# Recursion Lesson

- Pages:
**6** - Word count:
**1315** - Category: Service

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The first problem which is *the cost of tuition and parking in **Porterville College* is recursive since the nth value depends on the previous (or n-1) values. The recursive function for the problem is the following:

F(n) = F(n-1) + kn

where n = number of years and k = multiple factor of n per 5 years

Using the function, the questions raised can then be answered. Table 1 shows the solution for years 2010, 2015, and 2020.

Table 1. Cost of tuition and parking in Porterville College.

Year |
No. of Years since 1987 (n) |
Function F(n) |
Cost of Tuition and Parking |

1987 |
0 |
F(n) |
$1,200 |

1988 | 1 | F(n-1) + 8^{2}n |
$1,264 |

1989 | 2 | F(n-1) + 8^{2}n |
$1,392 |

1990 | 3 | F(n-1) + 8^{2}n |
$1,584 |

1991 | 4 | F(n-1) + 8^{2}n |
$1,840 |

1992 |
5 |
F(n-1) + 8^{2}n |
$2,160 |

1993 | 6 | F(n-1) + 9^{2}n |
$2,646 |

1994 | 7 | F(n-1) + 9^{2}n |
$3,213 |

1995 | 8 | F(n-1) + 9^{2}n |
$3,861 |

1996 | 9 | F(n-1) + 9^{2}n |
$4,590 |

1997 |
10 |
F(n-1) + 9^{2}n |
$5,400 |

1998 | 11 | F(n-1) + 10^{2}n |
$6,500 |

1999 | 12 | F(n-1) + 10^{2}n |
$7,700 |

2000 | 13 | F(n-1) + 10^{2}n |
$9,000 |

2001 | 14 | F(n-1) + 10^{2}n |
$10,400 |

2002 | 15 | F(n-1) + 10^{2}n |
$11,900 |

2003 | 16 | F(n-1) + 11^{2}n |
$13,836 |

2004 | 17 | F(n-1) + 11^{2}n |
$15,893 |

2005 | 18 | F(n-1) + 11^{2}n |
$18,071 |

2006 | 19 | F(n-1) + 11^{2}n |
$20,370 |

2007 | 20 | F(n-1) + 11^{2}n |
$22,790 |

2008 | 21 | F(n-1) + 12^{2}n |
$25,814 |

2009 | 22 | F(n-1) + 12^{2}n |
$28,982 |

2010 |
23 |
F(n-1) + 12^{2}n |
$32,294 |

2011 | 24 | F(n-1) + 12^{2}n |
$35,750 |

2012 | 25 | F(n-1) + 12^{2}n |
$39,350 |

2013 | 26 | F(n-1) + 13^{2}n |
$43,744 |

2014 | 27 | F(n-1) + 13^{2}n |
$48,307 |

2015 |
28 |
F(n-1) + 13^{2}n |
$53,039 |

2016 | 29 | F(n-1) + 13^{2}n |
$57,940 |

2017 | 30 | F(n-1) + 13^{2}n |
$63,010 |

2018 | 31 | F(n-1) + 14^{2}n |
$69,086 |

2019 | 32 | F(n-1) + 14^{2}n |
$75,358 |

2020 |
33 |
F(n-1) + 14^{2}n |
$81,826 |

2021 | 34 | F(n-1) + 14^{2}n |
$88,490 |

2022 | 35 | F(n-1) + 14^{2}n |
$95,350 |

The second problem which is *waiting in lines* states that:

As the school registrar, part of your job entails making the registration process as painless as possible for the students. One aspect of the registration process that influences the length of the time required to register is the amount of time students spend waiting in line. To get a better idea of the factors that influence the wait time, you have decided to model a few different situations.

(1) Determine the average wait time, when possible, for students in each of the following situations. For situations where an average wait time was determined, when will the registrar catch up so that there are no students waiting?

(2) What do you notice about the relationships among the arrival time, service time, and waiting time?

__Situation A__

Assume that no student is waiting in line. Each student needs two minutes to pick up his/her registration materials and students are scheduled to arrive every three minutes.

**ANSWER**:

**(1)** Let W = wait time, X = arrive time, S = service time

and m = number of students waiting in line, n = number of students arriving.

Since S=2mins ≤ X=3mins, then there will be no waiting.

**(2) ** Total service time is S_{total}(n) = 2n

__Situation B__

Assume that no student is waiting in line. Each student needs two minutes to pick up his/her registration materials and students are scheduled to arrive each minute.

**ANSWER:**

**(1)** Let W = wait time, X = arrive time, S = service time

Since S=2mins ≥ X=1mins, then there will be waiting of 1min à W=1min.

