HW-7
* Read Newnan Chap 7.
Work problems:
12th Ed. pp. 207ff: 6-1, 6-3, 6-8, 6-10, 6-21, 6-22, 6-37, 6-49, 6-54
Instructions
Do all your work below; use this as a text file or use the Excel functions if you need them.
List each problem number and do the work here.
Save this file to your computer as you work.
When you are finished, go back to the Assignments page where you found this and open the
Assignment.
You will find a section called “Assignment Materials.” Near the bottom of that section is “Attach File”
and a button to Browse for Local File.
Again do your work, save it, and them use the above to attach it.
Do your work here
hem.
open the
on is “Attach File”
12th Ed. pp. 211ff: 6-1, 6-3, 6-8, 6-10, 6-21, 6-22, 6-37, 6-49, 6-54
6-1
E = $6,000− $1,500 (A/G, 12%, 4)= $6,000 − $1,500 (1.359) = $3,962
6-3
ALTERNATIVE
p
n
int/yr
Salv
EUAC
Eff int
EUAC
First, compute A:
A = ($20,000 − $4,000) (A/P, 4%, 10) + $4,000 (0.04)
= $16,000 (0.1233) + $160= $2,132.80 per semiannual period
Now, compute the equivalent uniform annual cost:
EUAC = $2,132.80 (F/A, 4%, 2)= $2,132.80 (2.040)= $4,350.91
6-8
A = F[(er −1)/(em − 1)]
= $5 × 106 [(e0.15 − 1)/(e(0.15)(40) − 1)]
= $5 × 106 [0.161834/402.42879]
= $2,011
6-10
Solution at back of book is not correct.
1
2
3
4
5
6
1.4
1.5
1.6
1.7
1.8
1.9
7
8
9
10
npv=
EUAC=
Or =
2
2.1
2.2
2.3
$9.94
($1.76)
1.4 + .1 * (A/G,0.12,10)
1.7585
6-21
P = $1,000 (P/A, 9%, 5) + $3,500 (P/F, 9%, 4) + $1,500 (P/A, 9%, 5) (P/F, 9%, 5) +
$3,500 (P/F, 9%, 8)
= $1,000 (3.890) + $3,500 (0.7084) + $1,500 (3.890) (0.6499) + $3,500 (0.5019)
= $11,918
Equivalent Uniform Annual Amount = $11,918 (A/P, 9%, 10) = $1,857
6-22
This problem is much harder than it looks!
EUAC = {$600 (P/A, 8%, 5) + $100 (P/G, 8%, 5) + [$900 (P/A, 8%, 5) −
$100 (P/G, 8%, 5)][(P/F, 8%, 5)]}{(A/P, 8%, 10)}
= {$600 (3.993) + $100 (7.372) + [$900 (3.993) −
$100 (7.372)][0.6806]}{0.1490}
= $756.49
PVa(0)
PVb(0)
PV(0)
EUAC
3133
1944.1339
0.149
5077.1339
756.4929511
6-37
EUAC Comparison
Gravity Plan
Initial Investment: = $2.8 million (A/P, 10%, 40)
= $2.8 million (0.1023) = $286,400
Annual Operation and maintenance = $10,000
Annual Cost = $296,400
Pumping Plan
Initial Investment: = $1.4 million (A/P, 10%, 20)
= $1.4 million (0.1023) = $143,200
Additional investment in 10th year:
= $200,000 (P/F, 10%, 10) (A/P, 10%, 40)
= $200,000 (0.3855) (0.1023) = $7,890
Annual Operation and maintenance = $25,000
Power Cost: = $50,000 for 40 years = $50,000
Additional Power Cost in last 30 years:
= $50,000 (F/A, 10%, 30) (A/F, 10%, 40)
= $50,000 (164.494) (0.00226) = $18,590
Annual Cost = $244,680
Select the Pumping Plan.
6-49
Annual Cost of Diesel Fuel = [$80,000km/(16 km/liter)] × $0.88/liter = $4,400
Annual Cost of Gasoline = [$80,000km/(11 km/liter)] × $0.92/liter = $6,691
EUACdiesel = ($24,000 − $4,000) (A/P, 6%, 5) + $4,000 (0.06) + $4,400 fuel +
$900 repairs + $1,000 insurance
= $20,000 (0.2374) + $240 + $6,300
= $11,288
EUACgasoline = ($19,000 − $6,000) (A/P, 6%, 4) + $6,000 (0.06) + $6,691 fuel +
$700 repairs + $1,000 insurance
= $12,503
The diesel taxi is more economical.
