulate a
most economic replacement policy for replacing
faulty units or to take some remedial special action
to restore the efficiency of deteriorating units.
Introduction…
Following are the situations when the replacement of
certain items needs to be done:
• An old item has failed and does not work at all, or
the old item is expected to fail shortly.
• The old item has deteriorated and works badly or
requires expensive maintenance.
• A better design of equipment has been developed.
Introduction…
Replacement problems can be classified into the
following two categories:
• When the equipment deteriorate with time and the
value of he money
1. Does not change with time.
2. Changes with time.
• When the units fail completely all of a sudden.
Types of Failure
Gradual Failure
It is progressive in nature. That is, as the life of an
item increases, its operational efficiency also
deteriorates resulting in
1. increased running (maintenance & operating) cost
2. decrease in its productivity
3. decrease in the resale or salvage value.
Mechanical items like pistons, rings, bearings etc. and
automobile tyres fall under this category.
Types of Failure
Sudden Failure
This type of failure occurs in items after some
period of giving desired service rather than
deterioration while in service. The period of desired
service is not constant but follows some frequency
distribution which may be progressive, retrogressive
or random in nature.
1.Progressive failure: If the probability of failure of
an item increases with the increase in its life, then
such failure is called progressive failure. (light bulb or
tubes fails progressively)
Types of Failure
2.Retrogressive failure: If the probability of failure in
the beginning of the life of an item is more but as
time passes the chances of its failure become less,
then such failure is said to be retrogressive.
3.Random failure: In this type of failure, the constant
probability of failure is associated with items that
fail from random causes such as physical shocks,
not related to age. For example, vacuum tubes in air
born equipment have been found to fail at a rate
independent of the age of the tube.
Replacement of Equipment that deteriorates
gradually
Generally the cost of maintenance & repair of certain items
increases with time and a stage may come when these costs
become so high that it is more economical to replace the
item by a new one.
Replacement policy when value of money does not change
with time
The aim here is to determine the optimum replacement age
of an equipment/item whose running/maintenance cost
increases with time and the value of money remains static
during that period. Let
Replacement policy when value of money does not
change with time…
C : capital cost of equipment
S : Scrap (waste) value of equipment
n : number of years that equipment would be in use
f(t) : maintenance cost function
A(n) : average total annual cost.
Case 1: When t is a continuous variable. If the
equipment is used for n years, then the total cost
incurred during this period is an under:
T C  Capit alCost - Scrap Cost  Maint enanc
e Cost
n
 C - S   f(t )dt
0
Averageannual t ot alcost is
n
1
C -S 1
A(n)  T C 
  f(t )dt
n
n
n 0
For minimumcost , we must have
d
A(n)  0
dn
or
C-S
1
- 2  2
n
n
n
 f(t )dt 
0
1
f(n)
n
n
C -S 1
f(n) 
  f(t )dt  A(n)
n
n 0
Clearly,
d2
A(n)  0 at f(n) A(n)
2
dn
T hissuggest t hat t heequipmentshould be replacedwhen maint enanc
e cost equals
t heaverageannual t ot alcost .
Replacement policy when value of money does not
change with time…
Case 2: When t is a discrete variable. Here period of
time is considered as fixed and n, take the values
1,2,3,…. Then
C -S 1
A(n) 

n
n
n
 f(t )
t 1
Now, A(n) will be minimumfor t hat alue
v of n, for which
or
A(n  1)  A(n)
and
A(n - 1)  A(n)
A(n  1)  A(n) 0
and
A(n)  A(n - 1)  0
For t his,we can writ e
C-S
1 n 1
A(n  1) 

 f(t )
n  1 n  1 t 1
n
1 
1


C
S

f(t
)


 n  1 f(n  1)
n  1 
t 1

 1 

  n A(n)  f(n  1)

 n  1 
 1 
A(n  1) - A(n)  
  f(n  1) - A(n)

