HSC SC H HEAD ST START L T LECTU TURE Standard rd M - - PowerPoint PPT Presentation

HSC SC H HEAD ST START L T LECTU TURE

Standard rd M Mathematics

Presented by: KRYSTELLE VELLA

Satu turday 5 5th

th October 2

2019

WEL WELCO COME E TO Y YOUR HSC Y YEAR!

First o

• f a

all…

You will:

By the end of today’s lecture…

learn the first few topics of Standard Maths! q Financial M Maths q Measureme ment q Ne Netw twor

• rks

ks ask q questi tions about anything you are unsure of! learn stu tudy t techniques to prepare yourself for a successful year ahead

STUDY T TECHNI NIQUES

Although you don’t have exams for a while, it is a great i idea to learn s study t techniques f for S Standard M Maths n now!

How to study for Standard Maths

q Start the year with a fr fresh b book, , and keep it up-to-date and neat throughout. q If a concept confuses you now, seek extra h help before the year goes

• n (i.e. asking for extra help, YouTube videos etc!)

q Get into the habit of showing all y your w working. . q As you learn new formulas, create flashcards to help you practice and learn t them o

• ff b

f by hea heart. q Begin thinking about your exam technique to today.

Topic 1 1: Financial M Mathematics

In Year 11, you would have learned how to:

ü calculate simple i interest (I=Prn) ü calculate pe perce centag age in increa eases es and de decr creases ü use the straight-line method of de depr preci ciat ation ü calculate rates of pa pay (salaries, wages, overtime, commission etc!) ü calculate income t tax (PAYG) ü use bud budgets ts and calculate household e expenses

Sound f fami miliar??

Now in Year 12:

The two B BROAD t topics we cover include:

1.

• 1. Investments a

and L Loans 2.

• 2. An

Annuities

• 1. INVESTMENTS & LOANS

Investments are used to GR GROW money.

Future v value o

• f l

loans When calculating the future value of loans, compound i interest is calculated on the initial amount borrowed/invested plus any interest charged or earned. To calculate the future value of a loan:

𝐺𝑊 = 𝑄𝑊 (1 + 𝑠)*

wh where:

• FV

FV - future value of the investment (final balance)

• PV

PV - present value of the principal (initial quantity of money)

• r

r - rate of interest per compounding time period (note:

: this is expressed as a decimal)

• n - number of compounding time periods

FUTURE VALUE OF LOANS

Example q question:

Brendan invests $14,000 over 6 years compounding annually. What is the future v value of the investment if the compound interest rate is 5% p.a.?

So Solution:

Remember t the f figures r required t to f fulfil t the f future v value f formula!

• Present Value (PV) = $14 000
• Rate (r) = 5% p.a. (0.05)
• Number of periods (n) = 6

𝑮𝑾 = $𝟐𝟗 𝟖𝟕𝟐. 𝟒𝟓

(𝑢𝑝 2 𝑒𝑓𝑑𝑗𝑛𝑏𝑚 𝑞𝑚𝑏𝑑𝑓𝑡)

∴ 𝐺𝑊 = $14 000 (1 + 0.05)E

FUTURE VALUE OF LOANS

When solving future v value q questions, , it is so important that you correctly find the value of the RA RATE and NO. O . OF P PERIO IODS! For e example:

Jocelyn invests $5000 over 2 years compounding quarterly. Calculate the future value of the investment if the compound interest rate is 6% p.a.?

Wh What i t is th the interest r t rate per c compounding p period? How m many compounding p periods are th there?

𝐺𝑊 = $5632.46

PRESENT VALUE OF LOANS

We can also use the future-value formula to find the pr present va value of

• f an

an in inves estmen ent. 𝑸𝑾 = 𝑮𝑾 (𝟐 + 𝒔)𝒐 To To find p present v value: :

PRESENT VALUE OF LOANS

Example q question:

Calculate the present v value of an investment that has a future value of $25 000 after 8 years and earns 9.6% compound interest, compounding bi-annually.

So Solution:

Let’s break down the question to obtain our required values be befor

• re we

we insert them into the formula.

