CFA Level I Fixed Income formulas
Bond pricing, yield measures, duration and convexity, spreads, credit losses and securitization. After statements, this is the topic with the most to calculate.
167 formulas from the Fixed Income lessons (11–14% of the exam), 87 to memorize. Each says how deeply to learn it and opens to a quick worked example with the keystrokes for the TI BA II Plus and the HP 12C.
How deeply to learn each part
- Know the idea
- Understand it in general. It frames other topics, but you will not calculate with it.
- Use it
- Understand it well enough to solve a question with it. Recognize when it applies.
- Memorize
- Learn it by heart. The exam will not give it to you.
Module 1 Fixed-Income Instrument Features
The features of a bond
Quick example
NOK 300,000 par, 4.4% coupon, paid quarterly (m = 4). 300,000 × 0.044 ÷ 4 = NOK 3,300 a quarter.
- 300000×0.044÷4= 3300
- 300000ENTER0.044×4÷ 3300
ENTER enter (on the HP, puts the number on the stack)
Quick example
EUR 80m FRN, MRR 2.10% plus 95 bps, semiannual. (0.0210 + 0.0095) × 80m ÷ 2 = EUR 1,220,000.
- 0.021+0.0095×80000000÷2= 1220000
- 0.021ENTER0.0095+80000000×2÷ 1220000
ENTER enter (on the HP, puts the number on the stack)
Current yield, yield to maturity and the yield curve
Quick example
A 7% annual coupon bond trades at 108.50 per 100 par. 7 ÷ 108.50 = 6.45%.
- 7÷108.5= 0.0645
- 7ENTER108.5÷ 0.0645
ENTER enter (on the HP, puts the number on the stack)
Quick example
5 years, 4% annual coupon, price 96.00 per 100 par. Solve for r: 4.92%.
- 2NDCLR TVM with P/Y set to 1
- 5N96+|-PV4PMT100FV
- CPTI/Y 4.92
2ND second function (the shifted label above a key) · CLR TVM clear the time value of money keys · N number of periods · +|- change sign · PV present value · PMT payment · FV future value · CPT compute · I/Y interest rate per year (per period when P/Y is 1)
- fCLEAR FIN
- 5n96CHSPV4PMT100FV
- i 4.92
f gold shift key · CLEAR FIN clear the financial registers · n number of periods · CHS change sign · PV present value · PMT payment · FV future value · i interest rate per period
Quick example
Issuer 10-year yield 5.80%, government 10-year yield 3.95%. 5.80 − 3.95 = 1.85 points = 185 bps.
- 5.8−3.95=×100= 185
- 5.8ENTER3.95−100× 185
ENTER enter (on the HP, puts the number on the stack)
Bond indentures and covenants
Quick example
Cash 250m. Secured claims 100m, senior unsecured 200m. (250 − 100) ÷ 200 = 75% for senior unsecured.
- 250−100=÷200= 0.75
- 250ENTER100−200÷ 0.75
ENTER enter (on the HP, puts the number on the stack)
Module 2 Fixed-Income Cash Flows and Types
Bullet, amortizing and sinking fund bonds
Quick example
GBP 1,200,000 at 3% a year, three equal annual payments. 0.03 × 1,200,000 ÷ (1 − 1.03⁻³) = GBP 424,236.
- 2NDCLR TVM with P/Y set to 1
- 3N3I/Y1200000+|-PV
- CPTPMT 424,236.44
2ND second function (the shifted label above a key) · CLR TVM clear the time value of money keys · N number of periods · I/Y interest rate per year (per period when P/Y is 1) · +|- change sign · PV present value · CPT compute · PMT payment
- fCLEAR FIN
- 3n3i1200000CHSPV
- PMT 424,236.44
f gold shift key · CLEAR FIN clear the financial registers · n number of periods · i interest rate per period · CHS change sign · PV present value · PMT payment
Quick example
EUR 80 million at 5% a year, seven annual payments, EUR 25 million balloon at the end. EUR 10.755 million a year.
- 2NDCLR TVM with P/Y set to 1
- 7N5I/Y80+|-PV25FV
- CPTPMT 10.7551
2ND second function (the shifted label above a key) · CLR TVM clear the time value of money keys · N number of periods · I/Y interest rate per year (per period when P/Y is 1) · +|- change sign · PV present value · FV future value · CPT compute · PMT payment
- fCLEAR FIN
- 7n5i80CHSPV25FV
- PMT 10.7551 (with four decimals shown)
f gold shift key · CLEAR FIN clear the financial registers · n number of periods · i interest rate per period · CHS change sign · PV present value · FV future value · PMT payment
Floating, step-up, inflation-linked and zero-coupon bonds
Quick example
GBP 30 million, 1.2% real coupon, semiannual; the index rises from 250.0 to 254.0. Adjusted principal 30 × 1.016 = 30.48 million; 1.2% × 30.48m ÷ 2 = GBP 182,880.
- 254÷250×30000000×0.012÷2= 182,880
- 254ENTER250÷30000000×0.012×2÷ 182,880
ENTER enter (on the HP, puts the number on the stack)
Quick example
USD 100 million at a 10% PIK coupon owes 100 × 1.10 = USD 110 million after one year, and 133.1 after three.
- 1.1y^x3×100= 133.1
y^x power
- 1.1ENTER3y^x100× 133.1
ENTER enter (on the HP, puts the number on the stack) · y^x power
Callable, putable and convertible bonds
Quick example
Par EUR 1,000, conversion price EUR 80. 1,000 ÷ 80 = 12.5 shares per bond.
- 1000÷80= 12.5
- 1000ENTER80÷ 12.5
ENTER enter (on the HP, puts the number on the stack)
Quick example
Par EUR 1,000, conversion price EUR 50, shares at EUR 62. Ratio 20; 20 × 62 = EUR 1,240.
- 1000÷50×62= 1,240
- 1000ENTER50÷62× 1,240
ENTER enter (on the HP, puts the number on the stack)
Domestic, foreign and Eurobonds, and how bonds are taxed
Quick example
INR 80 million par, zero-coupon, issued at 78.50 per 100. 80m × (1 − 0.785) = INR 17.2 million.
- 1−0.785=×80= 17.2, in millions
- 1ENTER0.785−80× 17.2, in millions
ENTER enter (on the HP, puts the number on the stack)
Module 3 Fixed-Income Issuance and Trading
Bond indexes and how they differ from equity indexes
Quick example
Two bonds: 250m par at 108 and 750m par at 92. Market values 270 and 690, total 960. First bond: 270 ÷ 960 = 28.1% (not 25% by par).
- 250×1.08=STO1
- 750×0.92=STO2
- RCL1÷(RCL1+RCL2)= 0.28125
STO store · RCL recall a stored number
- 250ENTER1.08×STO1
- 750ENTER0.92×RCL1+1/xRCL1× 0.28125
ENTER enter (on the HP, puts the number on the stack) · STO store · RCL recall a stored number · 1/x reciprocal
Primary and secondary bond markets vs equity markets
Quick example
Bid 96.80, offer 97.05. Spread 0.25; mid 96.925. 0.25 ÷ 96.925 × 10,000 = 25.8 bps (25 bps of par).
- 97.05+96.8=÷2=STO1 96.925
- 97.05−96.8=÷RCL1=×10000= 25.79
STO store · RCL recall a stored number
- 97.05ENTER96.8+2÷STO1 96.925
- 97.05ENTER96.8−RCL1÷10000× 25.79
ENTER enter (on the HP, puts the number on the stack) · STO store · RCL recall a stored number
Module 4 Fixed-Income Markets for Corporate Issuers
Short-term funding for companies: credit lines, secured loans and commercial paper
Quick example
EUR 60m line, EUR 25m drawn all year at 5.10%, fee 0.20% on the unused part. 25m × 5.10% = 1,275,000; (60m − 25m) × 0.20% = 70,000. Total EUR 1,345,000.
- 25×0.051=STO1 interest 1.275 (EUR m)
- 60−25=×0.002+RCL1= 1.345
STO store · RCL recall a stored number
- 25ENTER0.051× interest 1.275 (EUR m)
- 60ENTER25−0.002×+ 1.345
ENTER enter (on the HP, puts the number on the stack)
Quick example
Total cost EUR 1,345,000 on EUR 25m drawn: 1.345 ÷ 25 = 0.0538, so 5.38%, above the 5.10% interest rate because of the fee.
- 1.345÷25= 0.0538
- 1.345ENTER25÷ 0.0538
ENTER enter (on the HP, puts the number on the stack)
How banks fund themselves short term
Quick example
Assets yield 7.15% and funding costs 3.60%. 7.15% − 3.60% = 3.55%.
