Client Profile:
PART 1
Client Profile:
Aneesas Financial Situation:
As of December 2021, Aneesa has saved $5,300 from her parents by doing her weekly household chores, as well as money from family members on special occasions (i.e. birthday, Eid etc.). She has begun saving this money when she started Year 7 in 2016, therefore she has saved $5,300 in 6 years. In December 2021, she also began working part-time as a Woolworths cashier where she earns $22.33/h, working from Monday-Thursday, from 6pm to 10 pm. Thus, she works a total of 16 hours a week as a cashier at Woolworths. Every Saturday and Sunday, she also works from 9am to 2pm (5 hours) as a tutor, where she earns $45/h. She tutors a group of high school students, specifically tutoring for maths and physics. She is expected to earn more money throughout the years as her salary will increase every year while working at Woolworths. As Aneesa is still currently living with her parents, she has limited expenses and bills to pay. However, she only has a budget of a maximum of $200 a week to spend on her expenses.
Under the General Retail Industry Award 2020, a part-time employee is entitled to the same rights as a full-time employee, including annual leave payments. Under this Award, a part-time and full-time level 1 employee is entitled to a minimum rate of $22.33/h. As Aneesa would progress throughout the levels, as she works every year, she is entitled to pay rises, such as $22.86/h in year 2, $23.22/h in year 3 etc. This is outlined in the table below:
https://awardviewer.fwo.gov.au/award/show/MA000004#P1078_88373
A table of expenses:
Expenses Weekly Cost (approximation) Yearly Cost (approximation) Factors to take into consideration
Entertainment $55 $2860 This figure may vary depending on special occasions such as birthdays, Eid, Ramadan etc.
Food $40 $2080 Since Aneesa is still living with her parents, she does not need to spend too much on food, as she is dependent on the food at home. However, this figure may change depending on specific occasions (e.g. birthday, Ramadan etc.)
Petrol $25 $1300 Aneesa usually catches public transport to and from university, using her school opal card (which is free). Thus, she does not spend too much money on petrol per week, as she barely travels to places that are far, other than her local Woolworths and tutoring for work.
Phone $25 (monthly)
$6.25 (weekly) $1300 This figure is taken from:
https://www.canstarblue.com.au/technology/tablets/student-phone-plans/This website provides the best phone plans for students, most of which were priced around $25/month
Personal care (clothes, skincare, etc.) $55 $2860 This figure is an approximation and may vary according to different circumstances, such as special events, or when running low on certain products etc.
= $181.25 = $10,400 Client Forecast Timeline:
Event Date Finances
Client graduates from Arthur Phillip High School in Paramatta. October 2021 Client has already saved $5,300.
Client begins working part-time as a cashier at Woolworths and a tutor. December 2021 Throughout her first year of work as a Woolworths cashier, she will earn $22.33/h. She is working from Monday-Thursday, from 6pm to 10pm (4 hours) as a cashier. She also works as a tutor on Sunday and Saturday, from 9am to 2pm (5 hours) where she earns $45/h.
Cashier:
$22.33 x 4 x 4 = $357.28
Tutor:
$45 x 5 x 2 = $450
Weekly income:
$807.00
Client begins studying at the university of Sydney, commencing her Bachelor of Project Management, majoring in construction for 3 years full time. February 2022 Client begins saving $607 weekly. She therefore saves $2640.45 monthly, which she puts into a savings account with an interest rate of 0.45% p.a. She begins saving with a starting deposit of $5,300. March 2022 Weekly income = $807.00
Budget = $200
Savings = $807 - $200
= $607.00
Monthly savings:
607 x 4.35 = $2640.45
Clients hourly rate increases every year working at Woolworths
(see table on page 3 for details) December 2022, December 2023, December 2024, December 2025 Client graduates from University of Sydney with a Bachelors in Project Management (Construction)
March 2025 Client finds a job as a construction project manager, earning a graduate project manager salary. She leaves her job as a part-time cashier at Woolworths, as well as her tutoring job. July 2025 Average graduate salary of a project manager: $67,500
Weekly income:
$67,500 52 = $1298.08
Client has saved $165612.95 in 5 years for a 30% deposit on an apartment, including the $5,300 she has saved before working. March 2027 Monthly savings: $2640.45
2640.451.000375(1.00037560-1)1.000375-1
$160252.4454
$160252.45
-----------------------------
5300(1.000375)60
$5420.578819
$5420.58
-----------------------------
$160252.45 + $5420.58
= $165673.03
Client puts in a 30% deposit for an apartment May 2027 30% x 488000 = $144000
Client is granted a home loan for the remaining 70% of the apartment May 2027 - 2052 Loan = $337,296
Rate = 2.29% p.a.
