Q1:

a): $S=0,P=50000,N=10$

$$D_n=\frac{50000-0}{10}$$ $$D_n=5000$$

We can use straight-line depreciation to find a depreciation rate of 5000 per year. BV and depreciation schedule is as follows:

b) $d=0.2,P=50000,N=10$

With declining-balance depreciation, the depreciation schedule and BV are here:

$$\text{Depreciation: }{D}_n=Pd(1-d{)}^{(n-1)}$$ $$B{V}_n=P(1-d{)}^n$$

Q2:

$C_0=2500000,O\&M=???,N=\infty ,i=10\%$ Since O&M occurs every 5 years, we need to get an adjusted $i_e$: $i_e=(1.1{)}^5-1$ $i_e=61.05\%$ every 5 years

The PW of the costs associated with this design (N = $\infty$) = 2500000 + ($\frac{P}{A}$, 61.05%, $\infty$) $\cdot$ 200000 = 2500000 + $\frac{200000}{.6105}$ = 2827600.33

If the pool’s service is reduced to 40 years, then we have $200,000 annuity.

= 2500000 + ($\frac{P}{A}$, 61.05%, 8) $\cdot$ 200000 = 2820361.66

Q3:

a) $C_0=2000000,B=820000,D=400000,O\&M=???$

Given MARR = 8% yearly, the effective interest rate for O&M every 3 years is

$i_e=(1.08{)}^3-1$ $i_e=25.97\%$ every 3 years

$$BCR=\frac{B-D}{C}$$ $$BCR=\frac{420000}{MARR}/\left(\frac{300000}{i_e}+2000000\right)$$ $$BCR=1.66>1$$

So by BCR, this project is viable.

b) Modified BCR: O&M counts as one of the disbenefits.

$$BC{R}_M=\frac{B-(D+O\&M)}{C}$$ $$BC{R}_M=\left(\frac{420000}{MARR}-\frac{300000}{i_e}\right)/200000$$ $$BC{R}_M=2.05>1$$

So by modified BCR, this project is viable.

Q4:

a) Equating to equal repeated lives, we have LCD(4, 6) = 12

For the CR1000

$$A=210,G=-10,i=12\%$$

(At n = 12, we don’t need to re-purchase, we just have to scrap it)

$$PW(CR1000)=\sum _{n=0}^{12}P{W}_n$$ $$=-60.65$$

Using AW, we don’t worry about repeated lives, so we just calculate with one lifecycle (N = 4):

$$AW(CR1000)=210$$ $$-680\left(\frac{A}{P},12\%,4\right)$$ $$+100\left(\frac{A}{F},12\%,4\right)$$ $$-10\left(\frac{A}{G},12\%,4\right)$$ $$=-6.54$$

For the CRX:

$$A=380,i=12\%$$

Again, at n=12, we only scrap it.

$$PW(CRX)=\sum _{n=0}^{12}P{W}_n$$ $$=887.4$$

Using AW, we don’t worry about repeated lives, just calculate it using one lifecycle (n = 6)

$$AW(CRX)=380$$ $$-1100\left(\frac{A}{P},12\%,6\right)$$ $$+250\left(\frac{A}{F},12\%,6\right)$$ $$=143.26$$

Better to pick CRX with its MARR of 12%.

b) Find the salvage value where CR1000 is a better option.

$$A{W}_{CR1000}>A{W}_{CRX}$$ $$210-680\left(\frac{A}{P},12\%,4\right)-10\left(\frac{A}{G},12\%,4\right)+s\left(\frac{A}{F},12\%,4\right)>143.26$$ $$-27.445+s(0.2092)>143.26$$ $$s>815.99$$

So the CR1000 needs to have a salvage value of 816 or more to make it the superior option.

