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Defining the Abatement Cost in Presence of Learning-by-Doing: Application to the Fuel Cell Electric Vehicle

A Correction to this article was published on 06 December 2017

This article has been updated


We consider a partial equilibrium model to study the optimal phasing out of polluting goods by green goods. The unit production cost of the green goods involves convexity and learning-by-doing. The total cost for the social planner includes the private cost of production and the social cost of carbon, assumed to be exogenous and growing at the social discount rate. Under these assumptions the optimization problem can be decomposed in two questions: (i) when to launch a given schedule; (ii) at which rate the transition should be completed that is, the design of a transition schedule as such. The first question can be solved using a simple indicator interpreted as the MAC of the whole schedule, possibly non optimal. The case of hydrogen vehicle (Fuel Cell Electric Vehicles) offers an illustration of our results. Using data from the German market we show that the 2015–2050 trajectory foreseen by the industry would be consistent with a carbon price at 52€/t. The transition cost to achieve a 7.5 M car park in 2050 is estimated at 21.6 billion € that is, to JEl 4% discount rate, 115 € annually for each vehicle which would abate 2.18 tCO\(_2\) per year.

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Fig. 1

Change history

  • 06 December 2017

    The correct affiliation for Anna Creti is: Université Paris-Dauphine, PSL Research University, LEDa, CGEMP, 75016 Paris, France.


  1. 1.

    A survey of learning-by-doing rates for different energy technology can be found in IEA (2000) and McDonald and Schrattenholzer (2001). Learning rates (cost reduction when production doubles) varies from 25% for photovoltaics, 11% for wind power, and 13% for fuel cell in the period 1975–2000.

  2. 2.

    Partial derivatives are denoted with indexes (except if it could be confusing) for instance \(C_X\) stands for \(\partial C/\partial X\).

  3. 3.

    According to the IPCC (2013) such a “Carbon Budget” is required to enforce a peak temperature target (see also Allen et al. 2009; Matthews and Caldeira 2008).

  4. 4.

    See Tol (2014) for a textbook derivation with several objectives. A targeted stabilization of the concentration of carbon in the atmosphere implies a growth rate of the CO\(_2\) price equal to the interest rate plus the decay rate of carbon in the atmosphere (Tol 2014, p. 53, the “shadow price” of CO\(_2\) \(\mu \) should be divided by \(U^\prime (C_t)\) to get the expression of the current price of CO\(_2\)).

  5. 5.

    Several recent contributions attempt to provide and test rule of thumb pricing. Golosov et al. (2014) propose a simple formula, based on a specified dynamic general equilibrium framework: the price of carbon should be proportional to world gross domestic product see also (Grimaud and Rouge 2014; Rezai and van der Ploeg 2015; Bijgaart et al. 2016).

  6. 6.

    The launching date and not the ending date should adjust to the change of duration reflecting that the DAC is a cost computed at the end of deployment.

  7. 7.

    Theoretically, both cases are possible: If the objective is to stabilize the concentration of CO\(_2\) in the atmosphere, the CO\(_2\) price should grow at a higher rate than the interest rate (e.g. Goulder and Mathai 2000). To slow the extraction of fossill exhaustible resource the growth rate of the carbon price should be lower than the intereste rate, with constant extraction costs (e.g. Grimaud and Rouge 2008).

  8. 8.

    For a complete discussion of our cost assumptions the reader is referred to Creti et al. (2015). The Excel model is available from the authors upon request.

  9. 9.

    Each cost component is obtained via Table 1 by adding the annualized capital cost, using the discount rate and the life time for the annualized factor, and the maintenance cost, using the capital cost and the percentage for maintenance. The Excel model is available from the authors upon request.

  10. 10.

    Estimates depend on various assumptions most notably the pure time discount rate and the convexity of the damage function. For instance Nordhaus (2011) computes 12$ /tCO\(_2\) (in 2005 US$) and Stern (2007) 68$/tCO\(_2\) while Golosov et al. (2014) find 15.5 $ and 135$ with their respective discount rate. At the country level, for France to reach its current policy objective, Quinet (2009) and Quinet (2013) suggest around 30 €/t for 2015; for the US, the EPA provides estimates (see the average of which is 56$ in 2015 with a 3% discount rate.