6-54
It is important to note that the customary “identical replacement” assumption is not
applicable here.
Alternative A
EUAB – EUAC = $1,500 − $5,000 (A/P, 15%, 10) = $1,500 − $5,000 (0.1993)
PS solution
Interest=
0.1
Years
Grav
0
2800000
10
Ann
yr 1 to 10
yr 11-40
EUAC
10000
$296,326.36
= +$504
Alternative B
EUAB – EUAC = $6,000 (P/A, 15%, 5) (A/P, 15%, 10) − $18,000 (A/P, 15%, 10)
= +$421
Choose A.
Check solution using NPW:
Alternative A: NPW = $1,500 (P/A, 15%, 10) − $5,000 = +$2,529
Alternative B: NPW = $6,000 (P/A, 15%, 5) − $18,000 = +$2,112
-20000
5
0.08
4000
0.0816 (annually)
$4,350.62
21753.08211
10
0.04
Alternative way
npv=
pmt=
600
700
800
900
1000
900
800
700
600
500
$5,076.75
($756.59)
Pump
$1,400,000.00 One-time
$200,000.00 One-time
$25,000.00
$50,000.00
$100,000.00
$77,108.66
$244,476.27
$307,228.36
$363,448.36
$2,392,261.64
$244,631.27
PV for 200k for 10 yrs
PV for 25,000 for 40 yrs
PV of 50,000 for 10 yrs
PV for yrs 11-40 which are then moved to yr 0
Total PV
12th Ed. pp. 207ff: 6-1, 6-3, 6-8, 6-10, 6-21, 6-22, 6-37, 6-49, 6-54
Classroom session 10-15-15
6-1
Int==
Year
0.12
0
1
2
3
4
6000
4500
3000
1500
using
PV
($5,357.14)
($3,587.37)
($2,135.34)
($953.28)
Note:
When seeking EUAB, EUAC or EU
the best method will be to use P
Using
Gradient
6000
6000
6000
6000
0
-1500
-3000
-4500
PW (NPV)= $12,033.13
PW (PV)
($12,033.13)
EUAB=
$3,961.72 using PMT
-1500(A/G,.12,4)
6000
-2038.5
3961.5
EUAW=EUAB
6-22
Int=
0
1
2
3
4
5
6
7
8
9
10
PV=
600
700
800
900
1000
900
800
700
600
500
$5,076.75
0.08
using
PV
($555.56)
($600.14)
($635.07)
($661.53) 661.5269
($680.58)
($567.15)
($466.79)
($378.19)
($300.15)
($231.60)
($5,076.75)
$756.59
-5
-4
-3
-2
-1
0
1
2
3
4
600
700
800
900
1000
900
800
700
600
500
When seeking EUAB, EUAC or EUAW
the best method will be to use PMT().
This was calulated using the
(1+i )^n method. It shows that
the result is the same as the
red numbers.
The red numbers uses PV().
Note:
To use PMT you must have
a value at either the beginning
of the time range (0, or in this
case, -6), or at the end (4 or 10 depending on which set of years
you are using.
You cannot use the 0 time when it is in the middle of the range.
NOTE: At the bottom of this window are labeled tabs.
TAB: HW-6 is this sheet with the new homework
HW - 6
Problems due the day BEFORE the next class.
* Read Newnan Chap. 5, 6
12th Ed. pp. 173ff: 5-1, 5-2, 5-3, 5-5, 5-9, 5-15, 5-32, 5-38, 5-37, 5-43, 5-61, 5-83
Instructions
Do all your work below; use this a text file or use the Excel functions if you need them.
List each problem number and do the work here.
Save this file to your computer as you work.
When you are finished, go back to the Assignments page where you found this and open the Assignment.
You will find a section called “Assignment Materials.” Near the bottom of that section is “Attach File” and a button
Again do your work, save it, and them use the above to attach it.
Do your work here
on is “Attach File” and a button to Browse for Local File.