- 7.15−3.6= 3.55
- 7.15ENTER3.6− 3.55
ENTER enter (on the HP, puts the number on the stack)
Repurchase agreements (repos)
Quick example
Borrow USD 15,000,000 for 45 days at a 2.40% repo rate. 15,000,000 × (1 + 0.024 × 45 ÷ 360) = USD 15,045,000.
- 0.024×45÷360+1×15000000= 15,045,000
- 0.024ENTER45×360÷1+15000000× 15,045,000
ENTER enter (on the HP, puts the number on the stack)
Quick example
Security worth 55 million, loan of 50 million: 55 ÷ 50 = 110%.
- 55÷50= 1.1
- 55ENTER50÷ 1.1
ENTER enter (on the HP, puts the number on the stack)
Quick example
Same trade: (55 − 50) ÷ 55 = 9.09%. Equivalent: 1 − 1 ÷ initial margin.
- 55−50=÷55= 0.0909
- 55ENTER50−55÷ 0.0909
ENTER enter (on the HP, puts the number on the stack)
Quick example
Margin 103%, loan with accrued interest 50,010,000, security now 50,600,000. 1.03 × 50,010,000 − 50,600,000 = +910,300: the borrower posts more collateral.
- 1.03×50010000−50600000= 910,300
- 1.03ENTER50010000×50600000− 910,300
ENTER enter (on the HP, puts the number on the stack)
Quick example
Lend 10,000,000 for 90 days at 2.00%, sell the bond for 10,000,000, buy it back at 9,700,000. Repurchase price 10,050,000, so profit 350,000 before costs.
- 0.02×90÷360+1×10000000−9700000= 350,000
- 0.02ENTER90×360÷1+10000000×9700000− 350,000
ENTER enter (on the HP, puts the number on the stack)
Long-term funding: investment-grade vs high-yield issuers
Quick example
A BBB bond for which the government yield is 3.20% and the credit spread 1.10% has YTM 4.30%. The spread is 1.10 ÷ 4.30 = 25.6% of the yield. For a B bond with a 4.80% spread, it is 4.80 ÷ 8.00 = 60%.
- 3.2+1.1=1/x×1.1= 0.2558
1/x reciprocal
- 3.2ENTER1.1+1/x1.1× 0.2558
ENTER enter (on the HP, puts the number on the stack) · 1/x reciprocal
Module 5 Fixed-Income Markets for Government Issuers
Government bond auctions and trading vs corporate bonds
Quick example
Offering 8.0 billion, non-competitive bids 1.4 billion. 8.0 − 1.4 = 6.6 billion must be filled by competitive bids.
- 8−1.4= 6.6
- 8ENTER1.4− 6.6
ENTER enter (on the HP, puts the number on the stack)
Quick example
C = 6.6 billion; bids below the cut-off total B = 5.0; bids at the cut-off A = 3.2. (6.6 − 5.0) ÷ 3.2 = 50% of each cut-off bid is filled.
- 6.6−5=÷3.2= 0.5
- 6.6ENTER5−3.2÷ 0.5
ENTER enter (on the HP, puts the number on the stack)
Module 6 Bond Valuation: Prices and Yields
Pricing a bond from its yield to maturity
Quick example
4 years left, 6% coupon paid annually, par 100, yield 5%. N = 4, I/Y = 5, PMT = 6, FV = 100 gives 103.55, a premium.
- 2NDCLR TVM with P/Y set to 1
- 4N5I/Y6PMT100FV
- CPTPV −103.5460
2ND second function (the shifted label above a key) · CLR TVM clear the time value of money keys · N number of periods · I/Y interest rate per year (per period when P/Y is 1) · PMT payment · FV future value · CPT compute · PV present value
- fCLEAR FIN
- 4n5i6PMT100FV
- PV −103.5460
f gold shift key · CLEAR FIN clear the financial registers · n number of periods · i interest rate per period · PMT payment · FV future value · PV present value
Quick example
5 years, 3% coupon paid twice a year, yield 4% stated annually. N = 10, r = 2%, PMT = 1.5. Price = 95.51.
- 2NDCLR TVM with P/Y set to 1
- 10N2I/Y1.5PMT100FV
- CPTPV −95.5087
2ND second function (the shifted label above a key) · CLR TVM clear the time value of money keys · N number of periods · I/Y interest rate per year (per period when P/Y is 1) · PMT payment · FV future value · CPT compute · PV present value
- fCLEAR FIN
- 10n2i1.5PMT100FV
- PV −95.5087
f gold shift key · CLEAR FIN clear the financial registers · n number of periods · i interest rate per period · PMT payment · FV future value · PV present value
Quick example
5 years, 6% coupon paid annually, price 104.30. Enter N = 5, PV = −104.30, PMT = 6, FV = 100 and compute I/Y: 5.01%.
- 2NDCLR TVM with P/Y set to 1
- 5N104.3+|-PV6PMT100FV
- CPTI/Y 5.0066
2ND second function (the shifted label above a key) · CLR TVM clear the time value of money keys · N number of periods · +|- change sign · PV present value · PMT payment · FV future value · CPT compute · I/Y interest rate per year (per period when P/Y is 1)
- fCLEAR FIN
- 5n104.3CHSPV6PMT100FV
- i 5.0066
f gold shift key · CLEAR FIN clear the financial registers · n number of periods · CHS change sign · PV present value · PMT payment · FV future value · i interest rate per period
Quick example
A 3-year zero priced at 103.00 per 100 of par. (100 ÷ 103)^(1/3) − 1 = −0.98%, a negative yield.
- 2NDCLR TVM with P/Y set to 1
- 3N103+|-PV100FV
- CPTI/Y −0.9805
2ND second function (the shifted label above a key) · CLR TVM clear the time value of money keys · N number of periods · +|- change sign · PV present value · FV future value · CPT compute · I/Y interest rate per year (per period when P/Y is 1)
- fCLEAR FIN
- 3n103CHSPV100FV
- i −0.9805
f gold shift key · CLEAR FIN clear the financial registers · n number of periods · CHS change sign · PV present value · FV future value · i interest rate per period
Accrued interest, flat price and full price
Quick example
A bond is worth 98.40 on its last coupon date. Yield 2% per period, settlement 110 days into a 182-day period. 98.40 × 1.02^(110/182) = 99.58.
- 1.02y^x(110÷182)×98.4= 99.585
y^x power
- 1.02ENTER110ENTER182÷y^x98.4× 99.585
ENTER enter (on the HP, puts the number on the stack) · y^x power
Quick example
Coupon of 1.85 per half-year, settlement 112 days into a 180-day period (30/360). (112 ÷ 180) × 1.85 = 1.151.
- 112÷180×1.85= 1.1511
- 112ENTER180÷1.85× 1.1511
ENTER enter (on the HP, puts the number on the stack)
Quick example
Full price 99.585, accrued interest 1.330. Flat price = 98.255.
- 99.585−1.33= 98.255
- 99.585ENTER1.33− 98.255
ENTER enter (on the HP, puts the number on the stack)
Quick example
12 April to 3 July in the same year: 30 × 3 + (3 − 12) = 81 days, out of 180 for a half-year.
- 7−4=×30+3−12= 81
- 7ENTER4−30×3+12− 81
ENTER enter (on the HP, puts the number on the stack)
How price, coupon, maturity and yield are related
Quick example
A 9-year 5% bond at par falls to 95.85 when its yield rises to 5.6%. 95.85 ÷ 100 − 1 = −4.15%.
- 95.847÷100−1= −0.0415
- 95.847ENTER100÷1− −0.0415
ENTER enter (on the HP, puts the number on the stack)
Quick example
A 10-year 6% bond priced at a constant 5% yield is 107.72 today and 106.46 two years later (8 years left). Same formula, fewer periods.
- 2NDCLR TVM with P/Y set to 1
- 8N5I/Y6PMT100FV
- CPTPV −106.4632
2ND second function (the shifted label above a key) · CLR TVM clear the time value of money keys · N number of periods · I/Y interest rate per year (per period when P/Y is 1) · PMT payment · FV future value · CPT compute · PV present value
- fCLEAR FIN
- 8n5i6PMT100FV
- PV −106.4632
f gold shift key · CLEAR FIN clear the financial registers · n number of periods · i interest rate per period · PMT payment · FV future value · PV present value
Matrix pricing
Quick example
Comparables yield 5.10% at 5 years and 5.55% at 8 years. At 7 years: 5.10 + (2 ÷ 3) × 0.45 = 5.40%.
- 2÷3×0.45+5.1= 5.4
- 2ENTER3÷0.45×5.1+ 5.4
ENTER enter (on the HP, puts the number on the stack)
Quick example
An issuer’s 6-year bond yields 4.78%. The interpolated 6-year government yield is 4.35%. The spread is 43 bps.
