Chapter3-Economics¶
Ex1-pg58¶
In [3]:
## Example 3.1
print('Example 3.1');
print('Page No. 58');
## given
P = 10000;## Principal Amount
i = 0.15;## Interest Rate
n = 4;##years
I = P*i*n;## Simple Interest
Ts= P+I;## The total repayment
print'%s %.2f %s'%('The total repayment is',Ts,' Euro')
Ex2-pg58¶
In [4]:
## Example 3.2
print('Example 3.2');
print('Page No. 58');
## given
P = 10000;## Principal Amount in Pound
i = 0.15;## Interest Rate
n = 4;##years
Tc = P*(1+i)**n;
print'%s %.2f %s'%('The total repayment after adding compond interest is ',Tc,' Pound')
Ex3-pg59¶
In [5]:
## Example 3.3
print('Example 3.3');
##Page No. 59
## given
P = 60000;##/ Principal Amount in Pound
i = 0.18;## Interest Rate
n = 10;##years
R = P*((i*(1+i)**n)/((1+i)**n -1));##Rate of Capital Recovery
print'%s %.2f %s'%('The annual investment required is',R,'Pound')
Ex4-pg61¶
In [6]:
import math
## Example 3.4
print('Example 3.4\n\n');
print('Page No. 61\n\n');
import numpy
## given
P = 100000.;##/ Principal Amount of boiler plant in Pound
n = 10.;## service life in years
S = 0.;##Zero Salvage value
nT = (n*(n+1.)/2.);##sum of years
z0=numpy.linspace(0,9)
leng=len(z0)
d_=numpy.zeros(leng)
for i in range (0,9):
d_[i+1] = ((P-S)/nT)*(n-i);
print'%s %.2f %s'%('The Annual depreciation for first year is ',d_[1],' Pound\n')
print'%s %.2f %s'%('The Annual depreciation for second year is ',d_[2],' Pound\n\n')
print'%s %.2f %s'%('The Annual depreciation for third year is ',d_[3],' Pound\n')
print'%s %.2f %s'%('The Annual depreciation for ten year is ',d_[9]/2,' Pound\n')
## Deviation in answer due to some .approximation of values in the book
Ex5-pg62¶
In [8]:
import math
## Example 3.5
print('Example 3.5\n\n');
print('Page No. 62\n\n');
## given
P = 40000.;##/ Principal Amount of boiler plant in Pound
nT = 10.;## service life in years
S = 4000.;## Salvage value
n = 6.;## years after which Asset value has to be calculated
##(a) Straight line method
d = ((P-S)/nT);## Depreciation
Aa = (d*(nT-n)) + S;
print'%s %.2f %s'%('The Asset value at the end of six years using Straight line method is ',Aa,' Pound\n')
## (b) Declining balance technique
f = 1.-(S/P)**(1./nT);## Fixed fraction of the residual asset
Ab = P*(1.-f)**n;
print'%s %.2f %s'%('The Asset value at the end of six years using Declining balance technique is ',Ab,' Pound\n\n')
## (c) Sum of the years digit
sum_nT = (nT*(nT+1.)/2.);##sum of 10 years
sum_n = 45.;##sum after 6 years
dc = ((sum_n/sum_nT)*(P-S));## Depreciation after 6 years
Ac = P-dc;
print'%s %.2f %s'%('The Asset value at the end of six years using Sum of the years digit is ',Ac,' Pound\n')## Deviation in answer due to direct substitution
##(d) Sinking Fund Method
r_i = 0.06;## Rate of interest
Ad = P-((P-S)*(((1.+r_i)**n-1.)/((1.+r_i)**nT-1.)));
print'%s %.2f %s'%('The Asset value at the end of six years using Sinking Fund Method is ',Ad,' Pound\n')## Deviation in answer due to direct substitution
