Safety Stock or Excess Capacity: -offs under Supply Risk

Aadhaar Chaturvedi • Victor Mart´ınez-de-Alb´eniz

IESE School, University of Navarra Av. Pearson 21, 08034 Barcelona, Spain [email protected] • [email protected]

June 2, 2009

This paper investigates the trade-offs between carrying safety stock and planning for excess capacity when a firm faces uncertain demand and supply. Using well-known results from queueing theory, we characterize a buyer’s optimal strategy regarding the amount of capacity that needs to be contracted with a supplier, and the corresponding inventory that it needs to hold. We typically find that both are higher when demand or supply variability are higher. We use our model to analyze multi-sourcing strategies: when multiple suppliers are available, we determine at optimality the size of the supply base, the amount of capacity to be installed, and the inventory levels to be maintained.

1. Introduction

Matching supply and demand is a difficult task for many companies. Indeed, demand is often unpredictable, and firms must plan for adequate resources to face it. Usually, two different approaches are used to adapt supply to demand. First, a company can carry a sufficiently large amount of inventory or safety stock, that will accommodate demand fluctuations. While this might increase expenses for maintaining this inventory, it also allows the firm to reduce the amount of capacity for production. Indeed, with higher inventory levels between demand and production, the utilization of production resources can be increased as inventory dampens any demand shock on production scheduling. In other words, production capacity can be used efficiently. An alternative approach is to carry low levels of inventory, and plan for a more flexible capacity, capable of being adjusted with demand changes. This can be achieved by planning a certain amount of excess capacity that will be used in periods of high demand. Of course, in this case, the expenses related to capacity installation run high, while inventory costs can be lowered. Figure 1 illustrates how a higher capacity level allows inventory to be reduced. The trade-off emerging from these two strategies is a delicate balancing act. Depending on the relative cost of capacity to inventory, firms will find the appropriate amount of capacity

1 3

2

1 Optimal inventory for capacity=3 Optimal inventory for capacity=2 0

Inventory Level −1

−2 0 10 20 30 40 50 60 70 80 90 100 Time Period

Figure 1: Example of inventory evolution, when capacity is limited (exponential demand with average 1). We observe that, when the installed capacity is lower, the number of instances where inventory levels drop below the safety stock increases, and hence the optimal safety stock required is higher. to be installed and inventory level to be maintained. This is specially important as some of these decisions may be irreversible. Indeed, increases in supply capacity may either be flexible and short-lived, e.g., when temporary employees are hired for seasonal increases of demand, or could be permanent, e.g., when a new line is installed or a new factory is opened. In such cases, large investments are required, which have a strong impact on the company’s fate. Given the impact and irreversibility of such capacity investments, it becomes necessary that adequate cost-benefit analysis supports the decisions taken. Specifically, one needs to ascertain that the additional capacity is beneficial in reducing the need for carrying inventory over a long time horizon. Furthermore, in taking these decisions, it is important to consider the possibility that installed capacity may fluctuate over time, due to machine break-downs, supplier disruptions, etc. Indeed, in recent times, supply risk has emerged as a priority for many industries and may have a strong impact on the capacity decision. The effect of limited supply capacity on inventory carrying cost is illustrated in apparel manufacturing. As shown by Raman and Kim [17], for apparel manufacturers solving suc- cessfully the capacity-inventory trade-off is key. First, they incur a high cost of carrying inventory due to high (for maintaining their inventory). Banks are reluctant to give them loans, for the lack of adequate collateral, since the salvage value of their prod- ucts is very low at the end of selling season. At the same time, they also face a high cost of stock-out since the margins on sales are high. Furthermore, demand is highly uncertain in this industry and demand shocks can severely deplete the inventory. Finally, the capacity of suppliers is limited and fluctuates, since a supplier of fabric caters to the demands of many other apparel manufacturers, which might be given higher priority. Given the high inven- tory carrying cost, together with the need to carry high inventory, it becomes imperative for

2 such apparel manufacturers to consider investing in a higher capacity from their suppliers, rather than carrying a high inventory. However, given the competitiveness of the industry, an inappropriate decision can severely affect the company’s future. Also, scarcity of capital resources requires that they optimally decide on the amount of supply capacity they need to procure. From an standpoint, an optimal decision would be to invest in supply capacity up to the point where the marginal increase in capacity investment is equal to the marginal decrease in inventory cost. This trade-off between capacity investment and inventory cost gets more involved when capacity is not only limited but also uncertain. A buyer who has invested in the capacity of an uncertain supplier might have too little or too much capacity at its disposal in a given period, resulting in an under-utilization of its investment. Given uncertain capacity, the buyer might also consider the alternative of diversifying its supply base. Indeed, the buyer can invest in the capacity of more than one supplier. This form of capacity investment has the added advantage that supply variability is reduced through diversification. However, supply diversification also has costs associated with managing a more complex supply chain. Thus the buyer not only needs to decide how much to invest in capacity but also how many suppliers to invest in, that is, the buyer needs to find the optimal diversification strategy, that balances the inventory carrying cost, capacity cost and the diversification cost. In other words the buyer needs to find the right balance between inventory levels it maintains and the capacity it installs at each of the potential suppliers. In this paper, we first investigate the trade-off between capacity investment and the cost of carrying inventory. We model an infinite-horizon periodic-review inventory system, with limited and uncertain capacity. The firm can make a single capacity investment into a single supplier, at the beginning of the model horizon, which determines the distribution of capac- ity available per period. After taking this decision, the firm relies on the supplier’s realized capacity to serve the demand. We analyze the optimal inventory policy and the optimal capacity level, which minimize the average operating cost of the buyer. Interestingly, to solve the limited capacity inventory problem, we use well-known results from the queueing theory literature on the waiting time in a single-server queue. This allows us to formulate closed-form first-order conditions for the optimal capacity investment and the correspond- ing optimal inventory levels. We then use our model to evaluate multi-sourcing strategies. Indeed, we provide a framework where a diversification strategy can be optimized, i.e., de- ciding the optimal size of the supply base, the amount of capacity to be installed, and the inventory levels to be maintained. Our model thus contributes to the literature on supply

3 too. The rest of the paper is organized as follows. In §2 we review the relevant literature. In §3 we present our basic model and find the optimal inventory level for a given capacity. In §4 we solve the optimal investment decision in capacity when a single supplier is present. In §5 we find the optimal capacity investment decision when sourcing from multiple suppliers is possible. We conclude in §6. All the proofs are contained in the appendix.

2. Literature Review

The literature relevant to this paper can be divided into three main groups, namely, capaci- tated inventory models, capacity investment models and multi-sourcing models. Capacitated inventory models deal with optimal inventory policies when supply is capaci- tated, and either fixed or random. Federgruen and Zipkin [10] show that modified base-stock policies are optimal under both average and discounted cost criterion. In other words, it is optimal to place orders up to a base-stock level when possible, and otherwise order full ca- pacity. Ciarallo et al [8] show that an order-up-to policy is optimal in an infinite-horizon periodic-review setting with uncertain supply capacity and uncertain demand. Tayur [18] further computes the optimal inventory policy for an infinite horizon problem with limited supply capacity, based on the average cost criterion. He uses the concept of shortfall (the amount by which the net inventory level falls short of the base-stock level) to model inventory and finds closed-form base-stock levels for exponentially distributed demand. Glasserman [12] develops bounds and approximations for setting base-stock levels with fixed limited ca- pacity, under the average cost criterion. He uses tail probabilities associated with random walks to obtain asymptotic approximations as the service level approaches 100%. We closely follow the modeling approach of Tayur and Glasserman and extend it to random capacity. The papers above focus on optimizing inventory policies when capacity is given. Much research has been done in jointly considering inventory and capacity decisions. Van Mieghem [15] provides a comprehensive review of the topic. We discuss the two main streams that analyze capacity investment problems. First, newsvendor-type capacity investment models analyze the capacity-inventory trade- off through a capacitated version of the multi-period discounted newsvendor model. However, since such models deal with the transient evolution of inventory (rather than focusing only on the steady state), it is not possible to find an optimal capacity level in closed-form, in a multi-period setting. Van Mieghem and Rudi [16] study newsvendor networks to jointly find optimal capacity and inventory with multiple products under the lost sales scenario.

4 Interestingly, they use three tools to optimize operating profits, namely capacity and inven- tory which are decided before demand is realized and activities (that are used to convert inventories into finished goods by utilizing capacity) which are decided after the demand is realized. However, for the backlog case they only solve the inventory problem for the unca- pacitated case. In contrast, we analyze a single-product capacity investment decision, with backlogging and average cost criterion. Angelus and Porteus [2] solve the capacity-inventory problem for a manufacturer in the case where both capacity and inventory decisions can be made in every period. That is, capacity can be added and removed at a certain cost in every period. They show that it optimal to bring capacity to a certain target interval and characterize these intervals for the lost-sales case. For the backlog case, they find that capacity and inventory act as economic substitutes, which we observe in our paper as well. However they do not characterize the optimal capacity levels (target intervals) for this case. Alternatively, queueing-type capacity investment models study manufacturing systems where both demand arrivals and production are treated as continuous processes (most often as Poisson processes). This results in continuous-review inventory policies where capacity, i.e., the rate of production, is a decision variable. Although such models can handle general demand distributions, most times they are limited to exponentially distributed production systems, that is, where the time taken to produce one unit in the production facility is expo- nentially distributed. In contrast, our model focuses on a periodic-review system where the capacity per period can be generally distributed. In this stream of literature, Bradley and Glynn [4] consider a manufacturer that faces a fixed installation and linear utilization cost structure for capacity. They develop heavy traffic and Brownian motion approximations for operational costs that yield insight into the capacity-inventory trade-off. However, they do not develop any closed-form solution for the optimal capacity decision, because the analysis becomes complex as they allow for correlated demand. Toktay and Wein [19] find optimal release policies (amount of goods released for production) in a capacitated production envi- ronment with updated demand forecasts in each period. De Kok [14] considers a capacitated packaging facility and discusses the optimal strategy of meeting service level through either outsourcing extra capacity needs or by postponing the extra capacity needs to the future. Caldentey and Wein [6] come closest to our paper in jointly solving the optimal capacity and inventory decision, with an average inventory cost objective. They consider that back-order costs are shared by both one supplier and one retailer, and solve the steady-state optimal decision on capacity (taken by the supplier) and inventory (taken by the retailer), for M/M/1 queues. However, the focus of their paper is to find equilibrium transfer payment contracts

5 between the supplier and the retailer to coordinate the supply chain. On a similar game theoretical approach, Cachon and Zhang [5] analyze a buyer-supplier equilibrium in which strategic suppliers select their capacity to minimize operational costs, and the buyer selects its allocation policy to them, in order to minimize the average delivery lead-time over an infinite horizon. They find that state-dependent allocation policies, i.e., policies that take into account the present state of workload on servers, deliver faster lead time compared to state-independent allocation policies. Finally, the literature on supply risk management through multi-sourcing is also related to our work. Tomlin [20] discusses mitigating supply risk through either sourcing from more than one supplier, compared to carrying safety inventory buffers. Indeed, diversification of suppliers is a possible solution to reduce supply disruptions. Federgruen and Yang [9] study the planning models used to determine quantities, in the presence of stochastic demand and supply default risk. Babich et al. [3] present a model for supplier pricing, where the buyer may buy from a single or from many suppliers. Chaturvedi and Mart´ınez- de-Alb´eniz[7] consider the design of optimal capacity auctions when suppliers are unreliable. In this paper, we jointly consider inventory, capacity and multi-sourcing decisions.

