# Replacement and Maintenance Analysis¶

## Example 8.1 Page 103¶

In [1]:
#initiation of variable
FC=4000.0;#in Rs.
i=0.0;#in % per annum
MC1=0;#in Rs.
print "Tabulation to determine economic life : ";
print"End of     Maintenance     Summation of     Avg cost of     Avg FC if     Avg total";
print" year      cost at end     maintenance      maintenance    replaced at    cost through";
print"             of year         costs          through year    year given    year given";
print"  A           B(Rs.)        C(Rs.)             D(Rs.)        E(Rs.)        F(Rs.)";
print"  1            0                0                 0          4000          4000";
print"  2          200              200               100          2000          2100";
print"  3          400              600               200       1333.33       1533.33";
print"  4          600             1200               300          1000          1300";
print"  5          800             2000               400           800          1200";
print"  6         1000             3000               500        666.67       1166.67";
print"  7         1200             4200               600        571.43       1171.43";
print"Economic life of the machine : 6 years";
print"Column C summarizes the summation of maintenance costs for each replacement period. The value corresponding to any end of year in this column represents the total maintenance costs of using the equipment till the end of that year. It gives the Economic life of the machine : 6 years";

#When i=12%
i=12.0;#in % per annum
FC=4000.0;#in Rs.
print"Tabulation to determine economic life : ";
print"End of     Maintenance         P/F       PW as of begin   Summation of    PW of        A/P     Annual equi";
print" year      cost at end        12%         of year of         PW of      cumulative     12%     total cost";
print"             of year           n          maintenance     maintenance   maintenance     n       of year";
print"  A           B(Rs.)        C(Rs.)             D(Rs.)        E(Rs.)        F(Rs.)    G(rs.)      H(rs.)";
print"  1            0           0.8929              0.00          0.00       4000.00     1.1200     4480.00";
print"  2          200           0.7972            159.44        159.44       4159.44     0.5917     2461.14";
print"  3          400           0.7118            284.72        444.16       4444.16     0.4163     1850.10";
print"  4          600           0.6355            381.30        825.46       4825.46     0.3292     1588.54";
print"  5          800           0.5674            453.92       1279.38       5279.38     0.2774     1464.50";
print"  6         1000           0.5066            506.60       1785.98       5785.98     0.2432     1407.15";
print"  7         1200           0.4524            542.88       2328.86       6328.86     0.2191     1386.65";
print"  8         1400           0.4039            565.46       2894.32       6894.32     0.2013     1387.83";
print"  9         1600           0.3606            576.96       3471.28       7471.28     0.1877     1402.36";
print" 10         1800           0.3220            579.60       4050.88       8050.88     0.1770     1425.00";
print"Economic life of the machine : 7 years";
print"For this problem, the annual equivalent total cost is minimum at the end of year 7. Therefore, the economic life of the equipment is 7 year.";

Tabulation to determine economic life :
End of     Maintenance     Summation of     Avg cost of     Avg FC if     Avg total
year      cost at end     maintenance      maintenance    replaced at    cost through
of year         costs          through year    year given    year given
A           B(Rs.)        C(Rs.)             D(Rs.)        E(Rs.)        F(Rs.)
1            0                0                 0          4000          4000
2          200              200               100          2000          2100
3          400              600               200       1333.33       1533.33
4          600             1200               300          1000          1300
5          800             2000               400           800          1200
6         1000             3000               500        666.67       1166.67
7         1200             4200               600        571.43       1171.43
Economic life of the machine : 6 years
Column C summarizes the summation of maintenance costs for each replacement period. The value corresponding to any end of year in this column represents the total maintenance costs of using the equipment till the end of that year. It gives the Economic life of the machine : 6 years
Tabulation to determine economic life :
End of     Maintenance         P/F       PW as of begin   Summation of    PW of        A/P     Annual equi
year      cost at end        12%         of year of         PW of      cumulative     12%     total cost
of year           n          maintenance     maintenance   maintenance     n       of year
A           B(Rs.)        C(Rs.)             D(Rs.)        E(Rs.)        F(Rs.)    G(rs.)      H(rs.)
1            0           0.8929              0.00          0.00       4000.00     1.1200     4480.00
2          200           0.7972            159.44        159.44       4159.44     0.5917     2461.14
3          400           0.7118            284.72        444.16       4444.16     0.4163     1850.10
4          600           0.6355            381.30        825.46       4825.46     0.3292     1588.54
5          800           0.5674            453.92       1279.38       5279.38     0.2774     1464.50
6         1000           0.5066            506.60       1785.98       5785.98     0.2432     1407.15
7         1200           0.4524            542.88       2328.86       6328.86     0.2191     1386.65
8         1400           0.4039            565.46       2894.32       6894.32     0.2013     1387.83
9         1600           0.3606            576.96       3471.28       7471.28     0.1877     1402.36
10         1800           0.3220            579.60       4050.88       8050.88     0.1770     1425.00
Economic life of the machine : 7 years
For this problem, the annual equivalent total cost is minimum at the end of year 7. Therefore, the economic life of the equipment is 7 year.