Ex6-pg67¶
In [9]:
import math
## Example 3.6
print('Example 3.6\n\n');
print('Page No. 67\n\n');
import numpy
## given
P = 9000.;## Capital Cost in Pound
n = 5.;## Project lifetime
Less_dep = 8000.;## Less Depreciation
##For Project A
d1 = numpy.array([[4500 ,3750 ,3000 ,1500 ,750 ]])## Saving in every year (before depreciation)
dT1 = sum(d1)
Net_S1 = dT1- Less_dep;## Total Net Saving
Avg1 = Net_S1/n;## Average net annual saving
R_R1 = (Avg1/P)*100.;
##For Project
d2 =numpy.array([[750 ,2250 ,4500 ,4500 ,1500 ]])## Saving in every year (before depreciation)
dT2 = sum(d2)
Net_S2 = dT2- Less_dep;## Total Net Saving
Avg2 = Net_S2/n;## Average net annual saving
R_R2 = (Avg2/P)*100.;
print(R_R1)
print(R_R2)
#print'%s %.2f %s'%('The percentage of Rate of Return on original investment for Project A is ',R_R1,'')
#print'%s %.2f %s'%('The percentage of Rate of Return on original investment for Project B is ',R_R2,'')
Ex7-pg68¶
In [10]:
import math
## Example 3.7
print('Example 3.7\n\n');
print('Page No. 68\n\n');
import numpy
## given
Pc = 10000.;## Capital cost for project C in Pound
Pd = 10000.;## Capital cost for project d in Pound
nc = 3.;## pay back period for C
nd = 3.;## pay back period for D
Ca = numpy.array([[4500, 3500, 2000, 2000 ,1000]]);## Annual Cash flow for C in Pound
Cc = numpy.array([[4500 ,8000, 10000, 12000 ,13000]])## Cumulative Cash flow for C in Pound
Da = numpy.array([[1500, 4000, 4500, 2200 ,1800 ,1000]]);## Annual Cash flow for D in Pound
Dc = numpy.array([[1500, 5500, 10000, 12200 ,14000, 15000]])## Cumulative Cash flow for D in Pound
Ac = 13000-Pc;## in Pound
Ad = 15000-Pd;## in Pound
print'%s %.2f %s'%('Additional amount from C after the pay back time is ',Ac,' Pound\n')
print'%s %.2f %s'%('Additional amount from D after the pay back time is ',Ad,' Pound\n')
Ex8-pg69¶
In [11]:
import math
## Example 3.8
print('Example 3.8\n\n');
print('Page No. 69\n\n');
import numpy
##Refer figure 3.6
## given
n = 5.;##years
C = 80000.;## COst of the project in Pound
S = 0.;## Zero Salvage Value
A_E = ([[10000 ,20000 ,30000 ,40000 ,50000]])## Annual Net cash flow for project E in Pound
C_E = ([[10000 ,30000 ,60000, 100000, 150000]])## Cummulative Net cash flow for project E in Pound
A_F = ([[50000 ,40000, 30000, 20000 ,10000]])## Annual Net cash flow for project F in Pound
C_F = ([[50000 ,90000 ,120000, 140000 ,150000]])## Cummulative Net cash flow for project F in Pound
##From the figure 3.6 (intercept of x-axis)
P_F = 1.75;## in years
P_E = 3.5;## in years
print'%s %.2f %s'%('The pay-back time of project F is ',P_F,'')
print'%s %.2f %s'%('The pay-back time of project E is ',P_E,'')
print('As the pay-back time is less for project F,\nProject F would always be choosen in practice\nsince prediction of savings in the early years are more reliable than long-term predictions.')