3. The Model

We model a buyer that faces the demand for a single product in every period. This demand, denoted by Dn in period n, is stochastic and exogenously given. It follows a c.d.f. FD(x) with mean d and is i.i.d across periods. The demand is served from inventory, which is replenished at the end of every period at a cost of c per unit, with zero lead-time. If the buyer’s inventory falls short of the demand in any period, then the unmet demand is backlogged to the next period at a penalty cost of p per unit1. On the other hand, if all the demand of a particular period is met, then any leftover units in the inventory are passed on to the next period at a holding cost of h per unit. The orders placed are shipped from one supplier (or more than one in §5). This supplier has a limited capacity Kn that is available to the buyer in period n, which may be insufficient to fill the buyer’s order. We model Kn as a random variable and follows a c.d.f. FK (x) which is i.i.d. across periods, with mean k. However, the buyer can control the capacity in place by making an initial one-time investment. As a result, the capacity distribution FK (x) depends on the buyer’s investment, e.g., its mean k is endogenized and can be varied.

1Since demand is backlogged, the assumption of zero lead-time could be relaxed easily with no impact on the structure of our results.

6 The buyer thus has two decisions to take that affect its operating cost: (i) it must make the necessary capacity investments, and (ii) it must decide on an appropriate inventory replenishment policy, that should consider the fact that supply is limited by capacity. The objective of the buyer is to minimize its average operating cost over an infinite horizon2. This includes the capacity cost per period, the inventory carrying cost which includes holding and back-ordering costs, and the variable purchasing cost. Since the average purchasing cost is fixed, equal to c · d per period, the problem of the buyer is to find the right balance between capacity and inventory costs. First, given a capacity level, consider the inventory decision. At the end of every period, the buyer observes the capacity of the supplier that is available. The buyer then places the order which arrives before the demand for the next period appears. It is well-known that a modified base-stock policy is optimal for the buyer facing such conditions with infinite horizon. In other words, if In is the inventory level at the beginning of period n (after receiving the order placed at the end of period n − 1), and y is the base-stock level, then it is optimal to order min{Kn, y − (In−1 − Dn)}. Thus, the inventory level at the beginning of period n + 1 is ½ y if y − (In − Dn) ≤ Kn In+1 = In−1 − Dn + Kn otherwise which can also be written as

In+1 = min(In − Dn + Kn, y). (1)

Similar to Tayur [18] and Glasserman [12] we introduce the shortfall Wn = y −In in period n as the amount by which the inventory falls short of the base-stock level. We can then write Equation (1) as

Wn+1 = max(Wn − Kn + Dn, 0) (2)

One can observe that the dynamics of the shortfall in Equation (2) are identical to the dynamics of waiting time in a queueing system with one server. In this queueing system, th the shortfall Wn represents the time that the n customer arriving into the system has to wait in the queue. The demand Dn represents the time taken by the server to serve the th n customer once it arrives at the server. Finally, the capacity Kn represents the inter- arrival time between the nth and the (n + 1)th customer arriving to the queue. Note that the interpretation of the shortfall as a waiting time is not very intuitive, since the shortfall in period n is interpreted as the waiting time of customer n in the queuing system. This is

2Since capacity decisions are involved, using an infinite horizon average cost criterion is reasonable, as transient effects should be secondary and thus need not be considered.

7 very different from manufacturing queues, where time is continuous and queues discrete. In our case, time is discrete and queues continuous. E(D ) d As a result, from standard queuing theory we obtain that when ρ = n = < 1 the E(Kn) k distribution of Wn converges to W as n → ∞ such that

W =d max(W − K + D, 0) (3) where =d indicates equivalence in distribution. For ρ ≥ 1 the shortfall increases over time (since on average demand is higher than capacity), and hence the distribution of shortfall does not converge to a steady state. The notation above allows us to formulate the average carrying cost as

¡ + +¢ ED,W p(D + W − y) + h(y − W − D)

Since the expression above is convex in y, we can therefore find the optimal base-stock level y through a first-order condition, which yields p P(D ≤ y − W ) = E F (y − W ) = (4) W D p + h The above result is similar to Tayur [18]. Thus, given a certain capacity distribution, we can optimize the base-stock level accordingly. Hence, we can write the average carrying cost of the buyer as ¡ + +¢ C(a) = min{ED,W p(D + W − y) + h(y − W − D) } (5) y where a represents the vector of investments in capacity, which in turn affects the capacity distribution FK . For example, if the supplier is fully reliable, a would be the capacity level in place; if the supplier’s capacity follows a Gamma distribution of shape ν and scale ξ, then a = (ν, ξ). Furthermore, in order to find the best balance between capacity investment H(a) and inventory cost C(a), the buyer’s problem can be written as

min {C(a) + H(a)} (6) a We are now in a position to find optimal capacity investment that would solve Equation (6). For this purpose, we follow three steps:

1. We first characterize the distribution of waiting time, with p.d.f., FW , in Equation (3) for different distributions of demand and capacity.

2. We evaluate the average carrying cost in Equation (5) for this particular distribution of waiting time.

8 3. We finally establish optimality conditions for the optimal capacity investment by char- acterizing buyer’s problem in Equation (6).

4. Single Supplier

In this section, we study the optimal balance between capacity and inventory when there is a single supplier. For this purpose, we analyze the waiting time distribution of a G/G/1 queue. Due to the complexity of this task, we first focus on a G/M/1 system, i.e., when demand is exponentially distributed, and derive closed-form solutions. We then consider the general case where we discuss analytical and numerical solutions.

4.1 Exponential Demand

Consider first that the demand is exponentially distributed with mean d. For a general distribution of supplier capacity, the distribution of shortfall in Equation (2) is given by the distribution of the waiting time in a G/M/1 queue. Using results from queueing theory, see Kleinrock [13], we can characterize the distribution of shortfall through its probability 3 density function fW (w). This p.d.f. is characterized by a parameter σ, and is given by · ¸ (1 − σ) − (1−σ) w f (w) = (1 − σ)δ(w) + σ e d (7) W d where δ(w) is the delta-dirac function. In other words, the distribution of the shortfall is (1 − σ) 0 with probability 1 − σ and an exponential of rate with probability σ. σ can be d found from the functional equation Z ∞ − (1−σ) x σ = e d dFK (x) (8) x=0 µ ¶ 1 − σ or equivalently as σ = L , where L represents the Laplace transform of the sup- K d K ply capacity distribution FK . In standard queueing theory, it has been shown that there exists a unique real solution for σ in the range 0 < σ < 1, when ρ = d/k < 1.

The probability distribution function for the shortfall enables us to characterize the base- stock level in Equation (4) as µ ¶ µ ¶ d h y(σ) = − ln ≥ 0. (9) 1 − σ p + h

3In a G/M/1 queue, the steady-state distribution of customers in the system, at arrival instants, is geometric. σ is the parameter that determines this geometric distribution, i.e., it is the ratio of steady-state probability of finding n + 1 customers in the system, to the probability of finding n customers in the system. However, there is no direct interpretation of σ as a probability ratio in our model.

9 To characterize the carrying cost in Equation (5) we introduce the random variable Z which is the sum of shortfall and demand. Then the distribution of Z can be found from the convolution of shortfall and demand. Specifically, for exponential demand, the distribution of Z is − (1−σ) x P(Z ≤ x) = 1 − e d , (10) (1 − σ) i.e., Z is exponentially distributed with rate . We can thus write the carrying cost d as ¡ + +¢ C(a) = EZ p(Z − y) + h(y − Z) (11) Both the optimal inventory y and the distribution of Z can be uniquely identified by σ. Furthermore, σ is determined by the mean of demand and the parameters of the capacity distribution, i.e., the vector of investments in capacity a. The following lemma provides some structural properties of the carrying cost.

Lemma 1 For an exponentially distributed demand, the cost of carrying inventory C(a) is convex in a if σ is convex in a.

Furthermore, if the capacity investment cost H is also convex in the vector of investments a, then the buyer’s problem in Equation (6) is convex and hence the optimal capacity investment is characterized by first-order conditions.

Theorem 1 For an exponentially distributed demand, if H(a) and σ(a) are convex in a, then the optimal vector of investment in capacity is given by jointly solving the system of equations µ ¶ ∂H ∂C hd h ∂σ = − = 2 ln (12) ∂ai ∂ai (1 − σ) p + h ∂ai The above theorem states that, at optimality, the marginal benefit from buying capacity, achieved as a reduction of carrying cost, should be equal to the marginal cost of buying that capacity. From Equation (9), we find that optimal inventory levels are increasing in σ. Also, if σ is decreasing in ai then from Equation (12), we find that a higher investment cost for ai would result in a lower ai at optimality. This implies that the optimal inventory level y and the capacity investment ai act as economic substitutes, as observed by Angelus and Porteus [2]. In the remainder of this section we characterize H(a) and σ(a) and show their convexity in a, for different distributions of supply capacity. We first focus on an exponential distribution of supply capacity, and then on general distributions of supply capacity that are commonly found in the literature: (i) fixed capacity, (ii) Bernoulli distributed capacity and (iii) Gamma distributed capacity.

