Question

Designing A New Arena :

You work for a company that has been tasked with designing a proposed seating plan for an NHL team’s new arena. Currently the NHL team plays in a rink similar to the one shown here:
Parameters of the Plan

• One ring of seats all the way around the rink is considered a row
• Row 1 is considered to be the row closest to the ice.
• The number of seats in the arena has to be between 18 000 and 22 500
• There cannot be more than 40 rows in the arena.
• The number of seats in each row must form an arithmetic sequence, increasing by the same number in each subsequent row.

PART A – SEATS AND ROWS

Using the parameters above, and your understanding of sequences and series:

1. Begin by selecting a total number of seats in the arena
2. Choose a realistic number of rows required for the entire arena
3. Choose a number of seats by which each row will increase
4. Based on your choices for 1 – 3, determine the number of seats in the first row
5. Based on the information from 1 – 4, determine the number of seats in the last row
6. Clearly summarize all of the information that you found in this part of your proposal in an organized list of a few bullet points.

Be sure to explain in detail any assumptions you make.

PART B – TICKET PRICES

The team’s current arena charges:

• $6000 for a season seat in rows 1-10
• $4000 for a season seat in rows 11-20
• $3000 for a season seat in rows 21-30
• $2000 for a season seat in rows 31-40.

The team has suggested modifying their current plan to use a geometric sequence, and decrease the price for a seat in each subsequent row by the same factor based on the price of a seat in the row in front of it.

Using the team’s current arena charges, your work from PART A, and your understanding of sequences and series:

1. Determine a reasonable price for a season ticket in the first row
2. Determine a reasonable factor by which the cost of each seat per game will decrease in each subsequent row behind row 1.
3. Determine the price per game of a season ticket in the last row

Answer 1

SEATING:

Let, Total no. of seats = 21000

No. of rows = 25

Increment in no. of seats between consecutive rows = 10

No. of seats in consecutive rows form an AP.

Let the no. of seats in 1st row be a

The increment in no. of seats between consecutive rows = d = 10

Seats in 2nd row = a+d; Seats in 3rd row = a+2*d..................... Seats in nth row = a+(n-1)d

Total no. of seats = a+(a+d)+.............+(a+24d) = 20000

i.e. 25a+(1+2+.........+23+24)*10 = 20000

i.e. 25a+((24)(24+1)/2)*10 = 20000 [since sum of n consecutive natural numbers is n(n+1)/2]

i.e. 25a+300*10 = 20000

i.e. a=680

The number of seats in the first row is 680 for a total seating of 20000 and a consecutive increment of 10 in each row.

a+24d= 680+24*10 = 920

The number of seats in the last row is 920 for a total seating of 20000 and a consecutive increment of 10 in each row.

Learning:

• To easily calculate the no. of total seats and the seating capacity of first row, take care that total capacity is a multiple of no. of rows.
• There is a bit of trial and error involved

PRICING:

Let the price of 1st bracket = p

The price of 2nd bracket = p*q

The price of 3rd bracket = p*q²

Since sum of 1st n terms of an AP is (n/2)*[2a+(n-1)d] and nth term is a+(n-1)d

Earlier revenue per game =

Revenue from 1-10 row +
Revenue from 11-20 row +
Revenue from 21-25 row

i.e. Earlier revenue per game =

((10/2)*[2*680+(10-1)*10])*6000 +
((10/2)*[2*(680+10*10)+(10-1)*10])*4000 +
((5/2)*[2*(680+20*10)+(5-1)*10])*3000

i.e. Earlier revenue per game = $ 90000000

The new revenue should also be at least this much

New Revenue per game = R = a*p+((a+d)*p*q)+((a+2d)*p*q²)+............((a+24d)*p*q²⁴)

On simplifying,

R = ap+apq+dpq+apq²+2dpq²+apq³+3dpq³+...................apq²³+23dpq²³+apq²⁴+24dpq²⁴ -----------------------------------(1)

R*q = apq+apq²+dpq²+apq³+2dpq³+apq⁴+3dpq⁴+...........apq²⁴+23dpq²⁴+apq²⁵+24dpq²⁵ ----------------------------------(2)

Subtracting equation (2) from (1)

R-R*q= ap +dpq+dpq2+dpq3+dpq4+..................+dpq²⁴-apq²⁵+24dpq²⁵

i.e. R(1-q) = ap -apq²⁵+24dpq²⁵ + dpq(q²⁴-1)/(q-1)

i.e, R = ((ap -apq²⁵+24dpq²⁵ + dpq(q²⁴-1))/(q-1))/(1-q)

Since we want the revenue to be at least more than previous revenue

90000000 <= (680(p -pq²⁵)+10(24pq²⁵ + pq(q²⁴-1)))/(1-q)

There are a lot of value of p and q that satisfy this equation.

By trial and error, a reasonable value of p = 2000 and q = 1.4

Reasonable price for first row = p = $ 2000

Reasonable factor for increment of cost of seat in each row = q = 1.4

Price per game of last row seat = 2000*(1.4^24) = $ 6,428,399.401