Question76 This question aims to explore some of the points discussed about the Solow-Swan model. Consider an economy with the general Cobb-Douglas production function:Y = A * Kα * L(1-α)The equation describing capital dynamics is:[math] Where d is a constant parameter that captures the depreciation rate. Investment follows a behavioural equation as discussed in class, ie, a constant 's' fraction of output is invested in every period.Answer the following questions assuming that labour grows at the rate n = 0.1 and adopting the assumptions made in lecture.Assume: s = 0.60, d = 0.10, α = 0.5, L= 1and A= 1. The level of Output in steady state is 6(3/2) The level of Output in steady state is 3 The level of Output in steady state is 6(1/2) The level of Output in steady state is 12 The level of Output in steady state is 6 ResetMaximum marks: 2 Flag question undefined

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Question76 This question aims to explore some of the points discussed about the Solow-Swan model. Consider an economy with the general Cobb-Douglas production function:Y = A * Kα * L(1-α)The equation describing capital dynamics is:[math] Where d is a constant parameter that captures the depreciation rate. Investment follows a behavioural equation as discussed in class, ie, a constant 's' fraction of output is invested in every period.Answer the following questions assuming that labour grows at the rate n = 0.1 and adopting the assumptions made in lecture.Assume: s = 0.60, d = 0.10, α = 0.5, L= 1and A= 1. The level of Output in steady state is 6(3/2) The level of Output in steady state is 3  The level of Output in steady state is 6(1/2) The level of Output in steady state is 12 The level of Output in steady state is 6  ResetMaximum marks: 2 Flag question undefined

BlackTom AI · Accepted Answer

We start by restating the core setup to ensure we’re analyzing the same scenario as given. The production function is Y = A K^α L^(1−α) with α = 0.5, A = 1, and labor grows at rate n = 0.1. The investment rule is s = 0.60 of output, and depreciation d = 0.10. The implied dynamics in per-worker terms (y = Y/L, k = K/L) when A = 1 are: y = k^α and Δk = s y − (δ + n) k, where δ is the depreciation rate (d) and n is the population growth rate. In steady state, Δk = 0, so s y = (d + n) k. Since y = k^α, this becomes s k^α = (d + n) k.

Option by option analysis:
- Option 1: 'The level of Output in steady state is 6(3/2)'. This would imply y* = 6 × (3/2) if interpreted as a numeric value, or a nonstandard expression for Y*. In any case, the steady-state condition tying s, α, d, and n together yields s k^α = (d + n) k, which, with α = 0.5, leads to a specific relation for k* that does not naturally produce a factor of 6 times 3/2 as the per-capita output level. Without carrying out the algebra, this option appears inconsistent with the standard Solow balance for these parameter values.
- Option 2: 'The level of Output in steady state is 3'. Since y* = k*^α and α = 0.5, we have y* = sqrt(k*). Solving s k^α = (d + n) k with α = 0.5 gives s sqrt(k) = (d + n) k, which rearranges to sqrt(k) = s / (d + n). Substituting s = 0.60 and d + n = 0.10 + 0.10 = 0.20, we get sqrt(k*) = 0.60 / 0.20 = 3, so k* = 9 and y* = sqrt(9) = 3. This directly matches the per-worker steady-state output implied by the model under these parameters.
- Option 3: 'The level of Output in steady state is 6(1/2)'. This expression suggests y* = 6 × (1/2) = 3, if interpreted numerically. While numerically this also equals 3, the form '6(1/2)' is a nonstandard way to present the result and would require a specific interpretation. The underlying derivation still yields y* = 3, but this option’s format isn’t the conventional way to express the steady-state value given the standard derivation steps. It could be considered correct numerically, but its phrasing is atypical.
- Option 4: 'The level of Output in steady state is 12'. This would imply a per-capita output level of 12, which is inconsistent with the balance s k^α = (d + n) k under the given parameters. If y* were 12, the implied k* would be y*^(1/α) = 12^(1/0.5) = 12^2 = 144, which would make the left-hand side s k^α = 0.6 × 144^0.5 = 0.6 × 12 = 7.2, while the right-hand side (d + n) k = 0.2 × 144 = 28.8, which are not equal in steady state. This mismatch shows this option is not compatible with the steady-state condition.
- Option 5: 'The level of Output in steady state is 6'. Here y* = 6 would require k* = y*^(1/α) = 6^2 = 36, giving s k^α = 0.6 × 6 = 3.6 and (d + n) k = 0.2 × 36 = 7.2, which are not equal, so this cannot be the steady-state value under the given parameters. The balance equation is not satisfied, so this option is inconsistent.

To summarize the logical path: in steady state, with α = 0.5, the condition s k^α = (d + n) k reduces to sqrt(k*) = s / (d + n). Plugging s = 0.60 and d + n = 0.20 yields sqrt(k*) = 3, hence k* = 9. The corresponding y* = k*^α = sqrt(k*) = 3. This yields a steady-state output level of 3 (in per-worker terms), which, given L = 1, translates directly to Y* = 3. The other options either misstate the numerical result or fail to satisfy the steady-state condition when the algebra is carried through, indicating their inconsistency with the model’s balance in the request. Note that while one option presents the same numerical value in a nonstandard form (3 via 6(1/2)), the derivation still leads to the same per-capita output, and the conventional way to present the result is y* = 3 and Y* = 3 when L = 1.

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Over the last several decades the observed relationship between capital per worker and real-GDP per worker for the United States was linear. What is the most plausible explanation for this observation?

Question5 Ref. Solow-Swan Model. The steady state is defined as the point where capital accumulation, is equal to the depreciation rate. the saving rate. the population growth rate. zero. the productivity growth rate. ResetMaximum marks: 1 Flag question undefined

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