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PROFESSOR AHMADI
PORTFOLIO THEORY
E X E R C I SE SE T
R E V I E W O F BA S I C
C O M PU TE R WO R K
CALIFORNIA STATE UNIVERSITY, SACRAMENTO
College of Business Administration
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Write program in Excel to solve the following problems.
Use Excel built-in functions to solve the following problems
1. For the 10-year period, the holding period yields of stock (X) and the Stock Market are provided
below:
Year
Stock X
1
Stock Market
20.2
3.5
15.8
34.6
51.5
7.3
30.6
37.7
26.9
33.5
2
3
4
5
6
7
8
9
10
23.6
-7.2
6.4
18.2
31.5
14.8
20.4
22.5
26.0
27.9
a. Estimate the beta.
b. Write the equation of the Characteristic Line.
2. Suppose that you observe the returns to Stock J and to the market portfolio over 5 months and
that you see this:
Month
1
2
3
4
5
Stock J
Market Portfolio
2
3
6
-4
8
4
-2
8
-4
4
a. Find the beta for Stock J
b. Write the equation of the Characteristic Line
c. Find the variance of the estimated line (residual variance).
4. The following is the rate of return on the market, and the rate of return on the IBM
Find the “beta” of IBM and the standard deviation of IBM’s beta.
Sum:
Markets
IBM
11%
9
10
12
42
10%
10
12
14
46
stock.
a. Practice Present Value with “Built-in Functions” in Excel
b. Find the corresponding “Financial Equation” for each “Function”
=PV(rate, nper, pmt, fv, type)
rate= interest rate per period; nper= total number of payment periods in an annuity
pmt= payment made each period; fv= present value
Type = 0 if payments are due the beginning of each period (can omit the 0 or enter) or 1 if payment is
due at the end of each period
Example 1: If you wanted to start a savings account that earns 6% annual interest, compounded
annually. How much would you have to deposit today (in a lump sum ), to have $5,000 in 5 years?
=PV(0.06, 5, 0, 5000)
($3,736.29)
If it is compounded semiannually the syntax will be:
=PV(0.06/2, 10, 0, 5000)
($3,720.47)
Example 2: (Present Value Annuity) How much is a bond that yields 8%, pays $90 for the next 14
years and $1,000 at the end of 14 years?
=PV(0.08, 14, 90, , 0)+PV(0.08, 14, , 1000)
($1,082.44)
TRY FUTURE VALUE FUNCTIONS AS WELL:
E.g.
. =FV(0.1,3,100)
=PMT(rate,nper,pv,fv,type)
rate=interest rate for the loan; nper=the total number of payments for the loan
pv=present value; fv=future value
type=0 indicates payments are due at the end of the period and 1 indicates payments are due at the
beginning of the period
Example 3: You take out a loan for $100,000 to buy a house. Mortgage rate is 12% a year, 30 years
fixed. What would monthly payments be?
=PMT(0.12/12, 30*12, 100000)
($1,028.61)
PPMT: Returns the payment on the principal for a given period
"=PPMT(rate, per, nper, pv, fv, type)"
rate=interest rate per period; per=specifies the period and must be in the range 1 to nper
nper= number of payment periods; pv=the present value; fv=the future value
type=0 if payment is at the end of each period(you can omit the 0 or enter it) or 1 if payment is at the
beginning of each period
Example 4: Use the PMT example above and find out the principal payment for the 1st period
"=PPMT(0.12/12, 1, 30*12, 100000)"
($28.61)
IPMT: Returns the interest payment for a given period for an investment
IPMT(rate, per, nper, pv, fv, type)
Example 5: Use the PMT example above and find the interest payment for the 1st period
"=IPMT(0.12/12, 1, 30*21, 100000)"
($1,000.00)
RATE
=RATE(nper, pmt, pv, fv, type, guess)
nper=the total number of pyment periods; pmt=the payment made each period
pv=the present value; fv=the future value
type=0 if payments are made at the end of each period or 1 if payments are made at the beginning of
each period.
Guess = is your guess for what the rate will be. If omitted guess will be assumed to be 10%
Example 6: You put $3,500 in a savings that is compounded semiannually and in 20 years you have
$11,417.13. What is the interest rate on that savings?