12th Ed. pp. 173ff: 5-1, 5-2, 5-3, 5-5, 5-9, 5-15, 5-32, 5-38, 5-37, 5-43, 5-61, 5-83
On some I added notes--PS
Also see the Recitation Excel Files on this assignment when posted.
5-1
NPW at start of year on 1/1/2018 is same as NWP at end of year 2017
NPW12/31/2017= –$14,000
NPW12/31/2015= –$14,000 (P/F, 9%, 2) = –$14,000 (0.8417) = –$11,784
0,09
($11.783,52)
5-2
P = $18,000 (P/A, 9%, 10) + $245,000 (P/F, 9%, 10)
= $18,000 (6.418) + $245,000 (0.4224)
= $219,012
P=
($219.008,49)
0,09
($115.517,84)
0,09
($103.490,65)
5-3
P = the first cost = $980,000
($98.000,00)
F = the salvage value = $20,000
0,12
AB = the annual benefit = $200,000
$1.130.044,61
Remember our convention of the costs being negative and the benefits being
0,12
positive. Also, remember the P occurs at time = 0.
$113.004,46
NPW = –P + AB (P/A, 12%, 10) + F (P/F, 12%, 10)
= –$980,000 + $200,000 (5.650) + $20,000 (0.3220)
= $156,440
Therefore, purchase the machine, as NPW is positive.
PS note: You can also do it these two ways. Doulble-click on the two cells to see how I did it.
Note the slightly different answer due to rounding in the tables used by the author.:
NPV=
$156.484,07
$156.484,07
5-5
P = ?, n = 36 months, i = 15%/12mo = 1.25%/month, A = $500
P = $500 (P/A, 1.25%, 36) = $500 (28.847) = $14,424
0,0125
($14.423,63)
5-9
NPW = -$600,000 - $100,000(P/A,15%,6) - $5,000(P/G,15%,6) +
$300,000(P/A,15%,6) -$15,000(P/G,15%,6) + $100,000(P/F,15%,6)
NPW = -$600,000 + ($300,000 - $100,000)(P/A,15%,6) +
(-$5,000 - $15,000)(P/G,15%,6) + $100,000(P/F,15%,6)
NPW = -$600,000 + $200,000(3.784) - $20,000(7.937) + $100,000(0.4323)
NPW = $41,290
Given a positive NPW, the plan should be accepted.
5-32
36
Compute the PW of Cost for a 25-year analysis period.
Note that in both cases the annual maintenance is $100,000 per year after 25 years.
Thus after 25 years all costs are identical.
Single Stage Construction
PW of Cost = $22,400,000 + $100,000 (P/A, 4%, 25)
= $22,400,000 + $100,000 (15.622)
= $23,962,000
Two Stage Construction
PW of Cost = $14,200,000 + $75,000 (P/A, 4%, 25) + $12,600,000 (P/F, 4%, 25)
= $14,200,000 + $75,000 (15.622) + $12,600,000 (0.3751)
= $20,098,000
Choose two stage construction.
5-37
NPW for 10 years of alternative A (identical equipment replace at year 5):
NPW(A) = -$4,500 + ($500 - $4,500)(P/F,6%,5) - $300(P/A,6%,10) +
$500(P/F,6%,10)
NPW(A) = -$4,500 - $4,000(0.7473) - $300(7.360) + $500(0.5584) = -$9,418
Alternatively, we could write this showing the two lives as:
NPW(A) = -$4,500 - $300(P/A,6%,5) + $500(P/F,6%,5) - $4,500(P/F,6%,5)
- $300(P/A,6%,5)(P/F,6%,5) + $500(P/F,6%,5)*(P/F,6%,5)
If you notice the second 3 terms are the same as the first three except for the 5 year
time difference. Therefore, we could rewrite this as:
NPW(A) = [-$4,500 - $300(P/A,6%,5) + $500(P/F,6%,5)][1 + (P/F,6%,5)]
NPW(A) = [-$4,500 - $300(4.212) + $500(0.7473)][1 + (0.7473)] = -$9,418
NPW for 10 years of alternative B:
NWP(B) = -$5,500 - $400(P/A,6%,10) = -$5,500 - $400(7.360) = -8,444
Alternative B is the better of the two, yielding a lower present worth of the costs.
