Question

Design a database to keep data about college students, their academic advisors, the clubs they belong to, the moderators of the clubs, and the activities that the clubs sponsor. Assume each student is assigned to one academic advisor, but an advisor counsels many students. Advisors do not have to be faculty members. Each student can belong to any number of clubs, and the clubs can sponsor any number of activities. The club must have some student members in order to exist. Each activity is sponsored by exactly one club, but there might be several activities scheduled for one day. Each club has one moderator, who might or might not be a faculty member. Draw a complete E-R diagram for this Database.
(a) All entities with their attributes must be represented, indicating all candidate keys. You must indicate and justify all assumptions you have made.

(b) Describe non-trivial domains for attributes where needed.

(c) Make a decision about the cardinality and participation constraints of all relationships, and add appropriate symbols to the E-R diagram.

Answer 1

Answer:

Complete design is attached below.please have a look.

Explanation:

Answer 2

Answer:

This is a typical example of a constraint max/min. The method used to solve this problem is called

the method of Lagrange multipliers. Let’s generalize the situation:

Given: A function: f(x, y, z) and a constraint that we can write as g(x, y, z) = 0.

Goal: Find min or max of f(x, y, z) for (x, y, z) satisfying g(x, y, z) = 0.

To have a “visual grasp” for the concept of Lagrange multipliers one can think about the following

problem:

Take a balloon (here approximated by a perfect sphere centered at the origin) and a box (think of

a cube for example). We want to find the maximum radius of the balloon (this is the function to

maximize) that can fit inside the box (this is the constraint). We start inflating the balloon and we

realize that the maximum radius is obtained when the balloon touches the box. At the touching

point(s) the surface of the balloon and the one of the box are tangent to each other!

This simple experiment is not a special case. In fact in general1

if P0 = (x0, y0, z0) is a point sitting

on the level surface given by the constraint where max/min for f occur, then at this point the level

surface of the constraint is tangent to the level surface of f passing through P0:

If the two surfaces are tangent, then all normal vectors to the two surfaces are parallel to each other.

In particular their gradients at P0 are parallel, that is

O~ f(P0) = λO~ g(P0) (3.1)

for some parameter λ. This parameter is called the Lagrange multiplier.

We discovered that the max/min points for a function f(x, y, z) constraint by g(x, y, z) = 0 are

found among the solutions (x, y, z, λ) for the system

O~ f(x, y, z) − λO~ g(x, y, z) = 0

g(x, y, z) = 0.

Notice that this system contains four equations and four unknowns:

∂

∂x

f(x, y, z) − λ

∂

∂x

g(x, y, z) = 0

∂

∂y

f(x, y, z) − λ

∂

∂y

g(x, y, z) = 0

∂

∂z

f(x, y, z) − λ

∂

∂z

g(x, y, z) = 0

g(x, y, z) = 0.

(3.2)

but in general it is not a linear system!

One can present the method of Lagrange Multipliers in a more efficient (but less illuminating) way.

Define in fact the new function

L(x, y, z, λ) = f(x, y, z) − λg(x, y, z).

The critical points of L solve the vector equation

O~ L(x, y, z, λ) = 0.

But remember that now the variables are (x, y, z, λ) so we need to take four partial derivatives for

L. If one does so then again (3.2) is obtained!