BUSINESS MATHEMATICS
FORECASTING
In forecasting, we try to establish and understand interrelationship between on one piece of information and another piece of information. In other words, the relationship between straight lines. For this purpose, we interpret data in lines on graphs.
Graph Presentation:
In graph,
X = independent variable (variable that is normally already given)
Y = dependents variable (variable that is to find)
y | |
0 | Level of Activity x |
Normally, we put given values on x-axis. Values to be determined are given in terms of y. We try to show that how much value of y changes with the change in value of x.
Relations may be: | |
X represent | Y represent |
Level of Activity > Units > Hours | Cost |
Advertising Cost | Sales |
Relationship between X and Y can be determined in following ways:
HIGH-LOW METHOD
In High-Low method, we only look at high level of activity and low level of activity. We ignore amounts between high level of activity and low level of activity. We look here only two pairs of data because we are easily able describe data in straight line method.
Though we use High-Low method to predict some value but this method is not best predictor because it only takes extremes into account and ignores values in between high and low.
Issues in Prediction under High Low Method
1. It only used two amounts for forecast. For a method to be a better predictor, it should use more values for prediction.
2. In High-Low Method, only absolute high and absolute low values are considered. Therefore, it can be said that they do not represent whole set of data itself.
Assumption in High-Low Method
Any change between high level of activity and low of activity will lead to a change in cost. It means if the cost changes, it is due to the change in level of activity. Changing cost is a variable cost.
Variable Cost /unit | = | Cost at High Level of Activity | – | Cost at Low Level of Activity |
High Level of Activity | – | Low Level of Activity |
By having above assumption, we can easily indentify the cost behaviours. It means we can identify cost into fixed cost and variable cost.
Example
The following cost information is available:
Output | 65,000 units | 105,000 units |
Cost | £133,000 | £210,000 |
Required:
Calculate the fixed and variable costs for the business and the total cost for 165,000 units.
Solution:
Variable Cost
Variable Cost /unit | = | Change in Cost due to change Level of Activity |
Change in Level of Activity |
Variable Cost /unit | = | Cost at High Level of Activity | – | Cost at Low Level of Activity |
High Level of Activity | – | Low Level of Activity |
Variable Cost /unit | = | £210,000 | – | £133,000 |
105,000 units | – | 65,000 units |
Variable Cost per unit = £77,000 / 40,000 units = £1.925 /units
Fixed Cost
At High level of Activity
Total Variable cost = Units at High level of activity x Variable cost per unit
Total Variable cost = 105,000 units x £1.925 per unit
Total Variable cost = £202,125
Total cost = £210,000
Total Cost = Fixed cost + Variable Cost
Fixed Cost = Total cost – Variable Cost
= £210,000 – £202,125
= £7,875
Total Cost for a Level of Activity of 165,000 units
£
Fixed Cost 7,875
Variable Cost (£1.925 per unit x 165,000) 317,625
325,500
REGRESSION ANALYSIS (LEAST SQUARE LINEAR REGRESSION)
In regression analysis, we try to develop a average relationship two variables (x,y) by considering more than two pairs of variables (x,y).
As in High-Low method, we just analyse High and low relationship values but in regression analysis we analyse more than two relationships between X and Y.
Regression Analysis Formulas
Variable Cost per unit (b)
b | = | n ∑ xy | – | ∑ x∑y |
n ∑ x2 | – | (∑ x)2 |
Fixed Cost in total (a)
a | = | ∑y | – | b | ∑x |
n | n |
To calculate above formula, we need:
∑x
∑y
∑xy
∑x2
∑y2
n
Cost (y)
y = a + bx
a = Fixed Cost
b = Variable cost / unit
x = Number of units
x = Independent variable (units)
y = Dependent variable (Cost)
Example
Months | Units | Cost | ||
January | 400 | 1,050 | ||
February | 600 | 1,700 | ||
March | 550 | 1,600 | ||
April | 800 | 2,100 | ||
May | 750 | 2,000 | ||
June | 900 | 2,300 |
Requried:
Using the following method, calculate the fixed and variable cost elements and forecast the cost for an output of 850 units.