Q5:

Find the value of $i^{\ast }$ such that NPW = 0.

$$0=-200000+20000\left(\frac{P}{F},i^{\ast },1\right)$$ $$+\frac{40000}{i^{\ast }}(\text{due to its indefinite revenue growth})$$ $$+15000\left(\frac{P}{F},i^{\ast },2\right)$$ $$+10000\left(\frac{P}{F},i^{\ast },3\right)$$ $$+5000\left(\frac{P}{F},i^{\ast },4\right)+0+0$$ $$0=-200000+20000\left(\frac{1}{1+i^{\ast }}\right)+\frac{40000}{i^{\ast }}+15000\left(\frac{1}{(1+i^{\ast }{)}^2}\right)+10000\left(\frac{1}{(1+i^{\ast }{)}^3}\right)$$ $$+5000\left(\frac{1}{(1+i^{\ast }{)}^4}\right)$$

Via linear interpolation:

$$\text{For }{i}_0=0.24,y_0=-89.097$$ $$i_1=0.23,y_1=7646.267$$ $$i^{\ast }=i_0-y_0\frac{(i_1-i_0)}{(y_1-y_0)}$$ $$i^{\ast }=.23988$$

The IRR is roughly 23.99%. Since our IRR (23.99%) > MARR (14%), this is a viable project.

Q6:

a): Find the monthly payment sent to the bank

$$6\%\text{ interest annually, compounded monthly}$$ $$i_s=\frac{.06}{12}$$ $$=.005$$

Monthly payment:

$$=15000000\left(\frac{i_s(1+i_s{)}^{120}}{(1+i_s{)}^{120}-1}\right)$$ = $166530.75 \text{ monthly payment}

b):

$$\text{Company pays 4 million}$$ $$\text{ + rest of the load remaining}$$ $$\text{ + future worth of 10 million after 4 years, i=7\% annually.}$$ $$=4000000+\left(15000000-166530.75\left(\frac{(1+i_s{)}^{48}-1}{i_s}\right)\right)+10000000(1.07{)}^4$$

$=23099007.53$

Q7:

a)

b)

$$\Delta IR{R}_{B-A}=(-8000-5000-(-14000-10000))-50000\left(\frac{A}{P},i^{\ast },8\right)$$ $$AW(B-A)=0=11000-50000\left(\frac{i^{\ast }(1+i^{\ast }{)}^8}{(1+i^{\ast }{)}^8-1}\right)$$ $$\text{For }{i}^{\ast }=.01=AW=4465.49$$ $$i^{\ast }=.01=AW=-2030.47$$

Using interpolation, we get:

$$i^{\ast }=0.1-4465.49\left(\frac{.01-.2}{A-B}\right)$$ $$i^{\ast }=14.06\%>\text{MARR}$$

Between not doing anything and option B, we choose option B since it’s greater than the MARR and option B has a higher initial cost.

$$\Delta IR{R}_{C_B}=(-6000-4000+8000+5000)$$ $$+(-68000-(-50000))\left(\frac{A}{P},i^{\ast },8\right)$$ $$AW(C-B)=0=3000-180000\left(\frac{A}{P},i^{\ast },8\right)$$ $$\text{For }{i}^{\ast }=.01=AW=647.57$$ $$i^{\ast }=.1=AW-373.99$$

Using linear interpolation:

$$i^{\ast }=x_0-i_0\left(\frac{x_1-x_0}{i_1-i_0}\right)$$ $$i^{\ast }=.01-647.57\left(\frac{.01-.1}{A-B}\right)$$ $$i^{\ast }=.067$$ $$6.7\%<MARR$$

We choose option B between B and C. This is because we’re less than the MARR, and option B has a lower initial cost. So the best option is B, based on $\Delta IRR$ comparisons.

