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Unformatted text preview: 22, x7, 25/ 2.2; 25;)?“ 512' 3552 712 (a) From Table 74, the GDS Recovery period is 5 years.
S/afz‘o it”; (h) In year 4 there was a $17,280 depreciation deduction.  DepreciatiOn Depreciation
Year Calculation Deduction Endof—Year
Book Value 1 $150,000 (0.2) $30,000 $120,000
2 150,000 (0.52) 48,000 72,000
3 150,000 (0.192) 28,800 43,200
4 150,000 (0.1 152) 17,280 25,920
5 150,000 (0.1152) 17,280 8,640
6 150,000 (0.0526) 8,640 0 (c) From (b), the book value at beginning ofyear 5 (end of year 4) is $25,920 (Bl (Q‘tM{BJ (0):“! {C} (E}=(A)+(D) 1117 141
EOY BTCF Depr. r1 T (40%) ATCF 0 $10M  $10M
1 $4M $2.5M $1.5M $0.6M $3.4M E
2 $4M $2.5M $1.5M —$0.6M $3.4M gr
3 $4M $2.5M $1.5M $O.61_Vt $3.4M L
4a $4M $2.5M $1.5M —$0.6M $3.4M I
4b 0 0 0  . _'__‘._ I :.. M Emillions of dollars
PW(15%) = ~$lOM + $3.4M (P/A,i%,4) = 6329304 IRR: 0 —"—' ~$lOM + $3.4M (PfA,i%,4); i' = 13.54%
Because PW(1[)%) < 0 and IR 4 15%, this project should not be recommended. 7~23 Assume repeatability. Purchase Option: From Table TZ, the ADS recovery period is 6 years (asset class
36.0). Applying the half year convention, depreciation deductions can be claimed over a 7—year period. C11: c17 = (0.5)($30,000/6) = $2,500 d2 = d3 = = d, = ($30,0001’6) = $5,000
(41 0:0 1c1=141  (B) 101=4tc1 (E1=141+(D1
EOY BTCF Depr. '11 T (40%) ATCF
0 "$3 0,000 '  83 0,000
1 0 $2,500 $2,500 $1,000 1,000
2 0 5,000 — 5,000 2,000 2,000
3 0 5,000  5,000 2,000 2,000
4 — 10,000 5,000 45,000 6,000 — 4,000:=—10,000+6,000
5 0 5,000 — 5,000 2,000 2,000
6 0 5,000 — 5,000 2,000 2,000
2 0 2,500 4 2,500 1,000 1,000
8 0 0 0 0 '1
l __lt‘ __ PW(12%) =  $30,000 + $1,000(P/A,12%,7) + $1,000(P/A,12%,5)(P/F,12%,1)
 $6,000(P/F,12%,4) =  $26,030.47 ' 1
_ [1/ AW(12%) = « $26,030.47 (NP,12%,8) =  $5,240 5 Leasing Option (MW
ATCF =  (1 — 0.40)(Leasing Cost) = AW(12%) . 2, For the leasing option to be more economical than the purchase option, ' ~ (0.6) (Leasing Cost) 4 — $5,240
Leasing Cost < $8,733
If Leasing Cost < $8,733 per year, lease the tanks; otherwise, purchase the tanks. 0.l8?4 (3) AW = $30,000(Ax?,10%,8) + 81
3.9847 (0) IR; use uglitative reasoiiinggnd A (BHA); by observation from
differences in PWS, both IRRS SMARR/AT = 10% and i'%3 > i‘%A Choose Machine B. $27 Assume repeatability of investments over a common multiple of lives. Design 31
(A) l (B) 1C1= (A) ~ 03) (D) = t (c) (E) = 141+1D)
OY BTCF Depr T (40%) A'I‘CF PW(10%)
0  $100,000 l 8100,000 0100,000
1 20,000 $20,000 50 20,000 18,182
2 20,000 32,000 4,800 24,800 20,495
3 20,000 19,200 ~ 320 19,680 14,286
4 20,000 11,520  3,392 16,608 11,343
5 20,000 11,520 9 3,392 16,608 10,312
6 20,000 5,760  5,696 14,304 8,075
7 20,000 0 — 8,000 12,000 6,158
7 30,000 ~ 1 2,000 18,000 9,238
PW(10%)= —$ 1,411 AW(10%) = PW (AIP,IO%,?) = 31,41 1 (0.2054) = 6% Desigg 82
""‘T (A) (B) (c) = (A)  03) (D) = 416‘) 1E1= (A) + (D) ‘l EOY BTCF Depr T (40%) PW 10%)
0 $200,000 $200,000 $200,000
1 40,000 $40,000 40,000 36,364
2 40,000 64,000 49,600 40,989
3 40,000 38,400 39,360 29,521
4 40,000 23,040 33,216 22,687
5 40,000 23,040 33,216 20,624
6 40,000 11,520 28,608 16,149
6 50,000 30,000 16,935 PW(10%) = 016,681 AW 10%) = PW(NP,10%,6) = — $16,681 (0.2296) = H 83% Therefore, choose Design Si since it has the greater annual worth (Note: Neither
design makes money  so if a system is not required, do not buy either 31 or $2.)