## Example 8.2 Page 105¶

In [2]:
#initiation of variable
FC=20000.0;#in Rs.
i=15.0;#in % per annum

#result
print "The other details are summarized in Table 8.3. It can be seen from the book.";
print "Total annual equivalent cost = [cumulative sum of PW as of beginning of year 1 of operation & maintenance cost + FC - PW as of beginning of year 1 of salvage]*(A/P,15,n)";
print "In column L, the annual equivalent cost is minimum for n=5. Therefore, the economic life of the machine is 5 years. ";

The other details are summarized in Table 8.3. It can be seen from the book.
Total annual equivalent cost = [cumulative sum of PW as of beginning of year 1 of operation & maintenance cost + FC - PW as of beginning of year 1 of salvage]*(A/P,15,n)
In column L, the annual equivalent cost is minimum for n=5. Therefore, the economic life of the machine is 5 years.


## Example 8.3 Page 107¶

In [3]:
#initiation of variable
FC=20000.0;#in Rs.
i=15.0;#in % per annum

#result
print "The details are summarized in Table 8.4. It can be seen from the book.";
print"Total annual equivalent cost = [summation of PW of maintenance cost + FC]]*(A/P,15,n)";
print"(column E + Rs. 6000)* Column G";
print "Column F * Column G";
print "In column H, the minimum annual equivalent cost occurs when n=8. Therefore, the economic life of the machine B is 8 years. ";
print "RESULT : Min annual equivalent cost for machine A : Rs. 2780";
print "Min annual equivalent cost for machine B : Rs. 3672.30";

The details are summarized in Table 8.4. It can be seen from the book.
Total annual equivalent cost = [summation of PW of maintenance cost + FC]]*(A/P,15,n)
(column E + Rs. 6000)* Column G
Column F * Column G
In column H, the minimum annual equivalent cost occurs when n=8. Therefore, the economic life of the machine B is 8 years.
RESULT : Min annual equivalent cost for machine A : Rs. 2780
Min annual equivalent cost for machine B : Rs. 3672.30


## Example 8.4 Page 109¶

In [6]:
#initiation of variable
#alternative 1
Pprice=200000.0;#in Rs
P=120000.0;#in Rs
F=25000.0;#in Rs
A=25000.0;#in Rs
i=12.0;#in % per annum
n=6.0;#in years

#calculation
AE1=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS

#result
print "The annual equivalent cost(AE(12%)) of this alternative in RS.  ",round(AE1,3);

#Alternative 2
P=150000.0;#in Rs
F=20000.0;#in Rs
A=14000.0;#in Rs
i=12.0;#in % per annum
n=6.0;#in years

#calcualion
AE2=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS

#result
print "The annual equivalent cost(AE(12%)) of this alternative in RS.  ",round(AE2,3);
print "Since, The equivalent cost of new machine is less than that of present machine, it is suggested that the present machine be replaced with the new machine.";

The annual equivalent cost(AE(12%)) of this alternative in RS.   51106.443
The annual equivalent cost(AE(12%)) of this alternative in RS.   48019.343
Since, The equivalent cost of new machine is less than that of present machine, it is suggested that the present machine be replaced with the new machine.


## Example 8.5 Page 111¶

In [7]:
#initation of variable
#alternative 1
Pprice=50000.0;#in Rs
P=15000.0;#in Rs
F=8000.0;#in Rs
A=14000.0;#in Rs
i=15.0;#in % per annum
n=5.0;#in years

#calcualtion
AE1=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS

#result
print "The annual equivalent cost(AE(15%)) of this alternative in RS.  ",round(AE1,3);

#Alternative 2
P=65000.0;#in Rs
F=13000.0;#in Rs
A=9000.0;#in Rs
i=15.0;#in % per annum
n=20.0;#in years

#calculation
AE2=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS

#result
print "The annual equivalent cost(AE(15%)) of this alternative in RS. : ",round(AE2,3);
print "Since, The equivalent cost of Old deisel Engine is less than that of New deisel Engine, it is suggested to keep the Old deisel Engine.";

The annual equivalent cost(AE(15%)) of this alternative in RS.   17288.209
The annual equivalent cost(AE(15%)) of this alternative in RS. :  19257.596
Since, The equivalent cost of Old deisel Engine is less than that of New deisel Engine, it is suggested to keep the Old deisel Engine.