Term of loan = 25 years (300 monthly repayments)
Monthly repayments = $1478
Below shows the average graduate project manager salary:
PART 2
Saving for a Deposit:
In order to attain a home loan for an apartment, the client must first put in a deposit for the apartment. The client is aiming to deposit a minimum of 20% for the home loan. To do this, Aneesa is hoping to save for 5 years for the loan. To increase Aneesas savings, she must invest her money into a savings account to earn interest on the money she deposits every week. This will result in a larger future value of her savings, which will then allow her to pay for a larger deposit on the apartment of her choice. This will also ultimately allow the client to reduce the amount of the home loan, and hopefully pay it off quicker. The client has already saved $5,300, which she will initially place into a savings account, and will further place $607 every week (607 x 4.35 = $2640.45) [$26450.45 monthly] into this account. 3 banks have been researched to provide the client with a savings account. The goal is to find a financial institution with a high interest rate to allow Aneesa to save more. This is shown below:
1) Westpac Bank
2) Bankwest Bank
3) Ubank
[The savings account and interest rate are highlighted in blue]
After researching these banks, it is recommended that the client saves her money into a savings account with Bankwest bank as it has the highest interest rate and will earn the client more money, thereby allowing her to make a bigger deposit for her home loan. If the client pays off a larger deposit, this would mean she would require a lesser home loan, meaning she would be able to pay it off quicker. After consulting with the client, she has decided to invest her money into a savings account with Bankwest Bank. The Bankwest website has a savings calculator which calculates the future value of an investment. This calculator was used to find out the future value of Aneesas savings and how much she would have for a deposit after 5 years of saving. This is exhibited on pages 7-8.
Therefore, the client will save $165,586 in 5 years by saving her money and investing it in a savings account with Bankwest Bank.
However, according to the spreadsheet, the client will save $165,673.02. The Bankwest calculations are probably lower as they most likely take into consideration factors such as bank fees, annual fees etc.
The graph below shows the growth of Aneesas savings:
Working Out:
p = principal value = $2640.45
r = interest rate (monthly)
r = (0.25% 12) + 1 = 1.000375
n = number of periods/months = 60 (5 years)
A = future value of savings
The client deposits $2640.45 into her savings account at the end of every month for 5 years. Since the interest is earnt at the end of the month, this repayment will earn interest for 59 months:
A1 = 2640.45 (1.000375)59
A2 = 2640.45 (1.000375)58
A3 = 2640.45 (1.000375)57
This trend continues until the clients last deposit in the 60th month. This deposit will then earn interest for 0 months:
A60 = 2640.45(1.000375)0
FV (future value) = A1 + A2 + A3 ++ A60
= 2640.45(1.0003750 + 1.0003751 + 1.0003752 + + 1.00037559)
This forms a geometric series:
a = 1.000375
r = 1.000375
n = 60
=
60.69133873
FV = 60.69133873 x 2640.45
$160252.4454 $160252.45
However, since the client had previous savings of $5,300 and had also deposited this money in the first month, this will also earn compound interest:
5300(1.000375)60 $5420.578819 $5420.58
Therefore, the future value of Aneesas savings will be:
$160252.45 + $5420.58 = $165673.03
Below is a table exhibiting the savings Aneesa will have at the end of every year on the course of her 5 years:
Year Compound Interest Working Out Total interest accumulated within that year Working Out
1 $37086.64 2640.451.000375(1.00037512-1)1.000375-1
$31762.73946
$31762.74
-------------------------------------
5300(1.000375)12
$5323.899252
$5323.90
-------------------------------------
$31762.74 + $5323.90
= $37086.64 $101.24 31762.74 - (2640.45 x 12)
= $77.34
--------------------
5323.90 5300
= $23.90
--------------------
$77.34+ $23.90
= $101.24
2 $69016.62 2640.451.000375(1.00037524-1)1.000375-1
$63668.70641
$63668.71
-------------------------------------
5300(1.000375)24
$5347.906273
$5347.91
-------------------------------------
$63668.71 + $5347.91
= $69016.62 $321.92 63668.71 (2640.45 x 24)
= $297.91
--------------------
5347.91 5323.90
= $24.01
--------------------
$297.91 + $24.01
= $321.92
3 $101090.57 2640.451.000375(1.00037536-1)1.000375-1
$95718.54671
$95718.55
-------------------------------------
5300(1.000375)36
$5372.021549
$5372.02
-------------------------------------
$95718.55 + $5372.02
= $101090.57 =$686.46 95718.55 (2640.45 x 36)
= $662.35
--------------------
5372.02 - 5347.91
= $24.11
--------------------
$662.35 + $24.11
= $686.46
4 $133309.16 2640.451.000375(1.00037548-1)1.000375-1
$127912.9091
$127912.91
-------------------------------------
5300(1.000375)48
$5396.245567
$5396.25
-------------------------------------
$127912.91 + $5396.25
= $133309.16 $1195.54 127912.91 (2640.45 x 48)
= $1171.31
--------------------
5396.25 - 5372.02
= $24.23
--------------------
$1171.31 + $24.23
= $1195.54
5 $165673.03 2640.451.000375(1.00037560-1)1.000375-1
$160252.4454
$160252.45
-------------------------------------
5300(1.000375)60
$5420.578819
$5420.58
-------------------------------------
$160252.45 + $5420.58
= $165673.03 $1849.78 160252.45 (2640.45 x 60)
= $1825.45
--------------------
5420.58 - 5396.25
= $24.33
--------------------
$1825.45 + $24.33
= $1849.78
The spreadsheets, tables, graphs, and calculations are based on the assumption that the interest rate does not change throughout the 5 years which Aneesa will be investing her money. They are also based on the assumption that Aneesa will consistently save the same amount of money ($2640.45) every month, regardless of her pay raises and bonuses as she grows older and gains experience. It is highly recommended that Aneesa invest her money into a savings account when saving for a deposit as she would be able to earn an extra $7246.03 (2640.45 x 60 = 158427, 165673.03 158427 = $7246.03) in 5 years on top of her savings. This could allow her to pay for a larger deposit, which could thereby lead to a reduced home loan which may be paid off quicker.
