Why a Saving-Rate Change Has a Transition
In the Solow model, a permanent change in the saving rate does not instantly change the capital already in place. It changes the share of current output devoted to investment, and that investment changes capital over time. The distinction matters: output per worker is initially unchanged, while investment and consumption can change immediately.
This tutorial traces those effects under the standard discrete-time Solow assumptions: output per worker is produced from capital per worker; a constant fraction of output is saved and invested; capital depreciates; and the labor force grows at a constant rate. There is no technological progress in the model considered here. The saving rate changes permanently at the start of period 0, while capital at that moment is inherited from the preceding period.
The denominator \(1+n\) accounts for the larger number of workers among whom next period's capital is shared. These are per-worker quantities, not aggregate totals. The equations presume that saving finances investment and that the production function and population growth rate do not change during the transition. If technology, population growth, or the saving rule also changes, the corresponding transition must account for that change.
What Changes Immediately, and What Changes Later?
Suppose the saving rate changes permanently from \(s_0\) to \(s_1\) at the beginning of period 0. Capital is predetermined: \(k_0\) is the capital inherited from before the change. Provided the production function has not changed, output in period 0 is still \(f(k_0)\). The new saving rate immediately changes investment and consumption:
Next period's capital depends on whether investment is enough to replace depreciation and provide capital for the growing workforce. In this model, break-even investment per worker is \((n+\delta)k_t\). Comparing investment with this amount identifies the direction of adjustment.
This expression follows by subtracting \(k_t\) from the law of motion and collecting terms. Since \(1+n>0\), the sign of the change in capital is the sign of investment minus break-even investment. Capital rises when \(s_t f(k_t)>(n+\delta)k_t\), falls when the inequality is reversed, and stays constant when they are equal.
If \(k_0\) was the old steady-state capital stock, its defining condition was \(s_0f(k_0)=(n+\delta)k_0\). Immediately after the saving-rate change, the difference between new investment and break-even investment is therefore
For positive output, a higher saving rate makes capital accumulation positive at the inherited old steady state; a lower rate makes it negative. This conclusion concerns the direction of the first adjustment, not its size or speed. Those require the production function and the model's parameters.
Worked Example: A Saving-Rate Increase Without Population Growth
Worked Example: A Saving-Rate Increase Without Population Growth
Consider \(f(k)=\sqrt{k}\), no population growth (\(n=0\)), and depreciation \(\delta=0.10\). Initially the economy is at its steady state with saving rate \(s_0=0.20\). The saving rate rises permanently to \(s_1=0.30\). Find the inherited capital and output, the immediate consumption and investment, and capital over the next three periods. Use the transition recurrence to four decimal places.
Method. First find the old steady state, then apply the new saving rate to the inherited capital and iterate the law of motion. The old steady state satisfies \(s_0\sqrt{k_0}=\delta k_0\). For positive \(k_0\), this gives \(0.20\sqrt{k_0}=0.10k_0\), hence \(\sqrt{k_0}=2\) and \(k_0=4\). Output is \(y_0=\sqrt{4}=2\).
At the new saving rate, investment is \(i_0=0.30(2)=0.60\), and consumption is \(c_0=(1-0.30)(2)=1.40\). Capital itself remains \(k_0=4\) during this period; the law of motion determines next period's capital:
For the next step, output is \(y_1=\sqrt{4.20}=2.049390\ldots\). Thus investment is \(0.30(2.049390\ldots)=0.614817\ldots\), and depreciated capital is \(0.90(4.20)=3.78\). Therefore \(k_2=0.614817\ldots+3.78=4.394817\ldots\), or \(4.3948\).
For the following step, \(y_2=\sqrt{4.394817\ldots}=2.096382\ldots\). Investment is \(0.30(2.096382\ldots)=0.628915\ldots\), and depreciated capital is \(0.90(4.394817\ldots)=3.955335\ldots\). Hence \(k_3=0.628915\ldots+3.955335\ldots=4.584249\ldots\), or \(4.5842\), to four decimal places.