 n  1 
T hus,
A(n  1) - A(n) 0  f(n  1)  A(n).
Similarly,it can be shown t hat
A(n)- A(n - 1)  0  f(n) A(n  1)
T hissuggest s t heopt imalreplacement policy:
Replace t heequipmentat t heend of n years,if t hemaint enanc
e cost in t he(n  1) t he year is more
t han t heavg. t ot alcost in t hent h year and t hent h yearsmaint enanc
e cost is less t han t heprevious
year's avg. t ot alcost .
Example:
A firm is considering replacement of a machine, whose cost
per year is Rs. 12,200 and the scrap value, Rs. 200. The
running (maintenance & operating) costs in rupees are
found from experience to as follows:
Year
1
Running Cost
200 500 800 1200 1800 2500 3200 4000
2
3
4
5
6
When should the machine be replaced?
7
8
We have given:
The running cost, f(n)
The scrap value (s) = 200
Cost of the machine (C)= 12,200
To find out the optimal time n when the
machine should be replaced, we need to
calculate an avg. Total cost per year
during the life of the machine.
(1)
Year of
service
n
(2)
Running
cost
f(n)
(3)
Cumulative
running cost
 f(n)
(4)
Depreciation
cost
C-S
(5)
(6)
Total cost
Avg. cost
TC=(3)+(4) A(n)=(5)/(1)
1
200
200
12000
12200
12200
2
500
700
12000
12700
6350
3
800
1500
12000
13500
4500
4
1200
2700
12000
14700
3675
5
1800
4500
12000
16500
3300
6
2500
7000
12000
19000
3167
7
3200
10200
12000
22200
3171
8
4000
14200
12000
26200
3275
Replacement policy when value of money
changes with time
When the time value of money is taken into
consideration, we shall assume that (1) the
equipment in question has no salvage value, and (2)
the maintenance costs are incurred in the beginning
of the different time periods.
• Since it is assumed that the maintenance cost
increases with time and each cost is to be paid just
in the start of the period, Let the money carry a rate
of interest r per year.
Replacement policy when value …
• Thus, a rupee invested now will be worth (1+r) after
a year, (1+r)2 after 2 years, and so on.
• In this way the Rs. invested today will be worth
(1+r)n after n years. Or, if we have to make a
payment of 1 Rs. In n years time, it is equivalent to
making a payment of (1+r)-n Rs. Today.
• The quantity (1+r)-n is called the present worth
factor (pwf) of 1 Rs. spent in n years time from now
onwards
Replacement policy when value …
• Thus, a rupee invested now will be worth (1+r) after
a year, (1+r)2 after 2 years, and so on.
• In this way the Rs. invested today will be worth
(1+r)n after n years. Or, if we have to make a
payment of 1 Rs. In n years time, it is equivalent to
making a payment of (1+r)-n Rs. Today.
• The quantity (1+r)-n is called the present worth
factor (pwf) of 1 Rs. spent in n years time from now
onwards
Replacement policy when value…
• (1+r)n is known as the payment compound amount
factor (Caf) of 1 Rs. spent in n years time.
• Let the initial cost of the equipment be C and let Rn
be the operating cost in year n.
• Let v be the rate of interest in such a way that
v= (1+r)-1 is the discount rate (pwf).
• Then the present value of all future discounts costs
Vn associated with a policy of replacing the
equipment at the end of each n years is given by
Replacement policy when value…
Then the present value of all future discounts costs Vn
associated with a policy of replacing the equipment
at the end of each n years is given by

 

Vn  C  R 0   vR1  v 2 R 2  ....... v n 1 R n 1  C  R 0 v n  v n 1 R 1  v n  2 R 2  ....... v 2n 1 R n 1  ....
n -1

  n
k
 C   v R k  *  v
k 0

 k 0
 
k
n -1
1


 C   v k R k 
n 1
k 0

 (1  v )
Now, Vn will be minimumfor t hat alue
v of n, for which
Vn 1  Vn  0 and Vn -1  Vn  0
For t his,we writ e
n -1


