Chapter14: Financial and Economic Evaluation

Ex14.1:pg-414

In [10]:
# given data
P=2000 # in rs
i=12 # interest rate in %
n=6 # time in years

F=P*(1+i/100.0)**n # Future value of investment

print "The amount will be Rs",round(F)

The amount will be Rs 3948.0

Ex14.2:pg-414

In [6]:
import math

# given data

P=10.0 # in lakh rs
i=12.25 # interest rate in %
F=20 # final amount in lakh rs

n=math.log(F/P)/math.log(1+i/100.0) # time in years

print "The number of years is ",round(n),"years"

The number of years is 6.0 years

Ex14.3:pg-415

In [11]:
#given data
F=100000 # final amount in rs
i=6 # interest rate in %
n=10 # time in years

P=F*(1/(1+i/100.0)**n) # initial amount

print "The initial value is Rs",round(P),

The initial value is Rs 55839.0

Ex14.4:pg-416

In [15]:
#given data
A=500 # annual amount invested each year in rs
i=9 # interest rate in %
n=6 # time in years

F=A*(((1+i/100.0)**n)-1)/(i/100.0) # future amount in rs

print "The Future amount will be Rs",round(F)

The Future amount will be Rs 3762.0

Ex14.5:pg-417

In [17]:
#given data
F=12000 # Total amount in rs
i=9 # interest rate in %
n=4 # time in years

A=F*(i/100.0)/(((1+i/100.0)**n)-1) # 

print "The amount deposited each year should be Rs",int(A)

The amount deposited each year should be Rs 2624

Ex14.6:pg-417

In [26]:
#given data
A=30000.0 # amount save each year in rs
i=10/100.0 # interest rate 
n=8 # time in years

P=A*(((1+i)**n)-1)/(i*((1+i)**n)) # amount spent on replacement in rs  
print "Amount spent on replacement is Rs",int(P)

Amount spent on replacement is Rs 160047

Ex14.7:pg-418

In [41]:
#given data
i=12/100.0 # interest rate 
n=10 # time in years

time=100.0 # days geyser is used in year
effi=0.9 # efficiency of geyser
w=100.0 # weight of water in kg
C=4.2 # heat capacity in kJ/kg-degree C
theta=60-15 # temperature difference in C
cost=4 # cost of electricity per kWh

Elec=(1/effi)*w*C*theta/3600.0 # electricity used in kWh/day
Elec=round(Elec,2)


A=Elec*time*cost # annual saving in Rs

P=A*(((1+i)**n)-1)/(i*((1+i)**n)) # final amount in rs

print "The final amount after 10 years is Rs",int(P)

The final amount after 10 years is Rs 13176

Ex14.8:pg-419

In [48]:
# given data

P=200000.0 #principal value in rs
i=10/100.0 # interest rate 
n=25.0 # time in years
L=2.0 # power produced  in kW

A=P*(i*(1+i)**n)/(-1+(1+i)**n) # annualised capital cost in rs

maintcost=P*0.05 # annual maintainence cost
Totalcost=A+maintcost #  total annual cost

Elec=L*0.25*10*365 # annual electricity production

Cost=Totalcost/Elec # unit cost of electricity production

print "unit cost of electricity production is Rs",round(Cost,1)

unit cost of electricity production is Rs 17.6

Ex14.9:pg-421

In [54]:
# given data

G=1 #gradient per period in lakh rs
i=12/100.0 # discount rate 
n=5 # time in years
A1=10 # payment at end of 1st year in lakhs rs
loan=40 # load applied for in lakhs


AGin=(1/i)-5*1/(-1+(1+i)**n) # gradient to uniform series conversion factor

Ag=A1+G*AGin # in lakhs Rs

Pg=Ag*(-1+(1+i)**n)/(i*(1+i)**n) # present worth in lakhs rs

print "The present worth is Rs",round(Pg,2),"Lakh"

if (Pg*0.85<loan):
    print "Loan is not given as amount is less than applied for loan"
else:
    print "Loan is given"

The present worth is Rs 42.44 Lakh
Loan is not given as amount is less than applied for loan

Ex14.10:pg-423

In [58]:
# given data

g=0.2 # annual gas price increase rate
i=10/100.0 # discount rate 
n=15 # time in years
A1=350*8 # payment at end of 1st year in lakhs rs

Pgg=(A1/(i-g))*(1-((1+g)/(1+i))**n) # present worth in Rs

print "The present worth of saving is Rs",round(Pgg)

