Managerial Accounting
20 May 2012
AC 505 Case Study II:
A. Break-even point in passengers and revenues per month = 35,000; $5,600,000
1) Per Passenger
Sales $160
Variable Expenses 70
Unit Contribution Margin $90

Fixed expenses ÷ Unit CM = $3,150,000 ÷ 90 = 35,000 passengers in break-even point

2) Contribution Margin Ratio (CM Ratio) = Contribution Margin ÷ Selling Price
= $90 ÷ $160 = .5625

Break-even point in dollars = Fixed costs ÷ CM Ratio = $3,150,000 ÷ .5625 = $5,600,000

B. Break-even point in number of passenger train cars per month = 556
Number of seats per train car = Average load factor × Number of seats per train car
= .70 × 90 = 63 passengers per train on average

Passengers in break-even point ÷ Number of seats per passenger train
= Number of passenger train cars per month to break-even

35,000 ÷ 63 = 555.5 or 556 train cars

C. Monthly break-even point in number of passenger cars when fare is raised to $190 = 486

Number of seats per train car = Average load factor × Number of seats per train
= .60 × 90 = 54 passengers per train on average Per Unit
Sales $190
Variable Expenses 70
Unit Contribution Margin $120

Fixed Expenses ÷ Contribution Margin = $3,150,000 ÷ $120 = 26,250 passengers to break even

Break-even point in passengers ÷ average passengers per train = Break-even point in train cars
= 26,250 ÷ 54 = 486.11 or 486 train cars to break-even
D. Break-even point in passengers and in number of passenger train cars when VC is $90
=45,000; 714

1) Per unit
Sales $160
Variable Expenses 90
Unit Contribution Margin $70

Fixed expenses ÷ Unit CM = $3,150,000 ÷ $70 = 45,000 passengers to break-even

2)
63 passengers on average per train car (.70 ×90 = 63)

Break-even point in passengers ÷ average passengers per train = Break-even point in train