#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.";
#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. ";
#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";
#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.";
#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.";
#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."
#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"
#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.";
#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."
#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."