10 4.1.1 Exponential Capacity

For exponentially distributed capacity with mean k, the distribution of shortfall can be characterized by the distribution of the waiting time of an M/M/1 queue. In that case, − x 1 F (x) = 1−e k , and hence the Laplace transform is L (s) = . As a result, Equation K K 1 + ks d (8) becomes σ = , which can be written as (1 − σ)(d − kσ) = 0. Since we know d + k(1 − σ) d that 0 < σ < 1, hence the only feasible solution of σ can be σ = = ρ, the utilization. k d Proposition 1 For exponentially distributed capacity with mean k, σ(k) = is decreasing k and convex in k.

Since exponential distribution is completely characterized by its mean k, we can define the capacity investment vector as a = k. Furthermore, consider that H(k) = Ccap · k is linear, with a capacity cost equal to Ccap per unit. Since σ is decreasing and convex in k, with Theorem 1 we can fully characterize the optimality conditions for the capacity investment: s µ ¶ k h h = 1 + − ln . d Ccap p + h

d In other words, at optimality, the utilization of capacity ρ = is increasing with the ratio k C of capacity to inventory cost cap , and decreasing in the ratio of back-ordering to inventory h p cost . h

4.1.2 Non-Exponential Capacity

In general one cannot find an explicit expression for σ from the functional equation (8), when supply capacity is not exponentially distributed. Hence, to show convexity of σ in a one has to implicitly differentiate the functional equation. For a capacity that is fully reliable and fixed at q, we can show σ to be convex in q.

− (1−σ) q Proposition 2 For a fixed capacity q, σ = e d , and σ(q) is decreasing and convex in q.

Here we define capacity investment vector a = q. Letting, H(q) = Ccap·q, we can characterize the optimal capacity decision from Equation (12). Note however that σ is implicitly defined, and therefore one needs to numerically find the optimal capacity investment. Next we consider the case where supply capacity is Bernoulli distributed, i.e., we consider supply capacity to be m with a probability α and 0 with a probability 1 − α. Note that when α = 1, we go back to the previous case, where capacity is fixed. When we define the

11 3.5 20

α=0.5 19 α=0.9 3 α=1 18

2.5 17 Optimal Mean Capacity

2 16 0.2 0.4 0.6 0.8 1 C cap 15

6 Average Cost 14 5 13 4

12 3 d=1 h=1

Optimal Inventory p=10 2 11 H(m,α)=5mα

1 10 0.2 0.4 0.6 0.8 1 1 1.2 1.4 1.6 1.8 2 C cap Mean Capacity

Figure 2: The supply capacity is Bernoulli distributed with parameters (m, α) and H(m, α) = Ccap · mα. We consider the reliability α to be fixed, and consider the impact of varying m on the right-hand side figure. On the left-hand side figure, we illustrate the sensitivity to the cost of capacity Ccap. capacity investment vector as a = (m, α), the buyer has two levers to invest in capacity: it can increase the reliability α, or increase the available capacity m when the supplier is reliable.

− (1−σ)m Proposition 3 For a Bernoulli distributed capacity with parameters m and α, σ = αe d + 1 − α and σ(m, α) is decreasing in m and α, and jointly convex in (m, α).

Assuming that H(m, α) is increasing and convex in (m, α), Proposition 3 allows us to find the optimal capacity decision from Equation (12). When we define the cost of capacity as

H(m, α) = Ccap · mα, although the capacity investment is not jointly convex in (m, α), we find numerically that the optimal capacity decision is still uniquely defined by the first-order conditions. To illustrate the proposition, Figure 2 presents, for α fixed, the optimal capacity investment m. We can observe in the figure that optimal inventory and mean capacity are economic substitutes. Finally, we consider the case where supply capacity is Gamma distributed with mean φ = νξ, where ν is the shape parameter and ξ is the scale parameter. This is a useful distribution that is unimodal, and usually occurs, for example, when the effective supply capacity is the sum of exponential variables. We use this property to model the effect of

12 multi-sourcing on capacity investment in §5. If a = (ν, ξ), the buyer has again two levers to invest in capacity, that is, it can either scale up the capacity by investing in ξ or it can invest in making the supply capacity more reliable by investing in ν.4 We find that σ is generally convex in (ν, ξ), however it is difficult to prove it. Instead we show in the following proposition the convexity of average carrying cost C in (ν, ξ).

Proposition 4 For a Gamma distributed capacity with shape parameter ν and scale param- eter ξ, the average carrying cost C is decreasing in ν and ξ, and jointly convex in (ν, ξ).

When H(ν, ξ) is increasing and jointly convex in (ν, ξ) Proposition 4 allows us to find the optimal capacity decision from Equation (12). Usually, the buyer in fact controls the average supplier capacity φ = νξ, as the cost is often expressed as Ccap · φ. If a = (φ), i.e., the buyer can only invest in the mean of supplier capacity, or if a = (φ, ν), i.e., the buyer can invest in both mean and the shape parameter, then σ is generally not convex in (φ, ν). However, we show in the following proposition convexity of average carrying cost with respect to the mean and the shape parameter of supply capacity.

Proposition 5 For a Gamma distributed capacity with mean φ and shape parameter ν, the average carrying cost C(φ, ν) is decreasing in φ and ν, and jointly convex in (φ, ν).

This proposition is quite relevant since it provides a well-behaved strucutre for our model with multiple sourcing in §5. Similarly as before, when the capacity cost H(φ) is increasing and convex in the mean of capacity, the optimal capacity investment is unique and can be found from Equation (12). Finally, taking ν given, and varying φ, Figure 3 illustrates the trade-off between capacity and inventory.

4.2 Non-Exponential Demand

When demand is non-exponential then the shortfall is distributed as the waiting time in a G/G/1 queue. No closed-form formulas exist for this case. In certain cases some results can be obtained, in particular when the supply capacity is exponential.

4 − (1−σ) k Note that for k = νξ fixed, when ν → ∞ we obtain σ → e d , which is equivalent to having a fixed supply capacity of k.

13 5 20 ν=0.5 ν 4 19 =1 ν=5 ν 3 18 =100

2 17 {Optimal Mean Capacity 1 16 0 0.2 0.4 0.6 0.8 1 C cap 15

5 Average Cost 14

4 13

12 d=1 3 h=1 11 p=10 {Optimal Inventory H(ν,ξ)=5νξ 2 10 0 0.2 0.4 0.6 0.8 1 1 1.2 1.4 1.6 1.8 2 C cap Mean Capacity

Figure 3: The supply capacity is Gamma distributed with mean φ and shape parameter ν and H(φ, ν) = Ccap · φ. We consider the shape ν to be fixed, and consider the impact of varying the mean capacity φ on the right-hand side figure. On the left-hand side figure, we illustrate the sensitivity to the cost of capacity Ccap.

4.2.1 Exponential Capacity

When supply capacity is exponential with mean k, the Pollaczek-Khinchine transform for- mulas characterize the Laplace transform of the waiting time in the steady-state, (k − d)s LW (s) = (13) LD(s) + ks − 1 where LW (s) is the Laplace transform of the waiting time distribution, LD(s) is the Laplace transform of the demand distribution, and d the average demand. Inverting the Laplace transform in Equation (13) can provide the distribution of the shortfall, and provides a numerical approach for optimizing the value of the capacity a = k to be installed. Recalling that Z = W + D, Equation (13) yields that

LD(s)k(1 − ρ)s LZ (s) = LD(s)LW (s) = (14) LD(s) + ks − 1 and hence the inventory carrying cost can be expressed as C(k) = pEZ−py+(p+h)E(y−Z)+. The Pollaczec-Khinchine formula provides the expected waiting time as µ ¶ µ ¶ ρ 1 + CV 2 EW = d D 1 − ρ 2

14 2.5 40 Gamma(CV2 =2) D Expoential(CV2 =1) D 35 Gamma(CV2 =1/2) 2 D Uniform(CV2 =1/3) D Fixed(CV2 =0) 30 D

Optimal Mean of Capacity d=1 1.5 1 2 3 4 5 h=1 C p=10 cap 25 C =4 cap

3.5 Average Cost

20 3

2.5 15 2 Optimal Inventory

1.5 10 1 2 3 4 5 1.2 1.4 1.6 1.8 2 2.2 2.4 C Mean of Capacity cap

Figure 4: On the left-hand side, optimal mean capacity and inventory level, as a function of the cost of capacity Ccap. On the right-hand side, total average cost as a function of mean capacity.

2 V arD where CVD = ³ ´2 . Hence, ED · µ ¶ µ ¶¸ ρ 1 + CV 2 EZ = EW + d = d 1 + D 1 − ρ 2 Unfortunately, E(y − Z)+ cannot be found in closed-form. We use a Fourier-series method to numerically invert the Laplace transform of Z, as described by Abate and Witt [1]. We proceed to show numerical results for the optimal capacity investment for different demand distributions, all with average 1: (a) fixed demand of 1, (b) uniform demand in [0,2], (c) 1 Gamma distributed demand with mean 1 and different coefficients of variation, of √ , 1 √ 2 (i.e., an exponential of rate 1) and 2. We compare the expected capacity at optimality, together with the inventory level. The right-hand side of Figure 4 shows that, given a certain level of mean capacity k, the higher the variability of the demand, the higher the inventory operating cost. Similarly the left-hand side of the figure shows that, given a certain cost of capacity, the higher the demand variability, the higher the mean capacity that the buyer installs and the higher is the corresponding safety stock y that it keeps. As a result, if H(k) = Ccap · k, the optimal investment in capacity is increasing as demand becomes more variable. This results in a

15 lower utilization, as one would expect. This implies that, when demand variability increases, the buyer reacts by both lowering the capacity utilization and increasing the safety stock.

