Ac 505 Case Study 2 Solution

476 WordsApr 14, 20122 Pages
Case Study 2 Solution 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 Contribution margin per passanger = $90 Break even point per passanger = $35,000 35000 Contribution margin ratio= $2 Break even point in dollars = $5,600,000 5600000 b. What is the break even point of passenger train cars per month 63 Compute # of seats per train car(remember load factor?) 555.5555556 Number of train cars rounded 556 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 the monthly break even point in the number of passenger cars? 54 Contribution Margin 120 Break even point in passengers 26250 Train cars (rounded) 486.1111111 Rounded 486 d. Contribution Margin= $70 Break even point in passenger 45000 Train Cars= 45,000/ 63 714.2 Rounded 714 e. How many passengers per month are needed to generate an after tax profit of $750,000 Contribution Margin= $205-$85 $120.00 After tax= $750,000/ .70 1071428.57 Number of Passengers $205- $85 = $1,071.429 + $3,600,000 $120 = 4671429 38928.57 Number of Passengers Rounded 38929 f. Number of discounted seats Contribution Margin= $120 - $70 $50 Passenger seats= 90 X.80 72 Train Cars= 72 X $ 50 X 50 $180,000 Train cars per day x

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