• Future Value (FV) =

$25 000

• Rate (r) = 9.6% p.a. compounding bi

bi-an annual nually Ø = 4.8% (0.048) per compounding period

• Number of periods (n) = 8 years compounding bi

bi-an annual nually Ø = 16 compounding periods

PRESENT VALUE OF LOANS

Solution co continued.. .. Therefore, the present value of this loan is equal to: 𝑄𝑊 = 25 000 (1 + 0.048)LE

𝑸𝑾 = $𝟐𝟐 𝟗𝟏𝟖. 𝟔𝟓

FINDING THE INTEREST EARNED

NOTE! Some questions may ask you to find t the i interest e earned only. In this case, we use the formula: Simply meaning that the interest earned is equal to the future value

• f the investment mi

minus the present value (principal) of the investment. Careful not to confuse this with finding the interest ra

rate.

𝑱 = 𝑮𝑾 − 𝑸𝑾

FINDING THE INTEREST RATE

To try and trick students, NESA may ask you to find the interest r rate used to calculate the present or future value of an investment. In these questions, we simply rearrange t the f future v value f formula.

Ex Example le qu question: The present value of Henry’s investment is $31 250. After 4 years, this has grown to $35 000. If interest compounds annually, calculate the compound interest rate at which Henry’s investment is growing, correct to two decimal places.

FINDING THE INTEREST RATE

So Solution: To begin, let’s insert the values we know into the future value formula. 𝟒𝟔 𝟏𝟏𝟏 = 𝟒𝟐 𝟑𝟔𝟏(𝟐 + 𝒔)𝟓 𝑮𝑾 = 𝑸𝑾(𝟐 + 𝒔)𝒐

In H Henry’s ’s i investment:

• Future Value (FV) = $35 000
• Present Value (PV) = $31 250
• Number of periods (n) = 4
• Rate (r) = ??

Now, we must use our algebraic knowledge to find the value of r.

FINDING THE INTEREST RATE

So Solution:

𝒔 = 𝟑. 𝟗𝟖%

Remove the $31 250 from both sides by division.

35 000 31 250 = 31 250 31 250 (1 + 𝑠)S 1. 1. 1.12 = (1 + 𝑠)S 2.

Simplify.

T 1.12 = T (1 + 𝑠)S

3. 3.

“Fourth-root” both sides to eliminate the 4.

1.028737345 = 1 + 𝑠

4.

Simplify.

𝑠 = 0.028737345 × 100%

5. 5.

Subtract 1 from either side, leaving the r on its own. Multiply by 100% to get a percentage answer.

SOLUTIONS (previous slide)

FINDING THE INTEREST RATE

Therefore, Henry’s investment grows at a compound interest rate of 2.87% p.a. (to two decimal places).

How c can w we t test o t our a answer t to m make s sure i it’s t’s r right? t?

APPRECIATION

Appreciation is the increase i in v value of goods. So Solution:

𝐺𝑊 = 𝑄𝑊(1 + 𝑠)*

𝑼𝒊𝒇 𝒘𝒃𝒎𝒗𝒇 𝒑𝒈 𝒖𝒊𝒇 𝒎𝒃𝒐𝒆 𝐛𝐠𝐮𝐟𝐬 𝟔 𝐳𝐟𝐛𝐬𝐭 𝐣𝐭 $𝟓𝟏𝟖 𝟐𝟒𝟓, 𝒖𝒑 𝒖𝒊𝒇 𝒐𝒇𝒃𝒔𝒇𝒕𝒖 𝒆𝒑𝒎𝒎𝒃𝒔.

Ex Example le qu question: : Flynn bought a block of land in 2014 for $319,000. If the land appreciates at 5% p.a., calculate it’s value after 5

• years. Answer to the nearest dollar.

𝐺𝑊 = 319 000 (1 + 0.05)m 𝐺𝑊 = $407 134

INFLATION

Inflation is a steady in increa ease e in in the p price o

• f g

goods a and s services in an economy over t time.

Inflation affects both the cost of living and the wages earned by individuals.

Example q question: : In 2010, the average cost of 1L of milk was $1.05. Accounting for an inflation rate of 6% p.a., determine the cost of 1L of milk in 2020.

Wh Which f formula d do w we u use t to c calculate inflati tion?

