Growing competition in electricity industry and the power source structure

Growing competition in electricity industry and the power source structure Hiroaki Ino Institute of Intellectual Property and Toshihiro Matsumura Institute of Social Science, University of Tokyo [Preliminary Not for Citation] Abstract We investigate how the strategic behaviors of electric-power producers work on their power source structure. We introduce the competitor who cannot use the technologies which requires an enormous set-up cost such as nuclear power into an electric-power market which is originally local monopoly. Does introducing such a competitor decrease the capacity of nuclear power in the market? We show that the capacity of nuclear power decreases if and only if nuclear plant is sufficiently efficient. 1 Introduction To generate electricity, we have various technologies such as nuclear, hydraulic, fossil and biomass power. Since 1990 s, the electric-power market has been reformed from local monopoly to competitive one by deregulation. Now we can observe that a variety of technologies compete in the electric-power market. Which technology is used to produce electric power (the power source structure) depends on the economic aspects as well as the technological, the environmental, and the policy aspects. Economically, the demand and cost structures are important. The demand of electricity frequently fluctuates between day and night or among seasons. Once the sunk cost of the power plant is paid, the technology such as nuclear or hydraulic power save little cost even if the quantity of power generation is reduced under its capacity, while the technologies such as fossil or biomass power requires the substantial cost due to the inputs for its additional supply. Thus, when a power company chooses to the technologies of their production, it is efficient for the 1

power company to use the technology like fossil power when they adjust their power supply to the volatile part of demand and the technology like nuclear power when they supply for the steady part of demand. This paper focus on this economic aspect to determine the power source structure. 1 The above explanation about choice of technologies only highlights the decision inside a firm and does not take the competition among firms into consideration. Although the explanation is appropriate if the markets is local monopoly of the power company, the power company now has the strategic interaction with new entrants. In particular, the entrants cannot usually use the technologies which requires an enormous set-up cost such as nuclear power. Does introducing such a competitor decrease the capacity of nuclear power in the market comparing to the monopoly case? We show that the capacity of nuclear power decreases if and only if the nuclear plants is sufficiently efficient. Furthermore, we investigate the case where the firm with nuclear plants specializes in generating the nuclear power and compete with the firm without nuclear plants. Such a market structure can be considered when the nuclear sector is separated from the the power company as a independent firm. We show that this specialization increases the capacity of nuclear power when the nuclear plant is not so efficient. However, the conclusion that the capacity of nuclear power decreases compared to the monopoly case if and only if it is sufficiently efficient is maintained. The paper is organized as follows. Section formulates the model. Section investigates the case of monopoly and shows that the model abstracts the feature explained in the second paragraph. In Section 4, we introduce the competitor who produces electricity without nuclear plants and investigates presents results. Section 5 examines the effect of specialization. Section 6 concludes the paper. 1 Holthausen (1976) investigates how the input of an ex ante control and that of an ex post control are chosen under uncertain demand in the monopoly market.

The model We consider an local market of electric-power whose structure is duopoly by firms 1 and or, as the benchmark case, monopoly by firm 1. The firms play the three-stage game where the power plants capacities are established under uncertain demand in the first stage, then the demand is identified in the second stage, and finally the quantities of power supplies are chosen in the third stage. There are two types of power plants to produce homogeneous electric power, that are referred as the nuclear plant and the natural-gas plant. The nuclear plant has the relatively high set-up cost of the power plant s capacity compared to the natural-gas plant; The natural-gas plant has the relatively high production cost of the power supply compared to the nuclear power. For simplicity, we adopt the following setups. In the first stage, firm i chooses the capacity of nuclear plant k i R + and the capacity of natural-gas plant l i R +. We suppose that the set-up cost of the nuclear plant s capacity is rk i where r > 0 while that of the natural-gas plant s capacity is 0. These costs are sunk in the following stages. The inverse demand of electric-power market is p = A Q where p is the market price and Q is the market demand. The scale of demand A is a random variable following the uniform distribution supported on the positive-valued interval [L, H] and this value is identified in the second stage. In the third stage, after observing A, firm i chooses the supply from nuclear plants x i (A) [0, k i ] and the supply from natural-gas plant y i (A) [0, l i ]. Denote firm i s aggregate supply q i (A) = x i (A) + y i (A). Note that since the demand becomes clear after the capacities are established and before the supplies are settled, firms decisions about the quantities of their supplies depend on A but capacities does not. We suppose that the production cost of the nuclear plant is 0 while that of the natural-gas plant is cy i (A) where c > 0. Note that in the final quantity-setting stage, these production costs are variable while the set-up costs of capacities become fixed costs. Since the set-up cost of natural-gas plant s capacity is 0, firm i always choose sufficiently large l i so that the y i (A) is not bounded upward by l i in the subgame perfect equilibrium. Thus, l i is not relevant to the other outcomes in the equilibrium. Furthermore, following assumptions are convenient to solve the problem algebraically. A.1. Var(A) > (L + c). A.. r < c. A.. c < L 48

