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# Recursion Lesson The whole doc is available only for registered users
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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) + 82n \$1,264 1989 2 F(n-1) + 82n \$1,392 1990 3 F(n-1) + 82n \$1,584 1991 4 F(n-1) + 82n \$1,840 1992 5 F(n-1) + 82n \$2,160 1993 6 F(n-1) + 92n \$2,646 1994 7 F(n-1) + 92n \$3,213 1995 8 F(n-1) + 92n \$3,861 1996 9 F(n-1) + 92n \$4,590 1997 10 F(n-1) + 92n \$5,400 1998 11 F(n-1) + 102n \$6,500 1999 12 F(n-1) + 102n \$7,700 2000 13 F(n-1) + 102n \$9,000 2001 14 F(n-1) + 102n \$10,400 2002 15 F(n-1) + 102n \$11,900 2003 16 F(n-1) + 112n \$13,836 2004 17 F(n-1) + 112n \$15,893 2005 18 F(n-1) + 112n \$18,071 2006 19 F(n-1) + 112n \$20,370 2007 20 F(n-1) + 112n \$22,790 2008 21 F(n-1) + 122n \$25,814 2009 22 F(n-1) + 122n \$28,982 2010 23 F(n-1) + 122n \$32,294 2011 24 F(n-1) + 122n \$35,750 2012 25 F(n-1) + 122n \$39,350 2013 26 F(n-1) + 132n \$43,744 2014 27 F(n-1) + 132n \$48,307 2015 28 F(n-1) + 132n \$53,039 2016 29 F(n-1) + 132n \$57,940 2017 30 F(n-1) + 132n \$63,010 2018 31 F(n-1) + 142n \$69,086 2019 32 F(n-1) + 142n \$75,358 2020 33 F(n-1) + 142n \$81,826 2021 34 F(n-1) + 142n \$88,490 2022 35 F(n-1) + 142n \$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.

(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 Stotal(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.

(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 Stotal(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.

(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 Stotal(n) = 2(3+n)

Wait time will be zero when arrive time X(n) equals start service time Sstart(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.

(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, Stotal(n)= 2(4+n)

Wait time will be zero when arrive time X(n) equals start service time Sstart(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>.

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