Answers to the problems

Supply & Demand

1. a)

      $D = S$  or

      $36-6P = 3P$ 

      $9P = 36$ 

      $P = 4$ 
 
      $S$  = 12 and  $D$  = 12


  b)

      $D = S$  or

      $15-4P = P$ 

      $5P = 15$ 

      $P$  = 3

      $S = 3$  and  $D = 3$ 

  c)

      $D = S$  or

      $15-3P = -3+3P$ 

      $6P = 18$ 

      $P = 3$ 

      $S = 6$  and  $D = 6$

2. Supplier bases his output decision on the fact he gets the market price plus the subsidy.

      $D = S$   or

      $36-6P = 3(P+3)$ 

     9 $P$  = 27

      $P$  = 3

      $S$  = 18 and  $D$  = 18

Supplier receives $6 for each unit produced. The supply curve has shifted to the right. 3. The consumer to buy an item must pay the market price plus the tax.

      $D = S$  or

      $15-3(P+2) = -3+3P$ 

     6 $P$  = 12

      $P$  = 2

      $S = 3$  and  $D = 3$

The consumer must pay $4 out of pocket for each unit consumed. The demand curve has shifted to the left.

4. Price dynamics

      $t$       $P(t)$    $S(t)$    $D(t) Þ(P(t)) 1      7     21    3      3
 
     2      4     12    12     0 
 
     3      4     12    12     0$

Period 2 is the equilibrium solution.

Index number problems

1. First what is the value of the market basket in each year?

     vmb(1985) = 10 x 1 + 10 x 1 + 10 x 1 = 30

     vmb(1990) = 10 x 2 + 10 x 2 + 10 x 2 = 60

     vmb(1995) = 10 x 1 + 10 x 3 + 10 x 2 = 60

Now use the formula CPI(a,b) = (vmb(a)/vmb(b)) x 100

     CPI(90,85) = (60/30) x 100 = 200

     CPI(95,90) = (60/60) x 100 = 100

     CPI(85,90) = (30/60) x 100 = 50

2. The nominal GNP in 1995 is

     (200 x 1500)/100 = 3000

The GNP deflator in 1990 is

     (1500/1000) x 100 = 150

Maoeconomic Models I

     a.   $Y = k(a + I)$  where  $k = 1/(1-b)$ 

          $k$  = 1/(1 - 0.75) = 1/(0.25)

           = 100/25

           = 4

          $Y$  = 4(200 + 200)

           = 4(400)

           = 1600

     b.  $Þ(Y)= kÞ(I)$ 

            = 4(20)

            = 80

     a.  $k$  = 1/(1 - 0.8)

          = 1/(0.2)

          = 10/2

          = 5

         $Y$  = 5(150 + 200)

          = 5(350)

          = 1750

     b.  $Þ(Y) = kÞ(I)$ 

             = 5(10)

             = 50

3. a. Initial equilibrium

Y

₀ :

        $Y$ ₀  = k(a + I₀ )

           = 4(250 + 150)

           = 1600

  b. Dynamics: Set up table

      $t$        $a + I$ _t      $bY$ _t-1      $Y$ _t
 
     0       400       1200     1600
 
     1       420       1200     1620 

     2       420       1215     1635
  
     3       420       1226     1646+
 

     Details for period three:

      $a + I$ ₃ = 420

      $bY$ ₂ = .75(1635) = 1226.25 approx 1226

      $Y$ ₃ = bY₂ + (a + I₃) = 1646.25 approx 1646

     c. Final equilibrium  $Y$ _oo:

        $Y$ _oo  = 4(420)

            = 1680

Maoeconomic Model II

     a.  $Y = k(a + I + G - bT)$ 

          = 4(200 + 200 + 200 - 0.75(200))

          = 4(600-150)