- 4.78−4.35=×100= 43 bps
- 4.78ENTER4.35−100× 43 bps
ENTER enter (on the HP, puts the number on the stack)
Module 7 Yield and Yield Spread Measures for Fixed-Rate Bonds
Periodicity: converting yields between compounding frequencies
Quick example
A bond earns 2.10% every six months. Periodicity 2 gives a semiannual bond basis yield of 2 × 2.10% = 4.20%.
Quick example
6.00% on a quarterly basis restated semiannually: 2 × [(1 + 0.06 ÷ 4)² − 1] = 2 × 0.030225 = 6.045%.
- 1.015y^x2=−1=×2= 0.06045
y^x power
- 1.015ENTER2y^x1−2× 0.06045
ENTER enter (on the HP, puts the number on the stack) · y^x power
Quick example
4.00% on a semiannual basis: (1 + 0.04 ÷ 2)² − 1 = 4.04%.
- 1.02y^x2=−1= 0.0404
y^x power
- 1.02ENTER2y^x1− 0.0404
ENTER enter (on the HP, puts the number on the stack) · y^x power
Quick example
A 10-year zero at 61.00 per 100. EAR = (100 ÷ 61)^(1/10) − 1 = 5.07%; semiannual basis = 2 × [(100 ÷ 61)^(1/20) − 1] = 5.01%.
- 2NDCLR TVM with P/Y set to 1
- 20N61+|-PV100FV
- CPTI/Y 2.5023 per half-year
- ×2= 5.0046
2ND second function (the shifted label above a key) · CLR TVM clear the time value of money keys · N number of periods · +|- change sign · PV present value · FV future value · CPT compute · I/Y interest rate per year (per period when P/Y is 1)
- fCLEAR FIN
- 20n61CHSPV100FV
- i 2.5023 per half-year
- 2× 5.0046
f gold shift key · CLEAR FIN clear the financial registers · n number of periods · CHS change sign · PV present value · FV future value · i interest rate per period
Current yield, simple yield and yield conventions
Quick example
A 3% annual coupon bond at a flat price of 94.00. 3 ÷ 94 = 3.19%. It ignores the 6.00 gain to par.
- 3÷94= 0.031915
- 3ENTER94÷ 0.031915
ENTER enter (on the HP, puts the number on the stack)
Quick example
A 3-year bond, 2.5% annual coupon, flat price 101.50. (2.5 + (100 − 101.5) ÷ 3) ÷ 101.5 = (2.5 − 0.5) ÷ 101.5 = 1.97%.
- 100−101.5=÷3+2.5=÷101.5= 0.019704
- 100ENTER101.5−3÷2.5+101.5÷ 0.019704
ENTER enter (on the HP, puts the number on the stack)
Quick example
A corporate yield of 6.00% on 30/360 is 6.00% × 365 ÷ 360 = 6.083% on actual/actual.
- 0.06×365÷360= 0.060833
- 0.06ENTER365×360÷ 0.060833
ENTER enter (on the HP, puts the number on the stack)
Yield to call and yield to worst
Quick example
A 5% semiannual bond at 102.00, callable in 3 years at 101. N = 6, present value (PV) = −102, PMT = 2.5, FV = 101 gives 2.2967% per half-year, so YTC = 4.59%.
- 2NDCLR TVM with P/Y set to 1
- 6N102+|-PV2.5PMT101FV
- CPTI/Y 2.2967 per half-year
- ×2= 4.5934
2ND second function (the shifted label above a key) · CLR TVM clear the time value of money keys · N number of periods · +|- change sign · PV present value · PMT payment · FV future value · CPT compute · I/Y interest rate per year (per period when P/Y is 1)
- fCLEAR FIN
- 6n102CHSPV2.5PMT101FV
- i 2.2967 per half-year
- 2× 4.5934
f gold shift key · CLEAR FIN clear the financial registers · n number of periods · CHS change sign · PV present value · PMT payment · FV future value · i interest rate per period
Quick example
Yields to first call, second call and maturity of 4.20%, 4.55% and 4.80%. YTW = 4.20%.
Quick example
A callable bond trades at 101.20 and its call is worth 1.35. Option-adjusted price = 101.20 + 1.35 = 102.55.
- 101.2+1.35= 102.55
- 101.2ENTER1.35+ 102.55
ENTER enter (on the HP, puts the number on the stack)
G-spread, I-spread, Z-spread and option-adjusted spread
Quick example
A bond yields 4.85% and the benchmark yields 3.95%. Spread = 90 bps.
- 0.0485−0.0395=×10000= 90
- 0.0485ENTER0.0395−10000× 90
ENTER enter (on the HP, puts the number on the stack)
Quick example
A 6-year bond yields 5.18%; governments yield 3.80% at 5 years and 4.45% at 10. y_gov = 3.80% + (1 ÷ 5) × 0.65% = 3.93%. G-spread = 125 bps.
- (6−5)÷(10−5)×(4.45−3.8)+3.8= 3.93, the government yield
- +|-+5.18=×100= 125
+|- change sign
- 6ENTER5−10ENTER5−÷4.45ENTER3.8−×3.8+ 3.93, the government yield
- CHS5.18+100× 125
ENTER enter (on the HP, puts the number on the stack) · CHS change sign
Quick example
A 5-year euro corporate bond yields 4.08% and the 5-year euro swap rate is 3.51%. I-spread = 57 bps.
- 0.0408−0.0351=×10000= 57
- 0.0408ENTER0.0351−10000× 57
ENTER enter (on the HP, puts the number on the stack)
Quick example
A 3-year 3.5% annual bond, spot rates 2.8%, 3.1%, 3.5%, Z = 0.80%. 3.5 ÷ 1.036 + 3.5 ÷ 1.039² + 103.5 ÷ 1.043³ = 97.84.
- 3.5÷1.036=STO1
- 3.5÷(1.039y^x2)=+RCL1=STO1
- 103.5÷(1.043y^x3)=+RCL1= 97.8400
STO store · y^x power · RCL recall a stored number
- 3.5ENTER1.036÷
- 3.5ENTER1.039ENTER2y^x÷+
- 103.5ENTER1.043ENTER3y^x÷+ 97.8400
ENTER enter (on the HP, puts the number on the stack) · y^x power
Quick example
Z-spread 210 bps, embedded call worth 55 bps a year. OAS = 210 − 55 = 155 bps.
- 210−55= 155
- 210ENTER55− 155
ENTER enter (on the HP, puts the number on the stack)
Module 8 Yield and Yield Spread Measures for Floating-Rate Instruments
Floating-rate notes: quoted margin and discount margin
Quick example
Semiannual FRN at MRR + 70 bps with MRR at 2.50%: (2.50% + 0.70%) × 100 ÷ 2 = 1.60 per 100 of par.
- 2.5+0.7=÷2= 1.6
- 2.5ENTER0.7+2÷ 1.6
ENTER enter (on the HP, puts the number on the stack)
Quick example
Two-year quarterly FRN, MRR 2.00%, QM 100 bps, DM 130 bps. PMT = 0.75; r = 0.825% per period; N = 8. Price = 99.42.
- 2NDCLR TVM with P/Y set to 1
- 8N
- 0.825I/Y
- 0.75PMT
- 100FV
- CPTPV −99.4217
2ND second function (the shifted label above a key) · CLR TVM clear the time value of money keys · N number of periods · I/Y interest rate per year (per period when P/Y is 1) · PMT payment · FV future value · CPT compute · PV present value
- fCLEAR FIN
- 8n
- 0.825i
- 0.75PMT
- 100FV
- PV −99.4217
f gold shift key · CLEAR FIN clear the financial registers · n number of periods · i interest rate per period · PMT payment · FV future value · PV present value
Quick example
Three-year semiannual FRN, MRR 3.00%, QM 70 bps, price 99.00. PMT = 1.85; N = 6; r = 2.0287% per period. DM = 2 × 2.0287% − 3.00% = 1.057% (106 bps).
- 2NDCLR TVM with P/Y set to 1
- 6N
- 99+|-PV
- 1.85PMT
- 100FV
- CPTI/Y 2.0287
- ×2−3= 1.057
2ND second function (the shifted label above a key) · CLR TVM clear the time value of money keys · N number of periods · +|- change sign · PV present value · PMT payment · FV future value · CPT compute · I/Y interest rate per year (per period when P/Y is 1)
- fCLEAR FIN
- 6n
- 99CHSPV
- 1.85PMT
- 100FV
- i 2.0287
- 2×3− 1.057
f gold shift key · CLEAR FIN clear the financial registers · n number of periods · CHS change sign · PV present value · PMT payment · FV future value · i interest rate per period
Money market rates: discount basis and add-on basis
Quick example
A 182-day bill with face value 2,000,000 at a 3.90% discount rate (360-day year): 2,000,000 × (1 − 182/360 × 0.039) = 1,960,567.