A.1 restricts our attention to the case where the demand has relatively high volatility. More concretely, this is the condition to exclude the case where the firm which can establish both the nuclear plant and the natural-gas plant does not supply positive amount of natural-gas power for all the situation satisfying the following assumptions. A. states that nuclear power is more efficient than natural gas if nuclear plants have no excess capacity. A. states that, for all the scale of demand, a firm can not be the monopolist if there is a rival firm. Monopoly In this section, as the benchmark case, we consider firm 1 is the monopolist. 4 Firm 1 can establish both nuclear and natural-gas plants..1 Quantity-setting stage In the final quantity-setting stage, the equilibrium outcomes for given k 1 and A are as follows. Note that by A., we have interior solutions with respect to q 1 for all A [L, H]. ( k1, A c ) (x m 1 (k 1 ; A), y1 m k 1 if k 1 [0, A c ] A [k 1 + c, ) (k 1 ; A)) = (k 1, 0) if k 1 [ A c, A ] A [k 1, k 1 + c] ( A, 0) if k 1 [ A, ) A (0, k 1] (1) A c q1 m if k 1 [0, A c ] A+c (k 1 ; A) = k 1 if k 1 [ A c, A ] p m if k 1 [0, A c ] (k 1 ; A) = A k 1 if k 1 [ A if k 1 [ A, ) A c, A ] () A if k 1 [ A, ) As seen in (1), firm 1 s optimal production plan reveals three patterns. If the scale of demand is large relatively to nuclear plant s capacity, i.e., A > k 1 + c, natural-gas power is supplied to make up for the lack of nuclear plant s capacity. If the scale of demand is not so large, i.e., A k 1 +c, all the supply is nuclear power. On top of that, if the scale of demand is sufficiently small, i.e., A < k 1, the nuclear plant have an excess capacity, i.e., k 1 > x m 1 (k 1; A). A.1 is equivalent to H L c > 0. This condition is always satisfied, e.g., if H L together with A.. More precisely, this is the necessary and sufficient condition for all the firm to have the interior solution of natural-gas power for some r (0, c) in Section 4 and sufficient condition in Section. In Section 5, A.1 is not used to induce the equilibrium since the firm with the nuclear plant cannot establish the natural-gas plant. 4 The analysis is same as in the case where firm 1 and firm decide their aggregate capacity and supply to maximize their collusive profit. 4

The final stage s monopoly profit without the set-up cost of capacities is given by π1 m (k 1 ; A) = k 1 (A k 1 ) (A c) 4 + ck 1 if k 1 [0, A c ] A [k 1 + c, ) if k 1 [ A c, A ] A [k 1, k 1 + c] A 4 if k 1 [ A, ) A (0, k 1] Note we can check π m 1 (k 1; A) smoothly connects as seen in figure 1. π m 1 A 4 (A c)(a+c) 4 (A c) 4 O A c A k 1 Figure 1: Graph of π m 1 (k 1; A). Capacity-setting stage In the first capacity-setting stage, since the firm establishes the capacities under uncertain demand, it maximizes the expected profit, i.e., 1 H max π1 m (k 1 ; a)da rk 1. () k 1 H L L Let k m 1 be the solution of this problem, that is, the equilibrium capacity of nuclear plants in the monopoly market. 5