4.2.2 Non-Exponential Capacity

When the capacity is not exponentially distributed, then we must resort to numerical sim- ulation to find the distribution of the shortfall. We present below our numerical results for different distributions of supply capacity, and different distributions of demand. We con- sider (i) fixed capacity, (ii) Bernoulli distributed capacity, with α = 0.8, and (iii) Gamma 1 distributed capacity with different coefficients of variation, of √ , 1 (i.e., an exponential √ 2 of rate 1) and 2; and (a) fixed demand of 1, (b) uniform demand in [0,2], (c) Gamma 1 distributed demand with mean 1 and different coefficients of variation, √ , 1 (i.e., an ex- √ 2 ponential of rate 1) and 2. Interestingly, the carrying cost is found to be convex in the average capacity, which guarantees that the optimization over capacity is well-behaved. We show the results in Table 1 (after the appendix). As in §4.2.1, we find that operating cost, inventory levels and capacity investment typically increase as demand or supply becomes more variable. We find that this trend to be consistent within a distribution, that is, for a given distribution, capacity and inventory levels increase with the coefficient of variation. However, this trend might not hold across distributions since the tail effects of different dis- tributions could considerably affect the shortfall distribution and consequently the inventory carrying cost. For example, cost and inventory are higher for a Bernoulli distributed capacity 1 (coefficient of variation ) compared to Gamma distributed capacity (coefficient of variation 2 1 √ , higher). 2 5. Multiple Suppliers

So far, we have considered the trade-off between capacity and inventory for a buyer that is served by a single supplier. In this section, we analyze the impact of multi-sourcing on the capacity-inventory relationship. Of course, when the capacity at a supplier is fully reliable, diversification will not help the buyer. In contrast, when capacity is unreliable, there exists an opportunity to diversify the supply base, which reduces the uncertainty of supply (its coefficient of variation). This allows the buyer to utilize capacity better, and reduce operating cost. At the same time, the supply chain will become more complex, as more suppliers are present, and this may create additional costs. We model this as a cost of diversification Cdiv to be paid for each supplier included in the supply base. When Cdiv = 0, it is clear that

16 the buyer would completely diversify its supply base and work with as many suppliers as possible, for a given total capacity. However, when Cdiv > 0, there is an optimal number of suppliers in the supply base, that should not be too large (because then the sourcing cost would be too high) nor too low (because then the variability of capacity may force the buyer to install a higher level of capacity). We investigate this trade-off and look at the optimal diversification strategy, that is, how many suppliers to be included in the supply base, how much capacity to be installed, and what inventory level to keep. For simplicity, we focus our analysis on the case where all suppliers are identical, i.e., their capacity investment cost, their purchasing cost and their individual capacity distribution are the same. In addition, in order to use the results of §4, we assume that all the suppliers’ capacity is Gamma distributed with the same shape parameter, e.g., when it is exponential. We also assume, as in section §4.1, that demand is exponentially distributed, with mean d. Essentially the buyer’s problem remains the same as in Equation (6), with the exception of the addition of an extra parameter, namely the number of suppliers s. We consider that 5 each supplier has a Gamma distributed i.i.d. capacity with shape νi = ν and scale ξi = ξ. As a result, the total available capacity to the buyer is Gamma distributed with shape sν and scale ξ. Thus, the average total capacity is φ = sνξ. From Proposition 5, we know that the inventory carrying cost is decreasing and jointly convex in (sν, φ) and also in (s, φ). Hence, we can determine the optimal inventory level and the optimal diversification strategy from Equation (9) and Equation (12) respectively. As before, we can use the capacity investment vector to define the cost structure for diversification. We take a = (s, φ). Assume first that H(s, φ) is increasing and convex. That is, we assume that marginal cost of contracting with each additional supplier (cost of diver- sification) is increasing and the marginal cost of buying higher mean capacity is increasing too. We can therefore characterize the optimal diversification strategy from Equation (12).6 From that equation, we see that the buyer would diversify less when the cost of diversifi- ∂H cation is higher. Similarly, the buyer would buy higher mean capacity from suppliers ∂s ∂H when the cost of capacity is lower. ∂φ It is worth analyzing the case where H is linear in (s, φ): we can define H(s, φ) =

Cdiv · s + Ccap · φ, where Cdiv is the cost of diversification and Ccap is the cost of capacity. Using the optimality equations of Theorem 1, the optimal supply base and the optimal mean

5Clearly, when choosing s identical suppliers, at optimality the buyer would install the same capacity in all of them. 6 Although s takes integer values, minφ{H(s, φ) + C(s, φ)} is convex in s, and hence can be optimized by simple search.

17 capacity are found by solving the equations

à !νs 1 σ = (1−σ)φ + 1 dνs µ ¶ hd h dσ(φ, s) (15) Cdiv = − ln (1 − σ)2 µp + h¶ ds hd h dσ(φ, s) C = − ln cap (1 − σ)2 p + h dφ dσ(φ, s) dσ(φ, s) where and can be found by differentiating implicitly the first equation. ds dφ These equations provide a value of s that might not be integer. Since the cost function is jointly convex in (s, φ), the optimal size of the supply base s∗ is bsc or dse. Finally, φ∗ is given by solving à !νs∗ 1 σ = (1−σ)φ + 1 dνs∗ µ ¶ (16) hd h dσ(φ, s∗) C = − ln cap (1 − σ)2 p + h dφ Once again, from Equation (12) we see that the optimal number of suppliers s is decreas- ing in Cdiv, that is, the buyer would diversify less when the cost of diversification is high.

Similarly the total capacity is decreasing and the inventory is increasing in Ccap. We show this optimal strategy in Figures 5 and 6. In the first figure, we see how the diversification strategy should vary as a function of the diversification cost. Interestingly, when the supply base is reduced, the average total capacity to be installed increases, as well as inventory. In the second figure, we observe the impact of the capacity cost. Here, we see that the number of suppliers tends to increase with Ccap. This is intuitive: when the capacity becomes more expensive, it becomes more cost-efficient to reduce its uncertainty, which can be achieved by increasing the number of suppliers. In summary, in a context of supply risk, our model can be used to jointly evaluate how many suppliers to use, and how much capacity and inventory are necessary.

6. Conclusion

The model developed in this paper solves the capacity-inventory trade-off of a buyer, in an infinite-horizon periodic-review setting, for the average cost criterion. We consider the general case where both demand and available capacity are stochastic. Using that a modified base-stock policy is optimal, we show that the shortfall, defined as the difference between base-stock level and current inventory, is distributed as the waiting time in a queuing system with one server. Thus, when demand is exponentially distributed we are able to obtain

18 10 d=1 h=1 p=10 C =1 5 cap

No. Of Suppliers 1 0.02 0.04 0.06 0.08 0.1 0.12 0.14 C div 2.3

2.2

2.1

2

Total Mean Capacity 0.02 0.04 0.06 0.08 0.1 0.12 0.14 C div

3.4

3.2

3 Optimal Inventory 0.02 0.04 0.06 0.08 0.1 0.12 0.14 C div

Figure 5: Optimal diversification strategy as a function of Cdiv, the cost to be paid per supplier.

6 d=1 4 h=1 p=10 2 C =0.1 div

No. of Suppliers 0 0 1 2 3 4 5 6 7 8 9 10 C cap 4

3

2

1

Total Mean Capacity 0 1 2 3 4 5 6 7 8 9 10 C cap 6

4

2 Optimal Inventory 0 1 2 3 4 5 6 7 8 9 10 C cap

Figure 6: Optimal diversification strategy as a function of Ccap, the capacity cost.

19 the first-order condition for the capacity decision. We provide closed-form solutions for several cases commonly used in the literature to model supply risk: when capacity is fixed, when it is Bernoulli distributed, and when it is Gamma distributed. When demand is not exponentially distributed, some analytical results can be obtained when capacity is exponential, and otherwise we discuss the optimal capacity decisions based on numerical experiments. We then apply the model to a situation where the buyer can choose to install capacity at multiple suppliers simultaneously, in order to diversify supply risk. We find that, as the cost of diversification increases, the supply base has fewer suppliers, and both the total installed capacity and the safety stock increase. As the cost of capacity increases, the supply base has more suppliers, and the total installed capacity decreases and the safety stock increases. Compared to the previous literature, the model provides a useful framework to evaluate supply risk management strategies. In particular, the application with multiple suppliers can be used to analyze more general supply base design problems, which are receiving increasing attention. This opens directions for future research. For example, while our model considers multiple suppliers, they are assumed to be identical. When they are asymmetric, the capacity decision must take into account that they might offer different supply reliability and different cost. As a result, suppliers must be prioritized, based on cost and reliability.

References

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[2] Angelus A. and E. L. Porteus 2002. “Simultaneous Capacity and Production Management of Short-Life-Cycle, Produce-to-Stock Goods Under Stochastic De- mand.”Management Science, 48(3), pp. 399-413.

[3] Babich V., A. N. Burnetas and P. H. Ritchken 2007. “Competition and Diversifica- tion Effects in Supply Chains with Supplier Default Risk.” Manufacturing & Service , 9(2), pp. 123-146.

[4] Bradley J. R. and P. W. Glynn 2002. “Managing Capacity and Inventory Jointly in Manufacturing Systems.” Management Science, 48(2), pp. 273-298.

[5] Cachon G. P. and F. Zhang 2007. “Obtaining Fast Service in a Queueing System via Performance-Based Allocation of Demand.” Management Science, 53(3), pp. 408-420.

20 [6] Caldentey R. and L. M. Wein 2003. “Analysis of a Decentralized Production-Inventory System.”Manufacturing & Service Operations Management, 5(1), pp. 1-17.

[7] Chaturvedi A. and V. Mart´ınez-de-Alb´eniz2008. “Optimal Procurement Design in the Presence of Supply Risk.” Working paper, IESE Business School.

[8] Ciarallo F. W., R. Akella and T. E. Morton 1994. “A Periodic Review, Production Planning Model with Uncertain Capacity and Uncertain Demand-Optimality of Ex- tended Myopic Policies.” Management Science 40(3), pp. 320-332.

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[10] Federgruen A. and P. Zipkin 1986a. “An Inventory Model with Limited Production Capacity and Uncertain Demands, I: The Average Cost Criterion.”Mathematics of Operations Research, 11, pp. 193-207.

[11] Federgruen A. and P. Zipkin 1986b. “An Inventory Model with Limited Production Capacity and Uncertain Demands, II: The Discounted Cost Criterion.”Mathematics of Operations Research, 11, pp. 208-215.

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[13] Kleinrock L. 1975. Queueing Systems. Wiley, 1975.

[14] de Kok T. G. 2000. “Capacity Allocation and Outsourcing in a Process Industry.” International Journal of Production Economics, 68, pp. 229-239.

[15] van Mieghem J. A. 2003. “Capacity Management, Investment and Hedging: Review and Recent Developments.” Manufacturing & Service Operations Management, 5(4), pp. 269-302.

[16] van Mieghem J. A. and N. Rudi 2002. “Newsvendor Networks: Inventory Management and Capacity Investment with Discretionary Activities.” Manufacturing & Service Op- erations Management, 4(4), pp. 313-335.