"=RATE(20*2, 0, -3500, 11417.13)"
3%
The annual interest rate will then be 3% X 2=6% since this account is compounded twice a year.
MIRR: Returns the Modified Internal Rate of Return for a series of periodic cash flows.
=MIRR(values, finance rate, reinvest rate)
values =an array of cells that contains the series of payments and incomes
finance rate =the interest rate you pay on the money used in the cash flows
reinvest rate =the interest rate you receive on the cash flows as you reinvest them
Example 7: Six months ago you opened a small hot dog stand by borrowing $8,000 at 10%. Each
month you manage to bring the following figures in profit and reinvest them in an account that earns
12% interest. What is your MIRR?
A
B
248
249
250
251
252
253
254
255
Month
0
1
2
3
4
5
6
Profit
-$8,000.00
$2,500.00
$3,500.00
$3,600.00
$4,200.00
$4,400.00
$5,000.00
=MIRR(B249:B255, 0.10, 0.12)
25%
IRR: Returns the internal rate of return for a series of cash flows
=IRR(values, guess)
values = an array of cells that contains the cash flows
guess = is your guess of what the IRR is (most of the time it can be omitted)
Example 8: Use the example for MIRR to find the IRR after the 4th month
"=IRR(B249:B253)"
24%
NPER: Returns the number of periods for an investment
=NPER(rate, pmt, pv, fv, type)
rate=the interest rate per period; pmt=the payment made each period
pv=the present value; fv=the future value
type=0 if payments are made at the end of each period or 1 if payments are made at the beginning of
each period
Example 9: If you decide to put $1,000 a month in an account that earns 12% interest a year, how
many periods will it take to earn $81,670?
=NPER(0.12/12, -1000, , 81670)
60.00018263
It will take approximately 60 periods or 60/12=5 years
NPV: Returns the net present value of an investment by using a discount rate and a series of
future payments and income.
"=NPV(rate, value1, value 2, ...)
rate=the rate of discount over the length of one period;
value 1, value 2, =represents payments and incomes
Example: Using the same example used for MIRR find the NPV
=NPV(0.10/12, 2500, 3500, 3600, 4200, 4400, 5000)+(-8000)
$14,474.37
DB: Returns the depreciation of an asset for a specified period
"=DB(cost, salvage, life, period, month)
cost=initial cost of the asset; salvage=value at the end of the depreciation
life= the number of periods over which the asset is being depreciated
period=period for which you want to calculate the depreciation
month=number of months in the first year
Example 10: A company buys a dump truck at $130,000 and in 10 years it is salvaged at $20,000. Show
the depreciation for the first 5 years
"=DB(130000, 20000, 10, 1, 12)" $22,230.00
"=DB(130000, 20000, 10, 2, 12)" $18,428.67
"=DB(130000, 20000, 10, 3, 12)" $15,277.37
"=DB(130000, 20000, 10, 4, 12)" $12,664.94
"=DB(130000, 20000, 10, 5, 12)" $10,499.23
DDB: Returns the depreciation of an asset for a specified period using the double-declining
balance method
=DDB(cost, salvage, life, period, factor)
cost=the initial cost of the asset
salvage=the value at the end of the depreciation
life=the number of periods over which the asset is being depreciated
period=the period for which you want to calculate the depreciation
factor=the rate at which the balance declines. 2 is for the double declining balance method
Example 11: Using the same example as the one used for DB, find the depreciation for the 1st month
and the first year
=DDB(130000, 20000, 120, 1, 2)
=DDB(130000, 20000, 10, 1, 2)"
$2,166.67
$26,000.00
SLN: Returns the straight line depreciation of an asset
=SLN(cost, salvage, life)
cost=the initial cost of an asset; salvage=the value at the end of the depreciation
life=the number of periods for which the asset is being depreciated
Example 12: Again using the example used in DB, find the depreciation allowance for each year.
