ECO102 Tutorial 10

• On test
  • Basic AE model, change in NX, offset with G
  • Solow growth model.
  • AS/AD question
• 1
  • $0$ coupon bond. $FV$ of $1000$
  • $r=0.1$
  • Price is $800$.
  • Yield is?
  • Face Value: How much they're lending to you.
    • How much that has to get repaid, minimum.
  • Matures in 1 year. So you have to pay $1000$ in 1 year.
  • You require interest rate as a compensation for lending money to someone else. In this market it's $r=0.1$ This doesn't correlate / influence price of bond.
  • Yield: $\frac{FV}{Price}=\frac{1000}{800}=\frac{5}{4}=1.25-1=0.25$ is the yield.
• 2
  • This adds a stipulation to the bond.
  • Coupon payments.
    • They used to come in pieces of paper with coupons at the bottom.
    • Then you used to go to the bank and cash those coupons in for money.
  • $FV=1000$
  • Matures in 2 years.
  • Let's say every year we need to pay $400$.
  • $CF$: Cash flow.
    • Year 1: $400$
    • Year 2: $400$ + $1000$ (face value)
  • Discount the cash based on interest rates and 'expected' rates of return basically.
    • $PV=\frac{400}{1.1}+\frac{400}{{1.1}^2}+\frac{1000}{{1.1}^2}=1520.66115702479$
  • Par value is Equilibrium price.
  • If face value is $1000$
  • And price is $1200$, then it's above par.
  • In the first question the price was $800$ and the face value was $1000$, so it was below par.
• 3
  • GDP growth is $0.02$
  • $MS=0.06$
  • Quantity theory of money
  • $\mathrm{\Delta }\mathrm{Inflation}=\mathrm{\Delta }\mathrm{MS}-\mathrm{\Delta }\mathrm{Y}$
  • $\mathrm{\Delta }\mathrm{Inflation}=0.06-0.02=0.04$
• 4
  • $m=0.7$
  • $CD=0.1$
  • $\text{Issues}=\mathrm{\$}12$
  • Cash drainage
  • And reserve ratio
  • Target reserve of $0.2$
  • Cash drainage is $0.1$
  • $\mathrm{\Delta }\mathrm{MS}=\frac{\text{Issuance}}{\text{Reserve Ratio}+\text{Cash Drainage}}$
    • $\frac{12}{0.2+0.1}=40.0$
  • Money supply isn't how much we issue into the economy
  • It depends on cash drainage and reserve ratio.
• 5
  • $MD=500-100r$
  • $\text{Issue}=\mathrm{\$}60$
  • $RR=0.1$
  • $CD=0.2$
  • Calculate the equilibrium interest rate.
  • $\mathrm{MS}=?$
  • $\mathrm{MS}=\frac{\text{Issuance}}{\text{Reserve Ratio}+\text{Cash Drainage}}$
  • $\mathrm{MS}=\frac{60}{0.1+0.2}=200.0$
  • Supply and demand for money
  • $MS=MD$
    • $200=500-100r$
    • $r=3$
    • $300\mathrm{\%}$ interest rate
    • Doesn't make sense, but this is the real solution.
• 6
  • $MD=1200-20r$
  • $\text{Issues}=\mathrm{\$}200$
  • $RR=0.25$
  • $CD=0$
  • Equilibrium interest rate are parity
  • Provide a pair of diagrams of MD and MS and interest rate parity to explain the relationship between the interest rate and the exchange rate. Point A
  • BOC buys $\mathrm{\$}50$ from the private sector, how does everything change? Point B
  • Point A:
    • $MD=1200-20r$
    • $MS=\frac{200}{0.25}=800.0$
    • $MS=MD$
      • $800=1200-20r$
      • $r=20$
      • $2000\mathrm{\%}$ interest rate ??!??!?!?!
      • Correct though
    • Exchange rate is determined by interest rate parity.
      • $E=\frac{1+r_{\text{foreign}}}{1+r_{\text{domestic}}}{E}_0$
      • Don't use decimals in this case.
      • $E=\frac{15}{20}=\frac{3}{4}=0.75$
      • So the exchange rate is $0.75$.
  • Point B:
    • We have to adjust the money supply because the BOC is buying $\mathrm{\$}50$ from the private sector.
    • This is quantitative easing. Taking money out of the market to stimulate the economy.
    • $\mathrm{MS}=\frac{200-50}{0.25+0}=600.0$
      • We do $-50$ because we've taken $50$ out of the market for $n$ years ($n$ is maturity on bonds).
    • $MS=MD$
      • $1200-20r=600$
      • $r=30$
      • $3000\mathrm{\%}$ interest rate ?!?!?!
    • Interest rate parity

      • $E=\frac{15}{30}=\frac{1}{2}=0.5$

      • So the exchange rate is $0.5$.

        left=-10; right=1300;
        top=70; bottom=-10;
        ---
        x=1200-20y
        x=600|dashed
        x=800|dashed
        y=20|dashed
        y=30|dashed
        (600,30)
        (800,20)
        

      • IRP always has a curve

        left=-2; right=10;
        top=40; bottom=-2;
        ---
        y = e^{-(x - (\ln(20) + 0.75))}
        y=20|dashed
        y=30|dashed
        x=0.75|dashed
        x=0.5|dashed
        (0.75,20)
        (0.5,30)