The present worth of saving is Rs 75273.0

Ex14.11:pg-424

In [64]:
# given data
Co=10000 # initial investment in rs
B=900.0 # net annual savings per year

nsp=Co/B # simple payback period

print "The simple payback period is ",round(nsp,2)
if nsp<20:
    print "proposal may be accepted"
else:
    print "proposal may not be accepted"

The simple payback period is  11.11
proposal may be accepted

Ex14.12:pg-425

In [42]:
# given data

import numpy

ProjectA=[-2400,600,600,600,600,600] 
ProjectB=[-2400,800,800,800,800,800]
ProjectC=[-2400,500,700,900,1100,1300]

ProjAcu=numpy.zeros(6)  # cumulative cash flow for project A
for i in range(0,6):
    
    ProjAcu[i]=ProjectA[i]+ProjAcu[i-1]
    
ProjBcu=numpy.zeros(6) # cumulative cash flow for project B
for i in range(0,6):
    
    ProjBcu[i]=ProjectB[i]+ProjBcu[i-1]

ProjCcu=numpy.zeros(6) # cumulative cash flow for project C
for i in range(0,6):
    
    ProjCcu[i]=ProjectC[i]+ProjCcu[i-1]
    
    
print "\t \t \t     Project A     \t \t \t \t  \t  \t  \t  \t  \t Project B \t \t \t \t \t \t  \t  \t  \t      Project C"
print "year\t \t current \t cummulative \t \t \t \t  \t current \t cummulative \t \t \t \t \t \t \t \t current \t  \t cummulative \t"
for i in range(0,6):
    
    print (i+1),"\t \t ",ProjectA[i],"\t \t \t  \t ",ProjAcu[i]," \t \t \t \t \t \t ",ProjectB[i]," \t ",ProjBcu[i],"\t \t \t  \t \t \t \t \t \t",ProjectC[i]," \t ",ProjCcu[i]," \t"

for i in range(0,6):
    
    if ProjAcu[i]==0:
        PA=i 
for i in range(0,6):
    
    if ProjBcu[i]==0:
        PB=i
for i in range(0,6):
    
    if (ProjCcu[i]<0 and ProjCcu[i+1]>0):
        PC=(i)-ProjCcu[i]/ProjectC[i+1]

print " The payback period for project 1 is ",PA,"\n The payback period for project 2 is",PB,"\n The payback period for project 3 is ",round(PC,2)


              

	 	 	     Project A     	 	 	 	  	  	  	  	  	 Project B 	 	 	 	 	 	  	  	  	      Project C
year	 	 current 	 cummulative 	 	 	 	  	 current 	 cummulative 	 	 	 	 	 	 	 	 current 	  	 cummulative 	
1 	 	  -2400 	 	 	  	  -2400.0  	 	 	 	 	 	  -2400  	  -2400.0 	 	 	  	 	 	 	 	 	-2400  	  -2400.0  	
2 	 	  600 	 	 	  	  -1800.0  	 	 	 	 	 	  800  	  -1600.0 	 	 	  	 	 	 	 	 	500  	  -1900.0  	
3 	 	  600 	 	 	  	  -1200.0  	 	 	 	 	 	  800  	  -800.0 	 	 	  	 	 	 	 	 	700  	  -1200.0  	
4 	 	  600 	 	 	  	  -600.0  	 	 	 	 	 	  800  	  0.0 	 	 	  	 	 	 	 	 	900  	  -300.0  	
5 	 	  600 	 	 	  	  0.0  	 	 	 	 	 	  800  	  800.0 	 	 	  	 	 	 	 	 	1100  	  800.0  	
6 	 	  600 	 	 	  	  600.0  	 	 	 	 	 	  800  	  1600.0 	 	 	  	 	 	 	 	 	1300  	  2100.0  	
 The payback period for project 1 is  4 
 The payback period for project 2 is 3 
 The payback period for project 3 is  3.27

Ex14.13:pg-426

In [48]:
# given data

import numpy

ProjAcu=[-2400.0,-1864.0,-1386.0,-959.0,-578.0,-238.0] # in Rs
ProjBcu=[-2400,-1686,-1048,-479.0,30,484] # in Rs
ProjCcu=[-2400,-1954,-1396,-755,-56.0,683] # in Rs