5-38
To equate the lives we will replace alternative A at the end of year 10 with an
identical process.
PW(A) = $18M [1 + (P/F, 9%, 10)] + $5M (P/A, 9%, 20)
− $4M[(P/F, 9%, 10) + (P/F, 9%, 20)]
PW(A) = $18M [1 + (0.4224)] + $5M (9.129) − $4M[(0.4224) + (0.1784)]
PW(A) = $68.8M
PW(B) =$25M + $3M (P/A, 9%, 20) − $6M (P/F, 9%, 20)
PW(B) =$25M + $3M (9.129) − $6M (0.1784)
PW(B) = $51.3M
PW(A) - PW(B) = $69.8M - $51.3M = $17.5M
Select Alternate B which saves $17.5M
5-43
To equate the lives we will replace alternative A at the end of year 2 with an identical
process.
PW(A) = $200,000 [1 + (P/F, 2%, 8)] + $10,000 (P/A, 2%, 16)
− $25,000 [(P/F, 2%, 8) + (P/F, 2%, 16)]
PW(A) = $200,000 [1 + (.8535)] + $10,000 (13.578) − $25,000 [(.8535) + (0.7284)]
PW(A) = $466,933
PW(B) =$250,000 + $15,000 (P/A, 2%, 16) − $40,000 (P/F, 2%, 16)
PW(B) =$250,000 + $15,000 (13.578) − $40,000 (0.7284)= $424,534
Select Alternate B.
5-61
Compute the present worth of the cash flow:
PW = ($725,000 + $143,000) + $260,000 (P/F, 9%, 25)
+ $12,000 (P/A, 9%, 40) - $32,000 (P/F, 9%, 40)
PW = $868,000 + $260,000 (0.1160) + $12,000 (10.757) - $32,000 (0.0318)
PW = $1,026,226.40
Compute the equivalent annual worth of this present value over 40 years:
AW = 1,026,226.40 (A/P, 9%, 40) = 1,026,226.40 (0.0930) = $95,439.06
Capitalized cost = $95,439.06/0.09 = $1,060,434
5-83
Assuming monthly compounding, we use a per period interest rate of 1.5%
Details below
int
costs
1
2
3
4
5
6
7
8
Sum:
Or:
30000
50000
210000
530000
620000
460000
275000
95000
$2.106.599,68
2
14000
10
18000
10
245000
10
200000
10
20000
Using PV()
e how I did it.
500
Using PV()
0,015
PW
29557
48533
200827
499358
575521
420689
247782
84333
$2.106.599,68
HW-9
Work problems:
12th Ed. 7-2, 7-6, 7-14, 7-18, 7-46.
NEXT EXAM DATE 10-27-2015. It will cover Chaps 5, 6 and 7
You may use a COMPUTER, and the textbook. No notes or homework.
Instructions
Do all your work below; use this as a text file or use the Excel functions if you need them.
List each problem number and do the work here.
Save this file to your computer as you work.
When you are finished, go back to the Assignments page where you found this and open the
Assignment.
You will find a section called “Assignment Materials.” Near the bottom of that section is “Attach File”
and a button to Browse for Local File.
Again do your work, save it to the computer, and then attach it to the Assignment.
Do your work here
hem.
open the
on is “Attach File”
12th Ed. 7-2, 7-6, 7-14, 7-18, 7-46.
7-2
(F/A, i, 35) = 10^6/5800 =
= 172.414 and is very close to 8% from tables.
(Exact = 8.003%)
PS alternative
A
-5800
Using Rate()
N
35
7-6
PS Alternative
$9,375 = $325 (P/A, i%, 36)
(P/A, i%, 36) = $9,375/$325 = 28.846
From compound interest tables, i = 1.25%
Nominal Interest Rate = 1.25 × 12 = 15%
Effective Interest Rate = (1 + 0.0125)12 − 1 = 16.08%
PV
A
N
Rate()=
Nominal
Effective
-9375
325
36
1,25%
15,003% =12*H17
16,079% =(1+H17)^12-1
7-14
The easiest solution is to solve one cycle of the repeating diagram:
PS alternatives:
0
1
80
-120
IRR()+
50%
Rate()
50%
Note that to use Rate() the
period is 1 and you use the 2nd
value as the pmt and the first
as pv.