- 182÷360×0.039=+|-+1=×2000000= 1960566.67
+|- change sign
- 182ENTER360÷0.039×CHS1+2000000× 1960566.67
ENTER enter (on the HP, puts the number on the stack) · CHS change sign
Quick example
A 126-day bill with face value 8,000,000 costs 7,900,000 (360-day year): 360/126 × 100,000 ÷ 8,000,000 = 3.571%.
- 8000000−7900000=÷8000000×360÷126= 0.035714
- 8000000ENTER7900000−8000000÷360×126÷ 0.035714
ENTER enter (on the HP, puts the number on the stack)
Quick example
30,000,000 placed for 270 days at a 4.20% add-on rate (365-day year): 30,000,000 × (1 + 270/365 × 0.042) = 30,932,055.
- 270÷365×0.042+1=×30000000= 30932054.8
- 270ENTER365÷0.042×1+30000000× 30932054.8
ENTER enter (on the HP, puts the number on the stack)
Quick example
A deposit will pay 5,000,000 in 60 days. At a 3.00% add-on rate (365-day year) it is worth 5,000,000 ÷ (1 + 60/365 × 0.03) = 4,975,463.
- 60÷365×0.03+1=1/x×5000000= 4975463.47
1/x reciprocal
- 60ENTER365÷0.03×1+1/x5000000× 4975463.47
ENTER enter (on the HP, puts the number on the stack) · 1/x reciprocal
Quick example
A 270-day deposit of 30,000,000 pays back 30,900,000 (365-day year): 365/270 × 900,000 ÷ 30,000,000 = 4.056%.
- 30900000−30000000=÷30000000×365÷270= 0.040556
- 30900000ENTER30000000−30000000÷365×270÷ 0.040556
ENTER enter (on the HP, puts the number on the stack)
Comparing money market instruments: bond equivalent yield
Quick example
A 91-day bill at a 3.70% discount rate (360-day year): 100 × (1 − 91/360 × 0.037) = 99.06.
- 91÷360×0.037=+|-+1=×100= 99.0647
+|- change sign
- 91ENTER360÷0.037×CHS1+100× 99.0647
ENTER enter (on the HP, puts the number on the stack) · CHS change sign
Quick example
A 150-day bill priced at 98.00 per 100: 365/150 × 2 ÷ 98 = 4.966%.
- 100−98=÷98×365÷150= 0.049660
- 100ENTER98−98÷365×150÷ 0.049660
ENTER enter (on the HP, puts the number on the stack)
Quick example
A CD at a 3.90% add-on rate on a 360-day year: 3.90% × 365/360 = 3.954%.
- 0.039×365÷360= 0.039542
- 0.039ENTER365×360÷ 0.039542
ENTER enter (on the HP, puts the number on the stack)
Quick example
A 75-day instrument has periodicity 365 ÷ 75 = 4.87.
- 365÷75= 4.8667
- 365ENTER75÷ 4.8667
ENTER enter (on the HP, puts the number on the stack)
Quick example
A 60-day BEY of 5.27%: m = 365/60 = 6.0833, so 2 × [(1 + 0.0527 ÷ 6.0833)^(6.0833/2) − 1] = 5.317%.
- 365÷60=STO1 m
- 0.0527÷RCL1+1=y^x(RCL1÷2)=−1=×2= 0.053167
STO store · RCL recall a stored number · y^x power
- 365ENTER60÷STO1 m
- 0.0527RCL1÷1+RCL1ENTER2÷y^x1−2× 0.053167
ENTER enter (on the HP, puts the number on the stack) · STO store · RCL recall a stored number · y^x power
Module 9 Term Structure: Spot, Par and Forward Curves
Spot rates and pricing a bond with the spot curve
Quick example
Spot rates 2.0%, 2.5%, 3.0% for 1, 2, 3 years. A 3-year bond with a 4% annual coupon: 4 ÷ 1.02 + 4 ÷ 1.025² + 104 ÷ 1.03³ = 3.922 + 3.807 + 95.175 = 102.904.
- 4÷1.02=STO1
- 4÷(1.025y^x2)=STO2
- 104÷(1.03y^x3)+RCL1+RCL2= 102.9036
STO store · y^x power · RCL recall a stored number
- 4ENTER1.02÷
- 4ENTER1.025ENTER2y^x÷+
- 104ENTER1.03ENTER3y^x÷+ 102.9036
ENTER enter (on the HP, puts the number on the stack) · y^x power
Quick example
A 2-year default-free zero trades at 93.00 per 100 of par. (100 ÷ 93)^(1/2) − 1 = 3.695%.
- 2NDCLR TVM with P/Y set to 1
- 2N93+|-PV100FV
- CPTI/Y 3.6952
2ND second function (the shifted label above a key) · CLR TVM clear the time value of money keys · N number of periods · +|- change sign · PV present value · FV future value · CPT compute · I/Y interest rate per year (per period when P/Y is 1)
- fCLEAR FIN
- 2n93CHSPV100FV
- i 3.6952
f gold shift key · CLEAR FIN clear the financial registers · n number of periods · CHS change sign · PV present value · FV future value · i interest rate per period
Par rates from spot rates
Quick example
The 3-year spot rate is 3.6%. d₃ = 1 ÷ 1.036³ = 0.8993: one unit paid in 3 years is worth 0.8993 today.
- 1.036y^x3=1/x 0.8993
y^x power · 1/x reciprocal
- 1.036ENTER3y^x1/x 0.8993
ENTER enter (on the HP, puts the number on the stack) · y^x power · 1/x reciprocal
Quick example
Spot rates 2.0% (1 year) and 2.6% (2 years). d₁ = 0.98039, d₂ = 0.94996. Par rate = (1 − 0.94996) ÷ (0.98039 + 0.94996) = 2.592%.
- 1.021/xSTO1
- 1.026y^x2=1/xSTO2
- 1−RCL2=÷(RCL1+RCL2)=×100= 2.5923
1/x reciprocal · STO store · y^x power · RCL recall a stored number
- 1.021/xSTO1
- 1.026ENTER2y^x1/xSTO2
- 1RCL2−RCL1RCL2+÷100× 2.5923
1/x reciprocal · STO store · ENTER enter (on the HP, puts the number on the stack) · y^x power · RCL recall a stored number
Implied forward rates: from spot to forward and back
Quick example
The 1-year spot rate is 2.0% and the 3-year spot rate is 3.0%. The 1y2y forward is (1.03³ ÷ 1.02)^(1/2) − 1 = 3.504%.
- 1.03y^x3=÷1.02=y^x0.5=−1= 0.0350
y^x power
- 1.03ENTER3y^x1.02÷0.5y^x1− 0.0350
ENTER enter (on the HP, puts the number on the stack) · y^x power
Quick example
One-year forwards today, in 1 year and in 2 years: 1.8%, 2.4%, 3.0%. The 3-year spot rate is (1.018 × 1.024 × 1.030)^(1/3) − 1 = 2.399%.
- 1.018×1.024×1.03=y^x(1÷3)=−1= 0.0240
y^x power
- 1.018ENTER1.024×1.03×31/xy^x1− 0.0240
ENTER enter (on the HP, puts the number on the stack) · 1/x reciprocal · y^x power
Quick example
A 3-year, 5% annual-coupon bond with one-year forwards of 1.8%, 2.4% and 3.0%: 5 ÷ 1.018 + 5 ÷ (1.018 × 1.024) + 105 ÷ (1.018 × 1.024 × 1.030) = 107.500.
- 105÷1.03+5=÷1.024+5=÷1.018= 107.5003
- 105ENTER1.03÷5+1.024÷5+1.018÷ 107.5003
ENTER enter (on the HP, puts the number on the stack)
How the spot, par and forward curves relate
Quick example
The 1-year spot is 6.0% and the 2-year spot is 5.2%. The 1y1y forward is 1.052² ÷ 1.06 − 1 = 4.41%, below both spot rates: an inverted curve.
- 1.052y^x2=÷1.06−1= 0.0441
y^x power
- 1.052ENTER2y^x1.06÷1− 0.0441
ENTER enter (on the HP, puts the number on the stack) · y^x power
Module 10 Interest Rate Risk and Return
The sources of return from a fixed-rate bond
Quick example
4 annual coupons of 5, each reinvested at 6%. 5 × (1.06⁴ − 1) ÷ 0.06 = 21.873: 20.000 of coupons plus 1.873 of interest on interest.