Q8:

a)

$$MARR=10\%$$ $$A=\frac{5}{h}\cdot \frac{1500h}{y}$$ $$A=\frac{7500}{y}$$ $$B{V}_{10}=P(1-d{)}^n$$ $$B{V}_{10}=21500(0.8{)}^{10}$$ $$B{V}_{10}=2308.54$$ $$AW=7500-21500\left(\frac{A}{P},10\%,10+2308.54\left(\frac{A}{F},10\%,10\right)\right)$$ $$AW=4146.70$$

b) For a 10% increase in MARR:

$$0.1+(0.1)(0.1)=0.11$$ $$AW=7500-21500\left(\frac{A}{P},11\%,10\right)+2308.54\left(\frac{A}{F},11\%,10\right)$$ $$AW=3987.32$$ $$AW=7500-21500\left(\frac{A}{P},9\%,10\right)+2308.54\left(\frac{A}{F},9\%,10\right)$$ $$AW=4315.3$$

c) Find the break-even (when PW = 0) If x is some number of hours:

$$A=\frac{5x}{y}$$ $$0=-21500+2308.54(0.3855)+5x(6.145)$$ $$x=\frac{21500-2308.54(0.3855)}{5(6.145)}$$ $$x=670.79\frac{hours}{year}\text{ if we want to break even.}$$

Q9:

a) For the PV system:

$$\frac{watts}{year}=\text{capacity}\times \text{capacity factor}\times \frac{hours}{year}$$ $$=25000\times .2\times \frac{8760hours}{year}$$ $$\frac{kilowatts}{year}=43800$$

For the wind system:

$$\frac{watts}{year}=\text{capacity}\times \text{capacity factor}\times \frac{hours}{year}$$ $$\frac{watts}{year}=35000\times .24\times \frac{8760hours}{year}$$ $$\frac{kilowatts}{year}=73584$$

b) For this PV system: Initial cost $=(25)(3500)$ $=87500$ O&M cost = $\frac{400}{year}$$ Cash of the excess power$=(43.8-20)(1000)(0.2)=$4760$

For the wind system: Initial cost = (35)(3000) = $105000

O&M cost = $\frac{1200}{year}$$

Cash of our excess power = (73.58 - 20)(1000)(.2) = $10716

c) For the PV system: Payback period = $\frac{\text{initial investment amount}}{\text{annual savings}}$ = $\frac{87500}{4760-400}$ $=20.068years$

For the wind system: Payback period = $\frac{105000}{10716-1200}$ $=11.03years$

d) Since our projects are mutually exclusive, we’re allowed to use the $\Delta IR{R}_{B-A}$ method

Option 1: Do nothing Option 2: Install the PV system Option 3: Install the wind system

Cost of not doing anything $=(.25)(20)(1000)$ = $$\frac{5000}{year}$

$\Delta IR{R}_{B-A}$ 0 = -87500 - (400 - 5000)($\frac{P}{A}$, $i$^*, 30) + 4760($\frac{P}{A}$,i^*, 30) = -87500 + 9360$\frac{(1+i^{\ast }{)}^{30}-1}{1^{\ast }(1+i^{\ast }{)}^{30}}$

$i^{\ast }$ = 0.101 0.101 < 0.12, so we choose to do nothing.

$\Delta IR{R}_{C-A}$ 0 = -105000 + (1200 - 5000)($\frac{P}{A}$, i^*, 30) + 10716($\frac{P}{A}$, i^*, 30) 0 = -105000 + 14516 $\cdot$ $\frac{(1+i^{\ast }{)}^{30}-1}{i^{\ast }(1+i^{\ast }{)}^{3}0}$ $i^{\ast }$ = .135 .135 > .12

Thus, we choose option C.

Happy farm should choose to go with the wind system.

Q10

a) ![[IMG_323103D91974-1.jpeg|]]

b) EV${}_6$ $=0.05(230000)+0.3(115000)+0.65(80000)$ $=$98000$

EV${}_5$ $=0.15(205000)+.45(90000)+0.4(55000)$ $=$93250$

EV${}_4$ = minimum of (EV${}_5$ and EV${}_6$) = $93250

EV${}_3$ =$0.55(150000)+.35(35000)$

EV${}_2$ $=0.66(EV$_4$)+0.34(50000)$

EV${}_1$ = minimum of ($E{V}_2$, $E{V}_3$)

Since EV${}_2$ < EV${}_3$, and $E{V}_5$ < $E{V}_6$, we choose to test them, then to make a minor modification if our result is unfavourable.

c) From b), we know EV${}_1$ = $78545$ and $E{V}_4$ = 93250.