1. High-Low Method
2. Regression analysis
Solution:
x = Units (Unit is an independent variable)
y = Cost (Cost is an dependent variable because it changes with change in number of units)
1. High-Low Method
High level of activity = 900 units
Low level of activity = 400 units
Cost at High level of activity = £2,300
Cost at low level of activity = £1,050
Variable Cost
Variable Cost /unit | = | Cost at High Level of Activity | – | Cost at Low Level of Activity |
High Level of Activity | – | Low Level of Activity | ||
Variable Cost /unit | = | £2,300 | – | £1,050 |
900 units | – | 400 units |
Variable cost /unit = (£2,300 – £1,050) / (900 – 400) = 2.5 /unit
Fixed Cost (at High level of activity)
Fixed cost = Total cost – Variable cost for High level of activity
= £2,300 – (£2.5 per unit x 900 units)
= £50
Cost for output of 850 units
Total cost = Fixed cost + Variable cost for 850 units
= £26 + (£2.5 x 850 units)
= £2,175
2. Regression analysis
| Months | Units (x) | Cost (y) | xy | x2 | y2 | |||||
| January | 400 | 1,050 | 420,000 | 160,000 | 1,102,500 | |||||
| February | 600 | 1,700 | 1,020,000 | 360,000 | 2,890,000 | |||||
| March | 550 | 1,600 | 880,000 | 302,500 | 2,560,000 | |||||
| April | 800 | 2,100 | 1,680,000 | 640,000 | 4,410,000 | |||||
| May | 750 | 2,000 | 1,500,000 | 562,500 | 4,000,000 | |||||
| June | 900 | 2,300 | 2,070,000 | 810,000 | 5,290,000 | |||||
| ∑ | 4,000 | 10,750 | 7,570,000 | 2,835,000 | 20,252,500 | |||||
| ∑ x | ∑y | ∑ xy | ∑ x2 | ∑y2 |
n = 6
Variable Cost
| b | = | n ∑ xy | – | ∑ x∑y |
| n ∑ x2 | – | (∑ x)2 |
| b | = | 6 x 7,570,000 | – | 4,000 x 10,750 |
| 6 x 2,835,000 | – | (4,000)2 |
| b | = | 45,420,000 | – | 43,000,000 |
| 17,010,000 | – | 16,000,000 |
| b | = | 2,420,000 |
| 1,010,000 |
| b | = | 2.396 |
So, the Variable Cost per unit (b) is £2.396
Fixed Cost
| a | = | ∑y | – | b | ∑x |
| n | n |
| a | = | 10,750 | – | 2.396 x | 4,000 |
| 6 | 6 |
| a | = | 1,792 | – | 2.396 | x 667 |
| a | = | 194.33 |
So, the Fixed Cost (a) in total is £194.33
Cost for Output of 850 units (x)
y = Cost
x = Units
a = Fixed cost (total)
b = Variable cost per unit
y = a + bx
= £194.33 + (£2.396 per unit x 850 units)
= £2,230.93
So the Cost (y) is £2,230.93
CORRELATION (r) “Correlation Coefficient”
Correlation explains the level of relationship between the data and line of best fit (or line of regression). It tells how well the line of best fit represents the given data. If the data supports the line of best fit very well, we are convinced that line that we have established represents the data. If the line does not support the date well, we will question the degree of relationship.
-1 ≤ r ≥ +1
Perfect positive correlation: when all data lies on the directly proportional line i.e. r = +1
Perfect negative correlation: when all data lies on the inversely proportional line i.e. r = -1
Near correlation: when data lies near the line i.e. r = -0.98, -0.90, +0.97, or +0.99 etc
+ 1 is perfect positive correlation for those values that are directly proportional to each other.