PART 3
Purchase and Loan Description:
Aneesa Malik is currently looking for an apartment in the Bankstown area or close by, that is priced around $500,000, with a maximum price of $550,000. It is required for the apartment to have a minimum of 2 bedrooms, minimum of 1 bathroom (preferably 2), and a minimum of 1 car space/garage. As she has saved $165,673 in 5 years, she needs an apartment that is priced accordingly so that she will be able to afford a minimum of 20% deposit with this money saved. Therefore, the home loan which the client will take will be needed to pay for the remaining of the apartment.
Below are a few apartment options which fit with the clients purchase requirements:
1)
2)
3)
After consulting with the client and considering these 3 options, it is recommended that Aneesa selects the first option of the apartment with a guide price of $480,000. Not only is this apartment priced less than she had targeted for ($500,000), but it also has the largest space out of all the apartments, with 2 bedrooms, 2 bathrooms and 2 car spaces. The client should be able to easily pay off a 30% deposit for this apartment, and acquire a home loan for the remaining 70%, which now, may be paid off across 25 years instead of 30 years, given the clients savings and salary. The client is not willing to spend all her savings on a deposit as she would like to be secure enough to still have some money left over to save for any emergencies. The 30% deposit which Aneesa would have to pay for this apartment would be: $480,000 x 30% = $144,000
However, when purchasing an apartment, there are additional costs which must be taken into consideration such as stamp duty, mortgage fees etc. These are outlined below:
From this table, it can be drawn that Aneesa will be making a 29% deposit plus upfront costs that equates to $144,000. Therefore, her loan will need to cover the remaining 71% of the loan. Thus, Aneesa will need to take a $337,296 loan. [$480,000 - $142,704 = $337,296]
Loan Details:
The client must apply for a home loan to purchase her required apartment. She has already saved for a 29% deposit and will need a loan of $337,296 to pay off the rest of the apartment. 3 banks have been investigated (Westpac, Commonwealth and NAB) for the client to select the most appropriate option, which would ultimately save costs in the long term with the lowest interest rate.
The details of the loan are as follows:
Loan amount = $337,296
Loan term = 25 years (300 monthly repayments)
Rate = variable rate
Payment type = principal and interest
Bank 1: Westpac Bank
Bank 2: ANZ Bank
Bank 3: NAB Bank
After taking into consideration these three options and after careful consultation with the client, it is recommended that Aneesa borrows a home loan from Westpac Bank as it has the lowest interest rate, thereby resulting in lower monthly repayments. This will benefit the client as it will help the client to save money by paying less interest. Below depicts a graph which represents the reduction of the loan as the client pays their monthly repayments:
Working Out:
M = monthly repayments
n = 300 (number of repayments)
Interest rate (monthly) = (2.29% 12) + 1 = 1.001908333
[throughout the working out, this rate will be rounded to 1.0019, however, within the actual calculations, the whole rate will be used]
A = amount owing after x months
A1 = 337296(1.0019)1 M
A2 = (337296(1.0019)1 M) (1.0019)1 M
When expanded, this equals:
= 337296(1.0019)2 M(1.01)1 M
When factorised, this equals:
= 337296(1.0019)2 M(1.0019 + 1)
A3 = (337296(1.0019)2 M(1.0019 + 1)) (1.0019)1 M
= 337296(1.0019)3 M(1.00192 + 1.0019 + 1)
This trend continues up to 300 months (25 years), ending when A360 = 0
0 = 337296(1.0019)300 M(1.0019299 + 1.0019298 + 1.0019297 ++ 1.00190)
337296(1.0019)300 = M (1.0019299 + 1.0019298 + 1.0019297 ++ 1.00190)
M = 337296(1.0019)300
(1.0019299 + 1.0019298 + 1.0019297 ++ 1.00190)
It is evident that a geometric series is being formulated:
a = 1.00190 = 1
r = 1.0019
n = 300
S300 = 1(1.001908333300-1)1.001908333-1 404.398087
M = 3372961.001908333300404.398087 $1477.742315
Rounded to the nearest dollar, this would equal to $1478, which is precisely what the Westpac online home loan monthly repayment calculator estimated Aneesas monthly repayment to be.
