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) ___%

Expert Solution

Total Rooms	=	335
Occupancy	=	66.60%
ADR	=	117.98
Revenue per available room (33566.60%117.98)	=	Occupancy*ADR
	=	26,323

Initial Fee {45000+300*(335-150)}	=	1,00,500
(+) Reservation Fee (3357.6512)	=	30,753
(+) Royalty Fees (26322.523355%)	=	4,40,902
(+) Advertising Fees (263223352.3%)	=	2,02,815
(+) Frequent Traveler Program (33566.60%$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%

ekkarill92 answered 2 years ago

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