SCN : 1. Inventory
1
1-Period Stochastic Demand: Newsvendor Model
Although we will work with a different cost notation, the initial results should be familiar to the reader. Let
D denote the demand with pdf and cdf f and F ; also let µ := E(D).
• c unit ordering cost
• cH unit holding cost
• cP unit shortage cost
Let x be the initial inventory level and y be the level after the order is received but before the demand
is materialized. For convenience, it is better to write the inventory costs in terms of y
L(y) := cH E(y − D)+ + cP E(D − y)+
Using the identity (y − D)+ = (D − y)+ + (y − D), the cost can be written in alternative forms
L(y) = cH (y − µ) + (cP + cH )E(D − y)+
We can also add the ordering costs to obtain the total costs
g(y) := cy + L(y) = cµ + (c + cH )(y − µ) + (cP + cH )E(D − y)+
Since (D − y)+ = max(D − y, 0), it is an upper envelope of convex functions so it is convex. E(D − y)+
is the expected value of a convex function so it is convex. Consequently, g(y) is a convex function and is
minimized at y ∗ which solves
F (y) =
cP − c
cP − c
underage cost
=
=
cP + cH
(cP − c) + (cH + c)
(underage cost)+(overage cost)
In the optimality condition ζ := (cP − c)(cP + cH ) is known as the critical fractile. Also note that the
critical fractile is the probability of not stocking out (sometimes known as Cycle Service Level). A similar
but different concept is the fill rate, defined as the ratio of expected unsatisfied demand to the expected
demand, i.e. E(D − y)+ /µ.
Since cµ is not a relevant cost, it can be dropped from the total cost expression. Note that cµ is the
cost of operating the inventory system if it were possible to choose the order quantity after the demand
observation. Because of this,g(y)−cµ is termed as the cost of not having perfect information on the demand.
Computational note: When the demand is N (µ, σ 2 ), the computation of expected shortages can be
streamlined as
E(D − y)+ = −(y − µ) (1 − Φ((y − µ)/σ)) + σφ((y − µ)/σ)
1
(1)
where Φ and φ are cdf and pdf for standard normal variate. After setting z = (y − µ)/σ, it follows that
E(D − y)+ /σ = −z (1 − Φ(z)) + φ(z) =: IN (z)
where IN (z) is known as the unit normal loss function and is tabulated at the end of the books. Manipulations
with normal density is not easy. For example, Φ cannot be put into a closed form expression. However,
it is possible to obtain it with numerical integration and all modern software -including Excel- has built-in
functions to compute these probabilities.
X ∼ N (µ, σ 2 )
Matlab
normpdf (x, µ, σ)
normcdf (x, µ, σ)
norminv(p, µ, σ)
Computing probabilities for
Functions
pdf, φ(x)
cdf, Φ(x)
inverse cdf, Φ−1 (p)
Excel
normdist(x, µ, σ, 0)
normdist(x, µ, σ, 1)
norminv(p, µ, σ)
For a given p (0 ≤ p ≤ 1), inverse cdf finds the outcome value x such that the probability of observing
smaller outcomes than x is exactly p. Note that if p = Φ(x) then x = Φ−1 (p). Note that Excel uses a single
function to compute both pdf and cdf, use the last argument of normdist to specify whether you want pdf
or cdf. ¤
Constraints: In general, there can be constraints on the inventory level y and the order size y − x
1. Inventory cannot be disposed off: y ≥ x
2. Storage capacity: y ≤ M
3. Limitations on order sizes, perhaps because of transportation economies: a ≤ y−x ≤ b where 0 ≤ a ≤ b
No matter how many restrictions are posed on y, WLOG we assume that there are only two: a lower lb and
an upper ub bound. Then we have a linearly constrained convex optimization problem
min g(y)
lb≤y≤ub
where we can apply the KKT conditions. There are three possibilities: i) arg miny g(y) ≤ lb, then set
y ∗ = lb. ii) arg miny g(y) ≥ ub, then set y ∗ = ub. iii) lb ≤ arg miny g(y) ≤ ub, then set y ∗ =arg miny g(y).
Multiple products with a single capacity constraint: When there are I products whose demands
and costs are indexed by i, we first compute the fractile solutions yi = F −1 ((cP i − ci )/(cP i + CHi )). If these
inventory levels satisfy the capacity constraint
I
X
yi ≤ M
i=1
we are done. Otherwise we know from KKT conditions that the constraint will be binding. Let λ ≥ 0 be
the dual variable for the constraint, KKT conditions require
gi0 (y) + λ = 0 for all i = 1 . . . I,
I
X
yi − M )λ = 0
(
i=1
2
Then it follows that λ should be increased starting from 0 until
I
X
i=1
µ
Fi−1
cP i − (ci + λ)
cP i + cHi
¶
=M
Note that now the underage cost is cP i − (ci + λ) and is smaller by λ units from the original underage cost.
The underage becomes chepaer once the dual price of the limited capacity is factored in.
