Question

In an effort to find alternate revenue streams for their athletic program, as well as to enhance the game day experience for fans, Eastern Michigan University recently ran a pilot program to sell beer at a home football game. Despite an approximate $3,000 net loss on the program, the pilot was deemed a success because it gave them valuable information needed about fan interest and required infrastructure if they decide to move forward with more football beer sales in the future. One potential reason for the net loss was the low attendance at the Sept. 26, 2015 game against Ball State University. Attendance for the game was 4,463, much lower than their normal attendance of approximately 6,000 fans. However, you should assume that the demand for beer (i.e., the percentage of fans that buy beer) has been correctly estimated by the pilot. In a normal season, EMU has six home football games. For this pilot program, EMU rented some infrastructure items (i.e., tables, tents, hedges and other equipment) that would be purchased if they decided to sell beer on an ongoing basis. One of the goals of the pilot program was to give administrators information on how much they could pay for this infrastructure and still earn a profit. Read the attached article for all other case data.

1. Identify the fixed and variable costs associated with this pilot program (as specified in the article). Assume that a 16-oz cup of beer is the cost object.

2. Write a profit equation for beer sales at EMU football games.

3. What attendance would have been required for the beer sales at the EMU v. Ball State game to break even? Given the current cost structure, is it likely that beer sales would be a good revenue stream? Why or why not?

4. The 2014 season’s average home attendance was 15,025 (though this number has been disputed by some). What would the net profit (loss) of the pilot have been at an average game in 2014? Assume that the demand for beer has not changed.

5. Assuming ‘normal’ game attendance and that the demand for beer has not changed, how much could EMU pay for infrastructure and still earn a profit over one six-game season?

Additional Information

EMU assistant athletic director for media relations Greg Steiner said the school paid Arbor Brewing Co. just under $900 for the beer it sold at the Sept. 26 game against Ball State at Rynearson Stadium, peddling 559 16‑ounce beers at $7 per cup. Sales brought in just under $4,000. But the university also incurred $1,200 in costs for extra security and an additional $5,000 in one‑time rental fees for tables, tents, hedges and other equipment, bringing the total overall cost to nearly $7,100. Taking into account only the cost of the beer and extra security — and not rental fees — EMU profited roughly $3.30 per beer sold or about $1,845. "(The rental fees) were expenditures we wouldn't have going forward if we were to continue in the future with beer sales because we would buy the equipment and not have those recurring costs," Steiner said

Answer 1

1]

fixed cost = rental fees and cost of extra security.  

variable cost = cost of purchasing the beer.

2]

profit = sale proceeds of beer - purchase cost of beer - cost of extra security.

Assuming that the tables, hedges etc. would be owned, the depreciation on those would be deducted from the above figure to calculate net profit

3]

to break even, sales must equal costs

(beers sold * sale price per cup) = cost of beer purchased + cost of extra security + cost of rental

(beers sold * $7) = $900 + $1,200 + $5,000

beers sold = $7,100 / $7 = 1,014

Actual attendance was 4,463. Beers sold = 559. Beers sold % = 559 / 4463 = 0.1253, or 12.53%

Attendance required to sell 1,014 beers = (1014 / 559) * 4463 = 8,096

4]

net profit/loss = sales - costs

sales (quantity) = average attendance * beers sold % = 15,025 * 12.53% = 1,883

sales value = 1,883 * $7 = $13,181

costs of beer purchased = (1883 / 559) * $900 = $3,032 (proportionate cost for 1883 beers, when the cost of 559 beers is $900)

cost of extra security = $1,200 * 6 = $7,200 (cost per game is $1,200)

profit = $13,181 - $3,032 - $7,200 = $2,949 (assuming the equipment is bought)