ProjAdis=[-2400,536,478,427,381,340] # in Rs
ProjBdis=[-2400,714,638,569,509.0,454] # in Rs
ProjCdis=[-2400,446,558,641,699,738.0] # in Rs

PA=0
PB=0
PC=0

for i in range(0,5):
     if (ProjAcu[i]<0 and ProjAcu[i+1]>0):
        PA=(i+1)-ProjAcu[i]/ProjAdis[i+1]
print "For project A"
if (PA==0):
    print "Net loss, Thus should be rejected"
else:
    print PA,"years is payback period" 
for i in range(0,5):
    
    if (ProjBcu[i]<0 and ProjBcu[i+1]>0):
        PB=(i)-ProjBcu[i]/ProjBdis[i+1]
print "For project B"

if (PB==0):
    print "Net loss, Thus should be rejected"
else:
    print round(PB,2),"years is payback period" 
    
for i in range(0,5):
    
    if (ProjCcu[i]<0 and ProjCcu[i+1]>0):
        PC=(i)-ProjCcu[i]/ProjCdis[i+1]
print "For project C"

if (PC==0):
    print "Net loss, Thus should be rejected"
else:
    print round(PC,2),"years is payback period" 

For project A
Net loss, Thus should be rejected
For project B
3.94 years is payback period
For project C
4.08 years is payback period

Ex14.14:pg-427

In [27]:
# given data

import numpy
i=12.0/100 # interest rate 
n=5.0 # years

ProjectA=[-2400,600,600,600,600,600] 
ProjectB=[-2400,800,800,800,800,800]
ProjectC=[-2400,500,700,900,1100,1300]


NPVA=ProjectA[0]+ProjectA[1]*(((1+i)**n) - 1)/(i*(1+i)**n)

print "NPV of Project A is ",round(NPVA)

NPVB=ProjectB[0]+ProjectB[1]*(((1+i)**n) - 1)/(i*(1+i)**n)

print "NPV of Project B is ",round(NPVB)

ProjectNPVc=0  # cumulative cash flow for project A
for i in range(0,5):
    
    ProjectNPVc=ProjectNPVc+(-ProjectC[0]+ProjectC[i])/(1+i)**(i+1)
    
print "NPV of Project C is ",round(ProjectNPVc)

# The answer for project C is wrong in the book

NPV of Project A is  -237.0
NPV of Project B is  484.0
NPV of Project C is  852.0

Ex14.15:pg-428

In [79]:
# given data
import numpy
Co=20000.0 # cost in Rs
B=3000.0 # annual benefit in rs
n=15.0 # time in years
i=15.0/100 # initial guess for rate
NPV=numpy.zeros(4)

NPV[0]=B*(((1+i)**n)-1)/(i*(1+i)**n)-Co
x=0
print "Iteration no. \t\t\t i*  \t \t \t \t  \t\t\t NPV(i*)" 
while NPV[x]<0:
    x=x+1  
    i=i-0.01
    NPV[x]=B*(((1+i)**n)-1)/(i*(1+i)**n)-Co
    
for z in range(0,4):
    print z+1," \t \t \t \t \t \t \t ",0.15-(z/100.0)," \t \t \t\t \t ",round(NPV[z])

IRR=i+(i+0.01-i)/(NPV[x]+NPV[x-1]) # using equation 14.28

print "The IRR is", round(IRR*100,1)

# the answer is slightly different in textbook due to approximation

Iteration no. 			 i*  	 	 	 	  			 NPV(i*)
1  	 	 	 	 	 	 	  0.15  	 	 		 	  -2458.0
2  	 	 	 	 	 	 	  0.14  	 	 		 	  -1573.0
3  	 	 	 	 	 	 	  0.13  	 	 		 	  -613.0
4  	 	 	 	 	 	 	  0.12  	 	 		 	  433.0
The IRR is 12.0

Ex14.16:pg-429

In [1]:
# given data

i=12.0/100 # interest rate
ProjAdisB=[0,536.0,478,427,381,340] # discounted benefit for A
ProjAdisC=[2400.0,0,0,0,0,0] # discounted cost for A
ProjBdisB=[0,714.0,638,569,509,454] # discounted benefit for B
ProjBdisC=[2400.0,0,0,0,0,0]  # discounted cost for B
ProjCdisB=[0,446.0,558,641,699,738] # discounted benefit for C
ProjCdisC=[2400.0,0,0,0,0,0] # discounted cost for C