Ex9-pg70¶
In [12]:
import math
## Example 3.9
print('Example 3.9\n\n');
print('Page No. 70\n\n');
## given
P = 1.;##/ Principal Amount in Pound
r_i = 0.1;## Compound interest rate
for i in range(1,4):
c = P*(1.+r_i)**i;
print'%s %.2f %s %.2f %s '%('compound intrest after year ',i,'' is 'equal to ',c,' Pound\n')
new_P = 1000.*P;##in Pound
new_c = 1000.*c;## in Pound
print'%s %.2f %s'%('The new amount at the compound interest after fourth year is ',new_c,' Pound\n\n')
## Discount rate
r_d = 0.10;## Discount rate
for j in range (1,4):
d = P*(1./(1.+r_d)**j);
print'%s %.2f %s %.2f %s '%('The amount receivable at discount in year ',j,'' is '',d,' Pound\n')
new_P1 = new_c;## in Pound
new_d = new_P1*d;## in Pound
print'%s %.2f %s'%('The new amount receivable at discount in fourth year is ',new_d,' Pound\n')
Ex10-pg71¶
In [14]:
import math
## Example 3.10
print('Example 3.10\n\n');
print('Page No. 71\n\n');
import numpy
## given
C = 2500.;## Cost of the project
P = 1000.;## Cash in flow
r_r = 0.12;## Rate of return
S = 0.;## Zero salvage value
n = 4.;##years
z0=numpy.linspace(1,4)
leng=len(z0)
d_=numpy.zeros(leng)
for j in range (1,4):
d_[j] = P*(1./(1.+r_r)**j);
P_v = d_[0]+d_[1]+d_[2]+d_[3];##Present value of cash inflow
N = P_v-C;
print'%s %.2f %s'%('Net present value is',math.ceil(N),'Pound\n')
if(P_v>C):
print('The project may be undertaken')
else:
print('The project may not be undertaken')
Ex11-pg72¶
In [15]:
import math
## Example 3.11
print('Example 3.11\n\n');
print('Page No. 72\n\n');
import numpy
## given
Cash_out = 80000.;## Present value of cash outflow for both projects E and F
r_r = .2;## Rate of return
n = 5.;## years
d = numpy.array([0.833, 0.694 ,0.579, 0.482 ,0.402])## Discount Factor for 20% of rate of return for 5 years
Ce = numpy.array([10000 ,20000, 30000 ,40000 ,50000])## Cash flow for project E in Pound
Pe = numpy.array([8330, 13880, 17370 ,19280 ,20100])## Present value for project E in Pound
Cf = numpy.array([50000 ,40000, 30000, 20000, 10000])## Cash flow for project F in Pound
Pf = numpy.array([41650 ,27760, 17370, 9640, 4020])## Present value for project F in Pound
Cash_inE = sum(Pe)##Present value of cash inflow in Pound
Cash_inF = sum(Pf)##Present value of cash inflow in Pound
Net_E = Cash_inE - Cash_out;## net present value for project E in Pound
Net_F = Cash_inF - Cash_out;## net present value for project F in Pound
if (Net_E>Net_F):
print('Project E is selected based on NPV')
else:
print('Project F is selected based on NPV')
Ex12-pg72¶
In [16]:
import math
## Example 3.12
print('Example 3.12\n\n');
print('Page No. 72\n\n');
## given
Cash_inG = 43000.;## Present value of cash inflow for project G in Pound
Cash_outG = 40000.;## Present value of cash outflow for project G in Pound
Net_G = Cash_inG - Cash_outG;## Net present value for G in Pound
PI_G = (Cash_inG/Cash_outG);## Profitability index for G
Cash_inH = 23000.;## Present value of cash inflow for project H in Pound
Cash_outH = 20000.;## Present value of cash outflow for project H in Pound
Net_H = Cash_inH - Cash_outH;## Net present value for H in Pound
PI_H = (Cash_inH/Cash_outH);## Profitability index for H
##The higher the profitability index the more desirable is the project.
if (PI_G>PI_H):
print('Project G is more attractive than Project H')
else:
print('Project H is more attractive than Project G')
Ex13-pg73¶
In [2]:
import math
## Example 3.13
print('Example 3.13\n\n');
print('Page No. 73\n\n');
import numpy
## given
Cash_out = 80000.;## Present value of cash outflow for project F in Pound
n = 5.;## years
Cash_in= numpy.array([50000, 40000 ,30000, 20000 ,10000])## Cashn in \flow for project F in Pound
NPV = 0.;##At the end of 5 years
##Let the unknown rate for project F be rm.