Another:
$120 = $80 (F/P, i%, 1)
$120 = $80 (1 + i)
(1 + i) = $120/$80 = 1.50
i* = 0.50 = 50%
Alternative Solution:
EUAB = EUAC
$80 = [$200 (P/F, i%, 2) + $200 (P/F, i%, 4) + $200 (P/F, i%, 6)] (A/P, i%, 6)
Try i = 50%
$80 = [$200 (0.4444) + $200 (0.1975) + $200 (0.0878)] (0.5481) = $79.99
Therefore i* = 50%
7-18
Year
0
1
2
3
4
5
6
Using IIR()
Cash flow
12500
-1000
-2000
-3000
-4000
-5000
-6000
13,18%
$12,500 = $1,000 (P/A, i%, 6) + $1,000 (P/G, i%, 6)
at 12%, $1,000 (4.111) + $1,000 (8.930) = $13,040
at 15%, $1,000 (3.784) + $1,000 (7.937) = $11,720
i* = 12% + (3%) ((13,040 – 12,500)/(13,040 −11,720)) = 13.23%
7-46
0
1
2
3
4
5
6
80
-120
80
-120
80
-120
50%
($0,00)
PS alternatives
PW of Benefits – PW of Cost = $0
$15,000 (P/F, i%, 4) − $9,000 − $80 (P/A, i%, 4) = $0
Try i = 12%
$15,000 (0.6355) − $9,000 − $80 (3.037) = +$289.54
Try i = 15%
$15,000 (0.5718) − $9,000 − $80 (2.855) = −$651.40
Performing Linear Interpolation:
i* = 12% + (3%) [289.54/(289.54 + 651.40)] = 12.9%
Price
Ann Taxes
N
Fut Val
Using Rate()
-9000
-80
4
15000
12,88%
years
EOY
0
1
2
3
4
-9000
-9000
-80
-80
-80
-80
-80
-80
14920 =15000-80
12,88% =IRR(C5:C9)
Future
1000000
8,00%
=(1+H17)^12-1
use Rate() the
nd you use the 2nd
pmt and the first
0,499999 =Int found by Goal seek
50%
HW-8
* Read Newnan Chap 7.
Work problems:
12th Ed. pp. 245ff: 7-1, 7-3, 7-4, 7-7, 7-15, 7-30, 7-32, 7-42, 7-44, 7-48, 7-54, 7-70.
NEXT EXAM DATE 10-27-2015. It will cover Chaps 5, 6 and 7
You may use a COMPUTER, and the textbook. No notes or homework.
Instructions
Do all your work below; use this as a text file or use the Excel functions if you need them.
List each problem number and do the work here.
Save this file to your computer as you work.
When you are finished, go back to the Assignments page where you found this and open the
Assignment.
You will find a section called “Assignment Materials.” Near the bottom of that section is “Attach File”
and a button to Browse for Local File.
Again do your work, save it to the computer, and then attach it to the Assignment.
Do your work here
4, 7-70.
hem.
open the
on is “Attach File”
12th Ed. pp. 249ff: 7-1, 7-3, 7-4, 7-7, 7-15, 7-30, 7-32, 7-42, 7-44, 7-48, 7-54, 7-70.
7-1
(F/A, i, 25) =
7
5
10
40
(F/A, 3.5%, 25) = 38.950 and (F/A, 4%, 25) = 41.646 bound the value. So we can
interpolate:
i = 3.5% + (0.5%) .
.. = 3.69%
7-3
We can solve for the rate of return directly:
F = P (1 + i)n
$20,000 = $5,000 (1 + i)3
(1 + i) = ($20,000/$5,000)1/3 = 1.587
i* = 0.587
Rate of Return = 58.7%
7-4
F = $5, P = $1, n = 5
F = P (1 + i)^n
$5 = $1 (1 + i)^5
(1 + i) = 5^0.20 = 1.38
i* = 38%
7-7
$5,000 = $1,500 (P/A, i%, 5)
(P/A, i%, 5) = $5,000/$1,500 = 3.333
Performing Linear Interpolation:
(P/A, i% 5)
i
3.352
15%
3.127
18%
i = 15% + (3%)((3.333 − 3.352)/(3.127 − 3.352))= 15.25%
7-15
Solve for the rate that yields a NPW of zero (ignore salvage value initially):
NPW = 0 = $14,000 (P/A, i%, 30) - $250,000
(P/A, i%, 30) = $250,000/$14,000 = 17.857
At i = 3.5%, P/A = 18.392; At i = 4%, P/A = 17.292.