Vn 1  Vn  C   v k R k  (1  v n ) 1  Vn
k 0


 v n R n  1  v Vn 
and similarly
1
(1  v n 1 )
Vn  Vn 1  V n 1 R n  1  v Vn 
1
(1  v n 1 )
Replacement policy when value…
Since v is the value of money, it will always be less
than 1, and therefore 1-v will always be +ve. This
implies that Vn/(1-Vn+1) will always be +ve.
Hence,
Vn 1  Vn  0  R n  (1 - v) Vn and Vn  Vn -1  0  R n 1  (1 - v) Vn
C  R 0  vR1  v 2 R 2  ...........  v n 1 R n 1
R n 1 
 Rn
2
n 1
1  v  v  ....... v
n 1
(1 - v )(1 v)   v k
n
1
k 0
The expression which lies between Rn-1and Rn is
called the weighted avg. cost of all the previous n
years with weights 1,v,v2,…..,vn-1 respectively.
Replacement policy when value…
• Hence the optimal replacement policy of the
equipment after n periods is:
(a) Do not replace the equipments if the next
period’s operating cost is less than the weighted avg.
of previous cost.
(b) Replace the equipments if the next period’s
operating cost is greater that the weighted avg. of
previous costs.
Selection of the best equipment amongst two
Step 1 Considering the case of 2 equipments (A & B),
we first find the best replacement age for both
equipments by making use of Rn-1<(1-v)Vn<Rn.
Let the optimal replacement age for A & B comes
out to be n1 and n2 respectively.
Step 2 Compute the fixed annual payment (or
weighted avg. cost) for A & B by using the formula
n 1
C  R 0  vR1  v R 2  ...........  v R n 1
Wn 
2
n 1
1  v  v  ....... v
2
Selection of the best equipment amongst two
and substitute n=n1 for A and n=n2 for B
Step 3
(i) If W(n1 )  W(n2 ), chooseequipmentA.
(ii) If W(n1 )  W(n2 ), chooseequipmentB.
(iii) If W(n1 )  W(n2 ), both equipmentsare equally good.
Example
• Let the value of money be assumed to be 10% per
year and suppose that machine A is replaced after 3
years whereas machine B is replaced after every 6
years. The yearly costs of both the machines are
given below:
Year
1
Machine
A
Machine
B
2
3
4
5
6
1,000 200
400
1,000
200
400
1,700 100
200
300
400
500
Determine which machine should be purchased (replaced).
Solution :
Since t hemoneycarries t herat eof int erest ,t hepresent wort h of t hemoney t obe spent
over a periodof one year is
100
10

 0.9091
100 10 11
t herefore,T he t ot aldiscount cost (present wort h)of A for 3 yearsis
v
 1000 200 X (0.9091) 400 X (0.9091)2  1512Rs. approx.
Again, t he t ot aldiscount edcost of B for six yearsis
 1700 100 X (0.9091) 200 X (0.9091)2  300 X (0.9091)3  400 X (0.9091)4 
 500 X (0.9091)5

2765Rs.
Avg. Yearlycost of MachineA  1512/3 504 Rs.
Avg. Yearlycost of MachineB  2765/6 461Rs.
T hisshows t hat t heapparentadvant ageis wit h machineB. But , t hecomparision is unfair since
t he periodsfor whicht hecost sare consideredare different So,
. if we consider 6 year period
for machineA also, t hen t het ot alcost of A will be
 1000 200 X (0.9091) 400 X (0.9091)2  1000X (0.9091)3  200 X (0.9091)4 
 400 X (0.9091)5

2647Rs.
T hisamountis Rs. 118less cost lier han
t
machineB over t hesame period.
Hence,machineA should be replecedor purchased.
Replacement of equipment that fails suddenly