Hospitality Property Management

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You are the manager of a 325 room hotel which has an 83% average occupancy rate and an ADR of $139.80. Twenty-two percent (22%) of your occupied rooms are occupied rooms are occupied by 2 registered guest. Seven percent (7%) of your rooms are occupied by 3 guests. Your hotel has a restaurant that seats 350 guest and you do 3.2 turns in the restaurant each day. The hotel has a small banquet room that seats 50 people. You have 3 banquets plan for the month of June, the first is for 27 guests, the second is for 12 guest, and the third is for 31 guest. In each banquet each guest will be served a meal. On average 13% of your occupied rooms order 1.2 meals through room service in your hotel each day. The A/C blower motor which serves you public areas is a 20 horse power motor. The A/C blower motor that supplies your restaurant and banquet room is a 15 horse power motor. You have 100,000 cubic feet of public space and the restaurant and banquet room have 40,000 cubic feet of space. You keep the inside temperature of the hotel set at 75 degrees. Each person staying in your hotel soils 5 ½ pound of linen per day and each cover served results in ¾ pound of soiled linen. The average outside temperature for the month June is 82 degrees and June has 30 days.

 

Factors: 0.125/mcf/cdd

0.75/Hp/cdd

Given that you pay the following cost information for electricity and water:

Electricity: $0.076/kWh

Water Rate: First 75 ccf @ $0.30 per ccf

Next 270 ccf @ $0.37 per ccf

All over 345 ccf @ $0.44 per ccf

Water Surcharge: A surcharge of 5.5% will be added to all water bills.

Sewer Rate: Commercial Sewer Rate: 255.5% of water bill (combined cost of water and water surcharge)

Sewer Surcharge: A surcharge of 5.5% will be added to all sewer bills

Given the following usage rates:

USAGE PER WATER ELECTRICITY

occupied room 225 gal 14 kWh

cover served 10 gal 1 kWh

pound soiled lined 4 gal 0.1 kWh

 

Answer the follow:

What is your forecasted electrical consumption (kWh and $) in the public areas and the restaurant for air conditioning for June using A/C horsepower to forecasted electrical consumption? __________ kWh $__________
What is your forecasted electrical consumption (kWh and $) in the public areas and the restaurant for air conditioning for June using square footage to forecasted electrical consumption? __________ kWh $__________
What is your foretasted electrical consumption (kWh and $) based on the number of occupied rooms, covers sold, and pounds of soiled linen?
__________ kWh $__________

What is your forecasted water consumption for the operation in hundreds of cubic feet and dollars? __________ ccf $__________
What is your forecasted sewer cost for the operation in dollars? $__________
What is your forecasted water and sewer cost for the operation in dollars? $__________

Sample Solution

Forecasted Electrical Consumption and Water Usage for June

Assumptions:

  • Number of occupied rooms per day = Average occupancy rate (%) * Total rooms
  • Number of guests = (Number of occupied rooms with 2 guests * 2) + (Number of occupied rooms with 3 guests * 3)
  • Total covers served per day = Restaurant turns * Restaurant seating capacity
  • Pounds of soiled linen per day = (Number of occupied rooms * Guests per occupied room * Linen per guest) + (Covers served * Linen per cover)

Electrical Consumption – Air Conditioning (Horsepower Method):

  1. Cooling Degree Days (CDD): We don’t have the CDD information, but it’s crucial for this calculation. You’ll need the average daily difference between the outside temperature (82°F) and the desired inside temperature (75°F) for June. Multiply this difference by the number of days (30) to get the total CDD.

  2. Public Areas:

    • kWh per CDD = 0.75 kWh/Hp/CDD (given)
    • Total HP = 20 HP (public area A/C motor)
    • Public Areas kWh per CDD = HP * kWh per HP/CDD = 20 HP * 0.75 kWh/Hp/CDD = 15 kWh/CDD

  3. Restaurant & Banquet Room:

    • kWh per CDD = 0.75 kWh/Hp/CDD (given)
    • Total HP = 15 HP (restaurant & banquet A/C motor)
    • Restaurant & Banquet kWh per CDD = HP * kWh per HP/CDD = 15 HP * 0.75 kWh/Hp/CDD = 11.25 kWh/CDD

Forecasted Electrical Consumption (Public Areas – Horsepower Method):

  • Public Areas kWh = Public Areas kWh per CDD * Total CDD (Requires CDD data)

Forecasted Electrical Consumption Cost (Public Areas – Horsepower Method):

  • Public Areas $ = Public Areas kWh * Electricity rate ($0.076/kWh)

Forecasted Electrical Consumption (Restaurant & Banquet – Horsepower Method):

  • Restaurant & Banquet kWh = Restaurant & Banquet kWh per CDD * Total CDD (Requires CDD data)

Forecasted Electrical Consumption Cost (Restaurant & Banquet – Horsepower Method):

  • Restaurant & Banquet $ = Restaurant & Banquet kWh * Electricity rate ($0.076/kWh)

Electrical Consumption – Air Conditioning (Square Footage Method):

Not recommended due to lack of specific information on A/C unit efficiency and building characteristics. This method requires a conversion factor that considers factors beyond horsepower.

Electrical Consumption – Rooms, Covers, Linen (Combined):

  1. Number of Occupied Rooms per Day:

    • Average occupancy rate (%) = 83%
    • Total rooms = 325 rooms
    • Number of occupied rooms per day = Occupancy rate * Total rooms = 0.83 * 325 rooms = 270.25 rooms/day (round to 270)

  2. Number of Guests:

    • Rooms with 2 guests = 22% of occupied rooms
    • Rooms with 3 guests = 7% of occupied rooms
    • Guests per day = (Rooms with 2 guests * Guests per room) + (Rooms with 3 guests * Guests per room)
    • Guests per day = (22% * 270 rooms) * 2 + (7% * 270 rooms) * 3 = 636 guests/day

  3. Total Covers Sold per Day:

    • Restaurant turns = 3.2 turns/day
    • Restaurant seating capacity = 350 guests
    • Total covers served per day = Restaurant turns * Restaurant seating capacity = 3.2 turns/day * 350 guests/turn = 1120 covers/day

  4. Pounds of Soiled Linen per Day:

    • Linen per guest = 5.5 pounds/guest/day (given)
    • Linen per cover = 0.75 pounds/cover (given)
    • Pounds of soiled linen per day = (Occupied rooms * Guests per room * Linen per guest) + (Covers served * Linen per cover)
    • Pounds of soiled linen per day = (270 rooms * Guests per room * 5.5 pounds/guest/day) + (1120 covers/day * 0.75 pounds/cover) = 3885 pounds/day

  5. Electrical Consumption from Rooms, Covers, Linen:

    • Electricity per occupied

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