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Correspondence to Guy Meunier.


Appendix A: proof of Lemma 1


To minimize the total cost (2), let us introduce \(\lambda _t\) the co-state variable associated to the relation (3), and \(\theta _t\) and \(\delta _t\) the Lagrange multipliers associated to the two constraints (4) on \(x_t\): it is positive (\(\theta _t\)) and smaller that the total fleet size (\(\delta _t\)). The first order conditions (together with the complementarity slackness conditions) are:

$$\begin{aligned} C_{x}(X_{t},x_{t})-c_{o}= & {} p_{t}^{CO2}+\lambda _{t}+\theta _t-\delta _t \end{aligned}$$
$$\begin{aligned} \dot{\lambda _{t}}-r\lambda _{t}= & {} C_{X}(X_{t},x_{t}). \end{aligned}$$

The main step of the proof consists in proving that \(x_{t}\) is increasing if \(x_{t}\in (0,N)\). This condition ensures that once \(x_{t}>0\) the number of green cars cannot come back to zero, and that \(x_{t}\) does not move when \(x_{t}=N\). If \(x_{t}\) is strictly positive ( \(\theta _{t}=0\)) and lower than the total car fleet (\(\delta _{t}=0\)), Eq. (19) becomes \( C_{x}(X_{t},x_{t})-c_{o} =p_{t}^{CO2}+\lambda _{t} \) and taking the time derivative:

$$\begin{aligned} C_{xX}\dot{X}_{t}+C_{xx}\dot{x}_{t}&=\dot{p}_{t}^{CO2}+\dot{\lambda }_{t} \\ C_{xX}x_{t}+C_{xx}\dot{x}_{t}&=\dot{p}_{t}^{CO2}+r\lambda _{t}+C_{X} \qquad&\text { thanks to Eq. } 20 \\ C_{xx}\dot{x}_{t}&=\dot{p}_{t}^{CO2}+r\lambda _{t}+[C_{X}8-C_{xX}x] \end{aligned}$$

The last term of the right hand side is positive because \(C_{X}(X,x)\) is concave with respect to x and \(C_{X}(X,0)=0\) (since \(C(X,0)=0, \forall X\)). Since \(C_{xx}\), \( p^{CO2}_t\) and r are all positive, \(\dot{x}\) is also positive so that \(x_t\) is increasing through time.

Then, since the CO\(_{2}\) price increases exponentially, \(x_{t}\) cannot be always null along an optimal trajectory. Then either \(x_{0}=0\) or \(x_{0}>0\). In the latter case \(T_{s}=0\), whereas in the former case \(T_{s}\) is the inf of the dates at which \(x_{t}>0\).

The ending date is finite, \(T_e<+\infty \): From the above proof, when \(x_t\) is positive its time derivative is bounded below by a positive number, so \(x_t\) necessarily reaches N in a finite time.

Finally, Eq. (5) is obtained by integrating Eq. (20), between t and \(+\infty \) (and using the boundary conditions \(\text {lim}_{t\rightarrow +\infty } e^{-rt}\lambda _t = 0\)):

$$\begin{aligned} \lambda _t=-\int _t^{+\infty }e^{-r(\tau - t)}C_X(X_\tau ,x_\tau )d\tau \end{aligned}$$

and injecting this expression into Eq. (19). \(\square \)

Appendix B: proof of Lemma 2


If \(C_{xx}=0\), given that \(C(X,0)=0\), \(C_{X}(X,x)=C_{Xx}(X,x)x\). Then, we resort by reductio ad absurdum assuming \(T_{s}<T_{e}\). Between the two dates the Eq. (5) is satisfied and taking its derivative with respect to t gives:

$$\begin{aligned} \dot{p}_{t}^{CO2}&=C_{Xx}\dot{X}_{t}+C_{xx}\dot{x}_{t}-\dot{\lambda }_{t}&\\&= C_{Xx} x_t +0 -[r\lambda _t+C_X] &\text {using Eq. } (20) \\&=C_{Xx}x_{t}-C_{X}+r\left[ \int _{t}^{+\infty }e^{-r(\tau -t)}C_{X}(X_{\tau },x_{\tau })d\tau \right]&\text { from } (21) \end{aligned}$$