## Example 8.6 Page 113¶

In [10]:
#intiation of variable
P=660000.0;#in Rs
F=400000.0;#in Rs
A=96000.0;#in Rs
i=10.0;#in % per annum
n=5.0;#in years

#caculation
AE1=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS

#result
print "The annual equivalent cost(AE(10%)) of this alternative in RS. : ",round(AE1,3);

#Alternative 2
P=150000.0;#in Rs
X=420000.0;#in Rs
i=10.0;#in % per annum
n=40.0;#in years

#calcualtion
AE2=(P-X)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS

#result
print "The annual equivalent cost(AE(10%)) of this alternative in RS. : ",round(AE2,3);
print "Since, The equivalent cost of alternative 2 is less than that of alternative 1, it is suggested that alternative 2 should be selected.";
print " calculations in the book is not accurate."

The annual equivalent cost(AE(10%)) of this alternative in RS. :  204587.345
The annual equivalent cost(AE(10%)) of this alternative in RS. :  108389.958
Since, The equivalent cost of alternative 2 is less than that of alternative 1, it is suggested that alternative 2 should be selected.
calculations in the book is not accurate.


## Example 8.7 Page 113¶

In [12]:
#initiation of variable
#alternative 1
P=10000.0;#in Rs
F=1500.0;#in Rs
A=1600.0;#in Rs
i=15.0;#in % per annum
n=7.0;#in years

#calcualtion
AE1=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS

#result
print "The annual equivalent cost(AE(10%)) of 10 hp motor  in RS. : ",round(AE1,3);

#alternative 1 part 2
P=10000;#in Rs
F=800.0;#in Rs
A=1000.0;#in Rs
i=15.0;#in % per annum
n=7.0;#in years

#calcualtion
AE2=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS

#result
print "The annual equivalent cost(AE(10%)) of 10 hp motor  in RS.  ",round(AE2,3);
print "Total annual equivalent cost of alternative in Rs. : ",round(AE1+AE2,3);

#Alternative 2
P=35000.0;#in Rs
F=4000.0;#in Rs
A=500.0;#in Rs
i=15.0;#in % per annum
n=7.0;#in years

#calcualtion
AE=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS

print "The annual equivalent cost of alternative 2 in RS.",round(AE,3);
print"Since, The equivalent cost of alternative 1 is less than that of alternative 2, it is suggested that the present 10 hp motor be augmented with an additional 5 hp motor.";
print " calculations in the book is not accurate"

The annual equivalent cost(AE(10%)) of 10 hp motor  in RS. :  3868.063
The annual equivalent cost(AE(10%)) of 10 hp motor  in RS.   3331.315
Total annual equivalent cost of alternative in Rs. :  7199.378
The annual equivalent cost of alternative 2 in RS. 8551.171
Since, The equivalent cost of alternative 1 is less than that of alternative 2, it is suggested that the present 10 hp motor be augmented with an additional 5 hp motor.
calculations in the book is not accurate


## Example 8.8 Page 115¶

In [22]:
#initiation of variable
P=10000.0;#in Rs
F=4000.0;#in Rs
A=500.0;#in Rs
i=12.0;#in % per annum
n=4.0;#in years

#calcualtion
AE=(P-F)*((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1)+F*i/100+A;#in RS
X= (AE-1000*i/100-750)/0.3292+1000;
#result
print "The annual equivalent cost(AE(15%)) of n in RSew machine : ",round(AE,3);
print "The comparative use value of old machine is Rs.", round(X,3), "which is less than the price(Rs. 8000) offered by the company which is supplying the new machine in the event of replacing the old machine by nw machine. ";
print"Therefore, it is advisable to replace the old machine with the new one.";

The annual equivalent cost(AE(15%)) of n in RSew machine :  2955.407
The comparative use value of old machine is Rs. 7334.771 which is less than the price(Rs. 8000) offered by the company which is supplying the new machine in the event of replacing the old machine by nw machine.
Therefore, it is advisable to replace the old machine with the new one.