Firm 1 s marginal expected revenue in the first stage is as follows. Note L < H c by A.1. c ] if k 1 [0, L c ] if k 1 [ L c, L ] d 1 H π1 m (k 1 ; a)da = dk 1 H L L [ 1 ch L c H L [ 1 ch c H L (L c)k 1 k1 ] ck 1 if k 1 [ L, H c [ ] ] 1 H H L Hk 1 + k1 if k 1 [ H c, H ] 0 if k 1 [ H, ] The line ABCD in the figure 6 depicts the shape of this expected marginal revenue. As we can see in this figure, firm 1 s expected marginal revenue is strictly decreasing when it is in the interval (0, c). Therefore, by A., the first order condition, MR m (k 1 ) d 1 H π1 m (k 1 ; a)da = r dk 1 H L L (4) is the necessary and sufficient condition for the problem (). Solving it yields the following result. Lemma 1 Suppose A.1-A.. Then, ( 1 L c + ) (c r)(h L) k1 m c(h c) r(h L) = ( 4c 1 H ) r(h L) [ ) if r c c [ (H L), c c if r ( (H L), c ] c if r 0, (H L). c (H L) ] (5) The first line of (5) corresponds to the region where k m 1 and the third line k m 1 ( L c, L ], the second line km 1 [ L, H c ] [ H c, H c ). From this result and (1), r > (H L) guarantees ym 1 (H) > 0 while y1 m (L) is always zero. In other words, unless the nuclear plant is drastically efficient, the natural-gas power is supplied when the demand happens to be high while the nuclear power always meets the steady demand. When firm 1 establishes nuclear plants, it faces a trade-off between cost and risk. If the demand is not volatile, since nuclear plants is less costly than natural-gas plants by A., the firms prefers nuclear plants. However, under the uncertain demand, establishing a large capacity of nuclear plants is risky since the part of its capacity can be an excess capacity when the demand happens to be small. The cost of this excess capacity can not be compensated since it becomes fixed sunk cost after the demand is identified. The firm balances this trade-off and, as a result, 6

the natural gas cope with the high demand unless r is so small that the effect of cost always dominate the effect of risk in the trade-off. Observe that the threshold which guarantees y m 1 (H) > 0 can be rearranged as c (H L) = c Var(A) Thus, if Var(A) increases, natural-gas power is supplied for lower r. 5 As well as high set-up cost of nuclear plants reduce the cost advantage of nuclear power, high volatility of demand enhance the risk to establish nuclear plants. In other words, r and Var(A) affect the trade-off in the same direction and similarly reduce the incentive to establish nuclear plants.. Example We provide a numerical example with H =, L = 1 and c = 0.4 and simulate how the expected ( x m market share of nuclear power E 1 (k1 m;a) A and the expected operating rate of nuclear plants ( ) x m E 1 (k1 m;a) A k1 m changes in r. See the Table 1. 6 The intuition of this result essentially follows the q m 1 (km 1 ;A) ) explanation of Lemma 1. As r gets low, the expected market share of nuclear power increases since the natural-gas power is less supplied because of the nuclear plant s cost advantage. Instead of that, the expected operating rate of nuclear plant decreases since the firm less care the risk of excess capacity. r market share operating rate 0.00 1.000 0.666 0.05 0.999 0.764 0.10 0.988 0.816 0.15 0.959 0.865 0.0 0.911 0.911 0.5 0.84 0.951 0.0 0.751 0.98 0.5 0.66 0.999 0.40 0.75 0.100 Table 1: Equilibrium management of nuclear plant (monopoly case) 5 Note that since y m 1 (A) is non-increasing in A by (1), y m 1 (A) = 0 for all A [L, H] if y m 1 (H) = 0. 6 From Lemma 1 and (1), the market share of nuclear power is below 1 if r > c /(H L) = 0.04 and the operating rate of nuclear plant is below 1 if r < c c /(H L) = 0.6. 7

4 Effect of the competition In this section, firm enters the market as the competitor of firm 1. We suppose that firm 1 can establish both nuclear and natural-gas plants and that firm can establish natural-gas plants but cannot use nuclear power 7. 4.1 Quantity-setting stage If all firm i s supply is produced by nuclear power, its reaction function in third stage is Ri N (q j ; A) = max[ A q j, 0]. If all firm i s supply is produced by natural gas, its reaction function in third stage is Ri G (q j ; A) = max[ A c q j, 0]. Thus, in the final quantity-setting stage, firm 1 s reaction function given k 1 is R1 G(q ; A) if q R G 1 1 (k1 : A) R 1 (q ; A) = k 1 if R1 G 1 (k1 ; A) q R1 N 1 (k1 ; A) R1 N(q ; A) if R1 N 1 (k1 ; A) q As seen in the figure, firm 1 s reaction function kinks at its nuclear capacity k 1. Firm s A q A-c R N 1 R G 1 k 1 A c A q 1 Figure : Reaction function of firm 1 given k 1 reaction function is always R G. The equilibrium outcome for given k 1 and A are obtained as the intersection of these reaction functions. Note that by A., we have interior solutions with respect to q 1 and q for all A [L, H]. 7 Dixit (1979) demonstrated that the incumbent yields an outcome in which the new entrant is inactive if there exists sufficiently large fixed cost to enter. 8