[17] Raman A. and B. Kim 2002. “Quantifying the Impact of Inventory Holding Cost and Reactive Capacity on an Apparel Manufacturer’s Profitability.” Production and Operations Management, 11(3), pp. 358-373.

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[19] Toktay L. B. and L. M. Wein 2001. “Analysis of a Forecasting-Production-Inventory System with Stationary Demand.” Management Science, 47(9), pp. 1268-1281.

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22 Appendix

Proof of Lemma 1

Proof. Equation (11) can be re-written as

Z y Z ∞ C(a) = h (y − z)g(z)dz + p (z − y)g(z)dz z=0 z=y where g(z) is the distribution function of Z = W + D. We then obtain

hd (p + h)d − (1−σ(a))y C(a) = hy − + e d (1 − σ(a)) (1 − σ(a)) ³ ´ h d ln p+h Substituting the value of y = − in the above equation, we get C(a) = hy(σ(a)). (1 − σ(a)) Hence the gradient of C(a) can be written as dC ∇ C(a) = ∇ σ(a) a dσ a where ∇aX denotes the vector³ of partial´ derivatives of X with respect to each element of h dC hd ln p+h the vector a and = − ≥ 0. The Hessian for C(y(a), a) can thus be written dσ (1 − σ)2 as d2C dC M + H (17) dσ2 dσ σ ∂σ ∂σ where Hσ is the Hessian matrix of σ and M is a symmetric matrix such that Mij = . ∂ai ∂aj dC d2C By construction M is positive semi-definite. Also ≥ 0 and ≥ 0. Hence if σ is jointly dσ dσ2 convex in a then the Hessian of C(y(a), a) is the sum of two positive semi-definite matrices and therefore C jointly convex in a.

Proof of Theorem 1

Proof. From³ Lemma´ 1 we can write the first-order conditions for Equation (6) using h dC hd ln p+h = − . dσ (1 − σ)2

Proof of Proposition 2

− (1−σ)q Proof. For fixed capacity, Equation (8) can be written as σ = e d or in other words

d ln(σ) q = − . (1 − σ)

23 q is monotonically decreasing in σ in the interval (0, 1). Indeed, taking the first with respect to σ we obtain dq d d ln(σ) = − − < 0 dσ σ(1 − σ) (1 − σ)2 1 because for all σ ∈ (0, 1), ln(σ) > 1 − . Taking the the second derivative yields σ d2q d (1 − 4σ + 3σ2 − 2σ2 ln(σ)) d(1 − σ)2 = ≥ ≥ 0 dσ2 σ2(1 − σ)3 σ2(1 − σ)3 Convexity of q in σ implies that σ is convex in q.

Proof of Proposition 3

Proof. From Equation (8) we get

− (1−σ)m σ = αe d + 1 − α. (18) d2σ We first show that σ is decreasing in m and α, then we show that ≥ 0 and finally we dm2 show that the determinant of Hessian of σ is non-negative.

From Equation (18) we can write m explicitly as a function of σ: µ ¶ d σ + α − 1 m = − ln . (1 − σ) α dm < 0, as in the previous proof. Similarly, from Equation (18) we can write α as an explicit dσ function of σ, 1 − σ α = (19) − (1−σ)m 1 − e d dα Again < 0 in the unit interval. dσ

Differentiating α with respect to σ in Equation (19) yields ³ ´ (1−σ)m m(1−σ) (1−σ)m − d − d dα − 1 − e − d e = ³ ´2 dσ − (1−σ)m 1 − e d

Differentiating again, d2α dσ2 ³ ´ ³ ´ m2(1−σ) (1−σ)m (1−σ)m (1−σ)m m(1−σ) (1−σ)m m (1−σ)m − d − d − d − d − d d2 e 1 − e − 2 1 − e − d e d e = ³ ´3 − (1−σ)m 1 − e d

24 m(1 − σ) d2α Letting x = , the numerator of is d dσ2 m £ ¤ m £ ¤ e−x x(1 − e−x) − 2(1 − e−x) + 2xe−x = e−x (x − 2) + (x + 2)e−x ≥ 0 d d because (x − 2) + (x + 2)e−x ≥ 0 for all x ≥ 0 (it is convex in x, takes value zero at x = 0 d2α and has slope zero at x = 0). Therefore ≥ 0. This implies that σ is convex in α. dσ2

To show joint convexity of σ in α and m, we need to show that the Hessian matrix of σ is positive semi-definite. For this purpose, we only need to show that the determinant of the matrix is non-negative (since we have already shown that σ is convex in α). We find the first derivatives as ∂σ (1 − σ) = − ³ ´ ≤ 0 ∂α m(σ+α−1) α 1 − d m(σ + α − 1) which implies that ≤ 1. Also, d ∂σ (1−σ) (σ + α − 1) = d ≤ 0 ∂m m(σ+α−1) d − 1 which implies that σ + α ≥ 1. We now evaluate the Hessian of σ: µ ¶ ∂2σ ∂2σ ∂m2 ∂m∂α ∂2σ ∂2σ ∂α∂m ∂α2

µ ¶  ³ ´ (1−σ)  2 2 −(1−σ)(σ+α−1)(2+ ) −2 (σ+α−1) (1−σ) − (σ+α−1)(1−σ) (σ+α−1)−1  d2 dm(σ+α−1)−d2 dα    d2   =  µ ¶ µ 2 ¶  2  −(1−σ)(σ+α−1) 2+ (1−σ) − m(1−σ)(2α+σ−1) + m(1−σ)  (d − m(σ + α − 1))  ( (σ+α−1)−1 ) d d−m(σ+α−1)   dα α2 

(20) We need to show that the determinant of the above matrix is positive. In doing so we ignore the common term outside the matrix, after which we need to show that µ ¶ m(2α + σ − 1) 2(σ + α − 1) − 2 ≥ 0. d m(2α + σ − 1) This is true if − 2 ≥ 0 (since σ + α ≥ 1). Substituting for the value of d 1 − σ α = , we need to show − (1−σ)m 1 − e d à ! − (1−σ)m m(1 − σ) 1 + e d ≥ 2. − (1−σ)m d 1 − e d

25 m(1 − σ) Taking x = , the above inequality is true if (x − 2) + (x + 2)e−x ≥ 0 which, as seen d above, is true for x ≥ 0. Thus the Hessian of σ is positive and σ is jointly convex in α and m.

Proof of Proposition 4

Proof. As in the previous proof, we first show that σ is decreasing in ν and ξ which would d2σ imply that average carrying cost too is decreasing in ν and ξ. We then show that ≥ 0 and dν2 finally we show that the determinant of the Hessian of average carrying cost is non-negative. For Gamma distributed capacity, σ can be found from Equation (8) as

à !ν 1 σ = (1−σ)ξ (21) d + 1 We can write the above equation as an explicit equation of ν, that is ln(σ) ν = − ³ ´. (22) (1−σ)ξ ln d + 1 Taking the derivative of the Taylor expansion of ν with respect to σ as σ → 1 yields dν −1 − d/ξ lim = < 0. σ→1 dσ 2 dν Since there is a unique solution to Equation (21) with 0 < σ < 1, and < 0, we get dσ σ→1 dν < 0 for all σ ∈ (0, 1). dσ Taking the first derivative of Equation (22) with respect to σ µ ¶ 1 dν 1 1 ξν = − . (23) ν dσ ln(σ) σ (1 − σ)ξ + d Taking the derivative of Equation (23) with respect to σ we get ³ ³ ´´ 2 (1−σ)ξ µ ¶2 2 1 dν 1 d ν (1 + ln(σ)) (ξ/d) 1 + ln d + 1 − + = − + h³ ´ ³ ´i2 . ν dσ ν dσ2 [σ ln(σ)]2 (1−σ)ξ (1−σ)ξ d + 1 ln d + 1

(1 − σ)ξ Letting i = + 1 ≥ 1, d 1 d2ν −(1 + ln(σ))(i ln(i))2 + (ξ/d)2(1 + ln(i))(σ ln(σ))2 + (ln(i))2(i2 + (ξσν/d)2 − 2ξσνi/d) = ν dσ2 (iσ ln(σ) ln(i))2 and substituting the value of ν from Equation (22) gives 1 d2ν − ln(σ)(i ln(i))2 + (ξσ ln(σ))2(2 + ln(i))/d2 + 2ξσi ln(i) ln(σ)/d = ν dσ2 (iσ ln(σ) ln(i))2

26 The numerator of the above expression can be written as

£ 2 2 2¤ −ξσ ln(σ£)[−i ln(i) − ξσ ln(σ)/d] /d − ln(¤ σ) (i ln(i)) − (ξσ) ln(i) ln(σ)/d − ln(σ) −iξσ ln(i)/d − (ξ)2σ2 ln(σ)/d2 > 0 ln(σ) d because each term within the square brackets is non-negative since ν = − > (in ln(i) ξ d d2ν steady-state the utilization ρ = < 1). Which implies that ≥ 0 and hence σ is convex ξk dσ2 in ν.

To show that σ is decreasing in ξ, we differentiate Equation (21) with respect to ξ: dσ ν(1 − σ)σ = − dξ (1 − σ)ξ + d − νξσ The fact that σ is decreasing in ν implies in Equation (23) that (1 − σ)ξ + d − νξσ ≥ 0. Therefore σ is also decreasing in ξ.