- 2NDCLR TVM with P/Y set to 1
- 4N6I/Y5PMT
- CPTFV −21.873, negative by the TVM sign convention
2ND second function (the shifted label above a key) · CLR TVM clear the time value of money keys · N number of periods · I/Y interest rate per year (per period when P/Y is 1) · PMT payment · CPT compute · FV future value
- fCLEAR FIN
- 4n6i5PMT
- FV −21.873
f gold shift key · CLEAR FIN clear the financial registers · n number of periods · i interest rate per period · PMT payment · FV future value
Quick example
Buy a 6-year, 4% annual bond at par. Yields drop to 3% at once and you sell after 2 years. Coupons 8.12, sale price 103.717. (111.837 ÷ 100)^(1/2) − 1 = 5.75%.
- 2NDCLR TVM with P/Y set to 1
- 2N100+|-PV8.12+103.717=FV
- CPTI/Y 5.753
2ND second function (the shifted label above a key) · CLR TVM clear the time value of money keys · N number of periods · +|- change sign · PV present value · FV future value · CPT compute · I/Y interest rate per year (per period when P/Y is 1)
- fCLEAR FIN
- 2n100CHSPV8.12ENTER103.717+FV
- i 5.753
f gold shift key · CLEAR FIN clear the financial registers · n number of periods · CHS change sign · PV present value · ENTER enter (on the HP, puts the number on the stack) · FV future value · i interest rate per period
Quick example
A 6-year, 3% annual bond bought at a 4% yield. After 2 years, 4 years remain. At 4%: N = 4, I/Y = 4, PMT = 3, FV = 100 gives 96.370.
- 2NDCLR TVM with P/Y set to 1
- 4N4I/Y3PMT100FV
- CPTPV −96.370
2ND second function (the shifted label above a key) · CLR TVM clear the time value of money keys · N number of periods · I/Y interest rate per year (per period when P/Y is 1) · PMT payment · FV future value · CPT compute · PV present value
- fCLEAR FIN
- 4n4i3PMT100FV
- PV −96.370
f gold shift key · CLEAR FIN clear the financial registers · n number of periods · i interest rate per period · PMT payment · FV future value · PV present value
Quick example
The bond above is sold for 94.80 while its carrying value is 96.370. 94.80 − 96.37 = −1.57, a capital loss.
- 94.8−96.37= −1.57
- 94.8ENTER96.37− −1.57
ENTER enter (on the HP, puts the number on the stack)
Investment horizon, price risk vs reinvestment risk, and duration gap
Quick example
A bond has a Macaulay duration of 4.6 years and the investor’s horizon is 6 years. 4.6 − 6 = −1.4: a negative gap, so falling rates are the risk.
- 4.6−6= −1.4
- 4.6ENTER6− −1.4
ENTER enter (on the HP, puts the number on the stack)
Macaulay duration: definition and calculation
Quick example
A 4-year, 3% annual bond at a 4% yield has price 96.370. Weighting each year by its share of that price gives 3.825 years.
- (1.04y^x4−1)×0.03+0.04=STO1 the denominator, 0.045096
- 4×(0.03−0.04)+1.04=÷RCL1= 22.1750
- +|-+(1.04÷0.04)= 3.8250
y^x power · STO store · RCL recall a stored number · +|- change sign
- 1.04ENTER4y^x1−0.03×0.04+STO1 the denominator, 0.045096
- 0.03ENTER0.04−4×1.04+RCL1÷ 22.1750
- 1.04ENTER0.04÷−CHS 3.8250
ENTER enter (on the HP, puts the number on the stack) · y^x power · STO store · RCL recall a stored number · CHS change sign
Quick example
A semiannual bond has a weighted average time of 7.64 half-year periods. 7.64 ÷ 2 = 3.82 years.
- 7.64÷2= 3.82
- 7.64ENTER2÷ 3.82
ENTER enter (on the HP, puts the number on the stack)
Quick example
6-year, 4% annual bond at a 5% yield, on a coupon date (t/T = 0). 21.0000 − 0.99 ÷ 0.063604 = 5.435 years.
- (1.05y^x6−1)×0.04+0.05=STO1 the denominator, 0.063604
- 6×(0.04−0.05)+1.05=÷RCL1= 15.5651
- +|-+(1.05÷0.05)= 5.4349
y^x power · STO store · RCL recall a stored number · +|- change sign
- 1.05ENTER6y^x1−0.04×0.05+STO1 the denominator, 0.063604
- 0.04ENTER0.05−6×1.05+RCL1÷ 15.5651
- 1.05ENTER0.05÷−CHS 5.4349
ENTER enter (on the HP, puts the number on the stack) · y^x power · STO store · RCL recall a stored number · CHS change sign
Quick example
A bond’s Macaulay duration is 6.10 years on its last coupon date (annual coupons). 0.4 years later, with the yield unchanged: 6.10 − 0.4 = 5.70 years.
- 6.1−0.4= 5.7
- 6.1ENTER0.4− 5.7
ENTER enter (on the HP, puts the number on the stack)
Module 11 Modified Duration, Money Duration and PVBP
Modified duration
Quick example
A semiannual bond yielding 5% a year has an annualized Macaulay duration of 7.45. Divide by the periodic factor, 1 + 0.05/2 = 1.025: 7.45 ÷ 1.025 = 7.268.
- 7.45÷1.025= 7.2683
- 7.45ENTER1.025÷ 7.2683
ENTER enter (on the HP, puts the number on the stack)
Quick example
A bond with an annualized modified duration of 4.80 sees its yield rise 30 bp. Estimated change: −4.80 × 0.0030 = −1.44%.
- 4.8×0.003=+|- −0.0144
+|- change sign
- 4.8ENTER0.003×CHS −0.0144
ENTER enter (on the HP, puts the number on the stack) · CHS change sign
Quick example
A bond is priced at 103.00. At a yield 5 bp lower it is 103.28 and at a yield 5 bp higher it is 102.77. (103.28 − 102.77) ÷ (2 × 0.0005 × 103.00) = 4.95.
- (103.28−102.77)÷(2×0.0005×103)= 4.9515
- 103.28ENTER102.77−2ENTER0.0005×103×÷ 4.9515
ENTER enter (on the HP, puts the number on the stack)
Quick example
An annual-pay bond yielding 4% has an approximate modified duration of 4.9515. Approximate Macaulay duration: 4.9515 × 1.04 = 5.15 years.
- 4.9515×1.04= 5.1496
- 4.9515ENTER1.04× 5.1496
ENTER enter (on the HP, puts the number on the stack)
Money duration and PVBP
Quick example
A holding of USD 10,000,000 par is priced at 98.00 (market value 9,800,000) with an annualized modified duration of 6.10. Money duration = 6.10 × 9,800,000 = USD 59,780,000.
- 10000000×0.98×6.1= 59,780,000
- 10000000ENTER0.98×6.1× 59,780,000
ENTER enter (on the HP, puts the number on the stack)
Quick example
The same position, with a money duration of 59,780,000, and a yield rise of 15 bp: −59,780,000 × 0.0015 = −USD 89,670.
- 59780000×0.0015=+|- −89,670
+|- change sign
- 59780000ENTER0.0015×CHS −89,670
ENTER enter (on the HP, puts the number on the stack) · CHS change sign
Quick example
A bond is priced at 100.0562 at a yield 1 bp lower and 99.9438 at a yield 1 bp higher. (100.0562 − 99.9438) ÷ 2 = 0.0562 per 100 of par.
- 100.0562−99.9438=÷2= 0.0562
- 100.0562ENTER99.9438−2÷ 0.0562
ENTER enter (on the HP, puts the number on the stack)
Quick example
A position has a money duration of 59,780,000. Its basis point value is 59,780,000 × 0.0001 = USD 5,978 per basis point.
- 59780000×0.0001= 5,978
- 59780000ENTER0.0001× 5,978
ENTER enter (on the HP, puts the number on the stack)
What makes duration higher or lower
Quick example
A 10-year zero-coupon bond yields 6% (annual). Macaulay duration is 10.00 years. Modified duration = 10 ÷ 1.06 = 9.434.
- 10÷1.06= 9.4340
- 10ENTER1.06÷ 9.4340
ENTER enter (on the HP, puts the number on the stack)
Quick example
A non-callable perpetual bond is priced to yield 3%. MacDur = 1.03 ÷ 0.03 = 34.33 years; ModDur = 1 ÷ 0.03 = 33.33.
- 1.03÷0.03= 34.3333
- 1.03ENTER0.03÷ 34.3333
ENTER enter (on the HP, puts the number on the stack)
Quick example
A quarterly-reset floater is 25 days into a 91-day period. (91 − 25) ÷ 91 = 0.7253 quarters, so 0.7253 ÷ 4 = 0.181 years.
- (91−25)÷91÷4= 0.1813
- 91ENTER25−91÷4÷ 0.1813
ENTER enter (on the HP, puts the number on the stack)
Module 12 Convexity and Portfolio Duration
Convexity and the convexity adjustment
Quick example
A 5-year, 4% annual coupon bond priced at 100.00 would be 101.1208 at a yield 25 bp lower and 98.8948 at a yield 25 bp higher. (101.1208 − 98.8948) ÷ (2 × 0.0025 × 100) = 4.452.