Repeatability is assumed. 728 (3) Straightline depreciation: Method 1
(C) = (A)  03) (D) =‘ i (C) (E) = (A) + (D)
T I T (40%) ATCF
$10,000   w$10,000
$14,150 $15,950 $6,380 $1770
3 1,000 — 0 0 $1,000 PW0(12%) = 010,000 ~ $?,770(P2’A,12%,5) +$1,000(P!F,12%,5) : $37,441,552 To have a basis for computation, assume that Method I is duplicated for years 610. Transform the additional PW5(12% = $3 7,449.68 t0 the present and get:
PW(12%) = PWD(12%)+(PJ’F,12%,5)PW5(12%)= $58,86?.10 Methodj (C) = (A)  (B) (D) = t {C} (E) (A) + (D) {13) Depr TI T £400 0) ATCF $40,000
$3,500 $10,500 $4,200 —$2,800 0 0 $5,000 PW(12%) = $40,000  $2,800(PHA,12%,10) +$5,000(PKF,12%,10) = $54,210.76
Thus, Method II is the better alternative. (0) Mgthod I
Assume that MVS is $1,000 The MACRS property class is 5 years. This means that the taxlife is 6 years
which is greater than the useful life of 5 years. PW (12%) = $37,311.71 13W(12%) over 10 years = 337,31l[l+(P/F,12%,5)] = $58,482
To get a ﬁgure for comparison, convert PW(12%) to annual worth over the
useful life ofS years: AW(12%) == $3?,311.11(A,1’P,12%,5) = $10,350.63 With straight line depreciation the annual worth is
1477510206) = $3 7',441.68(A/P,12%,5) = $10,368.69 } (41 181 671141  031 (D) = 4 6:1 181 = 141 + (D) I EOY BTCF Depr 71 T (40%) ATCF
0 $10,000 $10,000
1 $14,150 $2,000 $16,150 $6,460 $7,690
2 $14,150 $3,200 $17,350 $6,940 $7,210
3 $14,150 $1,920 $16,070 $6,428 $7,722
4 $14,150 $1,152 $15,302 $6,120.80 $8,029.20
5 «$14,150 $1,152 $15,302 $6,120.80 $8,029.20
5 $1,000 $1,000 $400 $600
6 0 $576 $576 $230.40 $230.40 We note that the annual worths are basically the same. This is due to the fact
that, whatever depreciation method we use, the depreciation deductions are small relative to the annual expenses. Method H The MACRS class life is 7 years. Assume that MV5 is $5,000 1—“ 1c)=14103) (D1=—11€1 (E1=141+031
EOY TI T (409/3) ATCF
0 $40,000 $40,000
1 — $7,000 $12,716 $5,086.40 $1,913.60
2  $7,000 $16,796 $6,718.40 $281.60
3 — $7,000 $13,996 $5,598.40 $1,401.60
4  $7,000 $11,996 $4,798.40 $2,201.60
5  $7,000 $10,572 $4,228.80 $2,771.20
6 $7,000 $10,568 $4,227.20 $2,772.80
7 $7,000 $10,572 $4,228.80 $2,771.20
8 $7,000 $8,784 $3,513.60 $3,486.40
9 $7,000 $7,000 $2,800 $4,200
10 $7,000 $7,000 $2,800 $4,200
_10 L $5,000 $5,000 $2,000 $3,000 11117112941 = $51,869.57 AW over the useful life of 10 years: AW(12%)$51,869.67(A1’P,12%,10)= $9,180.11 Thus, Method II is chosen also in this ease. AWSL(12%) = —$S4,210.?6(NP,12%,10)= $9,594.45
We notice a more signiﬁcant difference in this case. Here, the timing of the
depreciation deductions is of greater importance 729 Let X = Annual operating expenses
Note: The $4,000 shown in the BTCF column (A) represents the savings in rent.
Thus, Column A is the incremental BTCF between leasing and purchasing
the equipment.
