IB 2022 Regular exam Exercise 1 a. As seen in table 1 customer confidence πΆ0 decreases in all countries in the eurozone. ππ ππππ€ π‘βππ‘ πΆ = π0 + π1 (π − π) therefore, the decrease in customer confidence will lead to a negative effect on our Y as we know that π = πΆ + πΌ + πΊ (in a closed economy). In an IS-LM Diagram our IS curve will therefore shift left. This will lead to π < πΜ as we move leftwards along the PC curve. And therefore, increase unemployment. And a decrease in nominal wages. b. Central bank will implement an expansionary monetary policy and decrease the interest rate. By doing this output will go back to its initial level, and inflation will now move back rightwards along PC curve. This will increase investments which will increase demand which will increase output, decrease in unemployment, and increase wages. c. The new equilibrium will have lower consumptions but higher investments, as we decrease the interest rate. d. When m is increased, nominal wages is unaffected, however real wages will decrease as prices will increase, and our PS curve will shift down, and unemployment will increase, which leads do a decrease in output. Our rate of inflation at our natural level of output will therefore be higher than targeted. e. Contractionary moneytary policy Exercise 2 π°πΊ π¬πππππππ: π°πΊ = π = −ππππππ + ππππ πΏπ = π = π − πΜ πΏπ = π = 0.04 − 0.02 π³π΄ π¬πππππππ: π³π΄ = π = π. ππ ππ’π‘ππ’π‘ πππ’πππππππ’π = π π = −10000 · 0.02 + 2200 π = −200 + 2200 πΆπππππ π¬ππππππππππ = ππππ ππ· = π − π π π· = 2000 − 500 π«πππππππππ π°πππππ = ππ« = ππππ πΆ = π0 + π1 π − π1 π πΆ = 200 + 0.6 · 2000 − 0.6 · 500 πΆ = 200 + 1200 − 300 πͺππππππππππ = πͺ = ππππ πΌ = πΌ0 + π1 π − π2 (π + π₯) πΌ = 100 + 0.15 · 2000 − 2500(0.02 + 0) πΌ = 100 + 300 − 50 π°πππππππππ = π° = πππ 1 [π − π1 π + πΌ0 − π2 (π + π₯) + πΊ] 1 − π1 − π1 0 1 [200 − 0.6 · 500 + 50 − 2500(0.02 + 0) + 550] = 1 − 0.6 − 0.15 = 4[200 − 300 + 50 − 50 + 550] = 4[450] = 1800 π= π π π π Autonomous spending is now 50 lower. Output is 200 lower. This is due to the multiplier on 4. πΆ = π0 + π1 π − π1 π πΆ = 200 + 0.6 · 1800 − 0.6 · 500 πΆ = 200 + 1080 − 300 πͺππππππππππ = πͺ = πππ πΌ = πΌ0 + π1 π − π2 (π + π₯) πΌ = 50 + 0.15 · 1800 − 2500(0.02 + 0) πΌ = 50 + 270 − 50 π°πππππππππ = π° = πππ 2000 = −10000π + 2000 0 = −10000π π=0 πΌππ‘ππππ π‘ πππ‘π βππ π‘π ππ π‘π 0.02 π‘π ππ πππ’ππ π‘π ππ₯ππππ‘ππ ππππππ‘πππ π‘π ππ 0 yes, at zero lower bound π = π π π‘π ππ π = 0 When π π πππππππ ππ π‘π 0, π ππππ π‘βπ π πππ, πππ ππππ‘πππ ππππ πππ π‘βπππππππ πβππππ π π‘π πππ‘ π = 0 E No they can not keep