INFLATION

Let’s use the co compo pound d interest f formula. Therefore, the cost of 1L of milk in 2020 is $ $1.8 .88. 𝐺𝑊 = 𝑄𝑊(1 + 𝑠)* 𝐺𝑊 = 1.05 (1 + 0.06)Ln 𝐺𝑊 = $1.88

QUESTION TIME!!

Question 1 10 ( (Multiple C Choice)– 2017 H HSC E Exam:

QUESTION TIME!!

Question 1 17 ( (Multiple C Choice)– 2015 H HSC E Exam:

QUESTION TIME!!

Question 2 26 ( (d) – 2015 H HSC E Exam:

$369.17

DEPRECIATION

To calculate depreciation, we use the de decl clining-balance m method fo formula:

𝑻 = 𝑾𝟏 (𝟐 − 𝒔)𝒐

wh where:

• S is the salvage value of the asset after n periods
• V0 is the initial value of the asset
• r is the depreciation rate per period
• n is the number of period

This f formula i is a an a application o

• f t

the c compound i interest f formula!

DEPRECIATION

Example q question: : Rodney purchased a Hyundai i30 in 2014 for $19 000. The car depreciates at 16% per annum. Calculate t the v value o

• f t

the a asset a after 1 10 y years, c , correct t to t the n nearest do dollar ar. So Solution: 𝑇 = 19 000(1 − 0.16)Ln 𝑇𝑏𝑚𝑤𝑏𝑕𝑓 𝑤𝑏𝑚𝑣𝑓 = $𝟒𝟒𝟑𝟒

After 10 years, Rodney’s Hyundai is worth $3323, to the nearest dollar.

𝑇 = 𝑊

n (1 − 𝑠)*

SHARES AND DIVIDENDS

A share is when an individual/company owns p part o

• f a

another company. . Shares are bought and sold on the stock market, including the ASX (Australian Securities Exchange). To buy a and s sell s shares, , we use a broker. . A broker receives a a brokerage f fee for their work – a percentage of the transaction.

CALCULATING THE COST OF SHARES

To calculate the cost of shares, we must look at their or

• riginal

market v value and how m many are bought. Additionally, we must consider the brokerage f fees which apply. Example q question: : June purchased 100 shares at a market value of $12.64 each. She incurred brokerage fees at 4% of the price of the shares. Calculate the total cost of purchasing the shares.

So Solution: Cost of shares = $12.64 (market value) × 100 (quantity) = $126 $1264

Example q question: : June purchased 100 s shares at a market value of $12.6 .64 e

• each. She

incurred brokerage f fees a at 4 4% of the price of the shares. Calculate the total cost of purchasing the shares.

Brokerage fees = 4% × $1264 = $ $50.5 .56

Total c cost o

• f s

shares = = $ $1314.5 .56

DIVIDENDS

A shareholder is entitled to a share in the company’s profits – this i is k known a as a a di divi vide dend. d. To calculate the value of a dividend, we use the dividend y yield.

𝑒𝑗𝑤𝑗𝑒𝑓𝑜𝑒 𝑧𝑗𝑓𝑚𝑒 = 𝑏𝑜𝑜𝑣𝑏𝑚 𝑒𝑗𝑤𝑗𝑒𝑓𝑜𝑒 𝑛𝑏𝑠𝑙𝑓𝑢 𝑞𝑠𝑗𝑑𝑓 × 100% 𝑒𝑗𝑤𝑗𝑒𝑓𝑜𝑒 = 𝑒𝑗𝑤𝑗𝑒𝑓𝑜𝑒 𝑧𝑗𝑓𝑚𝑒 × 𝑛𝑏𝑠𝑙𝑓𝑢 𝑞𝑠𝑗𝑑𝑓

DIVIDENDS

Example q question: : The s share p price ce o

• f W

Wesfarmers i is cu currently $ $49.8 .84. a) The predicted dividend yield is 3.7%. What is the dividend?