(A) (B) 1 (C)={A)(B) {DP€03) (E)=(A)+ (D)
EOY BTCF Depr T1 T 40%) ATCF
0  $ 12,000  —~ —  $ 12,000
1 4,000  X $1,200 $2,800  X  $1,120 + 0.4X 2,880  0.6X
2 4,000  X 2,400 1,600  X ~ 640 + 0.4X 3,360 — 0.6X
3 4,000  X 2,400 1,600  X  640 + 0.4X 3,360  0.6X
4 4,000  X 2,400 1,600  X  640 + 0.4X 3,360  0.6X
5 4,000  X 2,400 1,600  X  640 + 0.4X 3,360  0.6X
6 4,000 » X 1,200 2,800  X ~ 1,120 + 0.4X 2,880 — 0.6X
7 4,000  X 0 4,000  X  1,600 + 0.4X 2,400  0.6X
8 4,000  X 0 4,000  X — 1,600 + 0.4X 2,400 w 0.6X
8 5,000 5,000 I 2,000 3,000
Solve for X:
PW(10%) = 0 =  $12,000 + $2,880('P1F,10%,1)+$3,360(P1’A,10%,4)(P1F,10%,1)
+ $2,880(P!F,10%,6) + $2,400(P/F,10%,7)
+ ($2,400 + $3,000)(PIF, 10%,8)  (0.6X)(P1‘A,10%,8)
X = $1 773.60
The added annual expense can be as high as $1,274 and the IRR will still exceed 10%.
Aft ‘  c 7—30 (3) Before—tax MARR — e‘ “X MARK = 0 5 = 0.25, or 25%
1  t 1— 0.40
(1)) Year Depreciation Year Depreciation
1 $12,861 5 $8,037
2 22,041 6 8,028
3 15,741 7 8,037
4 1 1,241 8 4,014 (c) BVg = 0, therefore T13 = $ 10,000 (Property having a 7—year recovery period
is fully depreciated after N+l=8 years.) 7—31 (d)
(A) (B) (C) = (A)  (B) (D) = 1 (C) (E) = (A) + (DJ I
EOY j BTCF Depr TI T (40%) 14101? 314/ (15%)
0 _ $90,000 [ $90,000 $90,000
1 15,000 $12,361 $2,139 —$856 14,144 12,299
2 15,000 22,041 7,041 2,316 17,316 13,472
3 15,000 15,741 741 296 15,296 10,053
4 15,000 11,241 3,759 1,504 13,496 7,717
5 15,000 3,037 6,963 2,735 12,215 6,073
6 15,000 3,023 6,972 2,739 12,211 5,279
7 15,000 3,037 6,963 2,735 12,215 4,592
3 15,000 4,014 10,936 4,394 10,606 3,467
3 10,000 10,000 4,000 6,000 1,961  1111711504) = 325,032, (e) No, reject the project because PW(ATCF) < 0 at MARR = 15%.
(A) ‘ 1B) 1C1=1411n) ] 001110 1E1=1A1+1D1
EOY ’ BTCF j Depr '11 T (40%) ATCF
0 $50,000  $50,000
1 14,000 $10,000 $4,000 —$1,600 12,400
2 14,000 10,000 4,000 1,600 12,400
3 14,000 10,000 4,000 1,600 12,400
4 14,000 10,000 4,000 1,600 12,400
5 14,000 10,000 4,000 1,600 12,400
6 14,000 0 14,000 5,600 3,400
14,000 0 14,000 5,600 3,400
14,000 0 14,000 "5,600 3,400
N 14,000 0 14,000 5,600 3,400 l Let X = N — 5 years. Set PW(10%) = O and solve for X: 0 = $50,000 + 312400111714, 1 0%,5) + 334001117413 0%,X)(P7’F,1 0%,5)
(P/A,10%,X) = 0.5741 X441 year. Thus,N=5+X=6ycars 732 Manufacturing designed for varying degrees of automation: Degrees Investment Cost Annuai Labor Cost Annual Power & Maintenance Cost A $ 10,000 $9,000 5 _ 500 B 14,000 7,500 300 C 20,000 5,000 1,000 D 30,000 3,000 1,500 (A)
(A) (B) (C) = (A)  (B) {D} = —t (C) (E) = (A) + (D)
EOY i Depr TI T (40%)  «$10,000
$2,000 $11,500 $4,900 Straight Line Depreciation: ($10,000~0)/5 = $2,000 (B) (C) = (A) — (B) (D) = 1(C) (E) = (A) + (D)
' T (40%) ATCF $14,000 ——~ $8,300 $11,100 Depreciation: ($14,000~0)r’5 = $2,800 (C)
(A; (B) (c) = (A)  03) (D) = t (c) (E): (A) + (u)
BTCF Depr T1 '1‘ (40%) ATCF
$20,000 $20,000
$6,000 $4,000 $10,000 $4,000 02,000 Depreciation: ($20,000OJES = $4,000 (D)
(A) 03) (010003) in) = t (c) (E) = (A) + (0)
EOY_ i_BTCF i Depr 11 T (40%) ATCF
0 030,000 I 030,000!
