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newnansm2 - Chapter 2 Engineering Costs and Cost Estimating...

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Chapter 2: Engineering Costs and Cost Estimating 2-1 This is an example of a ‘sunk cost.’ The $4,000 is a past cost and should not be allowed to alter a subsequent decision unless there is some real or perceived effect. Since either home is really an individual plan selected by the homeowner, each should be judged in terms of value to the homeowner vs. the cost. On this basis the stock plan house appears to be the preferred alternative. 2-2 Unit Manufacturing Cost (a) Daytime Shift = ($2,000,000 + $9,109,000)/23,000 = $483/unit (b) Two Shifts = [($2,400,000 + (1 + 1.25) ($9,109,000)]/46,000 = $497.72/unit Second shift increases unit cost . 2-3 (a) Monthly Bill: 50 x 30 = 1,500 kWh @ $0.086 = $129.00 = 1,300 kWh @ $0.066 = $85.80 Total = 2,800 kWh = $214.80 Average Cost = $214.80/2,800 = $129.00 Marginal Cost (cost for the next kWh) = $0.066 because the 2,801st kWh is in the 2nd bracket of the cost structure. ($0.066 for 1,501-to-3,000 kWh) (b) Incremental cost of an additional 1,200 kWh/month: 200 kWh x $0.066 = $13.20 1,000 kWh x $0.040 = $40.00 1,200 kWh $53.20 (c) New equipment: Assuming the basic conditions are 30 HP and 2,800 kWh/month Monthly bill with new equipment installed: 50 x 40 = 2,000 kWh at $0.086 = $172.00 900 kWh at $0.066 = $59.40 2,900 kWh $231.40 Incremental cost of energy = $231.40 - $214.80 = $16.60 Incremental unit cost = $16.60/100 = $0.1660/kWh
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2-4 x = no. of maps dispensed per year (a) Fixed Cost (I) = $1,000 (b) Fixed Cost (II) = $5,000 (c) Variable Costs (I) = 0.800 (d) Variable Costs (II) = 0.160 (e) Set Total Cost (I) = Total Cost (II) $1,000 + 0.90 x = $5,000 + 0.10 x thus x = 5,000 maps dispensed per year. The student can visually verify this from the figure. (f) System I is recommended if the annual need for maps is <5,000 (g) System II is recommended if the annual need for maps is >5,000 (h) Average Cost @ 3,000 maps: TC(I) = (0.9) (3.0) + 1.0 = 3.7/3.0 = $1.23 per map TC(II) = (0.1) (3.0) + 5.0 = 5.3/3.0 = $1.77 per map Marginal Cost is the variable cost for each alternative, thus: Marginal Cost (I) = $0.90 per map Marginal Cost (II) = $0.10 per map 2-5 C = $3,000,000 - $18,000Q + $75Q 2 Where C = Total cost per year Q = Number of units produced per year Set the first derivative equal to zero and solve for Q. dC/dQ = -$18,000 + $150Q = 0 Q = $18,000/$150 = 120 Therefore total cost is a minimum at Q equal to 120. This indicates that production below 120 units per year is most undesirable, as it costs more to produce 110 units than to produce 120 units. Check the sign of the second derivative: d 2 C/dQ 2 = +$150 The + indicates the curve is concave upward, ensuring that Q = 120 is the point of a minimum. Average unit cost at Q = 120/year: = [$3,000,000 - $18,000 (120) + $75 (120) 2 ]/120 = $16,000 Average unit cost at Q = 110/year: = [$3,000,000 - $18,000 (110) + $75 (120) 2 ]/110 = $17,523
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One must note, of course, that 120 units per year is not necessarily the optimal level of production. Economists would remind us that the optimum point is where Marginal Cost = Marginal Revenue, and Marginal Cost is increasing. Since we do not know the Selling Price, we cannot know Marginal Revenue, and hence we cannot compute the optimum level of output. We can say, however, that if the firm is profitable at the 110 units/year level, then it will be much more profitable at levels greater than 120 units. 2-6 x = number of campers (a) Total Cost = Fixed Cost + Variable Cost = $48,000 + $80 (12) x Total Revenue = $120 (12) x (b) Break-even when Total Cost = Total Revenue $48,000 + $960 x = $1,440 x $4,800 = $480 x x = 100 campers to break-even (c) capacity is 200 campers 80% of capacity is 160 campers @ 160 campers x = 160 Total Cost = $48,000 + $80 (12) (160) = $201,600 Total Revenue = $120 (12) (160) = $230,400 Profit = Revenue – Cost = $230,400 - $201,600 = $28,800 2-7 (a)
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