- 101.1208−98.8948=÷0.005÷100= 4.452
- 101.1208ENTER98.8948−0.005÷100÷ 4.452
ENTER enter (on the HP, puts the number on the stack)
Quick example
Same bond: PV₀ = 100.00, PV₋ = 101.1208, PV₊ = 98.8948, Δy = 0.0025. (101.1208 + 98.8948 − 200) ÷ (0.0025² × 100) = 0.0156 ÷ 0.000625 = 24.96.
- 101.1208+98.8948−200=÷0.0025÷0.0025÷100= 24.96
- 101.1208ENTER98.8948+200−0.0025÷0.0025÷100÷ 24.96
ENTER enter (on the HP, puts the number on the stack)
Quick example
Convexity 24.0 and a 150 bp yield move (either direction): ½ × 24.0 × 0.015² = +0.27% added to the duration estimate.
- 0.5×24×0.015×0.015= 0.0027
- 0.5ENTER24×0.015×0.015× 0.0027
ENTER enter (on the HP, puts the number on the stack)
Estimating price change with duration and convexity
Quick example
Modified duration 3.80, convexity 18.5, yield falls 75 bp. Duration: +3.80 × 0.0075 = 2.850%. Convexity: ½ × 18.5 × 0.0075² = 0.052%. Total +2.902%.
- 3.8×0.0075=STO1
- 0.5×18.5×0.0075×0.0075+RCL1= 0.02902
STO store · RCL recall a stored number
- 3.8ENTER0.0075×
- 0.5ENTER18.5×0.0075×0.0075×+ 0.02902
ENTER enter (on the HP, puts the number on the stack)
Quick example
AUD 10,000,000 par at a full price of 98.50 (market value 9,850,000) with convexity 40.0. MoneyCon = 40.0 × 9,850,000 = AUD 394,000,000.
- 10000000×0.985×40= 394,000,000
- 10000000ENTER0.985×40× 394,000,000
ENTER enter (on the HP, puts the number on the stack)
Quick example
MoneyDur 41,000,000, MoneyCon 394,000,000, yield +50 bp. −41,000,000 × 0.005 = −205,000; ½ × 394,000,000 × 0.005² = +4,925. Total −200,075.
- 41000000×0.005=+|-STO1
- 0.5×394000000×0.005×0.005+RCL1= −200,075
+|- change sign · STO store · RCL recall a stored number
- 41000000ENTER0.005×CHS
- 0.5ENTER394000000×0.005×0.005×+ −200,075
ENTER enter (on the HP, puts the number on the stack) · CHS change sign
Portfolio duration and convexity
Quick example
USD 7,000,000 par of a bond at 90.00 and USD 3,000,000 par of a bond at 70.00. Market values 6,300,000 and 2,100,000, so the first bond has weight 6,300,000 ÷ 8,400,000 = 0.75.
- 3000000×0.7=+6300000=STO1 8,400,000
- 6300000÷RCL1= 0.75
STO store · RCL recall a stored number
- 3000000ENTER0.7×6300000+STO1 8,400,000
- 6300000RCL1÷ 0.75
ENTER enter (on the HP, puts the number on the stack) · STO store · RCL recall a stored number
Quick example
Weights 50%, 30% and 20% with modified durations 3.0, 7.0 and 10.0: 0.50 × 3.0 + 0.30 × 7.0 + 0.20 × 10.0 = 1.5 + 2.1 + 2.0 = 5.6.
- 0.5×3=STO1
- 0.3×7+RCL1=STO1
- 0.2×10+RCL1= 5.6
STO store · RCL recall a stored number
- 0.5ENTER3×0.3ENTER7×+0.2ENTER10×+ 5.6
ENTER enter (on the HP, puts the number on the stack)
Quick example
Same weights (50%, 30%, 20%) with convexities 12, 60 and 110: 0.50 × 12 + 0.30 × 60 + 0.20 × 110 = 6 + 18 + 22 = 46.
- 0.5×12=STO1
- 0.3×60+RCL1=STO1
- 0.2×110+RCL1= 46
STO store · RCL recall a stored number
- 0.5ENTER12×0.3ENTER60×+0.2ENTER110×+ 46
ENTER enter (on the HP, puts the number on the stack)
Module 13 Effective Duration, Key Rate Duration and Empirical Duration
Effective duration and effective convexity
Quick example
Full price 100.80. A model gives 102.00 with the benchmark curve 15 bp lower and 99.55 with it 15 bp higher. (102.00 − 99.55) ÷ (2 × 0.0015 × 100.80) = 2.45 ÷ 0.3024 = 8.10.
- 102−99.55=÷2÷0.0015÷100.8= 8.10
- 102ENTER99.55−2÷0.0015÷100.8÷ 8.10
ENTER enter (on the HP, puts the number on the stack)
Quick example
Same prices: (102.00 + 99.55 − 2 × 100.80) ÷ (0.0015² × 100.80) = −0.05 ÷ 0.0002268 = −220.5. Negative, so the price gains less on the way down in rates than it loses on the way up.
- 102+99.55−201.6=÷0.0015÷0.0015÷100.8= −220.5
- 102ENTER99.55+201.6−0.0015÷0.0015÷100.8÷ −220.5
ENTER enter (on the HP, puts the number on the stack)
Price change with effective duration and convexity
Quick example
Effective duration 7.20, effective convexity −210, benchmark curve up 75 bp. Duration: −7.20 × 0.0075 = −5.400%. Convexity: ½ × (−210) × 0.0075² = −0.591%. Total −5.99%.
- 7.2×0.0075=+|-STO1
- 0.5×210+|-×0.0075×0.0075+RCL1= −0.059906
+|- change sign · STO store · RCL recall a stored number
- 7.2ENTER0.0075×CHS
- 0.5ENTER210CHS×0.0075×0.0075×+ −0.059906
ENTER enter (on the HP, puts the number on the stack) · CHS change sign
Key rate duration
Quick example
Shift only the 5-year benchmark rate by ±1 bp. Prices: PV₋ 100.0190, PV₀ 100.0000, PV₊ 99.9810. ΔPV ÷ Δr is estimated by (100.019 − 99.981) ÷ (2 × 0.0001), so KRD₅ = 0.038 ÷ (2 × 0.0001 × 100) = 1.90.
- 100.019−99.981=÷2÷0.0001÷100= 1.9
- 100.019ENTER99.981−2÷0.0001÷100÷ 1.9
ENTER enter (on the HP, puts the number on the stack)
Quick example
Key rate durations of 0.50 (2-year), 1.20 (5-year), 2.30 (10-year) and 1.65 (30-year) give an effective duration of 0.50 + 1.20 + 2.30 + 1.65 = 5.65.
- 0.5+1.2+2.3+1.65= 5.65
- 0.5ENTER1.2+2.3+1.65+ 5.65
ENTER enter (on the HP, puts the number on the stack)
Quick example
KRDs 0.60 (2-year), 2.20 (5-year), 3.40 (10-year). Rates move +15, 0 and −25 bp. −[0.60 × 0.0015 + 2.20 × 0 + 3.40 × (−0.0025)] = −[0.0009 − 0.0085] = +0.76%.
- 0.6×0.0015=STO1
- 3.4×0.0025+|-+RCL1=+|- 0.0076
STO store · +|- change sign · RCL recall a stored number
- 0.6ENTER0.0015×3.4ENTER0.0025CHS×+CHS 0.0076
ENTER enter (on the HP, puts the number on the stack) · CHS change sign
Empirical vs analytical duration
Quick example
Analytical duration 5.0. In a stress month the benchmark yield falls 60 bp and the credit spread widens 45 bp, so the bond’s own yield falls 15 bp and its price rises about 5.0 × 0.0015 = 0.75%. Against the benchmark move: 0.0075 ÷ 0.0060 = 1.25.
- 5×0.0015=÷0.006= 1.25, both yields fall so the signs cancel
- 5ENTER0.0015×0.006÷ 1.25, both yields fall so the signs cancel
ENTER enter (on the HP, puts the number on the stack)
Module 14 Credit Risk
Credit risk: probability of default and loss given default
Quick example
A GBP 3,000,000 secured bond with GBP 45,000 of accrued interest, backed by assets now worth GBP 1,100,000. 3,000,000 + 45,000 − 1,100,000 = GBP 1,945,000.
- 3000000+45000−1100000= 1,945,000
- 3000000ENTER45000+1100000− 1,945,000
ENTER enter (on the HP, puts the number on the stack)
Quick example
Expected exposure GBP 1,945,000 and a recovery rate of 45%. 1,945,000 × (1 − 0.45) = GBP 1,069,750.
