Marjorie Arevalo.pptx

Math Investment • • • • • • • • •

Simple Interest Compound Interest Simple Annuities Straight line Method Sinking fund Method Declining balancing Method Sum of year Digit (SYD) Method Break-even Capitalized cost and (Annual cost)

Simple Interest • 1-1 Definition of terms In a business transactions, Interest may be defined in two different ways 3. To Investor; Interest is an income derived from invested capital 4. To Debtor; Interest is money paid as rental for the use of money In short To an Investor, Interest is an incomes. To a debtor, Interest is an expense Interest is a fixed rated proportion as the rate of interest for any specified time unit. The ratio of the Interest earned in one time unit to the principal is called Interest rate (r). In the other words, r is the measure of the interest on one peso for time to time.

Principal Refers to the capital originally invested in a business transaction represented by letter P. Amount is referred to the full amount F which is the sum of the principal and the invest due at any time after the investment of the principal other word such as sum is also used instead of amount if it is met colloquially

Where in: P= Principal or the Capital R= Rate or percentage T= Time I= Interest

Example 1: Find the ordinary interest and the full amount on 3,000 pesos at 5% interest for two years . where in the formula of simple interest is I=Prt and the full amount is I=Prt I=3000(.05)(2) I=300 F=P+I = 3000 + 300 = 3,300

Example 2: Find the value of the principal P and the full amount F if the the Investment earns 200 in 18 months t at the rate of 8% P=I/rt

F=P+I

P=200/(0.8)(18/12)

F= 1,666.67 + 200

P=1,666.67

F= 1,866.67

Simple Annuities 1-1 Definition of the Terms

By definition annuity is a sequence of equal periodic payments classified inter two. 1. Annuity Certain 2. Contingent Annuity Annuity Certain – Refers to payment extended ever a fixed term of years the term of payment is fixed and known. One example is an equal payment for salary or housing load where the installment plan forms an annuity certain.

Contingent Annuity – refers to payments over a period of time whose length cannot be foretold accurately. One example is the equal sums paid as premium on life insurance policy where payments end at the death of the insured person. But as to when he is going to die is not known.

Depreciation Is the reduction of fall in the value of an asset or physical property during the course of its working life and due to the passage of the time. Types of Depreciation Physical Depreciation - is due to the reduction of the physical ability of an equipment or asset to produce results.

Functional Depreciation – is due to the reduction in the demand for the function that the equipment or asset was designed to render. This type of depreciation is often called Obsolescence.

Straight line method In this method of Computing depreciation, it is assumed that the loss in value is directly proportional to the age of the equipment or asset. Annual Depreciation charge, d d= Co – Cn n Where : Co = First cost Cn = Cost after “n” years

n = life of the property

Book value of the emd of “m” years of using, Cm Cm = Co – Dm Where : Dm – total depreciation after “m” years Dm = d(m) Example : A unit of welding machine cost 45,000 php. with estimated life of 5 yrs. Its salvage solve is 25,000 php. Find its annual depreciation charge d, total depreciation after 3 yrs. Dm, and Book value BV after 4 years.

Co = 45,000php

n = 5 yrs.

Cn = 25,000php

D3 = 3 yrs.

Dm = 4 yrs.

a.) d = Co - Cn n b.)

45,000 - 25,000 5

= 8,500php/ years

Dsub3 = 8,500/years (3 years) = 25, 500php c.)

Dm = 8,500/years (4 years) = 34,000php BV = Co - Dm BV = 45,000 – 34,000

BV = 11,000php Second hand price

Sinking fund Method In this method of Computing depreciation it is assumed that a sinking fund is established in which fund will accumulate for replacement purposes. Annual depreciation charge, d D =(Co-Cu) (i) n (1-i) -1

Where : Co = First cost Cn = Cost after “n” years n = life of the property

Book value at the end of “m” years of using, Cm Cm = Co - Dm Where Dm = total depreciation after “m” years n Dm = d[(1+i) - 1] i