i = 3.5% + 0.5% [(17.857 – 18.392)/(17.292 – 18.392)] = 3.7%
Including salvage value of $200,000:
NPW = 0 = $14,000 (P/A, i%, 30) + $200,000 (P/F, i%, 30) - $250,000
Try i = 5%: NPW = $14,000 (15.372) + $200,000 (0.2314) - $250,000 = $11,488
Try i = 6%: NPW = $14,000 (13.765) + $200,000 (0.1741) - $250,000 = -$22,470
Performing Linear Interpolation:
NPW
i
$11,488
5%
($22,470)
6%
i = 5% + (1%)((0 − 11,488)/(-22,470 − 11,488))= 5.34%
7-30
PIC
PW of Benefits – PW of Cost = $0
$30 (P/A, i%, 27) + $1,000 (P/F, i%, 27) − $875 = $0
Try i = 3 ½%
$30 (17.285) + $1,000 (0.3950) − $875 = $38.55 >$0
Try i = 4%
$30 (16.330) + $1,000 (0.3468) − $875 = −$38.30 < $0
i* = 3.75%
Nominal rate of return = 2 (3.75%) = 7.5%
7-32
a) Interest paid quarterly = $1,000 (0.06/4) = $15
NPW = 0 = $15 (P/A, i%, 40) + $1,000 (P/F, i%, 40) - $900
Try i = 1.75%
NPW = $15 (28.594) + $1,000 (0.4996) - $900 = $28.50
Try i = 2.00%
NPW = $15 (27.355) + $1,000 (0.4529) - $900= -$36.80
Interpolating we get: i = 1.859%/qtr
Annual effective rate = (1 + 0.0186)4 - 1 = 7.65%
b) NPW = 0 = $15 (P/A, i%, 20) + $950 (P/F, i%, 20) - $900
Try i = 1.75%
NPW = $15 (16.753) + $950 (0.7068) - $900 = $22.80
Try i = 2.00%
NPW = $15 (16.351) + $950 (0.6730) - $900= -$15.40
Interpolating we get: i = 1.899%/qtr
Annual effective rate = (1 + 0.01899)4 - 1 = 7.82%
7-42
a) The foregone cash rebate is like a hidden finance charge. You pay $12,000 for
the car but receive a car only worth $12,000 – $3,000 = $9,000. The monthly
payments = 12000/(4*12) = $250 for 48 months. NPW = 0 = 9000 – 250 (P/A, i, 48),
so (P/A, i%, 48) = 36.0 and interpolating
i = 1% + (0.25%)*(37.974 - 36.0)/(37.974 35.932)
= 1.242%, so r = (12) (1.242%) = 14.90%
and ia = (1 + 0.01242)12–1 = 0.15965 or 15.97%.
1.24% using Rate()
b) Worth of car = Cost – Rebate = $18,000 – $3,000 = $15,000.
The monthly payments = 18000/(4*12)
= $375 for 48 months.
NPW = 0 = 15,000 – 375 (P/A, i%, 48), so, (P/A, i%, 48) = 40.0 and interpolating
i = 0.75% + (0.25%)(40.185 - 40.0)/(40.185 37.974)
= 0.771%, so r = (12)(0.771%) = 9.65%
and ia = (1 + 0.00771)12–1 = 0.0965 or 9.65%.
c) Worth of car = Cost – Rebate = $24,000 – $3,000 = $21,000.
The monthly payments = 24000/(4*12)
= $500 for 48 months.
NPW = 0 = 21000 – 500 (P/A, i%, 48), so, (P/A, i%, 48) = 42.0 and interpolating
i = 0.50% + (0.25%)(42.580 - 42.0)/(42.580 - 40.185)
= 0.561%, so r = (12)(0.561%) = 6.73%
and ia = (1 + 0.00561)12–1 = 0.0694 or 6.94%.
