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Assignment 1 ADM 3351 By Yonathan Yirga 300148280 October 3, 2023
Problem 1 a)
(1+Spot rate (1)/2 )^2 = (1+r(0.5)/2) *(1+ r(1)/2) = 1.025 * 1.03 -1 = 1.05575 1+ Spot rate(1)/2 = sqrt(1.05575) = 1.027497 Spot rate (1) = 5.4994% b)
((1+R(0.5)/2) * (1+R(1)/2) * (1+R(1.5)/2) ) = (1+ spot(1.5)/2)^3 1.025 * 1.02749 * (1+R(1.5)/2) = 1.0265^3 (1+R(1.5)/2) = 1.0816/1.0532 = 1.02699 R(1.5) = 5.3979% c)
d(0.5)= (1+spot(0.5)/2)^-1 = (1.025)^-1 = 0.9756 d(1)= (1+spot(1)/2)^-2 = (1.027497)^-2 = 0.94719 d(1.5)= (1+spot(1.5)/2)^-3 = (1.0265)^-3 = 0.92453 d(2)= (1+spot(2)/2)^-4 = (1.0335)^-4 = 0.87651 Price = 5*d(0.5) + 5*d(1) + 5*d(1.5) + 105*d(2) Price = 5*0.9756 + 5*0.94719 + 5*0.92453 + 105*0.87651 Price = 4.878 + 4.7359 + 4.62265 + 92.03355 = $106.27 Price = $106.27 d)
((1+R(0.5)/2) * (1+R(1)/2) * (1+R(1.5)/2) * (1+R(2)/2 ) = (1+ spot(2)/2)^4 1.025 * 1.03 * 1.02699 * (1+R(2)/2 ) = 1.0335^4 1.084245 * (1+R(2)/2 ) =1.14088 (1+R(2)/2 ) = 1.052235 R(2) = 10.4469% We will use the zero-coupon bond. Price = FV/[(1+R(0.5)/2)+ (1+R(1)/2)+ (1+R(1.5)/2)+ (1+R(2)/2)+ (1+R(2.5)/2) ] 85 = 100/[1.025*1.02749*1.02699*1.052235*(1+R(2.5)/2)] (1+R(2.5)/2)=1.0337146 R(2.5)=6.7429% Problem 2 a)
Lets find d(2): d(2)=(1+spot(2)/2)^-4 = 1.041^-4 = 0.851524 Faire price(bond C) = FV*d(2) = $85.1524
b)
We solve for this equation (where x is number of bond A and y number of bond B: 102.5x + 103.5y = 100 2.5x + 3.5y = 0 We find x=3.5 and y=-2.5 Buy Number of Bonds: 3.5
Of Bond A Sell Number of Bonds: 2.5
Of Bond B Number of Bonds: Of Bond C Problem 3 a)
d(0.5) = (1+spot(0.5)/2)^-1 = (1.015)^-1 = 0.990099 d(1) = (1+spot(1)/2)^-2 = (1.02)^-2 = 0.961169 d(1.5) = (1+spot(1.5)/2)^-3 = 0.928599 Price (bond A) = 102.5*d(0.5) = 101.4851 Price (bond B) = 3.5*d(0.5) + 103.5*d(1) = 102.9463 Price (bond c) = 5*d(0.5) + 5*d(1) + 105*d(1.5) = 107.26369 b)
Price (bond A) = 102.5
(1+
?
2
)^1
101.4851 = 102.5
(1+
?
2
)^1
YTM(A)= 2% Price (bond B) = 3.5
(1+
?
2
)^1
+
103.5
(1+
?
2
)^2
102.9463 = 3.5
(1+
?
2
)^1
+
103.5
(1+
?
2
)^2
YTM(B)=3.965% Price (bond c) = 5
(1+
?
2
)^1
+
5
(1+
?
2
)^2
+
105
(1+
?
2
)^3
107.2637 = 5
(1+
?
2
)^1
+
5
(1+
?
2
)^2
+
105
(1+
?
2
)^3
YTM(C) = 4.9175%
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c)
Price (bond B) = 103.5*d(1) = 99.4809
on July 31, 2023 d)
Price (bond c) = 5*d(1) + 105*d(1.5) =
102.3087
on July 31, 2023 e)
Price (bond c) = 105*d(1.5) = 97.5029 on Jan 31, 2024 Problem 4 a)
Price (bond A) = ?
?
[1 − (
1
1+
?
2
)
2?
] +
?
(1+
?
2
)^2?
= 2.5[1-(1/(1+0.035))^12] + 100/[1+0.035]^12 = $946.713 Price (bond B) = $973.3506 Price (bond c) = $1000 Price (bond D)
= $1000 Price (bond E) = $1041.5832 Price (bond F) = $1083.16624
b)
If YTM>coupon rate, then the price of the bond is less than the face value. If YTM<coupon rate, then the price of the bond is greater than the face value. If YTM=coupon rate, then the price of the bond is the same as the face value. c)
As a bond comes to maturity, its price start to converge to its face value. If other things stay constant, the YTM of a bond will stay constant over time.
Problem 5 a)
Let x be the number of bonds A, y be the number of bonds B, and z be the bomber of bonds C: Z*105=103 Y*102+5z=3 101X+2y+5z=3 Therefore: z=0.981 , y=-0.0186 , x=-0.01849
So a portfolio composed 0.981 of bond C, short 0.0186 bond B, and short 1.867 bond A will replicate bond D. b)
The price of the new portfolio will be: -0.01849*103 –
0.0186*104.5 + 0.981*109.3 = 103.3748 Since the replica portfolio
’
s price is more than the price of Bond D, there is an arbitrage opportunity by shorting the replica portfolio and buying bond D. c)
buy Number of Bonds: 0.0184
Of Bond A buy Number of Bonds: 0.0186
Of Bond B sell Number of Bonds: 0.981
Of Bond C
Statement of Academic Integrity Individual Assignment Checklist & Disclosure Please read the disclosure below following the completion of your assignment. Once you have verified these points, hand in this signed disclosure with your assignment. 1. I acknowledge to have read and understood my responsibility for maintaining academic integrity, as defined by the University of Ottawa
’
s policies and regulations. Furthermore, I understand that any violation of academic integrity may result in strict disciplinary action as outlined in the regulations. 2. If applicable, I have referenced and/or footnoted all ideas, words, or other intellectual property from other sources used in completing this assignment. 3. A proper bibliography is included, which includes acknowledgement of all sources used to complete this assignment. 4. This is the first time that I have submitted this assignment or essay (either partially or entirely) for academic evaluation. 5. I have not utilized unauthorized assistance or aids including but not limited to outsourcing assignment solutions, and unethical use of online services such as artificial intelligence tools and course-sharing websites. Course code Adm 3351 Assignment # 1 Use of Plagiarism Detection Tools No Date of submission October 4, 2023 Name Yonathan Yirga Signature Y.Y.
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Question 8
If
Listen
==
1
tane√3 and 0 is in Quadrant II, what are sine and cose?
A
sine=
==
2
cose = 1/1/1
B
=
sine, cose=
√3
-
2
C
√√√3
sine=
cose = -
2
D
sine = -
12
√√3
cose =
2
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Question 8
Listen
1
tane = -
If
A
B
√3 and is in Quadrant II, what are sine and cose?
sine = - √, cose=
2'
ine = ½, cose =
=
12
√3
2
C
√3
sine =
2
"
cose=
==
12
D
√√3
sine = -, cose=
2
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Complete requirement 1
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