Question 5 of 8: Second-hand equipment resale — original price, current price, and profit

Question 5: Second-hand equipment resale — original price, current price, and profit (5/6)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two 54"×74" gyratory crushers (feed opening × mantle diameter, matching the appendix's $G\times D=3{,}996\ \text{in}^2$ primary-crusher line exactly); four 7 ft standard cone crushers. Appendix cost law $\text{Cost}=aX^b$ at M&S = 1,400: gyratory $a=271.25$, $b=1.189$, $X=G\times D$ (in²); standard cone $a=25{,}070$, $b=1.756$, $X=$ diameter (ft). M&S = 1,200 at construction; M&S = 1,800 now. Supplier paid 40% of original price/item, spent $1,109,570 total on dismantling/packaging, and sells at 60% of current price.

Find. (a) The approximate original (construction-era) price of each item. (b) The supplier's overall potential profit.

Approach. Compute each equipment type's base purchased cost at the appendix's own M&S = 1,400 reference, then scale linearly to M&S = 1,200 for the original (construction-era) price and to M&S = 1,800 for today's price; assemble the supplier's total investment (40% of original + dismantling) and total revenue (60% of current), and difference the two.

1. Base cost at M&S = 1,400. $$\text{Cost}_{\text{gyr},1400}=271.25\times3{,}996^{1.189}=\$5{,}196{,}400/\text{unit}$$ $$\text{Cost}_{\text{cone},1400}=25{,}070\times7^{1.756}=\$764{,}100/\text{unit}$$
2. Original (construction-era, M&S = 1,200) and current (M&S = 1,800) unit prices. $$\text{Cost}_{\text{gyr,orig}}=5{,}196{,}400\times\tfrac{1200}{1400}=\$4{,}454{,}100\qquad \text{Cost}_{\text{gyr,now}}=5{,}196{,}400\times\tfrac{1800}{1400}=\$6{,}681{,}100$$ $$\text{Cost}_{\text{cone,orig}}=764{,}100\times\tfrac{1200}{1400}=\$654{,}900\qquad \text{Cost}_{\text{cone,now}}=764{,}100\times\tfrac{1800}{1400}=\$982{,}400$$
3. Fleet totals (2 gyratory + 4 standard cone). $$\text{Total}_{\text{orig}}=2(4{,}454{,}100)+4(654{,}900)=\boxed{\$11{,}527{,}900}$$ $$\text{Total}_{\text{now}}=2(6{,}681{,}100)+4(982{,}400)=\boxed{\$17{,}291{,}800}$$
4. Supplier's total investment. Paid 40% of the ORIGINAL fleet price, plus dismantling/packaging: $$\text{Invest}=0.40\times11{,}527{,}900+1{,}109{,}570=4{,}611{,}100+1{,}109{,}570=\boxed{\$5{,}720{,}700}$$
5. Sale revenue and profit. Sells at 60% of the CURRENT fleet price: $$\text{Sell}=0.60\times17{,}291{,}800=\$10{,}375{,}100$$ $$\text{Profit}=\text{Sell}-\text{Invest}=10{,}375{,}100-5{,}720{,}700=\boxed{\$4{,}654{,}400}$$