7-44
The amount of cash paid will be $75,000 – $50,000 = $25,000 with $50,000
financed, so, the monthly payments will be 50,000 (A/P, 8%, 4) = (50,000) (0.3019)
= $15,095. The reduction in cost if one pays entirely in cash is $75,000 x 0.10 =
$7,500, so, a 100% cash payment would be $75,000 − $7,500 = $67,500 (true value
of equipment).
Year
Pay Cash Borrow from
Incremental
Manufacturer
Difference
0
-67,500
-25,000
-42,500
1
-15095
15095
2
-15095
15095
3
-15095
15095
4
-15095
15095
NPW = 0 = –42500 + 15095 (P/A, IRR, 4), so (P/A, IRR, 4) = 2.816. Interpolating
IRR = 15% + (3%)(2.855 - 2.816)/(2.855 - 2.690) = 15.72%.
Using the IRR() function:
7-48
Total Annual Revenues = $500 (12 months) (4 apt.) = $24,000
Annual Revenues – Expenses = $24,000 − $8,000 = $16,000
To find Internal Rate of Return the Net Present Worth must be $0.
NPW = $16,000 (P/A, i*, 5) + $160,000 (P/F, i*, 5) − $140,000
At i = 12%, NPW = $8,464
At i = 15%, NPW = −$6,816
15.69%
IRR= 12% + (3%) [$8,464/($8,464 + $6,816)]= 13.7%
7-54
The problem requires an estimate for n, the expected life of the infant. Seventy or
seventy-five years might be the range of reasonable estimates. Here we will use 71
years.
The purchase of a $1,000 life subscription avoids the series of beginning-of-year
payments of $64.50. Based on 71 beginning-of-year payments,
$1,000 – $64.50 = $64.50 (P/A, i%, 70)
(P/A, i%, 70) = $935.50/$64.50 = 14.50
6% < i*< 8%. By Calculator: i* = 6.83%
6.83% =RATE(75,-64.5,935.5)
7-70
Using incremental analysis, computed the internal rate of return for the difference
between the two alternatives.
Note: Internal Rate of Return (IRR) equals the interest rate that makes the PW of
costs minus the PW of Benefits equal to zero.
$9,000 – $3,000 (P/A, i*, 7) − $1,200 (P/F, i*, 8) = $0
Try i = 25%: $9,000 − $3,000 (3.161) − $1,200 (0.1678) = −$684.36< $0
Try i = 30%: $9,000 − $3,000 (2.802) − $1,200 (0.1226) = $446.88> $0
i* = 25% + (5%) [$684.36/($446.88 + $684.36)]= 28.0% (actual value is 27.9%)
The contractor should choose Alternative A and lease because 28% > 15% MARR.
OR:
$9,000 – $3,000 (P/A, i*, 7) − $1,200 (P/F, i*, 8) = $0
Amount years
Init
-9000
9000
Ann
3000
7 ($8,832.41)
Future
1200
8
($167.59)
Int
0.278988
PV
Goal seek:
($0.00)
Chapter 7 problems worked in class
7-6
Price
Down
month
N
Loan
Rate=
Nominal
Two ways
effective
interest
12375
3000
-325
36
9375
1.250%
15.003%
Pmt
N
PV
/mo
P/A=
28.84615
0.0125
0.1607858 Using Effect()
0.1607858 Using (1+int)^12) - 1
7-14
0
1
2
3
4
5
6
0.499999 =Int found by Goal seek
80
-120
80
-120
80
-120
50%
($0.00)
50%
7-42
Price
n
pmt
PV
Rate=
Effective
Case a
12000
48
3000
-250
9000
1.241%
15.96%
In words:
Case b
Case c
0% interest!!
18000
24000
48
48
Discount for cash
3000
3000
-375
-500
15000
21000
=RATE(B30,B32,B33)
0.770%
0.559%
9.64%
6.91%
=RATE(n,pmt,Price-discount)
years
EOY
7-46
0
1
2
3
4
-9000
-9000
-80
-80
-80
-80
-80
-80
14920 =15000-80
12.88128% =IRR(C5:C9)
int
-9000
-79.2079
-78.4237
-77.6472
14337.83
5102.548
0.01
-9000
($79.21)
($78.42)
($77.65)
$14,337.83
Years
Cash
0
1
2
3
4
-9000
-80
-80
-80
14920
INT=
0.01
-9000
Purchase answer to see full
attachment