##The amount standing at the end of 5 years is\n => 0 = 80000*(1+rm)^5 - 50000*(1+rm)^4 - 40000*(1+rm)^3 - 30000*(1+rm)^2 - 20000*(1+rm)^1 - 10000
## By taking (1+rm) = x\n =>8*x^5 - 5*x^4 - 4*x^3 - 3*x^2 - 2*x - 1 = 0\n\n')
from sympy import Eq, Symbol, solve
x = Symbol('x')
eqn= Eq(((x**5*8.0) - 5*x**4 - 4*x**3 - 3*x**2 - 2*x ), -1.0);
xres=1.3449924
rm=(xres-1)*100
rm = (xres - 1)*100;
print'%s %.2f %s'%('The value of rm for project F is ',math.ceil(rm),' per cent\n')
Ex14-pg74¶
In [4]:
import math
## Example 3.14
print('Example 3.14\n\n');
print('Page No. 74\n\n');
import numpy
import matplotlib
from matplotlib import pyplot
from sklearn import linear_model
## given
n = 5.;##years
C = 80000.;## Cost of the project in Pound
Cash_in = numpy.array([10000, 20000 ,30000 ,40000 ,50000])## Cash inflow in Pound
r_d1 = 15.;## Discount factor of 15%
r_d2 = 18. ;## Discount factor of 18%
r_d3 = 20.;## Discount factor of 20%
##At discount of 15%
df_1 = numpy.array([0.870, 0.756, 0.658 ,0.572 ,0.497])## Discount factor for every year
PV_1 = numpy.array([8700 ,15120, 19740 ,22880 ,24850])## Present value
Net_1 = sum (PV_1);## net present value
##At discount of 18%
df_2 = numpy.array([0.847, 0.718, 0.609 ,0.516 ,0.437])## Discount factor for every year
PV_2 = numpy.array([8470, 14360, 18270 ,20640, 21850])## Present value
Net_2 = sum (PV_2);## net present value
##At discount of 20%
df_3 = numpy.array([0.833 ,0.694 ,0.579 ,0.482, 0.402])## Discount factor for every year
PV_3 = numpy.array([8330 ,13880 ,17370, 19280, 20100])## Present value
Net_3 = sum (PV_3);## net present value
## f = N.P.V. cash inflow - N.P.V. cash outflow
##(1) By Numerical Method
ff = 2.*((sum (PV_2) - C)/(sum (PV_2) - sum(PV_3)));## in percentage
f = 18. + ff;
print'%s %.2f %s'%('the internal rate of return in percentage is ',f,' \n\n')## Deviation in answer due to direct substitution
##(2) By Graphical Interpolation
f_1 = (sum (PV_1) - C)/10**3;##At discount factor of 15%
f_2 = (sum (PV_2) - C)/10**3;##At discount factor of 18%
f_3 = (sum (PV_3) - C)/10**3;##At discount factor of 20%
%matplotlib inline
x = numpy.array([f_1 ,f_2 ,f_3]);
y = numpy.array([r_d1, r_d2, r_d3]);
import matplotlib.pyplot as plt
pyplot.title("Flow convergence loss in a conical nozzle")
pyplot.xlabel("Discount factor against f")
pyplot.ylabel("f ( *10^3 Pound)")
pyplot.plot(x,y)
pyplot.show()
import matplotlib.pyplot as plt
x = numpy.array([f_1 ,f_2 ,f_3]);
y = numpy.array([r_d1, r_d2, r_d3]);
pyplot.scatter(x,y)
pyplot.title("Discount factor against f")
pyplot.xlabel("f ( *10^3 Pound)")
pyplot.ylabel("Discount factor(%)")
Out[4]:
Ex15-pg77¶
In [5]:
import math
## Example 3.15
print('Example 3.15\n\n');
print('Page No. 77\n\n');
import numpy
## given
i_t = numpy.array([20, 40 ,60, 80, 100]);## Insulation thickness in mm
f_c = numpy.array([2.2, 3.5, 4.8, 6.1 ,7.4]);## Fixed costs in (10^3 Pound / year)
h_c = numpy.array([10.2, 6.5, 5.2, 4.6, 4.2]);## Heat costs in (10^3 Pound / year)