          = 1800

     b.  $200 = 4Þ(G)$ 

         $Þ(G)$ = 50

        200 = -0.75(4) $Þ(T)$ 

         $Þ(T)$ = -66.7

        200 = 4 $Þ G-0.75(4)Þ(T)$  with  $Þ (G) = Þ(T)$ 

        200 = (4 - 0.75(4)) $Þ(G)$ 

        200 =  $Þ(G)$  and  $Þ(T)$ = 200

2. The answer is obtained by substituting up the stack.

     2 into 1:  $Y = a + b(Y - T)  + I + G$ 

     3 into above:  $Y = a + b(Y - T) + I + T$ 

     4 into above:  $Y = a + b(Y - tY) + I + tY$ 

      $Y = bY - btY + tY + a + I$ 

      $(1 - b - t + bt)Y = a + I$ 

      $Y = k(a + I)$ 

     where  $k = 1/(1-b-t+bt)$

3. Note: Under exam conditions you will be given the equilibrium solution (in letters) the definition of

A

and

K

(in letters) and the numerical value of

K

     a.  $Y = K[A + (c/f)Ms/p]$ 

        where  $A = a + I + G - bT$ 

                = 200 + 600 + 200 - 0.75(100)

                = 925

        and  $K = k/[1 + (ekc/f)]$ 

              = 4/[1 + (0.25)(4)(2000)/1200]

              = 4/[1 + 10/6]

              = 4/(16/6)

              = 4(6/16)

              = 6/4

              = 3/2

         $Y$  = 1.5[925 + (2000/1200)(300/1)]

          = 1.5(1425)

          = 2137.5

     b.  $Þ(Y)= KÞ(G)$ 

         60 = 1.5 $Þ(G)$ 

         $Þ(G)$ = 40

     c.  $eÞ(Y) - Þ(i)$ = 0

         $Þ(i)= (e/f)Þ(Y)$ 

         $Þ(I)= -(ce/f)Þ(Y)$ 

            = -[(2000)(0.25)/1200]60

            = -25

     d.  $Þ(Y)= kÞ(G)$ 

         $Þ(Y)$ = 4(40)

            = 160

        160 =  $K(Þ()G + (c/f)Þ(Ms)/p)$ 

            = 1.5(40 + 2.5 $Þ(Ms)/p)$ 

         $Þ(Ms)/p$ = 40

  Substitute the 2nd eqn into the 1st

      $Y = a + bY - bT + I + G$ 

     Substitute 3rd eqn into the above

      $Y = a + bY - bT + I$ ₀  - ci + dY + G

     Collect terms

      $Y = bY + dY + a - bT + I$ ₀  - ci + G

      $Y = (b + d)Y + a - bT + I$ ₀  - ci + G

      $Y - (b + d)Y = a - bT + I$ ₀  - ci + G

      $(1 - (b + d))Y = a - bT + I$ ₀  - ci + G

      $Y = k$ ₀ (a - bT + I₀  - ci + G)

     where  $k$ ₀  = 1/[1 - (b + d)]

Bank expansion problems

1. Buy. Put high powered money into bank reserves.

2. Interest rates down. When FED buys at high price the effective yield is lowered.

3. Bank 1: Left-- FED buys $10M in governemt securities. Right-- Bank makes loan of $10M

    $A$          $L$            $A$          $L$  
  
   R  20    DD 100     10        100
  
   GS 20               20
   
   L  60               70

Bank 2: Left-- Customer deposits &10(paid by loan). Right-- Bank loans max

    $A$           $L$            $A$           $L$ 
 
 
   R  20     DD 110     11        110 
      
   GS 30                30 
 
   L  60                69

4. To make a profit banks convert reserves to interest paying assets.

5. Mathematically a contraction has a minus instead of a plus. In a contraction banks must contract the amount of loans to obtain the required reserves. They would do this by failing to lend out money when old loans are repaid.

6. $Þ(M)$ = (1/.1) x 10 = 100

7. In the real world there are leakges such as some money being kept as free reserves, in coin instead of demand deposits, and going abroad instead of into the US banking system.