## Example 8.9 Page 117¶

In [18]:
#Intiation of Variable
p1=0.07;#unitless
p2=0.11;#unitless
p3=0.12;#unitless
p4=0.18;#unitless
p5=0.21;#unitless
p6=0.20;#unitless
p7=0.11;#unitless
N0=100.0;#no. of transistors

#calcualtion
N1=N0*p1;#no. of transistors
N2=N0*p2;#no. of transistors
N3=N0*p3;#no. of transistors
N4=N0*p4;#no. of transistors
N5=N0*p5;#no. of transistors
N6=N0*p6;#no. of transistors
N7=N0*p7;#no. of transistors
Life=0;#in weeks
p=[p1, p2, p3, p4, p5, p6, p7];#Unitless
for i in range (7):
Life=Life+(i+1)*p[i];

#result
print "Expected life of each transistor in weeks : ",round(Life,3)
print "Average No. of failures/week :  ",round(100/Life);

#result of group replacement cost
print "Cost of transistor when replaced simultaneously = Rs. 3";
print "Cost of transistor when replaced individually = Rs. 9";
print "The cost of group replacement policy for seeral replacement periods are summarized in Table 8.6. This table can be seen from the book.";
print "From table it is clear that the avg cost/week is minimum for the 4th week. Hence, the group replacement period is 4 weeks.";
print "Individual replacement cost/week = Rs. 207";
print "Minimum group replacement ost/week = Rs. 196.50";
print "Since the min group replacement cost/week is less than the individual replacement cost/week, the group replacement policy is the best, and hence all the transistors should be replaced in 4 weeks."

Expected life of each transistor in weeks :  4.39
Average No. of failures/week :   23.0
Cost of transistor when replaced simultaneously = Rs. 3
Cost of transistor when replaced individually = Rs. 9
The cost of group replacement policy for seeral replacement periods are summarized in Table 8.6. This table can be seen from the book.
From table it is clear that the avg cost/week is minimum for the 4th week. Hence, the group replacement period is 4 weeks.
Individual replacement cost/week = Rs. 207
Minimum group replacement ost/week = Rs. 196.50
Since the min group replacement cost/week is less than the individual replacement cost/week, the group replacement policy is the best, and hence all the transistors should be replaced in 4 weeks.


## Example 8.10 Page 119¶

In [23]:
#initiation of variable
p1=(100.0-96)/100;#unitless
p2=(96.0-89)/100;#unitless
p3=(89-68)/100;#unitless
p4=(68.0-37)/100;#unitless
p5=(37.0-13)/100;#unitless
p6=(13.0-0)/100;#unitless
N0=1000.0;#no. of resistors

#calculation
N1=N0*p1;#no. of resistors
N2=N0*p2+N1*p1;#no. of resistors
N3=N0*p3+N1*p2+N2*p1;#no. of resistors
N4=N0*p4+N1*p3+N2*p2+N3*p1;#no. of resistors
N5=N0*p5+N1*p4+N2*p3+N3*p2+N4*p1;#no. of resistors
N6=N0*p6+N1*p5+N2*p4+N3*p3+N4*p2+N5*p1    ;#no. of resistors

Life=0;#in months
p=[p1, p2, p3, p4, p5, p6, p7];#Unitless
for i in range (7):
Life=Life+(i+1)*p[i];

#result
print "Expected life of each transistor in weeks : ",round(Life,3)
print "Average No. of failures/week :  ",round(100/Life);

#result of group replacement cost
print "Cost/transistor when replaced simultaneously = Rs. 4";
print "Cost/transistor when replaced individually = Rs. 10";
print "The cost of group replacement policy for several replacement periods are summarized in Table 8.7. This table can be seen from the book.";
print "From table it is clear that the avg cost/month is minimum for the 3rd month. Hence, the group replacement period is 3 months.";
print "Individual replacement cost/month = Rs. 2480";
print "Minimum group replacement ost/month = Rs. 2426.67";
print "Since the min group replacement cost/month is less than the individual replacement cost/month, the group replacement policy is the best, and hence all the transistors should be replaced in 3 months."

Expected life of each transistor in weeks :  4.17
Average No. of failures/week :   24.0
Cost/transistor when replaced simultaneously = Rs. 4
Cost/transistor when replaced individually = Rs. 10
The cost of group replacement policy for several replacement periods are summarized in Table 8.7. This table can be seen from the book.
From table it is clear that the avg cost/month is minimum for the 3rd month. Hence, the group replacement period is 3 months.
Individual replacement cost/month = Rs. 2480
Minimum group replacement ost/month = Rs. 2426.67
Since the min group replacement cost/month is less than the individual replacement cost/month, the group replacement policy is the best, and hence all the transistors should be replaced in 3 months.