To show that the average carrying cost C is jointly convex in ν and ξ we use the Hessian matrix, see Equation (17). Without loss of generality we assume d = 1 (showing convexity ξ in ξ or d is equivalent). The Hessian can be written as à ! ∂2C ∂2C ∂ν2 ∂ν∂ξ ∂2C ∂2C ∂ξ∂ν ∂ξ2  ¡ ¢ ¡ ¢  ³ ´ ∂2σ 2 ∂σ 2 ∂2σ 2 ∂σ ∂σ h ∂ν2 + 1−σ ∂ν ∂ν∂ξ + 1−σ ∂ν ∂ξ h ln p+h   = −   2  ¡ ¢ ³ ´2  (1 − σ) ∂2σ 2 ∂σ ∂σ ∂2σ 2 ∂σ ∂ξ∂ν + 1−σ ∂ν ∂ξ ∂ξ2 + 1−σ ∂ξ ∂σ For ease of notation we denote partial derivatives with subscripts, e.g., = σ and ∂ν ν ∂2σ = σ . We have already shown that σ is convex in ν, therefore to show that the average ∂ν2 νν cost is jointly convex in ν and ξ, we need to show that the determinant of the Hessian matrix defined above is positive. In other words we show below that µ ¶ 2 2 σξ σν (σννσξξ − σνξ) + σνσξ σνν + σξξ − 2σνξ ≥ 0 (24) 1 − σ σν σξ We evaluate the expression in Equation (24) term by term. The first term on the left-hand side of the equation is the determinant of the Hessian matrix of σ, that is à ! ∂2σ ∂2σ ∂ν2 ∂ν∂ξ det ∂2σ ∂2σ ∂ξ∂ν ∂ξ2

27 Taking i = (1 − σ)ξ + 1 the Hessian of σ can be written as

 ³ ´ ³ ´  2 −2ξiσ2 ln(i) + σi(ln(i)) (i2 + ν(σξ)2) σi −(νσ ln(i) + 1 − σ) + (1−σ)ν ln(i)(i+σξ)  i−νσξ i−νσξ    1  ³ ´ ³ ´  (i − νξσ)2  (1−σ)ν ln(i)(i+σξ) σνi(ν+1)(1−σ)2 2   σi −(νσ ln(i) + 1 − σ) + i−νσξ (i−νσµξ) − 2(νσ) (1 − σ) 

(25) The expression for the determinant of the above matrix is grouped into terms having common denominators, that is, terms having denominators (i − νσξ)2,(i − νσξ) and 1. The determinant can then be written as µ ¶ σ2i 2ν(1 − σ) ln(i)(νσ2ξ ln(i) + (1 − σ)i) + 4(νσ)2(1 − σ)ξ ln(i) (i − νξσ)4 +νi((1 − σ) ln(i))2 − i(νσ ln(i) + 1 − σ)2 which is equivalent to σ2i ¡ ¢ ν(1 − σ) ln(i)(2 + ln(i))(2νσ2ξ + i(1 − σ)) − i(νσ ln(i) + 1 − σ)2 (i − νξσ)4 Now we evaluate the second term of Equation (24). Ã ! 2 σν(1−σ) 2 2 σξ σν 1 −2σ ξν(1 − σ) + i−νξσ (i + ν(σξ) ) ln(i) σ + σ = 2 νν σ ξξ σ (i − νξσ)2 σi (ν+1)(1−σ) ln(i) 2 ν ξ i−νξσ − 2σ νi ln(i) µ ¶ 1 σ(1−σ) ln(i) (2νi2 + (νσξ)2 + i2) = i−νξσ (i − νξσ)2 −2σ2ν(i ln(i) + (1 − σ)ξ) Also µ ¶ 1 2σ(1 − σ) ln(i) −2σ = 2σ2νı ln(i) + 2σ(1 − σ)i − (νi2 + νσξi) νξ (i − νξσ)2 i − νξσ Adding the two up, we get µ ¶ σξ σν 1 σ(1 − σ) ln(i) 2 σνν + σξξ − 2σνξ = 2 2σ(1 − σ)(i − νξσ) + (i − νσξ) σν σξ (i − νξσ) i − νξσ 2 2 Multiplying the above expression with σ σ = νσ2i ln(i) yields 1 − σ ν ξ (i − νξσ)2

µ ¶ 3 µ ¶ 2 σξ σν 4(1 − σ)νσ i ln(i) ln(i) σνσξ σνν + σξξ − 2σνξ = 4 (i − νξσ) 1 + 1 − σ σν σξ (i − νξσ) 2 We can now write Equation (24) as µ ¶ 2 2 σξ σν (σννσξξ − σνξ) + σνσξ σνν + σξξ − 2σνξ µ 1 − σ σν σξ ¶ σ2i 2(1 − σ)νσ(i − νξσ)(2 ln(i) + (ln(i))2) = (i − νξσ)4 +ν(1 − σ) ln(i)(2 + ln(i))(2νσ2ξ + i(1 − σ)) − i(νσ ln(i) + 1 − σ)2

28 Removing the common term, we need to show that

2(1−σ)νσ(i−νξσ)(2 ln(i)+(ln(i))2)+ν(1−σ) ln(i)(2+ln(i))(2νσ2ξ+i(1−σ))−i(νσ ln(i)+1−σ)2 ≥ 0, that is, ν(1 − σ) ln(i)(2 + ln(i))(1 + σ) ≥ (νσ ln(i) + 1 − σ)2 or in other words

2ν(1 − σ) ln(i) + ν(1 − (ν + 1)σ2)(ln(i))2 ≥ (1 − σ)2

Since ν ln(i) = − ln(σ) and − ln(σ) ≥ 1 − σ for σ ≤ 1, we can bound ln(i) with 1 − σ. It is thus sufficient to show that

(1 − σ)2 + ν(1 − (ν + 1)σ2)(ln(i))2 ≥ 0 1 1 The above inequality is true for σ2 ≤ . For σ2 ≥ , we need to show that ν + 1 ν + 1 (1 − σ)2 ≥ −ν(1 − (ν + 1)σ2)(ln(i))2 x − 1 1 − σ Using the logarithmic inequality ln(x) ≥ and applying it to − ln(σ) ≤ , we get x σ 1 − σ that ln(i) ≤ . We then need to show that νσ (1 − (ν + 1)σ2) 1 ≥ − νσ2 which is indeed true since σ ≤ 1. Hence we have shown that the average carrying cost C is jointly convex in (ν, ξ).

Proof of Proposition 5

Proof. For Gamma distributed capacity we can write

† 1 σ (φ, ν) = ³ ´ν (26) (1−σ†(φ,ν))φ νd + 1

Where φ is the mean and ν the shape parameter of the capacity. Note that σ†(φ, ν) has been characterized similar to σ(ξ, ν) in Equation (21). Indeed, for a Gamma distribution, its mean is φ = νξ where ξ is the scale parameter. Without loss of generality we assume that d = 1 (since the analysis will remain same φ with φ as with ). This implies that in order to have a steady state distribution of the d shortfall we must have φ > d = 1. We find that σ† is not always convex in (φ, ν), therefore

29 to show convexity of the average carrying cost, we directly use Equation (17). The Hessian matrix for the average cost can be written as à ! ∂2C ∂2C ∂φ2 ∂φ∂ν ∂2C ∂2C ∂ν∂φ ∂ν2  ³ ´2 ³ ´  ³ ´ 2 † † † † 2 † ∂ σ + 2 ∂σ ∂σ ∂σ 2 + ∂ σ h ln h  ∂φ2 1−σ† ∂φ ∂φ ∂ν 1−σ† ∂φ∂ν  p+h   = − † 2 (1 − σ )  ³ ´ ³ ´2  ∂σ† ∂σ† 2 ∂2σ† ∂2σ† 2 ∂σ† ∂φ ∂ν 1−σ† + ∂φ∂ν ∂ν2 + 1−σ† ∂ν

We need to show that this Hessian matrix is positive semi-definite. For this purpose we φ formulate the matrix as a function of ν and the scale parameter is ξ = . Indeed, we can ν write σ† as 1 σ†(ξ(φ, ν), ν) = ((1 − σ†)ξ(φ, ν) + 1)ν We can then write the partial derivatives of σ† in terms of partial derivatives of σ(ξ, ν) so that we can use the results of the analysis of the proof of Proposition 4. For ease of notation ∂σ† ∂2σ† we denote partial derivatives with subscripts, e.g., = σ† and = σ† . ∂φ φ ∂φ2 φφ 1 σ† = σ φ ν ξ φ σ† = − σ + σ ν ν2 ξ ν 1 σ† = σ φφ ν2 ξξ 2φ 2φ φ2 σ† = σ − σ + σ + σ νν ν3 ξ ν2 νξ ν4 ξξ νν φ 1 1 σ† = − σ − σ + σ φν ν3 ξξ ν2 ξ ν ξν To show convexity of C in (φ, ν), we need to show that the Hessian matrix is positive semi-definite for all values of ξ and ν. Note that we no longer consider ξ as a function of (φ, ν) since we will show that the Hessian is positive semi-definite in the entire space of (ξ, ν) for ξ ≥ 0,ν ≥ 0 and νξ > 1. From here on, we prove convexity of the average carrying cost C in steps. The sketch of the proof is the following. In step 1 we show that the average carrying cost is decreasing in φ and ν. In step 2 we show that for C to be convex it is sufficient that a function R(ν, ξ, σ(ν, ξ)) is non-negative. Interestingly, this function is independent of φ. In step 3 we show that R ≥ 0 for all ξ ≥ 0 and for all ν ≥ 1. In the consequent steps 4, 6 and 5 we prove that R ≥ 0 for all ξ ≥ 0 and for all ν < 1 too.

Step 1 The average carrying cost C(φ, ν) is decreasing in φ and ν.

30 † We have already shown in the proof of Proposition 4 that σξ ≤ 0, which implies that σφ ≤ 0. dC ∂C We have also seen in the proof of Lemma 1 that ≥ 0 and hence ≤ 0. To show that dσ† ∂φ φ σ† ≤ 0, we need to show that −σ ≥ − σ . Letting i = (1 − σ)ξ + 1, this is equivalent to ν ν ν2 ξ i − 1 showing that ln i ≥ , which is indeed true for all values of i ≥ 1. i

Step 2 The average carrying cost C(φ, ν) is convex in (φ, ν) if for all ν, ξ ≥ 0

i (νσξ)2(1 − σ) R(ν, ξ) = [1 − (ν + 1)σ2](ln σ)2 + σξ[1 − (ν + 1)σ] ln σ − ≥ 0, 2ν(1 − σ) 2i where σ = σ(ν, ξ) and i = (1 − σ)ξ + 1.