Assumption:
All the spreadsheets, tables and graphs are based on the assumption that the interest rates remain the same and do not change. They are also based on the assumption that despite Aneesas pay raises and bonuses, she will consistently invest the same amount of money ($2640.45) every month into her deposit and home loan.
PART 4
Scenario:
This report will explore the effects of a certain scenario on the clients finances and her savings, deposit and loan. The scenario which will be examined is in regard to the clients future pay raises and bonuses.
The Deposit
According to the client forecast timeline (refer to pages 4-5) , the client has started to save her money in March 2022, starting with $607 monthly. 9 months later, in December of that year, she earns herself a pay raise, from $22.33/h to $22.86/h (refer to the hourly rate table on page 3). Every pay raise, she saves an extra $10 monthly, meaning she now saves $617 monthly. Consequently, every December when she earns a pay raise, she will save an extra $10 monthly. In July 2025, when the client has already graduated from university and has quit her part-time job, she has found a full-time job that correlates with her degree. As she begins working as a construction project manager, she earns the graduate project manager salary of $67,500 (refer to page 5). As she is earning a larger monthly income of $5625 (67500 12 = $5625), she is able to save an extra $300 monthly, which will raise her monthly deposits into her savings account to $2980.45.
If the client were to follow suit with this, she would end up with $169406 saved after 5 years. This is a $3,733 difference compared to her previous $165673 of savings. While it isnt much, it could still assist the client in saving up for a larger deposit, which would ultimately lead to a reduced loan which could be paid off quickly. With these savings of $169406, the client could afford a 34% deposit ($165920) on her $488,000 apartment, with $3486 to spare. This could then impact her home loan, which would be reduced to the remaining 66% of the apartment ($322080). Therefore, it is recommended that Aneesa takes this scenario into consideration and perhaps follow this plan which is in line with her pay raises.
The graphs below depict the changes which would occur to the clients savings in regard to this scenario:
As shown, the graph has become slightly steeper, meaning the client is saving more money as she earns pay raises and makes larger deposits into her savings account.
The Home Loan
The clients original monthly repayment is $1478 on a $337,296 loan with a variable interest rate of 2.29% p.a. Throughout the clients career and as she gains experience, she is bound to receive pay raises and bonuses. After 2 years of paying $1478 as a monthly repayment, she may decide to add an extra $350 every month, to make her monthly repayments $1828. Subsequently, she may repeat this cycle every 2 years.
Originally, the client would be able to pay off the $337,296 loan in 25 years. However, if the client were to follow the plan in the aforementioned paragraph, they would be able to pay off the loan in 162 months (13.5 years). With an average project manager salary of $128,265, the client would undoubtedly be able to pay an extra $350 on their monthly repayment every 2 years. Therefore, it is recommended that the client follow this plan in order to quickly pay off the loan for their apartment.
Average construction project manager salary:
The graphs below depict the changes which would occur to the home loan in regard to this scenario:
As demonstrated with the second graph, the figures become negative in a shorter period of time, showcasing that the loan would be paid off quicker. The first graphs shows that the line intercepts the x axis at 300 months, while with the second graph, the line intercepts the x axis at around 160 months.
References:
Westpac:
https://www.westpac.com.au/personal-banking/bank-accounts/savings-accounts/rates/https://www.westpac.com.au/personal-banking/bank-accounts/savings-accounts/savings-calculator/https://www.westpac.com.au/personal-banking/home-loans/all-interest-rates/
Bankwest:
https://www.bankwest.com.au/personal/bank-and-save/calculators/savings-term-deposit-calculatorhttps://www.bankwest.com.au/personal/bank-and-save/savings-accounts/hero-saver
Ubank:
https://www.ubank.com.au/banking-overview/calculators-and-rates
ANZ Bank:
https://www.anz.com.au/personal/home-loans/calculators-tools/calculate-repayments/
NAB:
https://www.nab.com.au/personal/home-loans/nab-base-variable-rate-home-loanhttps://www.nab.com.au/personal/home-loans/calculators/loan-repayments-calculator
