# Example

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```					 MBA             Quantitative Applications & Decision Making                   Lecture 1

Example

A small firm manufactures woolly socks. It has fixed costs of £3000 per month. (these
fixed costs remain constant for at least a few months and are 'fixed ' because they do
not vary with output. E.g rent, rates, depreciation and bank interest)
The variable costs are £2.40p per pair of socks produced being made up from labour
and materials and other direct costs. At the end of the month a wholesaler buys the
socks for £3.60p per pair.

The variable in this example are :-

(a) the monthly output and sales of socks

(b) the total costs incurred by the firm each month being the total of fixed and variable
Total Costs = Fixed Costs + Variable Costs

(c) the Sales revenue that the firm earns each month

We let X be the independent variable which represents the monthly output of pairs of
socks. The value of X will depend on availability of labour, material and machines

The Total Costs and Sales revenue both depend on output so these are the
dependent variables.

Cost Equation
If X is the number of socks output per month then

FC   +        VC

Total Costs (TC) = 3000 + 2.4 X

Revenue Equation
Each pair of socks sells for £3.60 so

Total Revenue (TR) = 3.6 X

Although these are relatively simple equations, they represent a useful model for the
firm

Profit Equation
Profit = Total Revenue - Total Costs

P    =        TR          -            TC

P    =    3.6 X          - (3000 + 2.4 X)

P = 1.2 X - 3000

We know from our earlier understanding that this is a straight line with intercept of -
£3000 and slope 1.2

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MBA               Quantitative Applications & Decision Making                Lecture 1

In practice these numbers are understandable in that the company will make a loss of
£3000 per month if no socks are produced and sold. The contribution to these fixed
costs is at the rate of £1.2 per pair of socks i.e for every pair of socks produced, the
company is moving towards profitability.

Now draw these three equations on a straight line graph.

We can estimate outcomes from the relationships which we have derived

(a) If the firm produces 1000 pairs of socks per month then the company will be
running at a loss of £1800 per month
Using P = 1.2 X - 3000              means      P = 1.2(1000) - 3000
P =     1200  - 3000
P = - 1800

(b) The firm would like to know how many socks are required to be made per month to
achieve break-even. Break-even is where Total Revenue equals Total Costs and
profits are zero

P    = 1.2 X - 3000 and break-even occurs when P = 0
1.2 X - 3000 = 0
X = 2,500

We can check this is true             P = TR - TC
When X = 2,500                        P = 3.6 (2,500) - (3000 + 2.4 (2,500))
P =     9000    - (3000 + 6,000)
P =     9000    -     9000 = 0

Check what assumptions you are making in your analysis?

What is the expected profit if the company makes 5000 pairs of socks per month?
What is the expected profit if the company makes 50000 pairs of socks per month?

What issues arise in these circumstances?

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MBA              Quantitative Applications & Decision Making                       Lecture 1

Differentiation and optimisation

In general a quadratic equation is of the form

Y = aX²     +   bX      + c

We need to see how Y changes with X changing.

If we allow X to change by a small amount δ X, then X moves from X to X + δX.
At the same time Y has also changed from Y to Y + δY. If this change is very
small we can see that we have a right angled triangle whereby we can use normal
notation to find slope of line. In reality we wish to find

δY /       δX

We know that    Y = X² - 3X - 4 so that we can find
Y + δY when X changes by δX

Y +   δY   =      (X +    δX)²       - 3(X + δX) - 4

Expanding        Y +   δY   =      X²   + 2X δX           +   δX²     - 3X - 3 δX   -   4

Original         Y          =      X²                                 - 3X          -   4

Subtracting            δY   =           + 2X δX       +       δX     - 3 δX

But if     δX is very small then        δX ² is approximately 0

δY   =       + 2X δX                         - 3 δX

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MBA               Quantitative Applications & Decision Making                      Lecture 1

Dividing by   δX           δY        =        + 2X                  - 3
δX

This is called the DIFFERENTIAL. It indicates the change in Y due to a change in X.
Please note that the differential has different values dependent on the value of X.

dY      = 1 when X = 2,            = -7 when X = -2
dX

There is one special case which interests us, a curve reaches an optimal point when
dy/dx = 0

An optimal point can be either           MAXIMUM or MINIMUM

By inspection we can normally find if Max or Min but always

dY / dX         < 0 means Maximum

dY / dX        > 0 means Maximum

In the above example, dY / dX can be found by differentiating again for the second
time.

