APPLY THE FOUR RULES
OF NUMERACY
Introduction to Quantitative Methods
T/616/0695
Tashkent International
University of Education
http://www.tiue.uz
Business Calculations
Quantitative Methods
Simple financial transactions
involving purchases, wages,
taxation, discounts
2.1 Gross Pay Calculations
Gross Pay is the total amount to which the employee is entitled.
Determining gross pay:
• Award or employment agreement conditions for minimum / maximum
hours, ordinary and overtime rates and leave entitlements;
• Rate per hour for part-time, casual or irregular employment;
• Rate per unit produced; or piece work method;
• Commission earned; where a percentage of sales achieved is paid to an
employee;
• Annual salary without any entitlement to overtime.
5
Payroll Accounting
2.2 Net Pay
Net pay is the Gross Pay less PAYG Withholding Tax and less any
other deductions. It is the amount actually paid to the
employee on pay day
6
Payroll Accounting
2.3 Ordinary Pay
Ordinary time is the normal hours worked each week, e.g. a 35
hour week.
Ordinary Pay = Ordinary Hours Worked x Rate per Hour
7
Payroll Accounting
2.4 Overtime
Overtime is paid for hours worked in excess of normal hours
Overtime is paid at one and a half times the normal rate for the
first 3 hours per week and thereafter at double the normal
rate.
Overtime pay = Overtime hours worked x Overtime rate per
hour
8
Payroll Accounting
2.5 Piece Rates
Employees under piece rate arrangements are paid per unit of
output.
EXAMPLE
Eric works in a clothing factory and is paid $4 for each shirt
stitched.
This week Eric stitched 100 shirts. How much is his gross pay?
9
Payroll Accounting
2.6 Commission
Commission is commonly paid to salespeople to motivate
employees to make sales.
Gross pay of salespeople usually consists of a retainer (a base
pay) plus commission.
10
Payroll Accounting
2.7 Annual Salary
A salary is remuneration usually quoted as a yearly figure, but
paid weekly, monthly or fortnightly. There is usually no payment
of overtime.
11
Payroll Accounting
2.8 Allowances
Allowances fall into two broad categories:
1 - By way of a reimbursement or compensation ( eg ) Car ,
Meal
2 – By way of Unconditional extra payment. ( eg ) Shift , First
aid, Travel
12
Payroll Accounting
2.9 Annual Leave Pay and Personal Leave Pay
Awards and employment agreements are required to provide a
minimum entitlement of 4 weeks annual leave per year and
10 days personal leave p.a. for full-time employees.
Some employees are also entitled to a leave loading of 17.5% (on
their annual leave payment)
13
Payroll Accounting
2.10 PAYG Withholding
PAYG Withholding (personal income tax) is deducted from an employee’s
pay and remitted by the employer to the ATO
Factors affecting the amount withheld:
 Gross Pay
 Residency
 Tax – Free threshold
 TFN Notification
 HELP debt
 Tax offsets
14
Payroll Accounting
Determining the Amount to Withhold
15
Payroll Accounting
Tax Offsets
16
Payroll Accounting
2.11 PAYG Withholding and HELP Repayments
HELP Tax Table is used to calculate the additional amount of tax
that must be withheld when the employee has a HELP debt
 See the Appendix for the HELP tax tables
17
Payroll Accounting
2.12 Leave Paid in Advance
Paid annual leave in advance =
Weekly PAYG Withholdings x the number of weeks leave being
taken
18
Payroll Accounting
2.13 Holiday Leave Loading and PAYG
Withheld
Some employees are entitled to annual leave loading (Holiday
Leave Loading or HLL), which is normally 17.5% of the gross
Holiday Pay
If using weekly tax tables, calculate the weekly wages plus
weekly leave loading, then use this amount to determine the
weekly PAYG Withheld
19
Payroll Accounting
2.14 Bonuses
PAYG Withheld on bonuses is calculated on a periodic (weekly,
fortnightly) basis as if an additional amount to normal weekly
pay.
20
Payroll Accounting
2.15 Salary Sacrifice
Salary sacrifice is an arrangement where an employee agrees to
forego part of his/her future salary or wages in return for the
