ECO101 Tutorial 12

ECO101 - Week 12 Monopoly.pdf

• 1
  • Monopolist
  • Eliminate the efficiency loss
  • What's the per-unit subsidy
  • $MC=Q$
  • $P=300-Q$
  • When $P=MC$
    • $Q=300-Q$
    • $Q=150$
  • Solve for revenue
    • Because monopoly
    • $R=PQ$
    • $R=(300-q)q$
    • $R=300q-q^2$
    • $MR=300-2q$
    • Just your $p$, like demand with double slope.
  • $MR=MC$
    • $300-2q=q$
    • $q=100$
  • Find $P$
    • $P=(300-100)$
    • $P=200$
  • Find subsidy
    • $MR=MC-s$
    • $300-2q=q-s$
    • $s=3q-300$
    • $s=(3)(150)-300=150$
    • Subsidy is now $150$
  • Find new $MC$
    • $MC-s$
    • $M{C}^{\prime }=Q-150$
  • Efficiency Loss
    • $\frac{bh}{2}=0.5((200-100)(150-100))=2500$

        left=-0.5; right=350;
        top=400; bottom=-0.5;
        ---
        y=x
        y=x-150
        y=300-2x
        y=300-x
        (150,150)|label:P=MC
        (100,200)|label:MC=MR
        (100,100)
        y=200|dashed|0<=x<=100
        y=100|dashed|0<=x<=100
        y=150|dashed|0<=x<=150
        x=100|dashed|0<=y<=200
        x=150|dashed|0<=y<=150
        100<=x<=150
    

• 2
  • Monopolist
  • $P=40-q$
  • $C=\frac{1}{2}{q}^2$
  • $MC=Q$
  • $P=MC$
    • $40-q=q$
    • $q=20⟹p=20$
  • Revenue function
    • $R=PQ$
    • $R=(40-q)q=-q^2+40q$
  • Marginal revenue
    • $MR=-2q+40$
    • See this is double the slope of the demand function $P=40=q$
  • Need to graph
  • Fee
    • $\frac{bh}{2}$
    • $\frac{(40-20)(20-0)}{2}=200$

        left=-0.5; right=50;
        top=50; bottom=-0.5;
        ---
        f(x)=40-x
        g(x)=40-2x
        h(x)=x
        y<=f(x)|20<=y|0<=x<=20
    

    • Shaded area is fee, consumer surplus
  • Profits before fee
    • $\pi =R-C$
    • $\pi =(pq)-c$
    • $\pi =(20)(20)-\frac{1}{2}(20{)}^2=200$
  • Profits after fee
    • $\pi =r-c+\text{fee}$
    • $\pi =pq-c+\text{fee}$
    • $\pi =(200)+\text{fee}$
    • $\pi =(200)+200=400$
• 3
  • Ordinary Price Discrimination
  • Sell to tourists
  • $p_1=14-\frac{1}{2}{q}_1$
  • $p_1=10$
  • $p_2=10-\frac{1}{2}{q}_2$
  • $p_2=?$
  • Tourists
    • Look at tourists $p_1$
      • $10=14-\frac{1}{2}{q}_1$
      • $q_1=8$
      • $(8,10)$
    • Revenue function for tourists
      • $R=PQ$
      • $R=(14-\frac{1}{2}{q}_1)q_1=-\frac{q_1^2}{2}+14q_1$
      • $MR=-q_1+14$
    • $MR(8)=-(8)+14=6$
  • Residents
    • Revenue
      • $R=PQ$
      • $R=(10-\frac{1}{2}{q}_2)q_2=-\frac{q_2^2}{2}+10q_2$
    • Marginal
      • $MR=-q_2+10$
    • $MR$ tourists = $MR$ residents
      • $6=10-q_2$
      • $q_2=4$
      • $P=10-\frac{1}{2}(4)=8$
  • Graph
    • $MR$ and $D$
  • Residents

        left=-2; right=30;
        top=20; bottom=-2;
        ---
        y=-x+10
        y=10-0.5x
        (10,0)
        (20,0)
        (0,10)
        y=6|0<=x|dashed
        x=4|0<=y<=8|dashed
        y=8|0<=x<=4|dashed
        (0,6)
        (4,0)
        (0,8)
    

  • Tourists

        left=-2; right=30;
        top=20; bottom=-2;
        ---
        y=-x+14
        y=14-0.5x
        (0,14)
        (14,0)
        (28,0)
        y=6|0<=x|dashed
        x=8|0<=y<=10|dashed
        y=10|0<=x<=8|dashed
        (0,6)
        (8,0)
        (0,10)
    

• 4
  • $C=32q-\frac{5}{100}{q}^2$
  • $P=40-\frac{15}{100}q$
  • Find $MC$
    • $MC=C^{\prime }=32-\frac{10}{100}q$
    • $MC=P$
    • $32-\frac{10}{100}q=40-\frac{15}{100}q$
    • $q=160$
  • $P$
    • $p=40-\frac{15}{100}(160)=16$
  • Fee
    • Consumer surplus
    • $0.5(40-16)(160)=1920$

        left=-0.5; right=600;
        top=50; bottom=-0.5;
        ---
    	y=40-(15/100)x
    	y=32-(10/100)x
    	(160,16)|label:(160,16)
    	y=16|0<=x<=160|dashed
    	x=160|0<=y<=16|dashed
    	(0,32)
    	(0,40)
    

  • Profit before fee
    • $\pi =r-c$
    • $\pi =(pq)-c$
    • $\pi =\left((40-\frac{15}{100}q)q\right)-(32q-\frac{5}{100}{q}^2)=-\frac{q^2}{10}+8q$
    • $\pi =-\frac{(160{)}^2}{10}+(8)(160)=-1280$
  • Profits after fee
    • $\pi =-\frac{q^2}{10}+8q+\text{fee}$
    • $\pi =-\frac{q^2}{10}+8q+1920$
    • $\pi =-\frac{(160{)}^2}{10}+(8)(160)+1920=640$