If k 1 [0, A c ] A [k 1 + c, ], [ x 1 (k 1 ; A) y1 (k ] 1; A) y (k 1; A) = [ ] A c k1 k 1 A c, [ ] q 1 (k 1 ; A) q (k = 1; A) [ A c ] A c, p (k 1 ; A) = A + c. (6) If k 1 [ A c, A+c ] A [k 1 c, k 1 + c], [ x 1 (k 1 ; A) y1 (k ] 1; A) y (k 1; A) [ k1 0 = A c k 1 ], [ ] [ q 1 (k 1 ; A) q (k = 1; A) k 1 A c k 1 ], p (k 1 ; A) = A c k 1. (7) If k 1 [ A+c, ) A (0, k 1 c], [ x 1 (k 1 ; A) y1 (k ] 1; A) y (k 1; A) = [ A+c ] 0 A c, [ ] [ q 1 (k 1 ; A) A+c ] q (k = A c, p (k 1; A) 1 ; A) = A + c. (8) Similarly to the monopoly case, firm 1 s optimal production plan reveals three patterns: when A > k 1 + c (high-demand case), natural-gas power is supplied; when A < k 1 c (low-demand case), the nuclear plant have an excess capacity; and otherwise (intermediate case), all the power supply is produced by nuclear plants without excess capacity. From these, we obtain firm 1 s profit in final stage without the capacity cost as (A c) π1(k 9 + k 1 c if k 1 [0, A c ] A [k 1 + c, ] k 1 ; A) = 1 (A+c k 1 ) if k 1 [ A c, A+c ] A [k 1 c, k 1 + c] (A+c) 9 if k 1 [ A+c, ) A (0, k 1 c] for given k 1 and A. The shape of π 1 is depicted in the figure. Note π 1 is not differentiable but we can check it is continuous in k 1. 4. Capacity-setting stage Firm 1 s profit maximization problem in the first capacity-setting stage is given by 1 H max π k 1 H L 1(k 1 ; a)da rk 1. (9) L Let k 1 be the solution of this problem, that is, the equilibrium capacity of nuclear plants under the duopoly with a new entrant. 9

π 1 (a+c) 9 (a+c)(a c) 9 (a c) 9 O a c a+c k 1 Figure : Graph of π 1 (k 1; A) Firm 1 s marginal expected revenue in the first stage is as follows. Note, by A.1, H c c [ ] if k 1 [0, L c ] 1 4cH (L+c) 1 H H L 4 + (L c)k 1 4 k 1 if k 1 [ L c, L+c ] π k 1 H L 1(k 1 ; a)da = 1 H L [ch [ ck 1] ] if k 1 [ L+c, H c ] L (H+c) > L+c. 1 H L 4 (H + c)k 1 + 4 k 1 if k 1 [ H c, H+c ] 0 if k 1 [ H+c, ] The line ABEF CG in the figure 6 depicts the shape of this expected marginal revenue. As we can see in the figure 6, though this expected marginal revenue goes up over c at first, it is strictly decreasing when in the interval (0, c). Therefore, by A., the first order condition, MR (k 1 ) d 1 dk 1 H L H is the necessary and sufficient condition for the problem (9). result. L (10) π 1(k 1 ; a)da = r, (11) Solving it yields the following Lemma Suppose A.1-A.. Then, ch r(h L) k1 = ( c 1 H + c ) (c + H) + 1r(H L) 10 [ if r c c(h L c) ( (H L) if r 0, c c(h L c) (H L) ), c ]. (1)

