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Answers and explanations to all the questions discussed for the exam

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Very clear document with all the types of question you can expect on your exam, with elaboration on how to calculate and etc.

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WEEK 1.......................................................................................................................................................... 1
Z(0,1) = 0.98, WHAT IS R2(0,1) AND WHAT IS R(0,1)?..................................................................................................2
WEEK 2.......................................................................................................................................................... 4
WEEK 3.......................................................................................................................................................... 7
WHICH CURRENCY IS TRADING AT A PREMIUM?..............................................................................................................7
WEEK 4.......................................................................................................................................................... 9
1. CALCULATE THE RISK-WEIGHTED ASSETS (RWA) OF THE BANK ACCORDING TO BASEL III.................................................11
FIND R..................................................................................................................................................................12
WEEK 5........................................................................................................................................................ 13




Week 1
Tutorial question 1:
A 1-year spot rate is at 1% and a 2-year spot rate is at 2% (all continuously compounded)
Calculate f (0,1.0,2 .0) ,i.e., a forward rate that applies to the period from 1 year to 2 years.
−1
Continuous compounding  Z=e−r ( 0 ,T 1 ) T1
, take logs  r = ln Z
T
( r ( 0,1 )=1%∧r ( 0,2 ) =2 % )

1

, −1 −1 Z ( 0,2 )
f ( 0,1.0,2 .0 )= ln F ( 0,1.0,2.0 )= ln
Δ Δ Z ( 0,1 )
−r ( 0,2) ∙2
1 e
¿− ln
Δ e−r ( 0,1) ∙1
1
¿− ¿
Δ
1
¿− (−r ( 0,2 ) ∙2−−r ( 0,1 ) ∙ 1)
Δ
−1
f ( 0,1.0,2 .0 )= (−0.02∙ 2−−0.01∙ 1 )=3 %
1
2
f ( 0,1.0,2 .0 )=r ( 0,1 )+ ( r ( 0,2 )−r ( 0,1 ) ) (other way of writing)
1
¿ 1 %+2 ( 2 %−1 % ) =3 %

Other way of calculating:
Z ( 0,1 ) =e−0.01∙ 1=0.9 9
−0.02∙ 2
Z ( 0,2 ) =e =0.96 1
Z ( 0,2 ) −f ( 0,1,2 )( T −T )
=e 2 1


Z ( 0,1 )
0.961 −f (0,1,2)(T −T ) 0.961
0.99
=e 2

0.99
1

(
=0.97 )
0.97=e−f (0,1,2)(T −T ) 2 1



ln 0.97=−f ( 0,1,2 ) (2−1)
f ( 0,1,2 )=3 %

Question from the slides
Z(0,1) = 0.98, what is r2(0,1) and what is r(0,1)?
−1
( 2 ∙1
r 2 ( 0,1 )=2 ∙ ( 0.98 ) −1 =0.0203=203b . p . ) (semi-annual)
Z ( 0,1 ) =e−r (0,1 ) ∙1 (continuous)
ln 0.98=¿−r ( 0,1 ) ∙ 1¿
r ( 0,1 )=−0.202=202 b . p .




Sample exam – Open question 1:
Continuously compounded 3M forward rates

2

,  f(0,0.25,0.5) = 1.24%,
 f(0,0.5,0.75) = 1.65%
 f(0,0.75,1.0) = 1.65%,
 r(0, 0.25) = 0.5% (The spot 3M rate)

Calculate a 1-year swap rate (assuming quarterly payments in a plain-vanilla fixed-for-floating swap).
Express the result in % per annum, round to the basis point.

−r 0 ,T i T i
Z ( 0 , T i )=e ( ) (general formula continuous compounding)
Z ( 0,0.25 ) =e−0.005 ∙0.025=0.99875
Z (0,0.50) −0.0124 ∙0.25
=e =0.9969
Z (0,0.25)
Z (0,0.75) −0.0165 ∙0.25
=e =0.99588
Z (0,0.50)
Z (0,1) −0.0165 ∙0.25
=e =0.99588
Z (0,0.75)
Z ( 0,0.50 )
=0.9969
Z ( 0.25 )
Z ( 0,0.50 )
=0.9969 → Z ( 0,0.50 )=0.99565
0.99875
Z ( 0,0.75 )
=0.99588 → Z ( 0,0.75 )=0.9915
0.99565
Z ( 0,1 )
=0.99588 → Z ( 0,1 )=0.9874
0.9915

1-year swap rate:
1−Z (0 , T )
c n ( 0 ,T )=n∙ m

∑ Z ( 0 ,T i)
i=1
1−0.9874
c 4 ( 0,1 ) =4 ∙ =0.0126=1.26 %
0.99875+ 0.99565+0.9915+0.9874

Multiple choice question 2, multiple choice question 3
2. A two-year discount factor is given by Z(0,2) = 0.8742. What is the interest rate r12(0,2) in decimals?
 if it were asked in percentages:
r i ( 0 , T i ) =n ∙ ¿
−1
r 12 ( 0,2 )=12∙ ( ( 0.8742 ) ¿ ¿ −1)=0.0674=6.74 % ¿
12 ∙2

3. You observe r(0,0.25) = 1.2% and f(0,0.25,0.5) = 3.0%. What is r4(0,0.5) in percent?
−0.012∙ 0.25
Z ( 0,0.25 ) =e =0.997
Z (0,0.5) −0.03 ∙0.25
=e =0.9925
Z (0,0.25)
Z ( 0,0.5 )
=0.9925 Z ( 0,0.5 )=0.9895
0.997

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