1. Copyright © 2000 Addison Wesley Longman Slide #22-1 Chapter Twenty Two RISK MANAGEMENT IN FINANCIAL INSTITUTIONS Part IV The Management of Financial Institutions

2. Copyright © 2000 Addison Wesley Longman Slide #22-2 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. Copyright © 2000 Addison Wesley Longman Slide #22-3 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. Copyright © 2000 Addison Wesley Longman Slide #22-4 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. Copyright © 2000 Addison Wesley Longman Slide #22-5 Duration of First National Bank's Assets and Liabilities

6. Copyright © 2000 Addison Wesley Longman Slide #22-6 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 DUR gap = DUR a - [L/A x DUR l ] = 2.7 - [(95/100) x 1.03] = 1.72 %ΔNW = - DUR gap x Δi/(1+i) = - 1.72 x.05/(1+.10) = -.078 = -7.8% ΔNW= -.078 x $100m = -$7.8m

7. Copyright © 2000 Addison Wesley Longman Slide #22-7 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. Copyright © 2000 Addison Wesley Longman Slide #22-8 Duration of Finance Company's Assets and Liabilities

9. Copyright © 2000 Addison Wesley Longman Slide #22-9 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 DUR gap =DUR a - [L/A x DUR l ] =1.16 - [90/100 x 2.77] = -1.33 years %ΔNW=- DUR gap X Δi /(1+i) =-(-1.33) x.05/(1+.10) =.061 = 6.1%

10. Copyright © 2000 Addison Wesley Longman Slide #22-10 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. Copyright © 2000 Addison Wesley Longman Slide #22-11 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 DUR gap = 0 1.Reduce DUR a = 0.98  DUR gap = 0.98 - [(95/100) x 1.03] = 0 2.Raise DUR l = 2.80  DUR gap = 2.7 - [(95/100) x 2.80] = 0