Proof See the Appendix. Q.E.D. k 1 [ L+c, H c ] if we are in the case of the first line of (1), and k 1 [ H c second line. Therefore, by (6), y 1, H+c ] if in the c(h L c) (H) > 0 if and only if r > c (H L). In other words, the natural-gas power is supplied for the high demand when the nuclear plant is not quite efficient. Again, the threshold can be rearranged as c c(h L c) (H L) = c(e(a) + c) Var(A) + c Thus, if Var(A) increases preserving the mean, natural-gas power is supplied for lower r. 8 These are because to establish nuclear plants yields the trade-off between cost and risk similarly to the monopoly case. 4. Comparison Does the capacity of nuclear power in the market decrease when a new entrant who produces without nuclear power is introduced? Proposition 1 Suppose A.1-A.. Then, there exists ˆr (0, c) such that k 1 k m 1 r ˆr. Proof See the Appendix. Q.E.D. This proposition indicates that the capacity of nuclear power decreases if and only if nuclear plant is sufficiently efficient relatively to the risk (volatility of demand). 9 The rationale behind this result is as follows. Due to the cost structure of nuclear and natural-gas plants, establishing 8 Note that since y 1(A) is non-increasing in A by (1), y 1(A) = 0 for all A [L, H] if y 1(H) = 0. 9 If A.1. is violated, i.e., Var(A) (L + c) /48, it is possible k 1 < k m 1 for all r (0, c). However, even in such a case, if we consider the case where c > r, we have k 1 > k m 1 if r > c unless r is so high that k 1 = k m 1 = 0. This is because when r > c, unless r is extremely high, nuclear plants are established under the duopoly to deter firm s production (see we have the region where MR > c), while no nuclear plants are established under the monopoly since they have no cost advantage but have risk of excess capacities. 11

nuclear plants, which yields the trade-off between cost and risk, yields another strategic effect if a competitor is introduced. Since nuclear plants have the relatively low production cost in the quantity-setting stage, larger capacity of nuclear plants commits larger production and deter the competitor s production, which promote the incentive to establish nuclear plants. This strategic effect is same as production deterrence of cost-reducing investment discussed in Dixit (1980): the investment cost and the reduced cost by the investment theoretically corresponds to the relatively high set-up cost and the relatively low production cost of nuclear plants in our context, respectively. On the other hand, by introducing the competitor, some production of firm 1 substitutes to the new competitor. Since firm supplies only natural-gas power even for the fundamental demand, this effect reduce the capacity of nuclear plants. When r is small, a lot of firm 1 s supply have already produced by nuclear even under monopoly. Therefore, the former effect of production deterrence (the latter effect of production substitution) is small (large) since it is effective when natural-gas (nuclear) plants is substituted for nuclear (natural-gas) plants. That is why the more the nuclear plant is efficient, the more likely the capacity of nuclear power decreases by introducing the competitor. When the policymaker introduces the competitor who produce without nuclear power in order to reduce the dependence on nuclear power for electricity, our result is pessimistic because it is an efficient technology that successfully decreases. Indeed, the following corollary of Proposition 1 further support this pessimism. Corollary 1 Suppose A.1-A.. Then, y 1 (A) = 0 for all A [L, H] if k 1 < km 1. Proof From the explanation just after Lemma, y1 (A) = 0 for all A [L, H] if we are in the case of the second line of (1). Obviously from the proof of Proposition 1, MR m and MR never intersect when k 1 [ L+c, H c ] (the first-line case of (1)). Therefore, ˆr < MR ( H c ). Q.E.D. Thus, the capacity of nuclear power decreases only when nuclear plant is so efficient that firm 1 no more supplies natural-gas power under the duopoly. 4.4 Market share of nuclear power We discuss how the competition influences the expected market share of nuclear power. 1

Proposition Suppose A.1-A.. Then, ( x 1 E (k 1 ; A) ) A q1 (k 1 ; A) + q (k 1 ; A) ( x m < E 1 (k1 m; A) ) A q1 m(km 1 ; A) if k 1 < k m 1. Proof See the Appendix. Q.E.D. This result is essentially follows the previous subsection. As in ordinary oligopoly model, even when the nuclear capacity decreases, we can show that the expected total output increases by the competition. Thus, if the capacity of nuclear plant is less under the competition than under monopoly, the market share is all the more less. This implies that the market share decreases when the nuclear plants are sufficiently efficient. As in Subsection., we also provide numerical examples of our two duopoly models with H =, L = 1 and c = 0.4. See the Table for the results. Figure 7 depicts the expected market share of nuclear power in this examples by the line labeled new entrant and the example in Subsection. by the line labeled monopoly. We can see that the expected market share is less under duopoly than under monopoly when the nuclear plants are sufficiently efficient. r market share operating rate 0.00 0.666 0.705 0.05 0.665 0.741 0.10 0.661 0.774 0.15 0.655 0.804 0.0 0.647 0.8 0.5 0.67 0.859 0.0 0.599 0.919 0.5 0.58 0.969 0.40 0.45 0.998 Table : Equilibrium management of nuclear plant (new-entrant case) 5 Effect of the specialization In this section, we investigate the case where the firm with nuclear plants specializes in the nuclear power supply. We suppose firm 1 generates its power production only by nuclear plants and firm only by natural-gas plants. 1