-1 is perfect negative correlation for those values that are inversely proportional to each other.
So the value of r should always lie between -1 and +1. If the value lies near to -1 or +1, there some degree of correlation.
Formula for Correlation
Now we put date calculate in above example into this formula.
r = (6 x 7,570,000 – 4,000 x 10,750) .
√(6 x 2,835,000 – 4,0002)(6 x 20,252,500 x 10,7502)]
r = 0.987
Here r is Positive and Near perfect correlation.
DETERMINATION (r2) “Coefficient of Determination”
In Determination, we just take the square of “Coefficient Correlation”.
Coefficient of Determination determines the amount of change in dependent variable (y) due to the change in independent variable (x).
In our above example, the y represents Cost and x represents Level of activity i.e. units
Example
From above data:
r2 = 0.9872 = 0.974 = 97.4%
It means there is 97.4% of change in cost due to the change in volume, the remaining 2.6% change is due to other unexplained fators.
EXPECTED VALUES (EVs)
Expected values are based on weighted averages. What is a weighted average as opposed to a simple average?
In Expected Value, we want to get a value that reflects the overall chance / likelihood / probability of an event occurring and the value associated with it. Therefore, it is a weighted average value.
In very simple term, it is the sum of all outcomes (i.e. positive or negative) resulted by multiplying positive (Sales, Profits, or gains) and negative (Losses) forecasts with all probabilities. Probabilities are the all percentages likelihood of all events / forecasts occurring in a situation and sum of all given probabilities should be 1 or 100%.
In order to calculate an expected value we multiply the probability of something happening (p) by the value of the outcome (x). The expected value is therefore (px).
To calculate the expected value of a series of things happening we simply add up the total of all the expected values. So the total expected value is equal to: ∑xp
Formula
Expected Value (EV) = ∑xp
Example
| Sale Forecast (x) | Probability (p) | xp | ||
| £100,000 | x | 0.10 | 10,000 | |
| £200,000 | x | 0.60 | 120,000 | |
| £300,000 | x | 0.30 | 90,000 | |
| 1.00 | 220,000 |
220,000 is the only the weighted average. It does not seem near to the original forecasts. We should analyse that if we want to get an individual outcome, is it possible to achieve value of or near to 220,000 in a period. Certainly we cannot.
So how to use the Expect value (EV) if could not achieve that result (220,000). We will use that figure as an average (weighted average value). By considering the above given example, let us suppose if we expect sale of £100,000 in one month, £200,00 in second month and £300,000 in third month, we can make an idea or we can expect that we are able to achieve sale round to £220,000. It may be less, it may more or it may near to £220,000.
OBSERVATION IN EXPECTED VALUE (EV)
1. There should not be same outcome in one set. We may have equal outcome but it is very rare.
2. It also ignores any risk related to any outcome. The reason is that the EV does not represent the range of initial forecasts (i.e. 200,000 = 300,000 – 100,000). So the range may go beyond 200,000.
In nutshell, our main of EV is calculated Weighted Average of different outcomes. We are not focusing on the risks. Expected values are also used in decision making situations. A series of expected value calculations can be made and the one with the best overall expected value would be chosen.
Example
A company has a choice of three alternative investments. If successful investment A gives a profit of £100,000, of which there is a 40% chance or probability, however if not it will yield loss of (£40,000). Investment B, if successful (60%), will yield a profit of £50,000; but if not, a loss of (£20,000). Investment C, if it is successful (80% chance), will yield profit of £40,000; however if not it will deliver loss of (£10,000).
Required:
Which investment should be considered?
Solution:
| Investments | | EV |
| A | 100,00 x 0.40 + (-40,000 x 0.60) | = 16,000 |
| B | 50,000 x 0.60 + (-20,000 x 0.40) | = 22,000 |
| C | 40,000 x 0.80 + (-10,000 x 0.20) | = 30,000 |
So, in A, B, C, we will choose. C