- 1−0.45=×1945000= 1,069,750
- 1ENTER0.45−1945000× 1,069,750
ENTER enter (on the HP, puts the number on the stack)
Quick example
A one-year probability of default of 2% on a loss given default of GBP 1,069,750. 0.02 × 1,069,750 = GBP 21,395.
- 1069750×0.02= 21,395
- 1069750ENTER0.02× 21,395
ENTER enter (on the HP, puts the number on the stack)
Quick example
POD 1.2% a year and LGD 45% give an expected loss of 0.54%, or 54 bp. A bond paying a 70 bp spread is more than fairly compensated (70 > 54).
- 0.012×0.45= 0.0054, so 54 bp
- 0.012ENTER0.45× 0.0054, so 54 bp
ENTER enter (on the HP, puts the number on the stack)
What drives credit spreads
Quick example
A 5-year zero-coupon bond priced at 82.10 per 100 of par. (100 ÷ 82.10)^(1/5) − 1 = 4.0235%.
- 100÷82.1=y^x51/x=−1= 0.040235
y^x power · 1/x reciprocal
- 100ENTER82.1÷51/xy^x1− 0.040235
ENTER enter (on the HP, puts the number on the stack) · 1/x reciprocal · y^x power
Quick example
A 5-year zero is quoted 82.10 bid and 82.60 offer. Bid yield 4.0235%, offer yield 3.8972%. Difference 0.1262%, or 12.6 bp.
- 100÷82.1=y^x51/x=STO1
- 100÷82.6=y^x51/x=+|-+RCL1= 0.001262
y^x power · 1/x reciprocal · STO store · +|- change sign · RCL recall a stored number
- 100ENTER82.1÷51/xy^xSTO1
- 100ENTER82.6÷51/xy^xCHSRCL1+ 0.001262
ENTER enter (on the HP, puts the number on the stack) · 1/x reciprocal · y^x power · STO store · CHS change sign · RCL recall a stored number
Quick example
A bond yields 4.62% against a 2.35% benchmark: a 227 bp yield spread. With a liquidity spread of 8 bp, the credit part is about 219 bp.
- 4.62−2.35=×100=−8= 219
- 4.62ENTER2.35−100×8− 219
ENTER enter (on the HP, puts the number on the stack)
Price impact of spread changes
Quick example
Modified duration 9.1 and a spread that tightens by 35 bp. −9.1 × (−0.0035) = +3.185%.
- 9.1×0.0035= 0.03185, so +3.185%
- 9.1ENTER0.0035× 0.03185, so +3.185%
ENTER enter (on the HP, puts the number on the stack)
Quick example
Modified duration 4.8, convexity 30 (already rescaled), spread widens 60 bp. −4.8 × 0.006 = −2.880%; ½ × 30 × 0.006² = +0.054%. Total −2.826%.
- 4.8×0.006=+|-STO1
- 0.5×30×0.006×0.006+RCL1= −0.02826
+|- change sign · STO store · RCL recall a stored number
- 4.8ENTER0.006×CHS
- 0.5ENTER30×0.006×0.006×+ −0.02826
ENTER enter (on the HP, puts the number on the stack) · CHS change sign
Quick example
A data service shows convexity 0.62 for a bond with modified duration 7.4. Used in the formula: 0.62 × 100 = 62, the same order as 7.4² ≈ 55.
- 0.62×100= 62
- 0.62ENTER100× 62
ENTER enter (on the HP, puts the number on the stack)
Module 15 Credit Analysis for Government Issuers
Sovereign credit analysis: the key ratios
Quick example
Debt of 195 and GDP of 300 (billions). 195 ÷ 300 = 0.65, so 65%. Higher is weaker.
- 195÷300= 0.65
- 195ENTER300÷ 0.65
ENTER enter (on the HP, puts the number on the stack)
Quick example
Debt of 195 and revenue of 78. 195 ÷ 78 = 2.5, so 250%. Higher is weaker.
- 195÷78= 2.5
- 195ENTER78÷ 2.5
ENTER enter (on the HP, puts the number on the stack)
Quick example
Interest of 7.8 on GDP of 300. 7.8 ÷ 300 = 0.026, so 2.6%. Higher is weaker.
- 7.8÷300= 0.026
- 7.8ENTER300÷ 0.026
ENTER enter (on the HP, puts the number on the stack)
Quick example
Interest of 7.8 on revenue of 78. 7.8 ÷ 78 = 0.10, so 10% of revenue goes on interest. Higher is weaker.
- 7.8÷78= 0.1
- 7.8ENTER78÷ 0.1
ENTER enter (on the HP, puts the number on the stack)
Quick example
GDP of USD 300 billion and 12 million people. 300,000 million ÷ 12 million = USD 25,000. Higher is stronger.
- 300000÷12= 25,000
- 300000ENTER12÷ 25,000
ENTER enter (on the HP, puts the number on the stack)
Quick example
Reserves of 45 and GDP of 300. 45 ÷ 300 = 0.15, so 15%. Higher is stronger.
- 45÷300= 0.15
- 45ENTER300÷ 0.15
ENTER enter (on the HP, puts the number on the stack)
Quick example
Reserves of 45 and external debt of 90. 45 ÷ 90 = 0.50: reserves cover half of what is owed abroad. Higher is stronger.
- 45÷90= 0.5
- 45ENTER90÷ 0.5
ENTER enter (on the HP, puts the number on the stack)
Quick example
Long-term external debt of 60 and GDP of 300. 60 ÷ 300 = 0.20, so 20%. Higher is weaker.
- 60÷300= 0.2
- 60ENTER300÷ 0.2
ENTER enter (on the HP, puts the number on the stack)
Quick example
External debt of 15 falls due within a year and GDP is 300. 15 ÷ 300 = 0.05, so 5%. Higher is weaker.
- 15÷300= 0.05
- 15ENTER300÷ 0.05
ENTER enter (on the HP, puts the number on the stack)
Agencies, supranationals and local government bonds
Quick example
Revenue 70, operating expenses 30, debt service (principal plus interest) 25. (70 − 30) ÷ 25 = 1.60×. Higher is stronger.
- 70−30=÷25= 1.6
- 70ENTER30−25÷ 1.6
ENTER enter (on the HP, puts the number on the stack)
Quick example
Operating expenses 30, debt service 25 and a 1.40× covenant. 30 + 1.40 × 25 = 65; revenue of 70 can fall 7.1% before a breach.
- 1.4×25+30= 65
- 1.4ENTER25×30+ 65
ENTER enter (on the HP, puts the number on the stack)
Module 16 Credit Analysis for Corporate Issuers
Financial ratios in credit analysis
Quick example
EBIT 84 on revenue 700. 84 ÷ 700 = 12.0%. Higher is better.
- 84÷700= 0.12
- 84ENTER700÷ 0.12
ENTER enter (on the HP, puts the number on the stack)
Quick example
EBIT 180 and interest expense 45. 180 ÷ 45 = 4.0×. Higher is better.
- 180÷45= 4
- 180ENTER45÷ 4
ENTER enter (on the HP, puts the number on the stack)
Quick example
EBIT 210 plus D&A 65. 210 + 65 = 275.
- 210+65= 275
- 210ENTER65+ 275
ENTER enter (on the HP, puts the number on the stack)
Quick example
Debt 1,320, EBIT 210, D&A 65. 1,320 ÷ (210 + 65) = 4.8×. Lower is better.
- 1320÷(210+65)= 4.8
- 1320ENTER210ENTER65+÷ 4.8
ENTER enter (on the HP, puts the number on the stack)
Quick example
Debt 700 and cash and marketable securities 100. 700 − 100 = 600.
- 700−100= 600
- 700ENTER100− 600
ENTER enter (on the HP, puts the number on the stack)
Quick example
Cash flow from operations 170 and dividends 40. 170 − 40 = 130.
- 170−40= 130
- 170ENTER40− 130
ENTER enter (on the HP, puts the number on the stack)
Quick example
CFO 170, dividends 40, debt 700, cash 100. 130 ÷ 600 = 21.7%. Higher is better.
- 170−40=÷(700−100)= 0.2167
- 170ENTER40−700ENTER100−÷ 0.2167
ENTER enter (on the HP, puts the number on the stack)
Quick example
Net income from continuing operations 70, D&A 30, deferred tax 6, other non-cash 4. 70 + 30 + 6 + 4 = 110.
- 70+30+6+4= 110
- 70ENTER30+6+4+ 110
ENTER enter (on the HP, puts the number on the stack)
Quick example
FFO 110 and debt 500. 110 ÷ 500 = 22.0%. Higher is better.