† We saw in the proof of Proposition 4 that σξξ ≥ 0, which implies that σφφ ≥ 0. Hence ∂2C ≥ 0. As a result, to show convexity of C in (φ, ν) we need to show that the determinant ∂φ2 of the Hessian matrix of C is non-negative, that is, we need to show that µ ¶ µ ¶ µ ¶ 2 2 2 2 σ† + (σ† )2 σ† + (σ†)2 ≥ σ† σ† + σ† φφ 1 − σ† φ νν 1 − σ† ν 1 − σ† φ ν φν which reduces to showing 2 2 4 σ† σ† + (σ†)2σ† + (σ† )2σ† ≥ (σ† )2 + σ† σ† σ† φφ νν 1 − σ† ν φφ 1 − σ† φ νν φν (1 − σ†) φν φ ν

After substituting the derivatives of σ† with derivatives of σ in the above inequality (1 − σ) (1 − σ) (1 − σ) (1 − σ) 2 σ σ + σ2σ + σ2σ ≥ σ2 + σ2 − σ σ − σ2σ + 2σ σ σ 2 ξξ νν ν ξξ ξ νν 2 ξν 2ν2 ξ ν ξ ξν ν ξ ν ξν ξ ν Interestingly, this inequality is independent of φ since σ can be defined as σ(ξ, ν). We therefore need to show that µ ¶ (1 − σ) (1 − σ) (1 − σ) 2 (σ σ − σ2 ) + (σ2σ + σ2σ ) ≥ σ σ − σ − σ σ + 2σ σ 2 ξξ νν ξν ν ξξ ξ νν ξ 2ν2 ξ ν ξν ν ξ ν ξν ν (27) To show the above inequality we first evaluate the individual terms. The first term on the left-hand side of the inequality in Equation (27) is similar to what we had evaluated in Proposition 4 to prove convexity of σ in (ν, ξ), that is, ³ ³ ´ ´ (1 − σ) 2 2 2 (σ σ −σ2 ) = σ (1−σ) νi ln(i)(2 + ln(i)) ξνσ2 + (1−σ)i − i (σν ln(i) + 1 − σ)2 ≥ 0 2 ξξ νν ξν (i−ξσν)4 2 2(1−σ)

31 The second term on the left-hand side of Equation (27) is

2 2 σνσξξ + σξ σνν 2 µ 2 ¶ (σi ln(i)) σνi(ν + 1)(1 − σ) 2 = 4 − 2(νσ) (1 − σ) (i − ξσν) µ(i − σξν) ¶ (1 − σ)2(νσ)2 σi(ln(i))2 + −2ξσ2i ln(i) + (i2 + ν(ξσ)2) (i − σξν)4 (i − σξν)  σνi3(ν + 1)(ln(i))2 (νσi ln(i))2 − 2 σ2(1 − σ)2  (i − σξν) (1 − σ)  =   (i − ξσν)4  σi(ν ln(i))2  −2ξ(νσ)2i ln(i) + (i2 + ν(ξσ)2) µ µ (i − σξν) ¶ ¶ σ2(1 − σ)2 i2 + 2νi2 + i(ξσν)2 (νσi ln(i))2 = iσν(ln(i))2 − 2 − 2ξ(νσ)2i ln(i) (i − ξσν)4 (i − σξν) (1 − σ)

The term on the right-hand side of the inequality in Equation (27) is µ ¶ (1 − σ) (1 − σ) 2 σξ 2 σξ − σξν − σξσν + 2σξνσν 2ν  ν ν µ ¶  (1 − σ)2σ (1 − σ)σi (1 − σ) − − −(νσ ln(i) + 1 − σ) + ν ln(i)(i + ξσ) (1 − σ)νσ  2ν(i − σξν) ν(i − σξν)2 i − σξν  = −  µ ¶  i − σξν  2(1 − σ)σ2i ln(i) 2σ2i2 ln(i) (1 − σ)ν ln(i)(i + ξσ)  − 2 − 3 −(νσ ln i + 1 − σ) +  (i − σξν) (i − σξν) µ i − σξν ¶  (1 − σ)(i − σξν)2 (1 − σ) + i(i − σξν) −(νσ ln(i) + 1 − σ) + ν ln(i)(i + ξσ) (1 − σ)2σ2  2 i − σξν  =  µ ¶  (i − σξν)4  2σνi2 ln(i) (1 − σ)ν ln(i)(i + ξσ)  +2σνi ln(i)(i − σξν) + −(νσ ln(i) + 1 − σ) + 1 − σ i − σξν

Since i + ξσ = ξ + 1 and (1 − σ)(ξ + 1) = i − σ, the above expression is   (1 − σ)(i − σξν)2 2 2 + σνi(i − σξν) ln(i) − (1 − σ)i(i − σξν) + (1 − σ)νi(ξ + 1) ln(i) (1 − σ) σ  2  4  2(σνi ln(i))2 2σ(νi ln(i))2(ξ + 1)  (i − σξν) − − 2σi2ν ln(i) +  µ 1 − σ ¶ i − σνξ  i − σξν 2 2 (1 − σ)2σ2  (1 − σ)(i − σξν) − i + νi(i − σ) ln(i) − σνi ln(i) − (σν) ξi ln(i)  =  2  (i − σξν)4  2(σνi ln(i))2 2σ(νi ln(i))2(ξ + 1)  − + 1 − σ i − σνξ  2 2  (1 − σ)(i − (σξν) ) 2 2 (1 − σ)2σ2  − + νi (1 − σ) ln(i) − νσi ln(i) − (σν) ξi ln(i)  = 2 4  2(σνi ln(i))2 2σ(νi ln(i))2(ξ + 1)  (i − σξν) − + 1 − σ i − σνξ

32 We now evaluate the term µ ¶ 2 2 (1 − σ) (1 − σ) 2 (σνσξξ + σξ σνν) − σξ 2 σξ − σξν − σξσν + 2σξνσν  2ν µ ν ν ¶  (ln(i))2i2σν (ξσν)2 (1 − σ)2σ2  i + 2νi + − 2ν(ξ + 1) +  =  i − σξν i  (i − σξν)4 (1 − σ) (i2 − (σνξ)2) − νi ln(i)(i(1 − σ) − σ + σ2νξ) 2 [using that 2νi = 2ν + 2νξ − 2νσξ]  µ ¶  (ln(i))2i2σν (ξσν)2 (1 − σ)2σ2  i + − 2νξσ +  =  i − σξν i  (i − σξν)4 (1 − σ) (i2 − (σνξ)2) − νi ln(i)((i − σ) − σ(i − σνξ)) 2  2  (ln(i)) iσν 2 (1 − σ)2σ2  (i − νξσ) +  =  i − σξν  (i − σξν)4 (1 − σ) (i2 − (σνξ)2) − νi ln(i)((i − σ) − σ(i − σνξ))  2 µ ¶  (1 − σ) (1 − σ)2σ2 (i − νσξ) σνi(ln(i))2 + (i + σνξ) =  2  (i − σξν)4  −νiµln(i)((i − σ) − σ(i − σνξ)) ¶  (1 − σ) (1 − σ)2σ2 (i − νσξ) σνi(ln(i))2 + νσi ln(i) + (i + σνξ) =  2  (i − σξν)4 −νi(i − σ) ln(i) Finally to show inequality in Equation (27) we need to show that µ ¶ (1 − σ) (1 − σ) (1 − σ) 2 (σ σ − σ2 ) + (σ2σ + σ2σ ) − σ σ − σ − σ σ + 2σ σ 2 ξξ νν ξν ν ξξ ξ νν ξ 2ν2 ξ ν ξν ν ξ ν ξν ν  µ ¶ 2  2 (1 − σ)i i 2  νi ln(i)(2 + ln(i)) ξνσ + − (σν ln(i) + 1 − σ)  (1 − σ)2σ2  µ 2 2(1 − σ) ¶  =  (1 − σ)  ≥ 0 (i − σξν)4  +(i − νσξ) σνi(ln(i))2 + νσi ln(i) + (i + σνξ)   2  −νi(i − σ) ln(i) (1 − σ)2σ2 Removing the common term and i, we need to show that (i − σξν)4 µ ¶ µ ¶ (1 − σ)i (1 − σ) ν ln(i)(2 + ln(i)) ξνσ2 + + (i − νσξ) σν(ln(i))2 + νσ ln(i) + (i + σνξ) 2 2i i ≥ (σν ln(i) + 1 − σ)2 + ν(i − σ) ln(i) 2(1 − σ) Replacing ln(i) = − ln(σ)/ν, and dividing both sides by 1 − σ, we need to show that µ ¶ µ ¶ µ ¶ ln(σ) ξνσ2 i σ(ln(σ))2 σ ln(σ) 1 σξν − ln(σ) 2 − + + (i − νσξ) − + + µ ν ¶ 1 − σ 2 (1 − σ)ν (1 − σ) 2 2i i σ ln(σ) 2 (i − σ) ≥ − + 1 − ln(σ) 2 (1 − σ) 1 − σ Which is equivalent to showing the inequality i(ln(σ))2 σi(ln(σ))2 (νξσ)2 iσ2(ln(σ))2 −νξσ2 ln(σ) + (1 − σ) + − (1 − σ) ≥ − iσ ln(σ) + σ ln(σ) 2ν ν 2i 2(1 − σ)

33 The above inequality can also be written as i(1 + σ) σ2i (νσξ)2(1 − σ) (ln(σ))2 − (ln(σ))2 + σξ ln(σ) − (ν + 1)σ2ξ ln(σ) − ≥ 0 2ν 2(1 − σ) 2i or in other words i ¡ ¢ (νσξ)2(1 − σ) R = 1 − (ν + 1)σ2 (ln σ)2 + σξ(1 − (ν + 1)σ) ln σ − ≥ 0 (28) 2ν(1 − σ) 2i

Step 3 R(ν, ξ) ≥ 0 for ν ≥ 1 and for all ξ.

ln(σ) After replacing ν with − in Equation (28), we need to show that ln(i) µ ¶ µ ¶ i σ2i σ2ξ (σξ)2(1 − σ) − ln(σ) (1 + σ) ln(i) − σξ + σ2ξ + (ln(σ))2 − + − ≥ 0 2 2(1 − σ) ln(i) 2i(ln(i))2

Removing the common − ln(σ) > 0 term and writing the second term on the left hand side ([i ln(i) − ξ(1 − σ)]σ ln(σ))2 of the above inequality as − , we would need to show that 2i(1 − σ)(ln(i))2

i [i ln(i) − ξ(1 − σ)]2σ2 ln(σ) (1 + σ) ln(i) − σξ + σ2ξ + ≥ 0 2 2i(1 − σ)(ln(i))2 x − 1 Using the logarithmic inequality ln(x) ≥ for x ≥ 0, we can substitute ln(σ) (using x 1 − σ − ln(σ) ≤ for σ ≤ 1) in the above inequality and thus we would need to show that σ i [i ln(i) − ξ(1 − σ)]2σ (1 + σ) ln(i) − σξ + σ2ξ − ≥ 0 2 2i(ln(i))2 which is similar to showing that