Example : An equipment costs 10,000 php with a salvage value of 500 php at the end of 10 years. Calculate the depreciation cost by sinking fund method at 4% interest. Co = 10,000php Cn = 500php i = 0.04 n = 10 yrs. (Co – Cn)(i) d= n (1 – i) -1

(10,000 - 500)(0.04) d= (1 – 0.04 10 - 1 ) d = 791.26php/yr.

n D[(1 – i) - 1] Dm = i a.) 2nd year 2 (791.76)[(1+0.04) - 1] Dm = 0.04 Dm = 1614.17 b.) 3rd year 3 (791.76)[(1+0.04) - 1] Dm = 0.04 Dm = 2470.00

Declining Balancing Method In this method of computing depreciation it is assumed that the annual cost of depreciation is fixed percentage of the Book value at the beginning of the year, This method sometimes known as constant percentage method of the mathesm on formula. Matheson formula : K=I–

n

Cn Co

Cm or K = I – n Co

The Value of K is the Constant percentage, hence K must be Decimal and a value less than 1. In this method the salvage or scrap value must not be zero.

Example : A machine costing 720,000php is estimated to have a book value of 40,545.73php when retired at the end of 10 yrs. Depreciation cost is computed using a constant percentage of the declining value. What is the annual rate of depreciation in %? Cn = 40,545.73php

n Cn K=1Co 10

K=1-

40,545.73 720,000

Co = 720,000php

n = 10 yrs.

BV = Co (1 –k)

n

40,545.73 = 720,000 (1 –k) K = 0.25 or 25 %

K = 0.25 or 25 %

10

Sum of the Years Digit (SYD) Method Respective depreciation charges. First year,

n year

D1 = (Co – Cn)

Second year,

D2 = (Co – Cn)

n year

n year Book value at the end of “m” years of using, Cm Third year,

D3 = (Co – Cn)

Cm = Co – (D1+ D2 +…….+ Dm) Sum of year digit

year

Year = n(n+1) 2

Example : A company purchases an asset for 10,000php and plans to keep it for 20 yrs. If the salvage value is zero at the end of 20 yrs. What is the Book Value at the 3rd year? n Co = 10,000php Cn = 0 n = 20 yrs. Dm = (Co – Cn) Year D1 = (10,000 – 0)(20/201) = 952.38 n(n+1) Year = 2 D = (10,000 – 0)(20-1/201)= 904.76 2 20(20+1) D3= (10,000 – 0)(20-2/201) = 857.14 Year = = 210 2 Dm = 2,714.28 BV = Co – Dm BV = 10,000 – 2,714.28 BV = 7285.72

Break – Even Refers to the situation where the sales generated (income) is just enough to cover the fixed and the variable cost (expenses) the level of production where the total income is equal to the total expenses is known as break-even point. Break-even Chart – Is a diagram which showns relationship between volume and fixed cost, variable cost, and income. The ff. is an example of even Chart

Example : The annual maintenance cost of a machine shop is 69,994php. If the cost of making a forging is 56php per units and its selling price is 135php per forced unit. Find the number of units to be forged to break-even. Expenses = Income

Prove!

69,994 + 56x = 135x 69,994 = 135x – 56x 69,994 = 79x x = 886

69,994 + 56(886) = 135(886) 119,610 = 119,610

Capitalized cost and Annual cost Capitalized cost - any structure or property is the sum of its first cost and the present worth of all costs for replacement, operation and maintenance for a long time or forever.

Annual cost – any structure or property is the sum of the annual depreciation cost

Example of Capitalized cost and annual cost : At 6% find the capitalized cost of a bridge 10 base cost is 250 million pess and life is 20 yrs. It the bridge must be practically rebuilt at a cost of 150 million pesos at the end of each 20 yrs.? Co = 250 million pesos Cn = 150 million pesos n = 20 years i = 0.06 Cc = Co + (Co-Cn)/(1+i)n -1 Cc = 250 + (250 – 150)/(1+0.06)20 -1 = 295.31 million pesos Ca = Co(i) + [(Co-Cn)(i)]/[(1+i)n -1] +A Ca = 250(0.06) + [(250 – 150)(0.06)]/[(1+0.06)20 -1] + 0 Ca = 17.72 million pesos