BCforA=sum(ProjAdisB)/sum(ProjAdisC)  # B mius C ratio
BCforB=sum(ProjBdisB)/sum(ProjBdisC) # B mius C ratio
BCforC=sum(ProjCdisB)/sum(ProjCdisC) # B mius C ratio
print "B - C for project A is ",round(BCforA,1)
print "B - C for project B is ",round(BCforB,1)
print "B - C for project C is ",round(BCforC,2)

B - C for project A is  0.9
B - C for project B is  1.2
B - C for project C is  1.28

Ex14.17:pg-431

In [95]:
# given data
Co=12000.0 # cost in Rs
Ca=200.0 # annual maintainence in Rs
C12=3000.0 # replacement cost in 12th year
S=1000.0 # salvage value in rs
n=20.0 # time in years
i=11/100.0 # interest rate

Cnet=Co-S*(1/(1+i)**n)+Ca*((((1+i)**n)-1)/(i*(i+1)**n))+C12*(1/(1+i)**12)

CR=Cnet*(i*(1+i)**n)/(((1+i)**n)-1)

print "The Capital Recovery cost is Rs",round(CR)

The Capital Recovery cost is Rs 1799.0

Ex14.18:pg-432

In [108]:
# given data
i=10/100.0 # rate
Acost=90000 # cost of A in Rs
Bcost=75000 #  cost of B in Rs
Acashfl=26000 # annual cash flow of A in Rs
Bcashfl=26000 # annual cash flow of B in Rs
nA=5 # useful life of A in years
nB=4 # useful life of B in years

NPVA=Acashfl*(((1+i)**nA) - 1)/(i*(1+i)**nA) - Acost # NPV for A
NPVB=Bcashfl*(((1+i)**nB) - 1)/(i*(1+i)**nB)- Bcost # NPV for B

print "The NPV for A is ",round(NPVA)

print "The NPV for B is ",round(NPVB)

AEA=i*NPVA/(1-(1+i)**(-nA))


AEB=i*NPVB/(1-(1+i)**(-nB))

print "The AE for A is Rs ",int(AEA)

print "The AE for B is Rs ",int(AEB)

print "The machine B will have higher profitability"

The NPV for A is  8560.0
The NPV for B is  7417.0
The AE for A is Rs  2258
The AE for B is Rs  2339
The machine B will have higher profitability

Ex14.19:pg-434

In [135]:
# given data
import numpy
Co=120000.0 # cost in Rs
N=5 # useful life
T=40/100.0 # tax rate 
i=9/100.0 # interest rate
Earning=[33000.0,35000.0,37000.0,39000,41000.0]
Depreciate=Co/N # depreciation in  Rs
pretax=numpy.zeros(5)
discshfl=numpy.zeros(5)
for x in range(0,5):
    pretax=Earning[x]-Depreciate
    tax=0.4*pretax
    ernng=pretax-0.4*pretax
    cashf=ernng+Depreciate
    discshfl[x]=cashf/(1+i)**(x+1)
netdiscntincm=sum(discshfl) # net discount income in Rs
NPV=netdiscntincm-Co # NPV
print "NPV of dryer is Rs",int(NPV)

# The answer in the book is wrong as the value of discounted cashflow is incorrect
    

NPV of dryer is Rs 2889

Ex14.20:pg-435

In [137]:
# given data

Co=300000 # cost in Rs
S=20000.0 # salvage value in Rs
N=15 # useful life

D=(Co-S)/N # Depreciation
BV=Co # Book Value
for i in range(0,N):
    BV=BV-D;
    print "The Book value at the end of ",i+1,"th year is Rs",round(BV)

The Book value at the end of  1 th year is Rs 281333.0
The Book value at the end of  2 th year is Rs 262667.0
The Book value at the end of  3 th year is Rs 244000.0
The Book value at the end of  4 th year is Rs 225333.0
The Book value at the end of  5 th year is Rs 206667.0
The Book value at the end of  6 th year is Rs 188000.0
The Book value at the end of  7 th year is Rs 169333.0
The Book value at the end of  8 th year is Rs 150667.0
The Book value at the end of  9 th year is Rs 132000.0
The Book value at the end of  10 th year is Rs 113333.0
The Book value at the end of  11 th year is Rs 94667.0
The Book value at the end of  12 th year is Rs 76000.0
The Book value at the end of  13 th year is Rs 57333.0
The Book value at the end of  14 th year is Rs 38667.0
The Book value at the end of  15 th year is Rs 20000.0