- 110÷500= 0.22
- 110ENTER500÷ 0.22
ENTER enter (on the HP, puts the number on the stack)
Quick example
Net income 120, working capital up 15, fixed-asset investment 60, net interest paid 10. 120 − 15 − 60 − 10 = 35.
- 120−15−60−10= 35
- 120ENTER15−60−10− 35
ENTER enter (on the HP, puts the number on the stack)
Seniority, recovery rates and notching
Quick example
Liquidation value V = 350. Secured claim S = 180 backed by collateral C = 130. Senior unsecured claims U = 220. Shortfall 50 joins U. (350 − 130) ÷ (220 + 50) = 220 ÷ 270 = 81.5%.
- (350−130)÷(220+180−130)= 0.8148
- 350ENTER130−220ENTER180+130−÷ 0.8148
ENTER enter (on the HP, puts the number on the stack)
Quick example
One-year POD 3%; the class recovers 25% in default. 0.03 × (1 − 0.25) = 2.25% a year.
- 1−0.25=×0.03= 0.0225
- 1ENTER0.25−0.03× 0.0225
ENTER enter (on the HP, puts the number on the stack)
Module 17 Securitization: How It Works and Why
Why securitize? Benefits for issuers, investors and markets
Quick example
Notes of 70% at +50 bp, 20% at +180 bp and 10% at +520 bp. 0.70 × 50 + 0.20 × 180 + 0.10 × 520 = 35 + 36 + 52 = 123 bp.
- (0.7×50)+(0.2×180)+(0.1×520)= 123
- 0.7ENTER50×0.2ENTER180×+0.1ENTER520×+ 123
ENTER enter (on the HP, puts the number on the stack)
Quick example
Corporate bond at +340 bp against a blend of 123 bp. 340 − 123 = 217 bp, or 2.17% a year on the amount raised.
- 340−123= 217
- 340ENTER123− 217
ENTER enter (on the HP, puts the number on the stack)
Module 18 Asset-Backed Securities, Covered Bonds and CDOs
Covered bonds
Quick example
A cover pool worth 2,142 backs covered bonds with a face value of 2,100. 2,142 ÷ 2,100 − 1 = 2.0% overcollateralization.
- 2142÷2100−1= 0.02
- 2142ENTER2100÷1− 0.02
ENTER enter (on the HP, puts the number on the stack)
Credit enhancement and credit tranching
Quick example
Loans worth 462 back notes of 420. 462 − 420 = 42 of overcollateralization.
- 462−420= 42
- 462ENTER420− 42
ENTER enter (on the HP, puts the number on the stack)
Quick example
Pool yield 8.4%, note coupon 5.1%, fees 0.6%. 8.4% − 5.1% − 0.6% = 2.7% a year.
- 0.084−0.051−0.006= 0.027
- 0.084ENTER0.051−0.006− 0.027
ENTER enter (on the HP, puts the number on the stack)
Quick example
Senior 200, mezzanine 50, junior 30; pool losses 65. The junior absorbs 30; the mezzanine bears 65 − 30 = 35 and keeps 15.
- 65−30= 35
- 65ENTER30− 35
ENTER enter (on the HP, puts the number on the stack)
Quick example
Mezzanine 50 and junior 30 sit below the senior tranche. The senior tranche loses nothing until pool losses exceed 50 + 30 = 80.
- 50+30= 80
- 50ENTER30+ 80
ENTER enter (on the HP, puts the number on the stack)
Credit card, auto loan and solar ABS
Quick example
Card ABS notes of 600 pay 5.4% a year. 600 × 0.054 ÷ 12 = 2.7 of interest a month.
- 600×0.054=÷12= 2.7
- 600ENTER0.054×12÷ 2.7
ENTER enter (on the HP, puts the number on the stack)
Quick example
A pool of 200 at 0.5% a month with a payment of 3.87. Interest is 200 × 0.005 = 1.00, so principal is 3.87 − 1.00 = 2.87.
- 200×0.005= 1
- +|-+3.87= 2.87
+|- change sign
- 3.87ENTER200ENTER0.005×− 2.87
ENTER enter (on the HP, puts the number on the stack)
Collateralized debt obligations and CLOs
Quick example
Loans of 400 yield 6.5% (26), debt of 360 costs 4.0% (14.4), the manager fee is 0.5% of the pool (2), no defaults. 26 − 14.4 − 2 = 9.6 left for equity.
- (400×0.065)−(360×0.04)−(400×0.005)= 9.6
- 400ENTER0.065×360ENTER0.04×−400ENTER0.005×− 9.6
ENTER enter (on the HP, puts the number on the stack)
Quick example
Cash left for equity of 9.6 on an equity tranche of 40. 9.6 ÷ 40 = 24%.
- 9.6÷40= 0.24
- 9.6ENTER40÷ 0.24
ENTER enter (on the HP, puts the number on the stack)
Quick example
Collateral of 690 against 600 of notes. 690 ÷ 600 = 1.15. If this falls below the trigger, cash goes to the senior notes.
- 690÷600= 1.15
- 690ENTER600÷ 1.15
ENTER enter (on the HP, puts the number on the stack)
Module 19 Mortgage-Backed Securities: RMBS, CMOs and CMBS
Prepayment risk and time tranching
Quick example
A tranche of 150 receives 50 in year 1, 40 in year 2, 35 in year 3 and 25 in year 4. (1 × 50 + 2 × 40 + 3 × 35 + 4 × 25) ÷ 150 = 2.23 years.
- (1×50)+(2×40)+(3×35)+(4×25)= 335
- ÷150= 2.2333
- 1ENTER50×2ENTER40×+3ENTER35×+4ENTER25×+ 335
- 150÷ 2.2333
ENTER enter (on the HP, puts the number on the stack)
Residential mortgage loans: the features that matter
Quick example
A loan of JPY 26,000,000 on an apartment worth JPY 40,000,000. 26,000,000 ÷ 40,000,000 = 65.0%.
- 26000000÷40000000= 0.65
- 26000000ENTER40000000÷ 0.65
ENTER enter (on the HP, puts the number on the stack)
Quick example
Mortgage payment 1,800 and other debt payments 350 a month; gross income 78,000 a year. 2,150 ÷ (78,000 ÷ 12) = 33.1%.
- 1800+350= 2,150
- ÷(78000÷12)= 0.3308
- 1800ENTER350+ 2,150
- 78000ENTER12÷ 6,500
- ÷ 0.3308
ENTER enter (on the HP, puts the number on the stack)
Mortgage pass-throughs and CMOs
Quick example
Loans of 250 at 3.2%, 350 at 3.8% and 400 at 4.7%. (250 × 3.2 + 350 × 3.8 + 400 × 4.7) ÷ 1,000 = 4.01%.
- (250×3.2)+(350×3.8)+(400×4.7)= 4,010
- ÷1000= 4.01
- 250ENTER3.2×350ENTER3.8×+400ENTER4.7×+ 4,010
- 1000÷ 4.01
ENTER enter (on the HP, puts the number on the stack)
Quick example
Loans of 400 with 280 months left and 600 with 180 months left. (400 × 280 + 600 × 180) ÷ 1,000 = 220 months.
- (400×280)+(600×180)= 220,000
- ÷1000= 220
- 400ENTER280×600ENTER180×+ 220,000
- 1000÷ 220
ENTER enter (on the HP, puts the number on the stack)
Quick example
WAC of 4.55% and fees of 0.48%. 4.55 − 0.48 = 4.07%.
- 4.55−0.48= 4.07
- 4.55ENTER0.48− 4.07
ENTER enter (on the HP, puts the number on the stack)
Quick example
Tranche A of 160 is first in line. Pool principal in year 1: 14 scheduled and 36 prepaid. 160 − 14 − 36 = 110.
- 160−14−36= 110
- 160ENTER14−36− 110
ENTER enter (on the HP, puts the number on the stack)
Commercial mortgage-backed securities (CMBS)
Quick example
Rent EUR 4.1 million, cash operating expenses 1.6 million, replacement reserves 0.3 million. (4.1 − 1.6) − 0.3 = EUR 2.2 million.
- 4.1−1.6−0.3= 2.2
- 4.1ENTER1.6−0.3− 2.2
ENTER enter (on the HP, puts the number on the stack)
Quick example
NOI of 2.2 and annual debt service (interest plus principal) of 1.8. 2.2 ÷ 1.8 = 1.22×.
- 2.2÷1.8= 1.2222
- 2.2ENTER1.8÷ 1.2222
ENTER enter (on the HP, puts the number on the stack)
Quick example
Three points to refinance a loan with GBP 6.5 million outstanding. 3 × 1% × 6,500,000 = GBP 195,000.
- 3×0.01×6500000= 195,000
- 3ENTER0.01×6500000× 195,000
ENTER enter (on the HP, puts the number on the stack)
No formulas match these filters.