Question

1.(50 pts.) A large bakery is considering three alternative investments for next year. Information on the three alternatives is shown below: Project A Project B Project C Cost of Equipment Installation Costs Expected Annual Saving Useful Life (Years) $80,400 $25,000 $22,600 $47,800 $20,000 $15,100 $75,450 $40,000 $24,000 - cannot choose from multiple 10 10 projects based If MARR is 12%, use rate-of-return analysis to determine which alternative should be selected. a on highest individual Here, ROR. Brajeet Investment Annual Savings ROR - 17% 105400 A + 22,600 -67800 18% +15,100 16.1% - 115450 +24000 ... Project B should be selected. For Project A, NPVA= -105400 + 32600 (P/A, 1, 10) 205-105400 + 22600 (P/A,ė,10) >> /4,é, 10) = 4663716814 si = 171. ✓ Project B. NPVB =-67800 + 15,100 PA, 10) 203-67800 +15,100(PA, s, 19) > PIA, , 10) = 4.4900 į = 187 For Project c, NPvc = -115450 + 24000 (P/A, 1, 10) 20c – {S So +27000 ( Atlo) PA, , 10) 74.810467 is i=16:17. From the rate of greturn for i livisual proje ander e For Wrong choice. TEMA 50130 V hot Pyyniec B has the highest

I need the solution for green colored text.

Answer 1

Here you are required to compare the projects on the basis of incremental rate of return. First of all determine the alternative with lowest initial investment. Project B has the lowest initial investment next costly project is Project A. Determine the incremental cash flow

	A	B	C
Cost + Installation	- 105400	-67800	-115450
Annual savings	22600	15100	24000
RoR	17%	18%	16.1%

Incremental cash flow between A and B

	A	B	A - B
Cost	-105400	-67800	-37,600
Savings	22600	15100	7,500

Now determine the NPW of the incremental cash flow

NPW(A-B) = -37,600 + 7,500(P/A,1%, 10)

$\large \implies 0 = - 37,600 + 7,500(P/A,i\%,10)$

7,500(P/A, 1%, 10) = 37,600

$\large \implies (P/A,i\%,10) = \frac{37,600}{7,500}$

$\large \implies (P/A,i\%,10) = 5.01333$

On solving for rate of return we get $\large \triangle RoR_{A - B} \approx 15\%$

MARR is 12% and the incremental RoR is greater than MARR thus select project A and reject Project B.

Now compare project A with project C

	A	B	A - B
Cost	-105400	-115450	- 10,050
Savings	22600	24000	1,400

Now determine the NPW of the incremental cash flow we get

$\large NPW_{(C-A)} = - 10,050 + 1,400(P/A,i\%,10)$

$\large \implies 0 = - 10,050 + 1,400(P/A,i\%,10)$

$\large \implies (P/A,i\%,10) =\frac{10,050}{1,400}$

$\large \implies (P/A,i\%,10) = 7.17857$

Solving for Rate of return we get 6.53%

The incremental RoR of C - A = 6.53% less than MARR. Therefore reject Project C and accept project A.

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