Project a Case 9-30
Ferdinand De Luna Prof. Suryoutomo
AC505 Case Study II Managerial Finance
Springfield Express is a luxury passenger carrier in Texas. All seats are first class, and the following data are available: Number of seats per passenger train car | 90 | Average load factor (percentage of seats filled) | 70% | Average full passenger fare | $160 | Average variable cost per passenger | $70 | Fixed operating cost per month | $3,150,000 |
a. What is the break-even point in passengers and revenues per month?
Break-even point in passengers = | Total Fixed Costs + Target Profit | | | | Contribution Margin per passenger | | | = | 3,150,000 + 0 | | | | 160-70 | | | = | 3,150,000 | | | | 90 | | | = | 35,000 is the break-even point in passengers | Break-even point in revenues = | Total Fixed Costs + Target Profit | | | | Contribution Margin Ratio | | | = | 3,150,000 + 0 | | | | (160-70) / 160 | | | = | 3,150,000 | | | | 0.5625 | | | = | $5,600,000 break-even point in revenues |
b. What is the break-even point in number of passenger train cars per month? Break-even point in train cars = | Total Fixed Costs + Target Profit | | | | Contribution Margin per train car | | | = | 3,150,000 + 0 | | | | (160-70) * (90*.70) | | | = | 3,150,000 | | | | 90*63 | | | = | 556 break-even point in train cars |
c. If Springfield Express raises its average passenger fare to $ 190, it is estimated that the average load factor will decrease to 60 percent. What will be the monthly break-even point in number of passenger cars? Break-even point in train cars = | Total Fixed Costs + Target Profit | | | | Contribution Margin per train car | | | = | 3,150,000 + 0 | | | | (190-70) * (90*.60) | | | = | 3,150,000 | | | | (120) * (54) | | | = | 486.1 or 487 train cars |
d. (Refer to original data.) Fuel cost is a
AC505 Case Study II Managerial Finance
Springfield Express is a luxury passenger carrier in Texas. All seats are first class, and the following data are available: Number of seats per passenger train car | 90 | Average load factor (percentage of seats filled) | 70% | Average full passenger fare | $160 | Average variable cost per passenger | $70 | Fixed operating cost per month | $3,150,000 |
a. What is the break-even point in passengers and revenues per month?
Break-even point in passengers = | Total Fixed Costs + Target Profit | | | | Contribution Margin per passenger | | | = | 3,150,000 + 0 | | | | 160-70 | | | = | 3,150,000 | | | | 90 | | | = | 35,000 is the break-even point in passengers | Break-even point in revenues = | Total Fixed Costs + Target Profit | | | | Contribution Margin Ratio | | | = | 3,150,000 + 0 | | | | (160-70) / 160 | | | = | 3,150,000 | | | | 0.5625 | | | = | $5,600,000 break-even point in revenues |
b. What is the break-even point in number of passenger train cars per month? Break-even point in train cars = | Total Fixed Costs + Target Profit | | | | Contribution Margin per train car | | | = | 3,150,000 + 0 | | | | (160-70) * (90*.70) | | | = | 3,150,000 | | | | 90*63 | | | = | 556 break-even point in train cars |
c. If Springfield Express raises its average passenger fare to $ 190, it is estimated that the average load factor will decrease to 60 percent. What will be the monthly break-even point in number of passenger cars? Break-even point in train cars = | Total Fixed Costs + Target Profit | | | | Contribution Margin per train car | | | = | 3,150,000 + 0 | | | | (190-70) * (90*.60) | | | = | 3,150,000 | | | | (120) * (54) | | | = | 486.1 or 487 train cars |
d. (Refer to original data.) Fuel cost is a