Question

In: Accounting

A 335-room hotel property recorded in 2004 a 66.6% occupancy and an ADR of $117.98. What...

A 335-room hotel property recorded in 2004 a 66.6% occupancy and an ADR of $117.98. What is the property’s franchise fee (1) on a per available room basis and (2) as a percentage of rooms revenue if the agreement required the hotel to pay a reservation fee of $7.65 per available room per month; a royalty fee of 5% of rooms revenue; an advertising fee of 2.3% of rooms revenue; and a frequent traveler program fee of $5.00 per occupied room. The hotel had frequent stay guests totaling 6% of the occupied rooms. The initial fee is a minimum of $45,000 plus $300 per room for each room over 150.

1. Please use the information from Question 1 to calculate the Total franchise fee.

              Total franchise fee (round to a whole number) $ ___

2. Please use the information from Question 1 to calculate the Franchise fee on PAR basis.

             Franchise fee on PAR basis (round to two decimal places) $___ PAR/yea

3.Please use the information from Question 1 to calculate the Franchise fee as a % of revenue.

             Franchise fee as a % of revenue (round to two decimal places) ___%

Solutions

Expert Solution

Total Rooms = 335
Occupancy = 66.60%
ADR = 117.98
Revenue per available room (335*66.60%*117.98) = Occupancy*ADR
=              26,323
Initial Fee {45000+300*(335-150)} =           1,00,500
(+) Reservation Fee (335*7.65*12) =              30,753
(+) Royalty Fees (26322.52*335*5%) =           4,40,902
(+) Advertising Fees (26322*335*2.3%) =           2,02,815
(+) Frequent Traveler Program (335*66.60%*$5) =                 1,116
(+) Frequent Travel Stay (26323*6%) =                 1,579
Total Franchise Fee           7,77,665
Franchise Fee per available room (777665/335)                 2,321
Room Revenue              26,323
Franchise cost                 2,321
% of Room Revenue (2321/26323)*100 8.82%

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