Question

TIUNIE 2.4 (25 points) Your company is upgrading the breakroom and kitchen. It is going to include an expresso machine, a fridge with compartments for each employee, a sink, microwave, toaster oven, tables chairs, a rock wall, snacks for everyone, and maybe some other bells and whistles. Your managers think that by updating this area employees will not take as long of lunches. They understand this purchase will be at a cost. You are tasked with considering two different options and presenting them to management. Use a 5% interest rate. E First Costa Annual Maintenance Cost Salvage Value Life Years Walmart Kit $40,000 $10,000 $12,000 3 Target $65,000 $12,000 $25,000 6 a. Using NPW (Net Present Worth Analysis) analysis determine which kitchen kit you should choose. (8 points) b. Using EUAW (Equivalent Uniform Annual Worth) analysis determine which kitchen kit you should choose. (8 points) C. You really want the Target kit because it looks nicer and has more bells and whistles. You are willing to keep these products around for longer and therefore extend the lives of these products. Perform the analysis to show that the Target option is the better choice. (5 points) d. Now from your analysis in part b think about how ethical presenting this information to management would be. Write 2-3 sentences about how you would present this information in a way that showed your bias. You will be graded on your ability to consider two options in an ethical comparison and how you perceive your bias. (4 points)

Answer 1

There are two alternatives available; W & T. The analysis period is 6 years. However, useful life of alternative-w is 3 years and alternative-T is 6 years. Thus, during our analysis period, alternative-W will be replaced once (that is, year-3) while there is no need of replacement of alternative-T. The rate of interest is 5% per year. Alternative-W Following is the formula to calculate the present worth: PW, = P+A(P/A,1,n)+F(P/F,1,n)+F(P/F,1,n) Here P is the initial cost (-$40,000) A is the annual cost (-$10,000) F1 is the cost of replacement (-$28,000) F2 is salvage value ($12,000) i is the interest rate (5%) Substitute the values in above equation. PW, = -$40,000 - $10,000 (P/A,5%,6) - $28,000 (P/F,5%,3)+$12,000 (P/F,5%,6) = -$40,000 - $10,000 (5.076) - $55,000(0.8638)+$12,000(0.7462) =-$40,000 - $50,076 - $47,509+ $8,954.40 = -$129,314.6 Thus, the present worth of the alternative-W is $129,314.6

Alternative-T Following is the formula to calculate the present worth: PW, = P +A(P/A,1,n)+F(P/F,1,n) Here P is the initial cost (-$65,000) A is the annual cost (-$12,000) Fis the salvage value ($25,000) i is the interest rate (5%) n is the time period (6 years) Substitute the values in above equation. PW, = -$65,000 - $12,000(P/A,5%,6)+$25,000 (P/F,5%,6) =-$65,000 - $12,000(5.076)+$25,000(0.7462) =-$65,000 – $60,912 +$18,655 = -$107,257 Thus, the present worth of the alternative-T is $107,257| Conclusion: It is clear from the above calculations that the present worth of the alternative-T is higher than that of alternative-W. Thus, alternative-T should be selected.

Following is the formula to calculate the EUAW: EUAW = P(A/P,1,n)+ A+F(A/F,1,n) Here P is the initial cost A is the annual cost F is the salvage value i is the interest rate n is the time period Alternative-W Substitute the values in above equation. EUAW, =-$40,000( A/P,5%, 3) – $10,000 +$ 12,000(A/F,5%, 3) =-$40,000(0.3672) - $10,000+ $12,000 (0.3172) =-$14,688 – $10,000+$3,806.40 =-$20,881.60 Thus, the EUAW is -$20,881.60

Alternative-T Substitute the values in above equation. EUAW, =-$65,000(A/P,5%,6)-$12,000+ $25,000(A/F,5%,6) =-$65,000(0.197) - $12,000 + $ 25,000(0.147) =-$12,805 - $ 12,000+ $3,675 = -$21,130 Thus, the EUAW is $21,130 Conclusion: It is clear from the above calculations that the EUAW of the alternative-W is higher than that of alternative-T. Thus, alternative-w should be selected.