As a check, the new steady state solves \(0.30\sqrt{k^*}=0.10k^*\), so \(k^*=9\). The calculated capital stocks \(4.20\), \(4.3948\), and \(4.5842\) are increasing and remain below 9, consistent with the direction of adjustment. A common mistake is to treat the higher investment as an immediate increase in \(k_0\); it instead raises \(k_1\), because capital is inherited at the start of the period.
Including Population Growth
With positive population growth, new investment must both replace depreciated capital and provide capital for additional workers. The exact recurrence divides the capital available next period by \(1+n\). Thus the net increase in capital per worker is the excess of investment over break-even investment, divided by \(1+n\). Simply adding that excess to current capital would omit the growth in the workforce.
Worked Example: A Saving-Rate Increase With Population Growth
Suppose \(f(k)=\sqrt{k}\), population grows at \(n=0.02\), and capital depreciates at \(\delta=0.08\). The economy begins at the steady state for \(s_0=0.20\), after which the saving rate rises permanently to \(s_1=0.25\). Calculate the inherited capital, period-0 consumption, and \(k_1\); then check the direction against the new steady state.
Method. The old steady-state condition is \(s_0\sqrt{k_0}=(n+\delta)k_0\). Here \(n+\delta=0.10\), so \(0.20\sqrt{k_0}=0.10k_0\). Dividing by \(\sqrt{k_0}\) for positive capital gives \(\sqrt{k_0}=2\), hence \(k_0=4\) and \(y_0=2\).
With the new rate, \(i_0=0.25(2)=0.50\), and \(c_0=(1-0.25)(2)=1.50\). Depreciated capital is \((1-0.08)(4)=3.68\). Capital available before adjusting for population growth is \(3.68+0.50=4.18\), so
Equivalently, break-even investment at \(k_0=4\) is \((0.02+0.08)(4)=0.40\). The excess investment is \(0.50-0.40=0.10\); dividing by \(1.02\) gives the increase \(0.098039\ldots\) in capital per worker. The new steady state satisfies \(0.25\sqrt{k^*}=0.10k^*\), giving \(\sqrt{k^*}=2.5\) and \(k^*=6.25\). The first-period increase from 4 to about 4.0980 is therefore in the direction of the new steady state. A common error is to add \(0.10\) directly to 4; the growing workforce means the increase is \(0.10/1.02\), not \(0.10\).
Consumption Can Move Differently from Capital
A saving-rate change affects consumption directly as well as through later changes in output. At the moment of an increase in saving, consumption falls at unchanged output because a smaller share is consumed. Later, as capital and output rise, consumption may recover and ultimately exceed its original level. The path is not described by capital alone: at every date, consumption is \((1-s_t)f(k_t)\).
A decrease in saving has the opposite immediate effect. Consumption rises at unchanged output, but lower investment can put capital on a downward path. In the standard Solow setting with a strictly increasing, concave production function and a unique stable steady state, capital moves toward the steady state associated with the new saving rate. Output moves with capital because \(f\) is increasing. The timing of a consumption recovery or decline depends on the parameters; it should be calculated rather than assumed.
Worked Example: A Saving-Rate Decrease and the Consumption Path
Let \(f(k)=\sqrt{k}\), \(n=0\), and \(\delta=0.10\). The economy begins in steady state with \(s_0=0.30\), then the saving rate falls permanently to \(s_1=0.20\). Calculate the initial output and consumption under both rates, \(k_1\), and consumption in period 1. Identify the new steady state.
Method. The old steady state solves \(0.30\sqrt{k_0}=0.10k_0\), so \(\sqrt{k_0}=3\) and \(k_0=9\). Output at the inherited capital stock is \(y_0=\sqrt{9}=3\), unchanged at the instant of the policy change.