5.1 Quantity-setting stage In final quantity-setting stage, firm 1 s reaction function given k 1 is { k 1 if q R N 1 1 (k1 ; A) R 1 (q ; A) = R1 N(q ; A) if R1 N 1 (k1 ; A) q As seen in the figure, firm 1 s reaction function kinks at its nuclear capacity k 1. Firm s A q A-c R N 1 R G 1 k 1 A c A q 1 Figure 4: Reaction function of firm 1 given k 1 reaction function is always R G. The equilibrium outcome for given k 1 and A are as follows. Note, by A., we have interior solution with respect to q 1 and q for all A [L, H]. { q 1 (k 1 ; A) = x 1 (k 1 ; A) = q (k 1 ; A) = y (k 1 ; A) = p (k 1 ; A) = k 1 A+c { A c k1 if k 1 [0, A+c ] A [k 1 c, ) if k 1 [ A+c, ) A [0, k 1 c] if k 1 [0, A+c ] A [k 1 c, ) if k 1 [ A+c, ) A [0, k 1 c] A c { A+c k1 if k 1 [0, A+c ] A [k 1 c, ) A+c if k 1 [ A+c, ) A [0, k 1 c] From these, we obtain firm 1 s profit in final stage without the capacity cost as { k1 (A+c k 1 ) π1 if k 1 [0, A+c (k 1 ; A) = ] A [k 1 c, ) (A+c) 9 if k 1 [ A+c, ) A [0, k 1 c] for given k 1 and A. The shape of π1 is depicted in the figure 5. Note π1 is not differentiable but we can check it is continuous in k 1. 5. Capacity-setting stage Firm 1 s profit maximization problem in the first capacity-setting stage is given by max k 1 1 H L H L π 1 (k 1 ; a)da rk 1. (1) 14

π 1 (a+c) 9 (a+c)(a c) 9 O a c a+c k 1 Let k 1 Figure 5: Graph of π 1 (k 1; A) be the solution of this problem, that is, the equilibrium capacity of nuclear plants under the duopoly of divided sectors. Firm 1 s marginal expected revenue in the first stage is as follows. H+L+c 1 H 4 [ k 1 ] if k 1 [0, L+c ] π1 1 (k 1 ; a)da = (H+c) k 1 H L H L 4 (H + c)k 1 + 4 L k 1 if k 1 [ L+c, H+c ] 0 if k 1 [ H+c, ] The line IF CG in the figure 6 depicts the shape of this expected marginal revenue. (14) As we can see in the figure 6, this expected marginal revenue is strictrly decreasing. Therefore, by A., the first order condition, MR (k 1 ) d 1 H π1 (k 1 ; a)da = r, (15) dk 1 H L L is the necessary and sufficient condition for the probrem (1). Solving it yields the following result. Note A.1 is not needed to derive this proposition. Lemma Suppose A.-A.. Then, k1 = 1 ( H + c ) (c + H) + 1r(H L). 15

Since firm 1 cannot use the natural-gas power, the solution which corresponds to the case where y 1 (k ; H) > 0 in Lemma disappears in this proposition. 5. Comparison Does the same principle occur when the competition is introduced by setting the nuclear sector up in independent business? Proposition Suppose A.1-A.. Then, there exists r (ˆr, c) such that k 1 > k 1 k 1 = k 1 if r r. if r > r and Proof See the Appendix. Q.E.D. When the firm 1 cannot use natural-gas power, it must cope with fluctuation of demand only by nuclear power. Thus, the firm establish nuclear plants for high demand even if that impose the risk of excess capacity. Therefore, as seen in the case r > r of Proposition, the nuclear capacity increases compared to the case where firm 1 can use both the technologies. However, this mechanism does not work when r r. This is because if the nuclear plant is efficient relatively to the risk (volatility of demand), firm 1 does not supply natural-gas power even when it can use both the technologies. In particular, since we can show that r > ˆr, we always have k 1 = k 1 in the case where the nuclear plant is so efficient that we have k 1 < km 1. Therefore, Proposition 1 is warrantable if k 1 is replaced by k 1. In other words, as for the conclusion that the capacity of nuclear power decreases by the competition if and only if nuclear plant is sufficiently efficient, we can apply the same principle as explained in Proposition 1. As for the market share of nuclear power, we also have the similar relation to Proposition. Proposition 4 Suppose A.1-A.. Then, E A ( q1 (k 1 x 1 (k 1 ; A) ; A) + q (k 1 ; A) ) ( x m < E 1 (k1 m; A) ) A q1 m(km 1 ; A) if k1 < k1 m. Proof As seen in Proposition, we always have k 1 = k 1 in the case where k 1 < km 1. Therefore, the similar proof to that of proposition is applicable if we replace the superscript * with the 16