**(2)** If students keep arriving, then the function W(n) which is the wait time of the nth student is:

where n = 1,2,3,…k

and where and n = 1,2,3,…k

This is also equal to the function X(n) which is the arrive time of the nth student:

where n = 1,2,3,…k

and where and n = 1,2,3,…k

and the total service time is S_{total}(n)= 2n

Student, n |
Arrive Time, X (mins) |
Service Time, S start (mins) |
Service Time, S end (mins) |
Wait Time, W (mins) |

n1 | 0 | 0 | 2 | 0 |

n2 | 1 | 2 | 4 | 1 |

n3 | 2 | 4 | 6 | 2 |

n4 | 3 | 6 | 8 | 3 |

n5 | 4 | 8 | 10 | 4 |

n6 | 5 | 10 | 12 | 5 |

n7 | 6 | 12 | 14 | 6 |

n8 | 7 | 14 | 16 | 7 |

n9 | 8 | 16 | 18 | 8 |

n10 | 9 | 18 | 20 | 9 |

Situation C

Assume that there are four students waiting in line as registration begins and that each new student arrives three minutes apart. Also assume that the service time is still two minutes.

**ANSWER:**

**(1)** Let W = wait time, X = arrive time, S = service time

Since S=2mins ≤ X=3mins, then eventually there will be no waiting. But since there are four students waiting in line, m=4, then there will be waiting for the first few students that are arriving. Based on the table shown below, the average wait time will be

where m = no. of students waiting, n = no. of students arriving

** (2)** If students keep arriving, then the function X(n) which is the arrive time of the nth student is:

where n = 1,2,3,…k

and where and n = 1,2,3,…k

and the total service time is S_{total}(n) = 2(3+n)

Wait time will be zero when arrive time X(n) equals start service time S_{start}(n):

Student, m & n |
Arrive Time, X (mins) |
Service Time, S start (mins) |
Service Time, S end (mins) |
Wait Time, W (mins) |

m1 | 0 | 0 | 2 | 0 |

m2 | 0 | 2 | 4 | 2 |

m3 | 0 | 4 | 6 | 4 |

m4 | 0 | 6 | 8 | 6 |

n1 | 0 | 8 | 10 | 8 |

n2 | 3 | 10 | 12 | 7 |

n3 | 6 | 12 | 14 | 6 |

n4 | 9 | 14 | 16 | 5 |

n5 | 12 | 16 | 18 | 4 |

n6 | 15 | 18 | 20 | 3 |

n7 | 18 | 20 | 22 | 2 |

n8 | 21 | 22 | 24 | 1 |

n9 | 24 | 24 | 26 | 0 |

n10 | 27 | 26 | 28 | 0 |

Situation D

Assume that five students are waiting and each transaction takes three minutes with arrivals four minutes apart.

**ANSWER:**

**(1) ** Let W = wait time, X = arrive time, S = service time

Since S=3mins ≤ X=4mins, then eventually there will be no waiting. But since there are five students waiting in line, m=5, then there will be waiting for the first few students that are arriving. Based on the table shown below, the average wait time will be

where m = no. of students waiting, n = no. of students arriving

**(2)** If students keep arriving, , then the function X(n) which is the arrive time of the nth student is:

where n = 1,2,3,…k

and where and n = 1,2,3,…k

and the total service time, S_{total}(n)= 2(4+n)

Wait time will be zero when arrive time X(n) equals start service time S_{start}(n):

Student, m & n |
Arrive Time, X (mins) |
Service Time, S start (mins) |
Service Time, S end (mins) |
Wait Time, W (mins) |

m1 | 0 | 0 | 2 | 0 |

m2 | 0 | 2 | 4 | 2 |

m3 | 0 | 4 | 6 | 4 |

m4 | 0 | 6 | 8 | 6 |

m5 | 0 | 8 | 10 | 8 |

n1 | 0 | 10 | 12 | 10 |

n2 | 4 | 12 | 14 | 8 |

n3 | 8 | 14 | 16 | 6 |

n4 | 12 | 16 | 18 | 4 |

n5 | 16 | 18 | 20 | 2 |

n6 | 20 | 20 | 22 | 0 |

n7 | 24 | 22 | 24 | 0 |

n8 | 28 | 24 | 26 | 0 |

n9 | 32 | 26 | 28 | 0 |

n10 | 36 | 28 | 30 | 0 |

References

Weisstein, Eric W. “Recursion.” *MathWorld. *1 May 2008. Wolfram Web Resource. 3 May 2008 <http://mathworld.wolfram.com/Pythagorean Theorem.html>.