Question 1:  Cost Concepts (11 marks)

a)   What is the estimated revenue for the tapered fastening pins?       Estimated annual revenue, R = 125,000 ($0.70) = $87,500 [1 mark]

b)   What are the estimated costs (to the nearest dollar) for one year for each of the alternative designs? Process A:

Costs A = [$18.60/hr X (16 hrs/1000 units) + $16.90/hr X (4.5hrs/1000units) + $0.10/unit] X 125,000 units [6 X ½ = 3 marks]

Costs A = $59,206 [½ mark]

Note: The “sunk” costs have already been spent and are not considered

Process B:

Costs B = [$18.60/hr X (7 hrs/1000 units) + $16.90/hr X (12hrs/1000units) + $0.10/unit] X 125,000 units [6 X ½ = 3 marks]

Costs B = $54,125 [½ mark]

Note: The “sunk” costs have already been spent and are not considered

c)    Which process should the company choose?

ProfitA  = R–CA  = $87,500 – $59,206 = $28,294 [1 mark]

ProfitB  = R–CB  = $87,500 – $54,125 = $33,375 [1 mark]

The company should produce the pins using process B [½ mark], because the profit will be higher [½ mark].

Question 2:  Cost Estimating (10 marks)

a)    Using a cost capacity factor of 0.65 and cost index values of 175 in 2010 and 210 in 2021 estimate cost of material for a new tent.

Cost of Material in 2010 for 1,050 m2  = 18.75 X 1050 = 19,687.50

Cost of Material in 2021 for 1,050 m2  = 19,687.50(210/175) = 23,625

Cost of Material in 2021 for 1,200 m2  = 23,625(1,200/1,050)0.65  = $25,767.16 [5 X ½ = 2.5 marks]

b)   Assuming a loaded labour rate of $120 per hour, a 91% learning curve what is the labour cost of the 50th tent? Cost of Labour for first 1200 sq metres = 120 X 98 = 11,760

Cost of Labour for 50th  unit = 11,760 X 50(LOG 0.91 / LOG 2)  = $6,906.26 [5 X ½ = 2.5 marks]

c)    Calculate the average labour cost per tent.

LOg 0.91

Total Labour cost = 11,760 X ∑ n  LOg 2      = $722,985.65 [3 X ½ = 1.5 marks]

Average labour cost per tent = 7,229.8565 ~ $7,229.86 [1 mark]

d)   What will the total price of the contract be to the client if you use a 20% profit markup?

Contract price = [100(25,767.16) + 722,985.65] X 1.2 = 3,959,641.98 ~ $3,959,642 [5 X ½ = 2.5 marks]

Question 3:  Time Value of Money (7 marks)

a)   What is the effective annual interest rate?

ieff  = (1 + iS )n  − 1 = (1 + )12  − 1 = 0.0616778118645 ~ 6.17% [1 mark]

b)   What is the future worth of the weekly payment plan after 20 years?

365.25 days/year X 20 years ÷ 7 days/week = 1,043.572 weekly time periods, but we need to round that up to the nearest whole time-period. There will be 1,044 weekly payments made. [1 mark]

Weekly interest rate can be calculated ...

ieff  = (1 + iS )n  − 1 = (1 + )7  − 1 = 0.001150464170287 ~ 0.115% / week [1 mark]

Or (will be close enough because of the numbers involved)

7 days X 0.06/365.25 days = 0.0011499 ~ 0.115% / week

FW = 1,000(F/A, 0.115%, 1044) = 1,000(2017.2337) = 2,017,233.7 ~ $2,017,234 +/– $600 [1 mark]

c)    What is the future worth of the single lump sum payment option after 20 years? FW = 650,000(F/P, 6.17%, 20) = 650,000(3.3116) = $2,152,540 +/– $900 [1 mark]

d)   Which option do you recommend your cousin take and why?

It is better for your cousin to take the lump sum payment now [½ mark], for two reasons; the future worth of the single lump sum is larger [¼ mark] and they might not live 20 more years [¼ mark].

e)    If the single lump sum payment option was taken, how many years would it take for the investment to reach 1

million dollars (provide answer in whole years)?

1,000,000 = 650,000(F/P, 6.17%, n)

(F/P, 6.17%, n) = 1.5384615, looking at the 6.17% table that number falls between 7 and 8 years. At 7 years it   won’t be a million, so in whole years we need to wait 8 years [1 mark] for the investment to reach $1,000,000.

Question 4:  Cash Flow Analysis (10.5 marks)

PW = –238,000 + (48,000 – 2,000)(P/A, 9%, 7) – 1,000(P/G, 9%, 5)(P/F, 9%, 2) + 25,000(P/F, 9%, 7) [9 X ½ = 4.5 marks] PW = –238,000 + 46,000(5.0330) – 1,000(7.1110)(0.8417) + 25,000(0.5470)

PW = 1,207.67 ~ $1,208 +/- 1 [½ mark]

AW = – 238,000(A/P, 9%, 7) + 48,000 – 2,000 – 1,000(P/G, 9%, 5)(P/F, 9%, 2)(A/P, 9%, 7) + 25,000(A/F, 9%, 7) [10 X ½ = 5 marks]

AW = – 238,000(0.1987) + 46,000 – 1,000(7.1110)(0.8417)(0.1987) + 25,000(0.1087)

AW = 237.62 ~ $238 +/- 2 [½ mark]

Question 5: Comparison Methods (15.5 marks)

a)    Hallard Co. (CFD – Not to Scale)

Initial cost, time line, years, MARR, 2 annuities, 1 gradient, salvage  [8 X ½ = 4 marks]

StartFilter Co. (CFD – Not to Scale)

Initial cost, time line, years, MARR, 2 annuities, salvage [7 X ½ = 3.5 marks]

b)   Because of the different service lives the best method for comparison is Annual Worth (AW). Otherwise you would need to take a repeated lives approach over 36 years.

AWH  = – 325,000(A/P, 6.5%, 18) + 37,500 – 30(A/G, 6.5%, 18) + 6,000(A/F, 6.5%, 18) [7 X ½ = 3.5 marks] AWH  = – 325,000(0.0959) + 37,500 – 30(6.8403) + 6,000(0.0309)

AWH  = 6,312.691 ~ $6,313 +/- 15 [½ mark]

AWS  = – 211,500(A/P, 6.5%, 12) + 29,500 + 7,000(A/F, 6.5%, 12) [5 X ½ = 2.5 marks]

AWS  = – 211,500(0.1226) + 29,500 + 7,000(0.0576)

AWs = 3,973.3 ~ $3,973 +/- 7 [½ mark]

Because AWH  > AWS  [½ mark], recommend Hallard Co. [½ mark]