Therefore, using that \(C_{X}(X,x)=C_{Xx}(X,x)x\) in that case,

$$\begin{aligned} 0<\dot{p}_{t}^{CO2} =r\left[ \int _{t}^{+\infty }e^{-r(\tau -t)}C_{X}(X_{\tau },x_{\tau })d\tau \right] \le 0 \end{aligned}$$

a contradiction.

Therefore, the number of green cars jumps from 0 to N at date \( T_{s}=T_{e}\), and the total discounted cost \(\Gamma \) could be written as a function of the date \(T_{s}\):

$$\begin{aligned} \Gamma =\int _{0}^{T_{s}}e^{-rt}\left[ (p_{t}^{CO2}+c_{o}).N\right] dt+e^{-rT_{s}}\Omega (0) \end{aligned}$$

Along the optimal trajectory, \(T_{s}\) should minimize this function. Taking the derivative with respect to \(T_{s}\) in the equation above and setting it equal to zero gives the Eq. (6). \(\square \)

Appendix C: proof of Lemma 3, Proposition 2

From Lemma 1, the optimal trajectory \((x_t^*)_{t\in [0,+\infty )}\) can be described as the launching of a deployment schedule. There is therefore no loss to minimize the cost over the set of trajectories defined as the launching (and ending) of a deployment schedule. The trajectory obtained from the optimal launching of the optimal deployment schedule coincides with the optimal trajectory described in Lemma 3.

Appendix C.1: proof of Lemma 3


Using the decomposition of the total discounted cost \(\Gamma \) provided by Eq. (11), the schedule \(\xi \) only influences the DAC and no other component of the cost. So that the \(\xi \) that minimizes \(\Gamma \) corresponds to the \(\xi \) that minimizes the DAC. A deployment schedule \((\bar{X},D,\xi )\) and its DAC are only defined for \(\xi \) such that \(\int _0^D \xi _\tau d\tau =\bar{X}\). The optimal \(\xi \), for a given \(\bar{X}\) and D, is then the solution of the optimization program:

$$\begin{aligned} \min _{\xi _{\tau }}\int _{0}^{D}e^{-r\tau }[C(X_{\tau },\xi _{\tau })-c_{0}\xi _{\tau }]d\tau \nonumber \\ \text { s.t. }\dot{X}_{t} =\xi _{\tau };\;0\le \xi _{\tau }\le N\text { and } X_{D}=\bar{X}. \end{aligned}$$

This result holds for any \(\bar{X}\), D and \(T_l\). \(\square \)

It is interesting to write the equations satisfied by \(\xi \) in order to recover the optimality conditions (5) satisfied by the optimal trajectory. Write the Lagrangian:

$$\begin{aligned} \mathcal {L}= e^{-r\tau }\left[ C(X_\tau , \xi _\tau )-c_o \xi _\tau \right] - \mu _\tau \xi _\tau - \theta _\tau \xi _\tau -\delta _\tau (N-\xi _\tau )-\alpha \xi _\tau \end{aligned}$$

in which \(\mu _\tau \) is the co-state variable of \(\dot{X}=\xi _\tau \); \(\delta _t\) and \(\theta _t\) are the Lagrange multiplier of \(\xi _\tau \le N\); and \(\xi _\tau \ge 0\) respectively, and \(\alpha \) is the Lagrange multiplier of the constraint \(\int _0^D \xi _\tau d\tau =\bar{X}\). The optimal \(\xi \) satisfies the equations:

$$\begin{aligned}&e^{-r\tau } \big [C_x(X_\tau ,\xi _\tau )-c_0\big ]+\delta _\tau -\theta _\tau =\mu _\tau +\alpha \\&\dot{\mu }_\tau =e^{-r\tau } C_X(X_\tau ,\xi _\tau ),\; \mu _D=0 \\&\delta _\tau (N-\xi _\tau )=0,\; \theta _\tau \xi _\tau =0 \end{aligned}$$