t_c = numpy.array([12.4, 10 ,10 ,10.7 ,11.6]);## Total costs in (10^3 Pound / year)
##(a) Graphical solution
##Refer figure 3.8
C_T = 9750.;## Minimum total cost in Pound
t = 47.;## Corresponding thickness of insulation in mm
print'%s %.2f %s'%('The most economic thickness of insulation is ',t,' mm \n')
##(b) Numerical solution
## The cost due to heat losses,C1, and the fixed costs,C2, vary according to the equations;-
## C1 = (a/x) + b and C2 = (c*x) + d
## Substituting the values of C1 and C2 together with the corresponding insulation thickness values , the following equations are obtained :-
## C1 = (150*10^3/x) + 2.7*10^3 and C2 = (65*x) + 0.9*10^3
##And to obtain the total costs
##CT = C1 + C2 = (150*10^3/x) + (65*x) + 3.6*10^3
## Differentiate to optimise, and put dCT/dx equal to zero
##dCT/dx =-((150*10^3)/x^2) + 65 = 0
##Let y = dCT/dx
from sympy import Eq, Symbol, solve
x = Symbol('x')
eqn= Eq(-((150*10**3)/x**2), 48.0);
print(eqn)
x=48.3
print'%s %.2f %s'%('The optimum thickness of insulation is ',x,'mm \n')
Ex16-pg79¶
In [6]:
import math
## Example 3.16
print('Example 3.16\n\n');
print('Page No. 79\n\n');
import numpy
## given
tb = numpy.array([36*10**3, 72*10**3, 144*10**3, 216*10**3]); ##operating time in s
U = numpy.array([971 ,863 ,727 ,636]);## Mean overall heat transfer rate in W/m^2-K
A = 50.;## area in m^2
dT = 25.;## temperature difference in degree celcius
ts = 54.*10**3;## Time in sec (h converted to sec)
##As Q = U*A*dT
z0=numpy.linspace(1,1,4)
leng=len(z0)
Q=numpy.zeros(leng)
Q_a=numpy.zeros(leng)
for i in range (1,4):
Q[i] = (U[i]*A*dT)/10**6;
Q_a[i] = ((tb[i]*Q[i]*10**6)/(tb[i] + ts))/10**6;
print'%s %.2f %s'%('the average heat transfer rate is ',Q_a[i],' *10^6 W \n')
##Refer figure 3.9
print('\n')
Q_max = 0.67*10**6;## Maximum value of Q in W
T_opt = 33.;## Time in h
print'%s %.2f %s'%('The maximum value of Q obtained is ',Q_max,' W \n')
print'%s %.2f %s'%('The most econnomic opertaing time for the heat exchanger to run is',T_opt,' h ')
Ex17-pg80¶
In [7]:
import math
## Example 3.17
print('Example 3.17\n\n');
print('Page No. 80\n\n');
## given
## C_T = 7*x + (40000/(x*y)) + 6*y + 10
##Differentiating C_T with respect to x and y:-
##dC_T/dx = 7 - (40000/(x^2*y))
##dC_T/dy = - (40000/(x*y^2)) + 6
##For optimum conditions :- dC_T/dx = dC_T/dy = 0
##dC_T/dx = 0 => 7 - (40000/(x^2*y)) = 0
##=> y = 40000/(7*x^2).......(1)
##dC_T/dy = 0 =>- (40000/(y^2*x)) +6 = 0
##=> y = (40000/(6*x))^0.5.......(2)
##From equation (1) and (2)
##=> 40000/(7*x^2) - (40000/(6*x))^0.5 = 0
from sympy import Eq, Symbol, solve
x = Symbol('x')
eqn= Eq( 40000./(7*x**2) - (40000./(6.*x))**0.5,20);
x=16.982634
##from equation (1)
y = 40000./(7*x**2);
##a = d^2C_T/dx^2 = 80000/(x^3*y)
##b = d^2C_T/dy^2 = 80000/(x*y^3)
a = 80000./(x**3*y);
b = 80000./(x*y**3);
if a > 0:
if b>0:
C= 7.*x + (40000./(x*y)) + 6.*y + 10.;## in Pound
##The optimum conditions must occur at a point of minimum cost- C_T_m
print'%s %.2f %s'%('The minimum cost is ',C,' Pound')