Cost structure

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University of Minnesota-Twin Cities *

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6035

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Industrial Engineering

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Oct 30, 2023

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pdf

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6

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Olivia Starr Take home exam #1 1. Cost Structure of Johnson’s Boat Company Key Activities Number of passengers Number of one-way trips per day Number of boats I. Costs that vary strictly with number of passengers: Ticket processing $0.40 per passenger Taxes $0.60 per passenger Total $1.00 per passenger II. Costs that vary strictly with number of one-way trips per day Gasoline $100.00 per one-way trip Crew salaries $70.00 per one-way trip Boat cleaning $20.00 per one-way trip Total $190.00 per one-way trip III. Costs that vary strictly with number of boats: Rent $2,200 per boat per month *Step costs: Total shore costs: 1 boat $11,000 per month 2 boats $ 13,500 per month 3 boats $14,200 per month 4 boats $16,000 per month – - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Total cost 1 boat $13,200 per month Total cost 2 boats $15,700 per month Total cost 3 boats $16,400 per month Total cost 4 boats $18,200 per month 2. Contribution Margins Contribution per passenger = fare - costs that vary strictly per passenger = 5-1 = $4.00 per passenger Contribution per one-way trip = (contribution per passenger)(number of passengers per trip) - costs that vary strictly per one-way trip Didn't add of boats times number of boats Should b 2 17,900 3 20 800 4 24 800
1-6 one-way trips: (4)(70) - 190 = $90.00 per one-way trip 7-12 one-way trips: (4)(66) - 190 = $74.00 per one-way trip 13-16 one-way trips: (4)(63) - 190 = $62.00 per one-way trip 17-20 one-way trips: (4)(58) - 190 = $42.00 per one-way trip 21-24 one-way trips: (4)(52) - 190 = $18.00 per one-way trip 3. How many one-way trips per day should Tom Johnson run with X number of boats? 1 Boat - 6 possible one-way trips per day Fixed cost for 1 boat for 1 month = $13,200 per month Profit = (contribution per trip)(number of trips) - fixed costs **Number of trips per day will be multiplied by 30 for the monthly profit 1 one-way trip: (90)(1)(30) - 13200 = - $10,500.00 2 one-way trips: (90)(2)(30) - 13200 = - $7,800.00 3 one-way trips: (90)(3)(30) - 13200 = - $5,100.00 4 one-way trips: (90)(4)(30) - 13200 = - $2,400.00 5 one-way trips: (90)(5)(30) - 13200 = $300.00 6 one-way trips: (90)(5)(30) - 13200 = $3,000.00 Tom should run 3 one-way trips per day with 1 boat. 2 Boats - 12 possible one-way trips per day Fixed cost for 2 boats for 1 month = $15,700 per month Profit = (contribution per trip)(number of trips) - fixed costs **Number of trips per day will be multiplied by 30 for the monthly profit 7 one-way trips: (74)(7)(30) - 15700 = - $160.00 8 one-way trips: (74)(8)(30) - 15700 = $2,060.00 9 one-way trips: (74)(9)(30) - 15700 = $4,280.00 10 one-way trips: (74)(10)(30) - 15700 = $6,500.00 11 one-way trips: (74)(11)(30) - 15700 = $8,720.00 12 one-way trips: (74)(12)(30) - 15700 = $10,940.00 Tom should run 12 one-way trips per day with 2 boats. 3 Boats - 18 possible one-way trips per day Fixed cost for 3 boats for 1 month = $16,400 per month Profit = (contribution per trip)(number of trips) - fixed costs **Number of trips per day will be multiplied by 30 for the monthly profit 13 one-way trips: (62)(13)(30) - 16400 = $7,780.00 14 one-way trips: (62)(14)(30) - 16400 = $9,640.00 15 one-way trips: (62)(15)(30) - 16400 =$11,500.00 16 one-way trips: (62)(16)(30) - 16400 =$13,360.00 17 one-way trips: (42)(11)(30) - 16400 = $5,020.00 18 one-way trips: (42)(12)(30) - 16400 = $6,280.00 Because the load factor is Breakeven load to an consideration passenger 190 4 47 S mama must be expressed in terms of load factors The load factors are the to averages for all trips 6 this was a typo calculations are off because don't need to calculate cost of boat need to compare the endpoints of each interval All the other choices are sub optimal of trips per day X Contributionper trip ibday
Tom should run 15 one-way trips per day with three boats. 4 Boats - 24 possible one-way trips per day Fixed cost for 3 boats for 1 month = $18,200 per month Profit = (contribution per trip)(number of trips) - fixed costs **Number of trips per day will be multiplied by 30 for the monthly profit 19 one-way trips: (42)(19)(30) - 18200 = $5,740.00 20 one-way trips: (42)(20)(30) - 18200 = $7,000.00 21 one-way trips: (18)(21)(30) - 18200 = - $6,6860.00 22 one-way trips: (18)(22)(30) - 18200 = - $6,320.00 23 one-way trips: (18)(23)(30) - 18200 = - $5,780.00 24 one-way trips: (18)(24)(30) - 18200 = - $5,240.00 Tom should run 20 one-way trips per day with 4 boats. 4. Breakeven Analysis Breakeven number of passengers for 2 boats, and 12 trips per day = fixed costs / contribution per passenger (per month) q = 15700/(12)(30) q=15700/360 q = 43.6112 passengers per one-way trip Breakeven number of passengers for 3 boats, and 15 trips per day = fixed costs / contribution per passenger (per month) q = 16400/(62)(15)(30) q=16400/450 q = 36.4445 passengers per one-way trip For a 2-boat business, Tom needs to be at 66.0774% (43.6112/66) capacity to breakeven. For a 3-boat business, Tom only needs to be at 58.7814% ( 36.4445/62) capacity in order to breakeven. Thus, the 2-boat business running 12 one-way trips per day to maximize profits is more risky than a 3-boat business running 15 one-way trips per day to maximize profits. Luckily. This is because the increase in fixed cost is not in a linear relationship with the increase in number of boats. Tom can run more trips and be satisfied that there will still be enough demand with three boats, and take on less risk because the cost of adding another boat is minimal compared to the benefits. with 3 boats should run 16 not 18 consider boats god given Don't factor in cost of boat total cont da O 6 90 540 12 74 888 16 62 992 18 42 756 with 4 boats should run I 20 42 840 24 18 432 To determine ofboats meal cont day x 30 days A boat monthly profit 130 540 13,200 3 000 need to decide frequency first
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