(Correct to two decimal places)

Dividend = 3.7 .7% (dividend yield) × $49.8 .84 (market price) = $ $1.8 .84 b) Wesfarmers decide to pay a dividend of $4.98. What is the dividend yield? (Correct to two decimal places) Dividend yield = $S.wx

$Sw.xS × 100% = 9.9

.99%

HSC QUESTION

Question 2 26 (e (e)– 2017 H HSC E Exam:

Shares: 500 × $3.20 = $1 $1600 600 Brokerage fees: 1.5% × $1600 = $ $24

CO COST

Total co cost: $1 $162 624

IN INCOME ME Dividends: 500 × $0.26 = $1

$130 30 Selling shares: 500 × $4.80 = $2 $2400 400 Total i inco come: $2530 $2530

PR PROF OFIT

Income - Expenses $2530 - $1624 = = $906 $906

NOT COVERED TODAY… q Credit cards q Annuities These harder topic areas will be covered in next year’s Head Start and Trial Revision lectures J

Topic 2 2: Meas Measuremen ement

In Year 11, you would have learned how to:

ü convert between metric u units (length, capacity, area etc) ü calculate errors i in m measurement ü use significant f figures and scientific n notation ü calculate perimeter, a , area & & v volume ü use Pythagoras’ t ’ theorem & & s similar f figures ü use Trapezoidal r rule ü work with ti time (latitude, longitude)

Now in Year 12:

The two B BROAD t topics we cover include:

1.

• 1. No

Non-ri righ ght-angled T Trigonometry 2.

• 2. Rates a

and r ratios

You may already know:

(S (SOH)

Hyp ypotenuse Oppo Opposit site Ad Adjacent 𝜾 𝑡𝑗𝑜 𝜄 =

{||{}~•€

• ‚|{•€*ƒ}€

𝑑𝑝𝑡 𝜄 =

„…†„‡€*•

• ‚|{•€*ƒ}€

𝑢𝑏𝑜 𝜄 = {||{}~•€

„…†„‡€*•

Trigonometr tric r rati tios – only apply to right-angled triangles (CA (CAH) (T (TOA)

FINDING UNKNOWN SIDES

When finding unknown sides in ri right-angled t triangles, , the question will give you one a angle other than the right angle, and

• ne l

labelled s side to help you answer the question. Example q question: : Find t the l length o

• f t

the u unknown s side x in t the t triangle b

• below. A

. Answer correct t to t two d decimal p places. .

𝟑𝟘° 20cm x

FINDING UNKNOWN SIDES

So Solution: First, let’s label the sides of the triangle according to the trigonometric ratios listed before. Remember, , we label the sides according to the other a angle provided in the triangle (not the right- angle).

𝟑𝟘° 20cm x

H A We are given the adjacent and hypotenuse, therefore we will use the cosine r ratio.

FINDING UNKNOWN SIDES

So Solution co continued: d:

cos 𝜄 = 𝑏𝑒𝑘𝑏𝑑𝑓𝑜𝑢 ℎ𝑧𝑞𝑝𝑢𝑓𝑜𝑣𝑡𝑓

𝒚 = 𝟐𝟖. 𝟓𝟘𝒅𝒏 (𝟑 𝒆𝒇𝒅𝒋𝒏𝒃𝒎 𝒒𝒎𝒃𝒅𝒇𝒕)

Now substitute our known values into the formula. To complete the equation, we need to rearrange the formula so x is the subject. With x as the subject, simplify the equation.

angle provided in the question hypotenuse length

𝐝𝐩𝐭 𝟑𝟘° = 𝒚 𝟑𝟏

unknown length

1. 1.

𝐝𝐩𝐭 𝟑𝟘° = 𝒚 𝟑𝟏 ×

2.

𝒚 = 𝟑𝟏 × 𝐝𝐩𝐭 𝟑𝟘°

3. 3.

FINDING UNKNOWN ANGLES

When finding unknown angles in ri right-angled t triangles, , the question will give you an unk unkno nown n an angle and two s sides relevant to solving the question. Example q question: : Find t the s size o

• f t

the u unknown a angle, t , to t the n nearest d degree.

17 cm 24 cm 𝜾

FINDING UNKNOWN ANGLES

So Solution: Again, let’s label the triangle according to the position of the unknown angle. We are given the adjacent and opposite sides, therefore we will use the tangent r ratio.