employer providing benefits of a similar value
Requirements for legitimate arrangement:
Not allowed for salary or wages already earned
• No access to sacrificed salary
• Contributions made to complying fund
21
Payroll Accounting
2.16 Deductions
Employers and employees may agree that other amounts be
deducted from the employee’s pay. These are personal expenses
of the employee
Examples include :
• Personal Superannuation contributions
• Private health insurance
• Union Fees
22
Payroll Accounting
2.17 Tax Calculator
As an alternative to using the tax tables the ATO provides an
electronic tax calculator
that can be used to determine withholding amounts.
23
Simple and compound interest
2 types of Interest
• Simple interest – interest is paid only on the principal
• Compound interest – interest is paid on both principal
and interest, compounded at regular intervals
Example
• a $1000 principal paying 10% simple interest after 3
years pays
.1  3  $1000 = $300
• If interest is compounded annually, it pays
•
•
•
•
.1  $1000 = $100 the first year, .
1  $1100 = $110 the second year
and .1  $1210 = $121 the third year
totaling $100 + $110 + $121 = $331 interest
Computing the Cost of Credit
• The cost of credit is determined by using the formula for
simple interest.
• Simple interest is computed on the amount borrowed
only and without compounding.
Simple Interest Formula
(continued)
• The cost is based on three elements:
1. A loan’s principal is the amount borrowed, or the unpaid
portion of the amount borrowed, on which the borrower
pays interest.
2. The rate is the percentage of interest you will pay on a
loan.
3. Time is the period during which the borrower will repay a
loan; it is expressed as a fraction of a year.
Simple Interest
• The length of time the borrower will take to repay a
loan is expressed as a fraction of a year—twelve
months, fifty-two weeks, or 360 days.
• Six months = ½
• Three months = ¼
• 90 days = 90/360 or 1/4
IMPLE INTEREST
FORMULA
Interest paid
Annual interest rate
I = PRT
Principal
(Amount of money
invested or
borrowed)
Time (in years)
Simple Interest Equation: Step 1
P
(Principal)
r
t
I
(Interest
Rate)
(Time
Period)
(Interest
Earned)
$1,000 invested at 7% interest rate for 5 years
1,000
.07
5
350.00
Simple Interest Equation: Step 2
P
(Principal)
I
(Interest
Earned)
A
(Amount
Investment
is Worth)
$1,000 invested at 7% interest rate for 5 years
1,000
350
$1,350.00
Time Value of Money Math Practice #1
Sara deposited $600.00 into a savings account
one year ago. She has been earning 1.2% in
annual simple interest. Complete the following
calculations to determine how much Sara’s
money is now worth.
Step One:
600.00
.012
1
7.20
Step Two:
600.00
7.20
607.20
Time Value of Money Math Practice #1
How much is Sara’s investment
worth after one year?
$607.20
Compound Interest
• Notice that the interest in our account was paid at
regular intervals, in this case every year, while our
money remained in the account. This is called
compounding annually OR one time per year.
COMPOUND INTEREST
FORMULA
Principal
(amount at
start)
amount at the
end
annual interest rate
(as a decimal)
 r
A  P 1  
 n
nt
time
(in years)
number of times per year
that interest in compounded
Compound Interest Equation – Single Sum
P (1 + r)n = A
Amount
Principal (1 + Interest Rate)Time Periods = Investment is
Worth
$1,000 invested at 7% interest rate
compounded yearly for 5 years
1,000 (1+ .07)5 = $1403.00
Compound Interest
• Suppose that instead of collecting interest at the end of
each year, we decided to collect interest at the end of
each quarter, so our interest is paid four times each
year. What would happen to our investment?
• Since our account has an interest rate of 5.5%
annually, we need to adjust this rate so that we get
interest on a quarterly basis. The quarterly rate is:
5.5 / 4  1.375%
Compound Interest
• So for our IRA account of $5000 at the end of a year looks like:
• After 10 years, we have:
0.055 