Then, \(\xi _\tau \) is increasing (similar reasoning than for the proof of Lemma 1), and integrating the second equation gives

$$\begin{aligned} e^{-r\tau }[ C_x(X_\tau ,\xi _\tau )-c_0]+\delta _\tau -\theta _\tau =\alpha -\int _\tau ^D e^{-rs} C_X(X_s,\xi _s)ds \end{aligned}$$

Together with the optimality conditions satisfied by \(\bar{X}^*\), \(D^*\) and \(T_l^*\), to be studied below, these first order conditions will coincide with 5. However, even for suboptimal \(\bar{X}\) and D the schedule \(\xi \) should satisfies these equations, which then gives the minimized deployment cost \(I^*(\bar{X},D)\) and the associated \(DAC(\bar{X},D)\). The derivative of the deployment cost are:

$$\begin{aligned} \frac{\partial I^*}{\partial \bar{X}}&=\alpha =e^{-rD}C_x(\bar{X},\xi _D)-c_0+\delta _D \end{aligned}$$
$$\begin{aligned} \frac{\partial I^*}{\partial D}&=e^{-rD}\big [C(\bar{X},\xi _D)-c_0\xi _D\big ] -\alpha \xi _D\nonumber \\&=e^{-rD}\big [C(\bar{X},\xi _D)-c_0\xi _D \big ] -\Big [e^{-rD}\big [C_x(\bar{X},\xi _D)-c_0\big ]+\delta _D\Big ] \xi _D \end{aligned}$$

Appendix C.2: proof of Proposition 2

There are two possible strategies to prove that \(\bar{X}^*\) and \(D^*\) are independent of the CO\(_2\) price \(p_0\). One is sketched in the main text. The other consists in looking at first order conditions and showing that the optimal duration and accumulated quantity satisfy a pair of equation independent from the CO\(_2\) price.


From the expression (11) of the total discounted cost, the optimal \(D^*\), \(\bar{X}^*\) and launching date satisfy the equations:

$$\begin{aligned} rI^*+\frac{\partial I^*}{\partial D}&=0 \nonumber \\ e^{rD}\frac{\partial I^*}{\partial X}+\frac{\partial \Omega }{\partial X}&=p_0e^{rT_e} \nonumber \\ p_0e^{rT_e}&=\frac{1}{N}\left[ re^{rD}I^*(\bar{X},D) +r\Omega (\bar{X})-c_0N\right] \text { using Eq. } (10) \end{aligned}$$

The first equation corresponds to Eq. (13), the third correspond to (15). And injecting the third into the second gives Eq. (14) satisfied by \(\bar{X}^*\).

The two Eqs.  (13) and (14) are independent of the CO\(_2\) price or the launching date, the couple \(\bar{X}^*\) and \(D^*\) is therefore independent of \(p_0\), and so is the optimal deployment schedule \((\xi _\tau ^*)_{\tau \in [0,D]}\) associated to them.

In addition, it is interesting to see how these equations together with Eq. (23) give back the Eq. (5).

From Eqs. (14) and (24):

$$\begin{aligned} \frac{\partial I^*}{\partial \bar{X}}=\alpha&=p_0e^{rT_e}-\frac{\partial \Omega }{\partial \bar{X}}=p_0e^{rT_e} - \int _{T_e}^{+\infty } e^{r(t-T_e)}C_X(\bar{X}+tN,N)dt \end{aligned}$$

so that for \(\xi _{\tau } \in (0,N)\) the Eq. (23) is equivalent to Eq. (5) with \(\xi _\tau =x_{T_s+\tau }\).

Furthermore, one can show that Eq. (14) leads to \(\xi _D=N\) and that Eq. (13) leads to \(\xi _0=0\). This relatively fastidious proof is available upon request.

Appendix D: growth rate of the CO\(_2\) price

Proof of Corollary 3.