17 cm 24 cm 𝜾

O A

FINDING UNKNOWN ANGLES

So Solution co continued: d:

𝑢𝑏𝑜 𝜄 = 𝑝𝑞𝑞𝑝𝑡𝑗𝑢𝑓 𝑏𝑒𝑘𝑏𝑑𝑓𝑜𝑢 Using our calculators, we can now solve this equation.

CALCULATOR STEPS:

ð SHIFT

𝑢𝑏𝑜 ta tan-1 will appear 1 2 4 =

ð ð ð

7

Wh What a t answer a appears o

• n y

your c calculator?

FINDING UNKNOWN ANGLES

𝜾 = 𝟒𝟔° 𝟐𝟗š𝟓𝟏. 𝟒𝟖"

𝑠𝑝𝑣𝑜𝑒𝑗𝑜𝑕 𝑢𝑝 𝑢ℎ𝑓 𝑜𝑓𝑏𝑠𝑓𝑡𝑢 𝑒𝑓𝑕𝑠𝑓𝑓:

𝜾 = 𝟒𝟔°

Hopefully you got a decimal answer of 35.3 .31121344. We must round our final answer to degrees.

°′ ′′

PROPERTIES OF NON-RIGHT-ANGLED TRIANGLES

A B C a b c

The s sides o

• f t

the t triangle a are n named a according t to the o

• pposite a

angle.

AREA OF NON-RIGHT-ANGLED TRIANGLES

𝑩 = 1 2 𝒃𝒄 𝐭𝐣𝐨 𝑫

The area of a non-right-angled triangle is equal to ha half the p product o

• f t

two s sides multiplied b by t the s sine o

• f t

the included a angle.

si side an angle si side

AREA OF NON-RIGHT-ANGLED TRIANGLES

Question 2 27– 2009 H HSC E Exam:

What i is t the a area o

• f t

this t triangle? (Correct to two decimal places) 2.3 .34km2

SOLUTION (previous slide)

SINE RULE

𝒃 𝐭𝐣𝐨 𝑩 = 𝒄 𝐭𝐣𝐨 𝑪 = 𝒅 𝐭𝐣𝐨 𝑫 The s sine r rule r relates th the s sides a and a angles i in a a n non- ri right ght-angled tr triangle.

To u use t the s sine r rule, , you must be given either: § two angles and one side or § two sides and an angle opposite one of the given sides

SINE RULE

The sine rule can be used to find both si sides es and an angles. To f find a a si side: 𝒃 𝐭𝐣𝐨 𝑩 = 𝒄 𝐭𝐣𝐨 𝑪 = 𝒅 𝐭𝐣𝐨 𝑫 To f find a an an angle:

𝐭𝐣𝐨 𝑩 𝒃

=

𝐭𝐣𝐨 𝑪 𝒄

=

𝐭𝐣𝐨 𝑫 𝒅

SINE RULE

Example q question – Finding a side Find t the v value o

• f x

x correct t to t the n nearest w whole n number.

x 12cm 43° 33°

Do w we h have t the c correct i information t to u use t the s sine r rule?

SINE RULE

𝑏 sin 𝐵 = 𝑐 sin 𝐶

Substitute the known values into the sine rule formula.

Solution: :

Make x the subject of the formula. Solve x to the nearest whole number.

𝑦 sin 43 = 12 sin 33

1. 1. 𝑦 sin 43 × sin 43 = 12 sin 33 × sin 43 𝑦 = 12 sin 43 sin 33 2. 𝒚 = 𝟐𝟔𝒅𝒏 3.

SINE RULE

Example q question – Finding an angle Find t the v value o

• f t

the m missing a angle, c , correct t to t the n nearest d degree.

39cm 42cm 𝜾 98°

Do w we h have t the c correct i information t to u use t the s sine r rule?

SINE RULE

Solution: : sin 𝐵 𝑏 = sin 𝐶 𝑐

Substitute the known values into the sine rule formula.

sin 𝜄 39 = sin 98 42

Make sin 𝜄 the subject of the formula.

sin 𝜄 = 39 × sin 98 42

Insert equation into calculator. 39 × sin(98) 42

𝟏. 𝟘𝟐𝟘𝟔𝟒𝟓𝟕𝟒𝟔𝟒

Use calculator to find value of sin 𝜄.