F1  50001 

4 

0.055 

F10  50001 

4 

41
 $5280.72
410
 $8633.85
Compound Interest Formula
• P dollars invested at an annual rate r, compounded n times
per year, has a value of F dollars after t years.
• Think of P as the present value, and F as the future value of
the deposit.
nt
 r
F  P  1  
 n
Compound Interest
Period
Interest
Credited
Times
Credited
per year
1
Rate per
compounding
period
R
Annual
year
Semiannual
6 months 2
R
2
Quarterly
quarter
4
R
4
Monthly
month
12
R
12
Compound Interest
• Number of times interest is compounded has effect on return
• Interest compounding frequently will yield higher returns
$1,000 invested at 7% for 5 years
Compounding Method
Daily
Monthly
Quartely
Semi-Annually
Annually
Amount Investment is
Worth
$1,419.02
$1,417.63
$1,414.78
$1,410.60
$1,402.55
Example 1
• Example: $800 is invested at 7% for 6 years. Find the
simple interest and the interest compounded annually
Simple interest:
I  PRT  $800  .07  6  $336
Compound interest:
M  P(1  i ) n  $800(1.07)6  $1200.58
I  M  P  $1200.58  $800  $400.58
Example 2
• Example: $32000 is invested at 10% for 2 years. Find the
interest compounded yearly, semiannually, quarterly, and
monthly
yearly:
M  P(1  i ) n  $32000(1.10) 2  $38720
I  M  P  $38720  $32000  $6720
semiannually:
M  P(1  i ) n  $32000(1.05) 4  $38896.20
I  M  P  $38896.20  $32000  $6896.20
Example 2 (cont.)
• Example: (continued)
quarterly:
M  P(1  i ) n  $32000(1.025)8  $38988.89
I  M  P  $38988.89  $32000  $6988.89
monthly:
i  1012%  .833%, n  12  2  24
M  P(1  i) n  $32000(1.00833) 24  $39052.20
I  M  P  $39052.20  $32000  $7052.20
Depreciation
49
Depreciation
Fixed assets like plant and machinery etc.
are used in the business for the purpose of
production or providing services. With the
passage of time and utilisation, value of such
fixed assets decreases.
50
Value of portion of fixed assets utilized for
generating revenue must be charged during a
particular accounting year to ascertain the true
cost. This portion of cost of fixed asset allocated
is called depreciation.
Depreciation means reduction in value of asset
or in the utility due to passage of time, natural
wear and tear, exhaustion of the subject matter.
51
Causes of Depreciation
Lapse of time
natural wear and tear
exhaustion of the subject matter
Obsolescence of technology
52
Objectives of Providing for Depreciation
To ascertain the true results of operations
To present true and fair value of the fixed asset
To accumulate funds for the replacement of the
asset
53
Factors in measurement
Estimation of exact amount of depreciation is
not easy.
Generally following factors are considered in
calculation of depreciation.
54
1. Cost of asset including expenses for
installation etc.
2. Estimated useful life of the asset.
3. Estimated scrap value (if any) at the time
of useful life of the asset.
55
Methods of providing depreciation
1) Straight Line method (SLM)
2) Reducing Balance Method RBM
3) Machine Hour Method
4) Production Units Method
56
Straight Line method
In this method, an equal amount is
written off every year during the
working life of the asset to nil or its
residual value at the end of its
useful life.
57
Straight Line method
SLM: The underlying assumption of this method is
that the particular asset generates equal utility
during its lifetime.
Cost of AssetDepreciation= Scrap Value
Useful Life
58
Straight Line method
Example
Cost of machinery: 18000
Installation Charges:2000
Useful Life of Asset: 5 Years
Calculate Depreciation as per SLM
59
Straight Line method
20000-0
Depreciation=
5 years
Depreciation = 4000 p.a.
60
Reducing Balance Method
Under this method, a fixed percentage of
diminishing value of the asset is written off
each year.
The annual charges of the
depreciation decrease from year to year.
61
Written Down Value (WDV)= (Acquisition Cost – Depreciation)
Depreciation = WDV*Depr Rate
Reducing Balance Method
62
Reducing Balance Method
RBM: The main advantage of this method is
that total charge to total revenue is uniform
when the depreciation is high, repairs are
negligible and as the repairs increases the
burden of depreciation gets lesser and lesser.
63
Reducing Balance Method
RBM:
For First Year Depreciation
=Acquisition value* Rate For Second
Year on words
Depreciation=Written down value*
Rate
64
Reducing Balance Method
Example
Cost of machinery: 50000
Scrap Value of machine:5000
Useful Life of Asset: 10 Years
Depreciation %: 15% p.a.
65
Machine Hour Method
Where it is possible to keep a record of the actual
running hours of each machine, depreciation may
be calculated on the basis of hours worked.
The machine hour rate of depreciation is
calculated after estimating the total numbers of
hours that machine would work during its whole
life.
Under machine hour method Depreciation is
calculated for each hour the machine works.
66
Machine Hour Method
Example
Cost of machine: 500000
Estimated working hours: 40000
Scrap Value: 10000
The pattern of distribution of effective working
hours:
67
Machine Hour Method
Year
hours
1-2: 5000 per year
3-5: 7000 per year
6-8: 3000 per year
Compute depreciation p.a.
68
Machine Hour Method
Solution:
1-2
5000
X (500000-10000)
40000
=61250 p.a.
69
Machine Hour Method
Solution:
3-5
7000
X (500000-10000)
40000
=85750 p.a.
70
Machine Hour Method
Solution:
6-8
3000 X (500000-10000)
40000
=36750 p.a.
71
Production Unit Method
Under this method depreciation is
determined by comparing annual
production with the estimated
total production.
72
Production Unit Method
The amount of depreciation is computed
by the using following formula:
Depreciation for the period=
depreciable
Production during the period
amount
Estimated total production
X
73
Production Unit Method
Example
Cost of machine: 30000
Estimated total production: 4000
Scrap Value: 2000
Pattern of distribution of production:
74
Production Unit Method
Year
1:
2:
3:
Units
2000
1500
500
75
Production Unit Method
Solution:
Year 1
2000
X (30000-2000)
4000
=14000
76
Production Unit Method
Solution:
Year 2
1500
X (30000-2000)
4000
=10500
77
Production Unit Method
Solution:
Year 3
500
X (30000-2000)
4000
=3500
Thank You
Quantitative
Methods