superscript **. Q.E.D. As in Subsection., we also provide numerical examples of our two duopoly models with H =, L = 1 and c = 0.4. See the Table. Figure 7 depicts the expected market share of nuclear power in this example by the line labeled specialization. We can see that the expected market share is less than the previous section unless the nuclear plants are sufficiently efficient. r market share operating rate 0.00 0.666 0.705 0.05 0.665 0.741 0.10 0.661 0.774 0.15 0.655 0.804 0.0 0.647 0.8 0.5 0.67 0.859 0.0 0.65 0.884 0.5 0.611 0.906 0.40 0.596 0.97 Table : Equilibrium management of nuclear plant (devided-sector case) 6 Conclusion We have investigated how the growing competition in an electric-power market affects the power source structure. We consider the volatile demand and two types of technologies: the nuclear power and the natural-gas power. The nuclear power is supposed to have the relatively high set-up cost compared to the natural-gas power and the natural-gas power is supposed to have the relatively high production cost compared to the nuclear power. We have shown that that introducing a competitor with the natural-gas plant into the monopoly market decreases the capacity of nuclear power of the original monopolist if and only if the nuclear plant is sufficiently efficient. References Dixit, A. (1979). A Model of Duopoly Suggesting a Theory of Entry Barriers. The Bell Journal of Economics, 10, 0-. 17

Dixit, A. (1980). The Role of Investment in Entry-Deterrence. The Economic Journal, 90, 95-106. Holthausen, D.M. (1976). Input Choices and Uncertain Demand. The American Economic Review, 66(1), 94-10. APPENDIX Proof of Lemma. First, note that MR (k 1 ) c when k 1 L+c and that MR (k 1 ) = 0 when k 1 H+c. The former fact is because MR (k 1 ) is quadratic in k 1 when k 1 [ L c, L+c ] and ( ) ( ) L c L + c MR = c, MR c(l c) = c + (H L) > c. Therefore, we must have k1 ( L+c H+c ) by A. and (11). Next, observe that MR is strictly decreasing when k 1 [ L+c, H+c L+c ]. Decreasing in [, H c ] is obvious from (10). Decreasing in [ H c, H+c ] is implied by the facts MR (k 1 ) is quadratic in k 1 when k 1 [ H c, H+c ] and MR ( H c ) > 0 and MR ( H+c ) = 0. Note that MR ( H c ) > 0 is implied from the second equality of following expression: MR ( H c ) = c(h + c) (H L) c(h L c) = c. (16) (H L) Therefore, the second order condition is globally satisfied in ( L+c, H+c ). From the third term of the expression (16), MR ( H c ) < c by A.1, which is equivalent to H L c > 0 since Var(A) = (H L) 1. Hence, when r [MR ( H c ), c), we have k1 ( L+c, H c ] and when r (0, MR ( H c )], k 1 [ H c, H+c ). Notice this is because MR is strictly decreasing on [ L+c ]. Therefore, solving the first order condition (11) yields the result., H+c Q.E.D. Proof of Proposition 1. Lemmas 1 and imply lim k1 = H + c < lim k1 m = H r 0 r 0. where the inequality holds by A.. In other words, k1 < km 1 when r is sufficiently close to 0. Furthermore, when A.1 is satisfied, i.e., V ar(a) > (L+c) 48, Lemmas 1 and imply lim k1 = L r c > lim r c km 1 = L c. In other words, k 1 > km 1 when r is sufficiently close to c. 18