The total discounted cost should be written:

$$\begin{aligned} \Gamma (T_l)=&\int _0^{T_l+D} e^{-rt}(c_o+p^{CO2}_t)N dt + e^{-rT_l}I + e^{-r(T_l+D)}\Omega (\bar{X}) \\&-p_0\int _{T_l}^{T_l+D}e^{(\rho -r)t}\xi _{t-T_l}dt \end{aligned}$$

which is similar to (8) except that the second line above, the value of interim abatement, replaces \(p_0\bar{X}\). The derivative of the second line with respect to \(T_l\) is (after an integration by parts):

$$\begin{aligned} -p_0 \int _{T_l}^{T_l+D}(\rho -r)e^{(\rho -r)t}\xi _{t-T_l}dt=-(\rho -r)p^{CO2}_{T_l+D} e^{-r(T_l+D)} \int _{0}^{D}e^{(\rho -r)(t-D)}\xi _tdt \end{aligned}$$

so that the derivative of the discounted cost with respect to \(T_l\) is

$$\begin{aligned} \frac{\partial \Gamma }{\partial T_{l}}= & {} e^{-r(T_{l}+D)}\left[ (c_{o}+p_{T_{l}+D}^{CO2})N-rIe^{rD}-r\Omega (\bar{X} )- p^{CO2}_{T_l+D} (\rho -r) \right. \\&\times \left. \int _{0}^{D}e^{(\rho -r)(t-D)}\xi _t \right] dt \end{aligned}$$

This derivative is equal to zero at the optimal launching date (the second order condition is satisfied) which gives Eq. (17). \(\square \)

The following result is proved:

Result The ending (resp. launching) CO\(_2\) price of scenario \((\xi _\tau )_{\tau \in [0,D]}\) is increasing (resp. decreasing) with respect to \(\rho \) if \(\xi _0=0\) and \(\xi ^\prime _\tau >0\).


We denote \(\xi _\tau ^\prime \) the derivative with respect to time of \(\xi _\tau \).

The ending CO\(_2\) price is given by Eq. (17). The derivative of its denominator with respect to \(\rho \) is:

$$\begin{aligned}&\int _0^D ((\rho -r)(D-\tau )-1)e^{-(\rho -r)(D-\tau )} \xi _\tau d\tau \\&\quad =\underbrace{\phantom {\int _0^D} \left[ (D-\tau )e^{-(\rho -r)(D-\tau )} \xi _\tau \right] _0^D}_{=0}\underbrace{-\int _0^D (D-\tau )e^{-(\rho -r)(D-\tau )} \xi ^\prime _\tau d\tau }_{<0} \end{aligned}$$

so the ending CO\(_2\) price is increasing with respect to \(\rho \).

The launching CO\(_2\) price is:

$$\begin{aligned} p^{CO2}_{T_l^*}= \frac{r I +(r\Omega (\bar{X})-c_0N) e^{-rD}}{Ne^{(\rho -r)D}-(\rho -r)\int _0^De^{(\rho -r)\tau } \xi _\tau d\tau } \end{aligned}$$

The derivative of the denominator with respect to \(\rho \) is

$$\begin{aligned}&DNe^{(\rho -r)D}-\left[ \tau e^{(\rho -r)\tau } \xi \right] _0^D+\int _0^D \tau e^{(\rho -r)\tau } \xi ^\prime _\tau d\tau \\&\qquad =\,D(N-\xi (D))e^{(\rho -r)D}+\int _0^D \tau e^{(\rho -r)\tau } \xi ^\prime _\tau d\tau >0 \end{aligned}$$

so, the launching CO\(_2\) price is decreasing with respect to \(\rho \). \(\square \)

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Creti, A., Kotelnikova, A., Meunier, G. et al. Defining the Abatement Cost in Presence of Learning-by-Doing: Application to the Fuel Cell Electric Vehicle. Environ Resource Econ 71, 777–800 (2018).

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  • Dynamic abatement costs
  • Learning by doing
  • Fuel cell electric vehicles

JEL Classification

  • Q55
  • Q42
  • C61