Before the change, consumption at that output was \((1-0.30)(3)=2.10\). Under the new saving rate, period-0 consumption is \((1-0.20)(3)=2.40\), while investment falls from \(0.30(3)=0.90\) to \(0.20(3)=0.60\). Depreciated capital is \(0.90(9)=8.10\). Therefore \(k_1=8.10+0.60=8.70\).
Period-1 output is \(y_1=\sqrt{8.70}=2.949576\ldots\), so period-1 consumption is \(c_1=0.80(2.949576\ldots)=2.359661\ldots\), or \(2.36\) to the nearest cent in these output units. The new steady state solves \(0.20\sqrt{k^*}=0.10k^*\), giving \(k^*=4\). The results check the distinction: consumption first rises from 2.10 to 2.40 because the saving rate falls, then declines to about 2.36 as output falls. A common mistake is to assume the immediate consumption increase means consumption must keep rising; it does not, because the capital stock starts adjusting downward.
A Reliable Transition Calculation
For a multi-period problem, keep the timing consistent: use the current capital stock to find current output, apply the saving rate to that output, and then use the law of motion to obtain next period's capital. If the saving rate is permanently changed, use the new rate in every subsequent period. The following sequence prevents most timing and unit errors.
Use the given \(k_0\), or solve the old steady-state condition if the economy is stated to begin there. Do not reset capital when the saving rate changes.
Find \(y_t=f(k_t)\), investment \(i_t=s_t y_t\), and consumption \(c_t=(1-s_t)y_t\). Separate the immediate allocation change from capital adjustment.
Apply \(k_{t+1}=[s_t f(k_t)+(1-\delta)k_t]/(1+n)\). Carry enough precision through repeated calculations and round only the requested final values.
Compare investment with \((n+\delta)k_t\), and, when needed, solve the new steady-state condition \(s_1f(k^*)=(n+\delta)k^*\). Confirm that the computed path moves in the expected direction.
These calculations are aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus. The transition analysis is useful because it distinguishes a change in the allocation of current output from the slower accumulation of productive capital. It also makes clear why a long-run comparison alone does not describe what happens between the initial and new steady states.
Check Your Understanding
For each problem, apply the stated discrete-time Solow assumptions and keep the timing of the saving-rate change explicit.
- Let \(f(k)=\sqrt{k}\), \(n=0\), and \(\delta=0.05\). An economy starts at its old steady state with \(s_0=0.15\), then permanently changes to \(s_1=0.20\). Find \(k_0\), period-0 consumption, and \(k_1\). (Answer: \(k_0=9\), \(c_0=2.40\), \(k_1=9.15\).)
- Let \(f(k)=\sqrt{k}\), \(n=0.06\), and \(\delta=0.04\). Starting at the steady state for \(s_0=0.20\), the saving rate rises to \(s_1=0.25\). Find \(k_0\), new period-0 consumption, and \(k_1\). (Answer: \(k_0=4\), \(c_0=1.50\), \(k_1=4.0943\), to four decimal places.)
- Let \(f(k)=\sqrt{k}\), \(n=0\), and \(\delta=0.10\). An economy starts at the steady state for \(s_0=0.40\), then changes to \(s_1=0.25\). Find the initial output, new period-0 consumption, and \(k_1\). (Answer: \(y_0=4\), \(c_0=3.00\), \(k_1=15.40\).)
- Let \(f(k)=\sqrt{k}\), \(n=0\), and \(\delta=0.10\). An economy starts at the steady state for \(s_0=0.20\) and raises its saving rate to \(s_1=0.30\). At what capital stock does consumption under the new rate equal its original period-0 level? (Answer: \(k=256/49\approx5.2245\).)
- Let \(f(k)=\sqrt{k}\), \(n=0\), and \(\delta=0.20\). The inherited capital is \(k_0=1\), and the saving rate is permanently \(s=0.40\) from period 0 onward. Calculate \(k_1\), \(k_2\), and period-1 consumption. (Answer: \(k_1=1.2000\), \(k_2\approx1.3982\), and \(c_1\approx0.6573\), to four decimal places.)