CALCULATOR STEPS:

ð

sin

ð

SHIFT

=

SINE RULE

Solution c continued: :

Convert to degrees and minutes.

°′ ′′

𝜾 = 𝟕𝟖°

Round to required unit (nearest degree).

𝜄 = 66° 51š29.32”

The c cosine r rule ca can be u used t to s solve p problems involving th three s sides a and o

• ne a

angle. .

COSINE RULE

To find the third s side given two sides and the included angle: 𝒅𝟑 = 𝒃𝟑 + 𝒄𝟑 − 𝟑𝒃𝒄 𝐝𝐩𝐭 𝑫 To find an unknown angle given three sides: 𝐝𝐩𝐭 𝑫 = 𝒃𝟑 + 𝒄𝟑 − 𝒅𝟑 𝟑𝒃𝒄

COSINE RULE

Question 2 26 – 2013 H HSC E Exam:

Which f formula a are w we u using h here? cos 𝐷 = 𝑏¯ + 𝑐¯ − 𝑑¯ 2𝑏𝑐

C c a b cos 𝐷 = 53¯ + 66¯ − 98¯ 2 × 53 × 66 cos 𝐷 = 110°

COSINE RULE

13cm 14cm 𝟒𝟑°

x

Example q question – Finding a side Find t the v value o

• f x

x correct t to t two d decimal p places. Try a and s solve t this q question a and w we’l ’ll w work t through t the s solutions to together!

SOLUTIONS (previous slide)

SINE RULE or COSINE RULE??

*Note: these questions are from the Cambridge General Maths textbook.

Cosine r rule Sine r rule

Could y you s solve t these o

• n y

your o

• wn?

NOT COVERED TODAY… q Bearings q Angles of elevation and depression q Radial surveys These harder topic areas will be covered in next year’s Head Start and Trial Revision lectures J

Topic 3 3: Ne Networks

NETWORK TERMINOLOGY

TERMIN INOLOGY DEFIN INIT ITIO ION ne network a set of objects/tasks connected together to create a sequence ve vertices the objects of a network (normally drawn as points) ed edges es connect the objects of a network together pa path a sequence of events through a network connecting the starting vertex to the finishing vertex directed n network when the edges in a network point in only one direction undirected n network when the edges in a network are bidirectional (or undirected) degree o

• f a

a v vertex the number of edges which protrude from each vertex weighted e edge when the connection/edge have weights (numbers) assigned to them

WHAT IS A NETWORK DIAGRAM?

A ne network diagram has several objects connected together to create a sequence.

degree of a vertex – as seen above, the vertex C has two edges protruding from it; therefore, its degree is 2 weighted edge - this connection has a value of two assigned to it edge – connecting points C and E together vertex –

• bject E

denoted by a point

2

8 3 4 2 9 7 A E B D C

2

1

2

1

This network is di directed d – each e edge o

• nly p

points i in one direction (i.e. the edge connecting A to B only travels from A to B).

DRAWING NETWORK DIAGRAMS

Let’s begin this topic by learning how to draw simple n network d diagrams from information given in a table. This activity will help us understand the need for specific sequences of tasks. In the table below, we have objects A, B, C and D and an explanation of the connections between them.

A B C D A

• 4

2

• B

4

• 5
• C

2 5

• 7

D

• 7

DRAWING NETWORK DIAGRAMS

Before drawing a network diagram from this table, we need to understand a few key points evident:

§ the same object (vertex) does not connect to each other i.e. object A does not connect to itself § the values in the boxes represent the weighted edge between those two vertices i.e. the edge between A and B is assigned a value of 4 § not all vertices connect to eachother i.e. vertex D does not connect to vertex B § the network is undirected – i.e. the edge connecting A to B is presented two times (A, B) (B, A) indicating the insignificance

• f the edge

DRAWING NETWORK DIAGRAMS

Now we can attempt to represent this information in a network diagram. There are, of course, many different ways to interpret and visualise this

• information. If this explanation doesn’t work for you, take an approach which

best suits your learning! If you haven’t already, quickly sketch down this information so we can draw the network.

