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Chapter Twenty Two

Part IV The Management of Financial Institutions. Chapter Twenty Two. RISK MANAGEMENT IN FINANCIAL INSTITUTIONS. Managing Credit Risk. Solving Asymmetric Information Problems 1. Screening 2. Monitoring and Enforcement of Restrictive Covenants 3. Specialize in Lending

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Chapter Twenty Two

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  1. Part IV The Management of Financial Institutions Chapter Twenty Two RISK MANAGEMENT IN FINANCIAL INSTITUTIONS

  2. Managing Credit Risk • Solving Asymmetric Information Problems • 1. Screening • 2. Monitoring and Enforcement of Restrictive Covenants • 3. Specialize in Lending • 4. Establish Long-Term Customer Relationships • 5. Loan Commitment Arrangements • 6. Collateral and Compensating Balances • 7. Credit Rationing

  3. Managing Interest-Rate Risk • First National Bank • Assets Liabilities • --------------------------------------------------------------------------------------------------------------------- • Reserves and cash items $ 5 m | Checkable deposits $ 15 m • | • Securities | Money market deposit accounts $ 5 m • less than 1 year $ 5 m | • 1 to 2 year $ 5 m | Savings deposits $ 15 m • greater than 2 year $ 10 m | • | CDs: Variable-rate $10 m • Residential mortgages | less than 1 year $ 15 m • Variable rate $ 10 m | 1 to 2 year $ 5 m • Fixed rate (30 year) $ 10 m | greater than 2 year $ 5 m • | • Commercial Loans | Fed funds $ 5 m • less than 1 year $ 15 m | • 1 to 2 year $ 10 m | Borrowings: less than 1 year $10 m • greater than 2 year $ 25 m | 1 to 2 year $ 5 m • | greater than 2 year $ 5 m • Physical capital $ 5 m | • | Bank capital $ 5 m

  4. Income Gap Analysis • Rate-Sensitive Assets = $5m + $ 10m + $15m + 20% x $20m • RSA = $32 m • Rate-Sensitive Liabs = $5m + $25m + $5m+ $10m + 10% x $15m • + 20%x$15m • RSL = $49.5 m • i  5%  • ΔAsset Income = + 5% x $32.0m = + $ 1.6m • ΔLiability Costs = + 5% x $49.5m = + $ 2.5m • ΔIncome = $ 1.6m - $ 2.5 = - $0.9m • If RSL > RSA, iNIM, Income • GAP = RSA - RSL • = $32.0m - $49.5m = -$17.5m • ΔIncome = GAP x Δi • = - $17.5m x 5% = -$0.9m

  5. Duration of First National Bank's Assets and Liabilities

  6. Duration Gap Analysis • %ΔP - DUR x Δi/(1+i) • i 5%, from 10% to 15%  • ΔAsset Value = %ΔP x Assets • = -2.7 x .05/(1+.10) x $100m • = -$12.3m • ΔLiability Value = %ΔP x Liabilities • = -1.03 x .05/(1+.10) x $95m • = -$4.5m • ΔNW = -$12.3m - (-$4.5m) = -$7.8m • DURgap = DURa - [L/A x DURl] • = 2.7 - [(95/100) x 1.03] • = 1.72 • %ΔNW = - DURgap x Δi/(1+i) • = - 1.72 x .05/(1+.10) • = -.078 = -7.8% • ΔNW = -.078 x $100m • = -$7.8m

  7. Example of Finance Company • Friendly Finance Company • Assets Liabilities • ------------------------------------------------------------------------------------------------- • Cash and Deposits $ 3 m | Commercial Paper $ 40 m • | • Securities | Bank Loans • less than 1 year $ 5 m | less than 1 year $ 3 m • 1 to 2 year $ 1 m | 1 to 2 year $ 2 m • greater than 2 year $ 1 m | greater than 2 year $ 5 m • | • Consumer Loans | Long-Term Bonds • less than 1 year $ 50 m | and other long-term • 1 to 2 year $ 20 m | debt $ 40 m • greater than 2 year $ 15 m | • | Capital $ 10 m • Physical capital $ 5 m |

  8. Duration of Finance Company's Assets and Liabilities

  9. Gap and Duration Analysis • If i 5% • Gap Analysis • GAP = RSA - RSL = $55 m - $43 m = $12 million • ΔIncome = GAP x Δi = $12 m x 5% = $0.6 million • Duration Gap Analysis • DURgap = DURa - [L/A x DURl] • = 1.16 - [90/100 x 2.77] = -1.33 years • %ΔNW = - DURgap X Δi /(1+i) • = -(-1.33) x .05/(1+.10) • = .061 = 6.1%

  10. Managing Interest-Rate Risk • Problems with GAP Analysis • 1. Assumes slope of yield curve unchanged and flat • 2. Manager estimates % of fixed rate assets and liabilities that are rate sensitive

  11. Managing Interest-Rate Risk • Strategies for Managing Interest-Rate Risk • In example above, shorten duration of bank assets or lengthen duration of bank liabilities • To completely immunize net worth from interest-rate risk, set DURgap = 0 1. Reduce DURa = 0.98 DURgap = 0.98 - [(95/100) x 1.03] = 0 2. Raise DURl = 2.80 DURgap = 2.7 - [(95/100) x 2.80] = 0

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