Question

state consumption per worker. 5. Excel Problem. Suppose that you have a standard Solow model with a Cobb-Douglas production function. The central equation of the model can be written: ka+1 = sAk +(1-3)ka. Output per worker is given by: y = Ak Consumption per worker is given by: a = (1 - sy. (a) Suppose that A is constant at 1. Solve for an expression for the steady state capital per worker, steady state output per worker, and steady state consumption per worker. 105 (b) Suppose that a = 1/3 and d = 0.1. Create an Excel sheet with a grid of values of s ranging from 0.01 to 0.5, with a gap of 0.01 between entries (i.e. you should have a column of values 0.01, 0.02, 0.03, and so on). For each value of s, numerically solve for the steady state values of capital, output, and consumption per worker. Produce a graph plotting these values against the different values of s. Comment on how the steady state values of capital, output, and consumption per worker vary with s. (c) Approximately, what is the value of s which results in the highest steady state consumption per worker? Does this answer coincide with your analytical result on the previous question?

Answer 1

Ket= SAKE +1-8) ke Yt = Akta , et =(1-5) ift A- Qu steady state KE+1=kt = k* - k*= sket SR*& + (1-8) k* SKA = SKAL kta s kt r d = - se strady state y* = kxd steady state (*) (* = (-5) yt fe= 6-9472

For part b , the excel cannot be attached here. Here are the snaps of the working.

Savings rate	capital per worker	output per worker	consumption per worker
0.01	0.0316	0.3162	0.3131
0.02	0.0894	0.4472	0.4383
0.03	0.1643	0.5477	0.5313
0.04	0.2530	0.6325	0.6072
0.05	0.3536	0.7071	0.6718
0.06	0.4648	0.7746	0.7281
0.07	0.5857	0.8367	0.7781
0.08	0.7155	0.8944	0.8229
0.09	0.8538	0.9487	0.8633
0.1	1.0000	1.0000	0.9000
0.11	1.1537	1.0488	0.9334
0.12	1.3145	1.0954	0.9640
0.13	1.4822	1.1402	0.9920
0.14	1.6565	1.1832	1.0176
0.15	1.8371	1.2247	1.0410
0.16	2.0239	1.2649	1.0625
0.17	2.2165	1.3038	1.0822
0.18	2.4150	1.3416	1.1001
0.19	2.6190	1.3784	1.1165
0.2	2.8284	1.4142	1.1314
0.21	3.0432	1.4491	1.1448
0.22	3.2631	1.4832	1.1569
0.23	3.4881	1.5166	1.1678
0.24	3.7181	1.5492	1.1774
0.25	3.9528	1.5811	1.1859
0.26	4.1924	1.6125	1.1932
0.27	4.4366	1.6432	1.1995
0.28	4.6853	1.6733	1.2048
0.29	4.9385	1.7029	1.2091
0.3	5.1962	1.7321	1.2124
0.31	5.4581	1.7607	1.2149
0.32	5.7243	1.7889	1.2164
0.33	5.9947	1.8166	1.2171
0.34	6.2693	1.8439	1.2170
0.35	6.5479	1.8708	1.2160
0.36	6.8305	1.8974	1.2143
0.37	7.1171	1.9235	1.2118
0.38	7.4076	1.9494	1.2086
0.39	7.7019	1.9748	1.2047
0.4	8.0000	2.0000	1.2000
0.41	8.3019	2.0248	1.1947
0.42	8.6074	2.0494	1.1886
0.43	8.9167	2.0736	1.1820
0.44	9.2295	2.0976	1.1747
0.45	9.5459	2.1213	1.1667
0.46	9.8659	2.1448	1.1582
0.47	10.1894	2.1679	1.1490
0.48	10.5163	2.1909	1.1393
0.49	10.8466	2.2136	1.1289
0.5	11.1803	2.2361	1.1180

Graphs:

capital per worker 12.0000 10.0000 8.0000 6.0000 4.0000 capital per worker 2.0000 0.0000 Savi ngs rate

output per worker 2.5000 2.0000 1.5000 1.0000 -output per worker 0.5000 0.0000 to Savi ngs rate

consumption per worker 1.4000 1.2000 1.0000 0.8000 0.6000 consumption per worker 0.4000 0.2000 0.0000 FTTTTTTTTTTT Savi ngs rate

From the graphs we see, as savings rate increases, the capital per worker increases at an increasing rate, output per worker increases at a decreasing rate while and consumption per worker increases , reaches a maximum (approximately at 1.2) and then declines as savings rate increases.

Below are the formulas used in excel

DELTA Pg No x= 1 szol then formulae: k* = 15 312 c) to maximise stedy state consumption per worker. c = (-os dc =, Ri-s). I us 1 ds Voll 205 dc=0 = as is =rs ars 1-5 - 25 35=1 ) s=1/3 > 0.33. 25 -s- l a s 1-o l 35+1 Voi I u srs usus / Voil y sus) AS arc Lo hence it s = Astituirs is the 0.33 is a maximum point. s= 0.33 Yes the aus. coincides with our previous finding where we reach max of consumph perworker (1.2) at S=0.33