A B C D A

• 4

2

• B

4

• 5
• C

2 5

• 7

D

• 7

DRAWING NETWORK DIAGRAMS

• 1. Begin by creating a

a v vertex f for o

• bject A. Note which vertices also connect to

A, and use edges to connect these vertices together. In this case, object B and C also connect to object A. Ensure you label the weighted edges of these connections!

A B C 4 2

DRAWING NETWORK DIAGRAMS

2. Continue to connect t the v vertices t together a according t to t the t table. For example, C connects to B with a weighted edge of 5, and C connects to D with a weighted edge of 7.

A B C D 5 4 1 2

DRAWING NETWORK DIAGRAMS

Remember, there is no one way to draw this diagram. Have a look at the network diagram below – it is representing the same information!

A B C 4 2 5 7 D

MORE TERMINOLOGY!!!

WALK: : A sequence of vertices and the edges between them. PATH: : A path is a walk that doesn’t ’t v visit a any v vertex m more t than

• n
• nce.

CYCLE: : A cycle is a walk with the same s start a and e end v vertex, which doesn’t ’t v visit a any v vertex m more t than o

• nce.

PUTTING TERMINOLOGY INTO PRACTICE

A B C D E F Id Identify t the f following a as a a w walk, p , path, c , cycle o

• r o
• ther,

, giving r reasons f for y your c choice: a) ABCDED b) DEFA c) ABCDEFA d) AD

WALK: : Follows the edges of the network. Why can this not be a path?? PATH: : Follows the edges of the network and doesn't visit any vertex more than once. CYCLE: : Follows the edges of the network and has the same start and end vertex. NEIT ITHER: : There is no edge connecting A and D.

THE SHORTEST PATH

The shortest path is a path between two v vertices in a network for which the sum of the weights o

• f i

f its e edges i is m minimised. Example Q Question: : Consider the following network.

E F A C D B H G 4 6 1 6 5 1 4 6 2

a) W What i is t the s shortest p path fr from A A t to B B? b) W What i is t the s shortest p path fr from F F t to G G? ACDB ( (11) FBHG ( (9)

WHY IS THE MINIMUM SPANNING TREE NOT ALWAYS THE SHORTEST PATH ?

The minimum spanning tree does not always provide the shortest path between two points. For example, the network diagram below shows the minimum spanning tree which has a path ABC, whereas the shortest path between vertices A and C is AC.

A B C 4 5 3 Minimum s spanning t tree: : AB ABC Shortest p path f from A A-C: : AC AC

MINIMUM SPANNING TREES

Us Using al algorit ithms A minimum spanning tree connects all t the v vertices of a network together without any cycles and with the mi minimu mum m possible total edge weight. When solving minimum spanning trees, there are two algorithms that can be used:

Prim’s ’s Al Algorith thm Prim’s algorithm suggests that to find the minimum spanning tree, we fi first select t the s shortest e edge, and then continue b by s selecting t the s shortest attached e

• edge. This continues until all

vertices are included, and helps avoid any possible loops. Kruskal’s ’s Al Algorith thm Kruskal’s algorithm suggests that to find the minimum spanning tree, we first s select t the s shortest e edge, and then select t the n next s shortest e edge, , regardless o

• f w

whether i it i is a attached t to the p previous e edge o

• r n
• not. We only add

connections that connect part of the network that was not previously joined.

USING PRIM’S ALGORITHM

Let’s ’s u use P Prim’s ’s A Algorithm f first t to f find t the m minimum s spanning t tree o

• f

f this n network:

Total: 9 : 9

USING KRUSKAL’S ALGORITHM

Now, l , let’s ’s u use K Kruskal’s ’s A Algorithm t to f find t the m minimum s spanning t tree o

• f t

f this

• network. N

. Note: It : It s should p produce t the e exact s same w weight!!

LET’S DO SOME PRACTICE QUESTIONS

LET’S DO SOME PRACTICE QUESTIONS

LET’S DO SOME PRACTICE QUESTIONS

SOLUTIONS (previous slide)

THANK NKYOU Y YEAR 1 12!

See y you n next y year J