=SLN(130000, 20000, 10)
$11,000.00
Find the depreciation allowance for each month
=SLN(130000, 20000, 120)
$916.67
POWER: Returns the result of a number raised to a power
=POWER(number, power)
Example 13: Find 1.08 raised to the power of 60
=POWER(1.08, 60)
101.2570637
PRODUCT: Multiplies numbers given
=PRODUCT(number 1, number 2, ...)
number 1, number 2=numbers you want to multiply
Example 14: Multiply the following numbers:
20, 30, 33, 44, 21
=PRODUCT(20, 30, 33, 44, 21) or =PRODUCT(A392:A396)
18295200
SQRT: Returns a positive square root
=SQRT(number)
or
=SQRT(A23)
Example 15: The standard deviation is equal to the square root of a variance. If the variance is 55 find
the standard variation using the SQRT function
=SQRT(55)
7.416198487
Average
"=Average(Number 1, Number 2....)
Example: Find the average of the following AT&T stock prices
49
51
58
49
56
"=Average(49,51,58,49,56)" or
"=Average(B55:B59)"
52.6
Note: Recognize that there is a difference between leaving a cell blank and entering a "0". If a cell is
left blank, it will be ignored. If a 0 is entered, it will be counted as a zero value.
Variance
=Var(Number 1,Number 2...)
Example: Find the variance to the AT&T stock prices listed above
"=Var(49,51,58,49,36)"
17.3
*Covariance
=Covar(array1, array2)
Example: Find the Covariance of IBM and AT&T using their stock prices listed below
IBM AT&T
100
49
95
51
98
58
113
49
115
56
"=Covar( {100,95,98,113,115}, {49,51,58,49,56} )" or
"=Covar(b82:b86, c82:c86)"
0.28
Standard Deviation
"=STDEV(number1,number2,....)"
Example: Use the numbers for AT&T above and find the standard deviation
"=STDEV(49,51,58,49,56)"or
"=STDEV(C82:C86)"
4.159326869
MMULT
=MMULT(Array 1,Array2)
Note: The number of columns in array 1 must be the same as the number of rows in array 2
Example: Multiply the two matrices below
B
C
D
E
F
108 0.1
-0.1
0.0
2 0
109 -0.1
0.4
0.3
0 2
110
0.0
0.3
0.7
0
G H
0
0
0 2
=MMULT(B108:D110,F108:H110)
and instead of pressing return press CTRL+SHIFT+RETURN
0.20
-0.2
0
-0.2
0.8
0.6
0
0.6
1.4
Make sure you highlight the complete array where the answers will show up. The result is an array with
the same number of rows as array 1 and the same number of columns as array 2
MINVERSE
=MINVERSE(Array)"
Example: Find the inverse Matrix of the matrix below:
0.2
-0.2
0
1
0.05
-0.2
0.8
0.6
1
0.07
0
0.6
1.4
1
0.3
1
1
1
0
0
0.05
0.07
0.3
0
0
Make sure you higlight a complete array with the same number of columns and rows when entering the
function
"=MINVERSE(B123:F127)" CTRL+SHIFT+RETURN
0.676990018
-0.735858715
0.058868697
0.789096493
-1.433324802
-0.735858715
0.799846429
-0.063987714
0.446634246
-2.789864346
0.058868697
-0.063987714
0.005119017
-0.23573074
4.223189148
0.789096493
0.446634246
-0.23573074
-0.094599437
0.522139749
-1.433324802
-2.789864346
4.223189148
0.522139749
-15.86895316
TRANSPOSE
This function changes rows into columns and the reverse
=Transpose(array)
then press CTRL+SHIFT+RETURN
Example: Change the following column of numbers to a row of numbers
Make sure you highlight the complete array when entering functions
"=Transpose(B145:B149)" then press CTRL+SHIFT+RETURN
1
2
3
4
5
1
2
3
4
5
Logarithms
=Ln(number)
Example: You start with $100, the rate is 8% and at the end of so many periods we get $158.69. How
many periods did it take to get $158.69?
$100(1+8%)to the power of n=$158.69
100(1.08)to the power of n=158.69
(1.08) to the power of n=1.5869
log(1.08) to the power of n=log (1.5869)
n log (1.08)=log (1.5869)
n=log(1.5869)/log(1.08)
"=ln(1.5869)/ln(1.08)"
6.000210246
It took 6 periods to get $158.69