Thus for             Y    = X²           - 3X - 4

dY/dX =          2X - 3

d²Y / dX²        = +2          Since this is positive must be a Minimum

Using another example                    Y = - X²    + 5X - 10

dY / dX = -2X + 5

At Max or Min dY / dX = 0 means -2X + 5 = 0 means 2X = 5
means X = 2.5

But     d ² Y / dX ²      =       -2   which means this is a maximum.

We can apply this methodology to other equations at higher level than X squared.

Y = X3           + 3X 2         - 9X + 12

Since this equation has an X cubed term we know that there are two curves.

dY / dX = 3X2 + 6X - 9

At Max or Min dY / dX = 0 which means 3X2 + 6X - 9 = 0

Similarly dividing by 3              3(X2 + 2X - 3) = 0           hence X2 + 2X - 3 = 0

This can be factorised               (X + 3)(X - 1) = 0

This has two solutions                         X + 3 = 0     or      X - 1 = 0

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MBA             Quantitative Applications & Decision Making                Lecture 1

X = -3       or     X = 1

If we differentiate again            d2 Y / dX2   = 6X + 6

When X = -3 then       d2Y / dX2 = -18 + 6 = - 12           Maximum

When X = 1 then        d2Y / dX2 =   6 + 6 = + 12Minimum

We can use the idea of calculus to consider production problem. In Break-even, as a
company increases the number of product made, the profit increases. The larger the
production of X then the larger the profit. This is somewhat unrealistic. A company is
constrained from making an ever larger profit because it needs to not only produce the
product but also sell the product.

Example
A firm makes a product for which fixed costs are £800 per month, with variable costs of
£80 per product. Records show that quantity sold depends on price and has been
ascertained that quantity sold meets the demand equation Q = 20 (100 - p) where
p is the selling price to customers.

Information provided

Fixed Cost    FC =      £800
Variable Cost VC =      £80
Selling Price SP =      £p we do not know what price to charge
Number of products =    Q we do not know how many to make or sell

Total Revenue      TR = price x quantity
=   p x     Q

But we have a demand function which relates Q to p          Q = 20 (100 - p)

Q/20 = 100 - p

We can rearrange this formula such that                     p = 100 - Q/20

Substituting this value of p into TR formula we obtain

Total Revenue          TR = (100 - Q/20) Q

TR = 100Q - Q2 /20

Total Cost             TC = Fixed Cost + Variable Coast

TC =        800     +       80Q

Profit                 P = Total Revenue - Total Cost

P = 100Q - Q2 /20 - (800 + 80Q)

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MBA              Quantitative Applications & Decision Making                     Lecture 1

P =     - Q2 /20 + 20Q - 800

We need to find when this Profit function is a Maximum. To do this we differentiate P
with respect to Q. i.e. How does Profit vary with changes in quantity.

dP / dQ = - 2Q /20 + 20

= - Q /10     + 20

Maximum occurs when dP / dQ = 0           hence -Q / 10 + 20 = 0

Q    = 200

The optimal quantity to make and sell is 200. Substituting this value of Q into price p
equation we find

Price p = 100 - Q / 20

= 100 - 200 / 20         = 90

The optimal position for the company is to set a selling price of £90 and sell 200 units
every month.

The maximum profit P =         P =     - Q2 /20 + 20Q - 800

And when Q = 200               P =      - 200.200/20 + 20.200 - 800

=    - 2000           + 4000    - 800

Max P     =    £1200 per month.

Please note that the company, with this information cannot do better. If it wants to sell
more than 200 units per month then it will need to reduce the price and hence the
profits. If the company tries to increase profits by increasing price then it will sell less
than 200 units.

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