ξ(1−σ) i [i − ]2σ (1 + σ) ln(i) − σξ + σ2ξ − ln(i) ≥ 0 2 2i i − 1 (1 − σ)ξ Using the inequality ln(i) ≥ , we can show that i− ≥ 0 (since i = (1−σ)ξ +1). i ln(i) Also using the logarithmic inequality ln(1 + x) ≤ x ∀x ≥ 0, we get that ln(i) ≤ (1 − σ)ξ. Using these inequalities it is sufficient to show that i (i − 1)2σ (1 + σ) ln(i) − σξ + σ2ξ − ≥ 0 2 2i which is similar to showing i (1 − σ)2ξ2σ (1 + σ) ln(i) − σξ + σ2ξ − ≥ 0 2 2i

34 (1 − σ)ξ After substituting for ln(i) in the above inequality with ln(i) ≥ we would need to i show that 1 + σ (1 − σ)ξσ − σ − ≥ 0 2 2i ∂σ or simply i ≥ σξ. Since ≤ 0, therefore i ≥ νξσ, hence the above inequality is definitely ∂ξ true for ν ≥ 1. Therefore R(ν, ξ) ≥ 0 when ν ≥ 1. ∂R(ν, ξ) ∂2R(ν, ξ) 3 Step 4 lim R(ν, ξ) = 0, lim = 0 and lim = −2ξ + 2 + + ν > 0, for 1 1 1 2 ν→ ξ ν→ ξ ∂ν ν→ ξ ∂ν ν all values of ξ. 1 From Equation (22) we see that when d = 1, ν = . Due to uniqueness of solution of σ→1 ξ 1 1 0 < σ < 1, we can also say that σ → 1 as ν → . Similarly i = (1 − σ)ξ + 1 → 1 as ν → . ξ ξ Using the Taylor approximation for ln σσ→1− = −(1 − σ), we get µ ¶ 1 − σ 1 − σ lim R(ν, ξ) = lim − + (1 − σ) − = 0 1 σ→1− ν→ ξ 2 2 The first derivative of R with respect to ν can be written as ∂R −ξ(ln σ)2(1 − (ν + 1)σ2) ∂σ i ln σ ∂σ iσ2(ln σ)2 = + (1 − (ν + 1)σ2) − ∂ν 2ν(1 − σ) ∂ν νσ(1 − σ) ∂ν 2ν(1 − σ) i(ν + 1)σ(ln σ)2 ∂σ i(ln σ)2(1 − (ν + 1)σ2) i(ln σ)2(1 − (ν + 1)σ2) ∂σ − − + ν(1 − σ) ∂ν 2ν2(1 − σ) 2ν(1 − σ)2 ∂ν ∂σ ∂σ ∂σ +ξ(ln σ)(1 − (ν + 1)σ) + ξ(1 − (ν + 1)σ) − σ2ξ ln σ − (ν + 1)σξ(ln σ) ∂ν ∂ν ∂ν νσ2ξ2(1 − σ) σν2ξ2(1 − σ) ∂σ (νσξ)2 ∂σ (νσξ)2(1 − σ)ξ ∂σ − − + − i i ∂ν 2i ∂ν 2i2 ∂ν Taking Taylor expansion of the logarithm in Equation (23) we find that ∂σ i ln i lim = lim − = −1 1 i→1 ν→ ξ ∂ν i − 1 and hence ∂R(ν, ξ, σ(ν, ξ)) lim ν→ 1 ∂ν ξ µ ¶ i(1 − (ν + 1)σ2) i(1 − (ν + 1))σ2 (νσξ)2 = lim − − − ξ(1 − (ν + 1)σ) − 1 ν→ ξ νσ 2ν 2i 1 1 = −1 + + 1 − 2 2 = 0. From Equation (25), we can find the second derivative of σ with respect to ν. Again, taking the Taylor expansion of the logarithm, we find ∂2σ lim = lim σ(i − 1)(i2 + ν(σξ)2) = 0. 1 2 i→1 ν→ ξ ∂ν

35 The second derivative of R with respect to ν gives a long expression, which in the limits can be written as ∂2R(ν, ξ, σ(ν, ξ)) lim 2 ν→ 1 ∂ν ξ    −ξ(ln σ)2(1 − (ν + 1)σ2) ∂σ i ln σ ∂σ iσ2(ln σ)2   + (1 − (ν + 1)σ2) −   2ν(1 − σ) ∂ν νσ(1 − σ) ∂ν 2ν(1 − σ)     i(ν + 1)σ(ln σ)2 ∂σ i(ln σ)2(1 − (ν + 1)σ2) i(ln σ)2(1 − (ν + 1)σ2) ∂σ  ∂  − − +  = lim ν(1 − σ) ∂ν 2ν2(1 − σ) 2ν(1 − σ)2 ∂ν ν→ 1 ∂ν  ∂σ ∂σ ∂σ  ξ  2   +ξ(ln σ)(1 − (ν + 1)σ) + ξ(1 − (ν + 1)σ) − σ ξ ln σ − (ν + 1)σξ(ln σ)   ∂ν ∂ν ∂ν   νσ2ξ2(1 − σ) σν2ξ2(1 − σ) ∂σ (νσξ)2 ∂σ (νσξ)2(1 − σ)ξ ∂σ   − − + −  ½ i i ∂ν 2i ∂ν 2i2 ∂ν ¾ ξ 2 1 1 1 ξ −( 2 ) + (−ξ + 1 + ν ) − ( 2ν ) + (1 + ν ) + ( 2ν ) + ( 2 + ν + 1) − (1) − (1) + (ξ) − (1 + ξ) = lim ξ ξ ν→ 1 −(ξ) + 1 + (1 − ) + ( ) ξ µ ¶ 2 2 3 = lim −2ξ + 2 + + ν 1 ν→ ξ ν > 0 for all ξ. ³ ´ ³ ´ 2 i ln σ Step 5 The equation 1−(ν+1)σ z2 + 2 1−(ν+1)σ z − 1 = 0, where z = ≤ 0 ν ν (1 − σ)νσξ can have at most one (real) negative root.

After some manipulations, R(ν, ξ) can also be written as µ ¶ µ ¶ µ ¶ µ ¶ i ln σ 2 1 − (ν + 1)σ2 i ln σ 1 − (ν + 1)σ R(ν, ξ) = + 2 − 1 (1 − σ)νσξ ν (1 − σ)νσξ ν

i ln σ Letting z = ≤ 0 in the above equation results in a quadratic equation (1 − σ)νσξ µ ¶ µ ¶ 1 − (ν + 1)σ2 1 − (ν + 1)σ z2 + 2 z − 1 ν ν The roots of this equation are r ³ ´ ³ ´2 1−(ν+1)σ 1−(ν+1)σ 1−(ν+1)σ2 − ν ± ν + ν 1−(ν+1)σ2 ν The following scenarios cover the entire space of 0 < σ < 1 and ν, that is, 1 1. σ2 ≥ ν + 1 1 2. σ ≤ ν + 1 1 1 3. σ ≥ and σ2 ≤ ν + 1 ν + 1

36 The first case gives both positive roots, which are indeed not valid roots since z ≤ 0. The second case gives one positive and one negative root and the third case also gives one positive and one negative root.

i ln σ Step 6 z = ≤ 0 is monotonic in ν for all values of ξ. (1 − σ)νσξ

Substituting − ln σ = ν ln i and i = (1 − σ)ξ + 1 into z, we get i ln i z = − σ(i − 1) µ ¶ d i ln i ∂i ∂σ = i − 1 − ln(i) ≥ 0. Also = −ξ ≥ 0, for all ξ ≥ 0, since σ is decreasing di σ(i − 1) ∂ν ∂ν i ln i in ν. Therefore is increasing in ν for all values of ξ ≥ 0. Which implies that z is σ(i − 1) decreasing in ν for all values of ξ ≥ 0. End of the proof. From step 2 we get the sufficient condition R(ν, ξ) ≥ 0 for all ξ and ν. From step 3 we know that this condition is satisfied for ν ≥ 1 and for all values of ξ. 1 From step 4 we know that R is positive and increasing in the limit ν → . Steps 5 and 6 ξ together imply that equation R(ν, ξ) = 0 can have at most one real root along ν ≥ 0 for all values of ξ ≥ 0. However if R were to be less that 0 then it would have to have at least two 1 real roots between ≤ ν ≤ 1 (follows from steps 3 and 4). Hence R ≥ 0 when ν ≤ 1 too ξ and for all ξ. This completes the proof.

37 Table 1: G/G/1 Results with C = 1 and unit mean demand. cap √ Demand → Fixed (CV=0) Uniform (CV= 1 ) Gamma (CV= √1 ) Exponential (CV=1) Gamma (CV= 2) 3 2 Capacity ↓ Cost=1 Cost=2.47 Cost=3.73 Cost=5.27 Cost=7.77 Fixed (CV=0) Cap=1 Cap=1.43 Cap=1.78 Cap=2.27 Cap=3 Inven=1 Inven=1.94 Inven=2.24 Inven=2.75 Inven=3.51 Cost=3.2 Cost=3.79 Cost=4.53 Cost=5.94 Cost=8.39 1 Bernoulli (CV= 2 ) Cap=1.6 Cap=1.66 Cap=1.89 Cap=2.35 Cap=3 38 Inven=2 Inven=2.89 Inven=2.91 Inven=3.31 Inven=3.95 Cost=2.85 Cost=3.66 Cost=4.46 Cost=5.88 Cost=8.31 Gamma (CV= √1 ) Cap=1.78 Cap=1.93 Cap=1.91 Cap=2.17 Cap=2.86 2 Inven=1.81 Inven=2.39 Inven=2.94 Inven=3.61 Inven=4.28 Cost=3.83 Cost=4.52 Cost=5.18 Cost=6.44 Cost=8.88 Gamma (CV=1) Cap=2.03 Cap=2.12 Cap=2.05 Cap=2.31 Cap=2.89 Inven=2.43 Inven=3.02 Inven=3.51 Inven=4.16 Inven=4.92 √ Cost=5.27 Cost=5.81 Cost=6.3 Cost=7.47 Cost=9.8 Gamma (CV= 2) Cap=2.46 Cap=2.46 Cap=2.4 Cap=2.47 Cap=2.94 Inven=3.41 Inven=4.05 Inven=4.42 Inven=5.25 Inven=6.25