Therefore, the proof is complete if MR m and MR cross exactly at a single point on (0, c) (See the figure 6). In absolute value, the slope of MR m is smaller than or equal to c H L for all k 1 by (4) and that of MR is lager than or equal to c H L when k 1 > L+c by (10). Observe that the former fact comes since the slope of MR m equals c H L when k 1 ( L, H c ) and we have the relations MR m (k 1 ) = (L c) + 4k ( ) 1 > MRm L = c (17) k 1 H L k 1 H L MR m (k 1 ) = H 4k ( ) 1 k 1 H L > MRm H c = c (18) k 1 H L when k 1 ( L c, L ) in (17) and when k 1 ( H c, H ) in (18). Notice that the inequalities of (17) and (18) hold because MR m is quadratic and decreasing in each area. Observe that the latter fact comes since the slope of MR equals c H L when k 1 ( L+c MR (k 1 ) = (H + c) k 1 k 1 (H L) < MR k 1 ( ) H + c, H c ) and we have the relation = H + c (H L) < c H L when k 1 ( H c, H+c ). Notice that the first inequality holds because MR is quadratic and decreasing when k 1 ( H c ) and that the final inequality makes use of A.1. Moreover, if, H+c both the slopes of MR and MR m are equal to c H L for some k 1, (19) MR (k 1 ) = ch ck 1 > MR m (k 1 ) = ch c ck 1 by (10) and (4). Hence, MR and MR m never cross at multiple points. Q.E.D. Proof of Proposition. First, we will show that E A (q 1 (k ; A) + q (k ; A)) > E A (q m 1 (km ; A)) when k < k m. If k < k m, y (H) = 0 by Corollary 1, and thus we must have H c < k < H+c from the explanation just after Lemma. Combined with L+c < k (See the first paragraph of Lemma s proof), we find the relation L < k c < H < k + c. Thus, by (6)(7)(8), we obtain the expected total output under duopoly with the new entrant: E A (q 1(k ; A) + q (k ; A)) = k c L a c H da + k c = 4 (k ) + H + c a c + k da k + H 4L c 6cH + 4cL. 1 Since this is increasing in k when H c < k < H+c, we have the relation ( ) ( )) H c H c E A (q1(k ; A) + q(k ; A)) > E A (q1 ; A + q ; A = H ch L + cl c. (0) 19

Meanwhile, the expected total output under monopoly E A (q m 1 (k 1; A)) is non-decreasing in k 1 since q m 1 (k 1; A) is non-decreasing in k 1 for all A [L, H] by (). Thus, using k m 1 < H from Lemma 1, we obtain ( )) H H E A (q1 m (k 1 ; A)) E A (q1 m ; A a = L da = H L. (1) 4 By (0) and (1), we are done that E A (q 1 (k ; A) + q (k ; A)) > E A (q m 1 (km ; A)) if 0 < H ch L + cl c H L 4 = H 4cH L + 4cL 4c. 1 The numerator of right hand side is increasing in H since differentiating it with respect to H yields (H c) > 0 by A.1 and A.. Thus, using H > L + c from A.1, we obtain H 4cH L + 4cL 4c > (L + c) 4c(L + c) L + 4cL 4c = 4c(L c). Finally, we find that 4c(L c) > 0 by A.. Next, we will show that E A (x 1 (k ; A)) < E A (x m 1 (km ; A)) if k < k m. For that, it is sufficient to show that x 1 (k ; A) < x m 1 (km ; A) for all A [L, H]. When k < k m, we must have H c < k as above. Thus, L+c < H c < k by A.1 and this implies that x 1 (k ; L) = L+c < k1 by (8). In the mean time, x m 1 (km ; L) = min[ L, km 1 ] by (1). Therefore, x 1 (k ; L) < x m 1 (km ; L) since we have the facts that L+c < L by A. and thatk < k m. Moreover, x 1 (k 1 ; A)/ A = 1 when x 1 (k 1 ; A) < k 1 xm 1 (km 1 ; A)/ A = 1 and xm 1 (km 1 ; A) < km 1. Therefore, x 1 (k ; A) < x m 1 (km ; A) for all A [L, H]. Q.E.D. Proof of Proposition. Let r = c(h + c) (H L) c(h L c) = c. (H L) Then, r (ˆr, c) since H L c > 0 by A.1, and ˆr < MR ( H c ) = r as seen in the proof of Corollary 1. Then, from Lemma and Lemma, k1 = k 1 for r (0, r]. As for the case where r ( r, c), k1 and k1 L+c must be in (, H c ) by MR ( H c ) = MR ( H c ) = r and MR ( L + c ) = c + H + (H L) 10c 1 Note the last inequality holds since H > 8c by A.1 and A. and H L > c by A.1. Since when k 1 ( L+c, H c ), the slope of MR equals c H L and that of MR is smaller than c H L similarly to the relation (19), MR (k 1 ) < MR (k 1 ) on ( L+c, H c ). Therefore, k 1 > k1 if r > r. Q.E.D. > c. 0

MR I c + c(l c) (H L) c A c c (H L) B E c c(h L c) (H L) F ˆr C c (H L) O L c L c L+cL H c ˆk G H+c H c D H k 1 Figure 6: Expected marginal revenues of nuclear plant s capacity 1

Figure 7: Market share of nuclear power