16 104,500 I $6,000 010,500 $4,200 —$300__ Depreciation: ($30,000~0)75 = $6,000 (51) Annual Worth AWA = $10,000(AIP,15%,5) — $4,900 = 07,833
AWE = $14,000(AIP,15%,5) ~ $3,860 = $8,036.20
ch — $20,000{A}?,15%,5)  $2,000 = 457,966
AWE): —$30,000(NP,15%,5)  $300 = $9,249 Select to automate to Degree A. (b) Present Worth
PWA = ~$ 10,000  $4,900(P;’A,15%,S) = $26,425.78
PWB = $14,000  $3,860(P!A,15%,5) = $26,939.49
PWc = $20,000 ~ $2,000(P/’A,15%,5) = $26,704.40
FWD: —$30,000  $300(PKA,15%,5) = $31,005.66 Select to automate to Degree A. (0) [RR
Dearsze
A B C D
Investment $10,000 $ 14,000 $20,000 $30,000
Annual Cost $4,900 3,860 2,000 300 Because all numbers are costs, we cannot 00mpare with the alternative of do nothing,
thus select A as our base comparison. Bi
—$4,000 + $1,040(P/A, i'%,5) = 0, i'% 2 9.431%; since i*% < 15%, reject B C—_A
—$10,000 + $2,900CPKA, i'%,5) = 0, i'% E 13.8%; since i'% < 15%, reject C M
$20,000 + $4,600(P!A, i'%,5) = 0, 1%
Select A. 4.8%; since i‘% < 15%, reject D and Ill 735 Depreciation schedule (MACRS Syear property class). The cost basis (B) is
assumed to be $345,000. $345,000 0.2000 $69,000 $276,000
276,000 0.3200 110,400 165,600
165,600 0.1920 66,240 99,360
99,360 0.1152 39,744 59,616
59,616 0.1152 39,744 19,872
19,872 0.0576 19,872 0 (3) Economic Value Added (EVA): 1 = 05
T(50%) 55 43,000
1,600
45,760 $1 12,000a 1 10,400
66,240 39,744 72,256
39,744 72,256
112,000 19,872 92,128
J 120,000b 120,000c a BTCFk = $120,000 6 $8,000 = $112,000
b MW = $120,000
c Gain on disposal = MV6 H BVG = $120,000  0 = $120,000 PW(10%)
$11,818 $21,500  0.10 ($345,000) =  $13,000 800 0.10( 276,000) a  26,800 —22,148
22,880  0.10( 165,600) = 6,320 4,748
36,128 —0.10( 99,360) = 26,192 17,888
36,1280.10( 59,616): 30,166 18,730
46,064 ~0.10( 19,872) = 44,077 24,880 60,000 33,870  "16161: $66,150 The present equivalent ofthc EVA = $66 150. (b) Afl‘er—tax cash flow (ATCF): _ BOY BTCF T(50%) ____ PW(10%) 0 $345,000 0 $345,000 —$34S,000
1 112,000 — $21,500 90,500 82,274
2  800 111,200 91,896
3 ~ 22,880 89,120 _ 66,95?
4  36,128 75,872 51,822
5  36,128 75,872 47,109
6 112,000  46,064 65,936 37,222
6 120,000 — 60,000 60,000  33,870 Total: $66,150 Yes; the PW( 10%) of the ATCF (mm) is the same as the present
equivalent of the annual EVA amounts. 736 Beginningof—year book values: Year BVH l dk BVk 1 $ 180,000 $ 36,000 $ 144,000
2 144,000 5?,600 86,400
3 86,400 34,560 51,840
4 51,840 20,736 31,104
5 ' 31,104 20,736 10,368
6 10,368 10,368 0 NOPAT“ (0.10)Bv,_. EV?” ] 1 0 $ 18,000  8 18,000
2 $ 13,392 14,400 — 27,720
3 893 8,640  7,747
4 9,464 5,184 4,280
5 9,464 3,110 6,354
6 15,892 1,037 14,855
7 22,320 0 22,320
8 22,320 0 22,320
9 22,320 0 22,320
10 22,320 0 22,320
10 _1 8,600 18,600 a From Table 26: Column (C) algebraically added 10 Column (D).
b Equation 723; EVAk = NOPATk 4 1 a 13v“ PWU 0%) =— 318,000 (P/F,10%,1) + $14.:‘55(P/F,1'0%,6) ...
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 Depreciation, PW, BTCF Depr

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