1.1
Pricing in the Newsvendor Problem
The basic principle of inventory management is matching the demand and the stocks. Traditionally the
stock levels were the only decision variables. Recently, with ideas borrowed from economics theory, prices
also become decision variables. By adjusting the price p, the demand can be regulated to better match
the stocks. Naturally the demand mean becomes a function of theprice and we write µ(p). In almost all
practical contexts, the demand mean decreases in p. Some examples are
1. µ(p) = a − bp with a > 0 and b > 0.
2. µ(p) = ap−b with a > 0 and b > 1.
There generally are two ways to introduce randomness into the demand
1. Additive Randomness: D(p) = µ(p) + ξ where µ(p) = a − bp.
2. Multiplicative Randomness: D(p) = µ(p)ξ where µ(p) = ap−b .
In order to keep D(p = 0) ≥ 0, ξ is lower bounded by −a in the additive case and by 0 in the multiplicative
case. In general, we assume that ξ ∈ [A, B] 1 .
Whether the demand randomness is additive or multiplicative, the revenue is pE(y ∧ D(p)). The costs
are still g(y) so the expected profit becomes
Π(y, p) = pE(y ∧ D(p)) − g(y, p)
where g(y, p) is the obvious extension of g(y), obtained by replacing D with D(p) in the expressions.
Additive Randomness: We use the additive form to simplify Π(y, p). First let define the safety stock
z := y − µ(p). With this transformation and using E((µ(p) + ξ) ∧ (µ(p) + z)) = µ(p) + E(ξ) − E(ξ − z)+ ,
the profit can be manipulated
Π(z, p) = pE((µ(p) + ξ) ∧ (µ(p) + z)) − cH E(z − ξ)+ − cP E(ξ − z)+ − c(µ(p) + z)
= p(µ(p) + E(ξ)) − cH E(z − ξ)+ − (cP + p)E(ξ − z)+ − c(µ(p) + z + E(ξ) − E(ξ))
= (p − c)(µ(p) + E(ξ)) − (cH + c)E(z − ξ)+ − (cP + p − c)E(ξ − z)+
= Ψ(p) − L(z, p)
1
When we compare a random variable against a real number, we think of inequalities in the almost sure sense, e.g. P (A ≤
D(p = 0) ≤ B) = 1
3
where Ψ(p) := (p − c)(µ(p) + E(ξ)) is the profit function with perfect information. L(z, p) is almost the
same function as we defined before except that the overage and underage costs are increased by c and p − c
in the current profit maximization framework. Finally, our problem is
max Π(z, p).
z,p
As a first cut, consider the problem under perfect information i.e. maxp Ψ(p). This problem has a
concave objective function (∂ 2 Ψ/∂p2 = −2b < 0) and the optimal price is
p0 =
a + bc + E(ξ)
2b
The objective function Π(z, p) is concave in z and in p separately but not jointly. Thus, when p is fixed
we obtain the newsvendor problem. When z is fixed, we set ∂Π(z, p)/∂p = 0 to find
p(z) = p0 −
E(ξ − z)+
2b
Substituting this price back into the profit, we obtain a one dimensional optimization problem maxz Π(z, p(z)).
The objective function is not necessarily concave or not even unimodal. But it can be made unimodal with
the following restrictions (Petruzzi and Dada 1999)
1. Condition 1: 2r(z)2 + dr(z)/dz > 0 for A ≤ z ≤ B where r(z) is the failure rate function for ξ.
2. Condition 2: a − b(c − 2cP ) + A > 0.
If the second condition fails, there could be multiple critical z values but the largest maximizes Π(z, p(z)).
Multiplicative Randomness: We follow the same steps as in the additive randomness but with transformation z := y/µ(p)
ΠM (z, p) = pE((µ(p)ξ) ∧ (µ(p)z)) − cH µ(p)E(z − ξ)+ − cP µ(p)E(ξ − z)+ − cµ(p)z
= (p − c)µ(p)E(ξ) − (cH + c)µ(p)E(z − ξ)+ − (cP + p − c)µ(p)E(ξ − z)+
= ΨM (p) − LM (z, p)
where superscript M is used to differentiate current functions from their additive counterparts which take
superscript A later on. Our problem is to maximize the profit. The profit is concave in z and unimodal in
p. We first find the optimal price for a fixed z by setting ∂ΠM /∂p = 0.
µ
¶
(c + cH )E(z − ξ)+ + cP E(ξ − z)+
b
M
0,M
p (z) = p
+
b−1
µ − E(ξ − z)+
where p0,M is the optimal price under perfect information and p0,M = bc/(b − 1). It is easy to establish that
pM (z) ≥ p0,M .
Similar to the additive randomness case, optimization over p, z is reduced to optimization over z by
inserting p(z) instead of p: maxz ΠM (z, p(z)). Once more ΠM (z, p(z)) can be made unimodal with extra
conditions. Indeed Condition 1 of the additive case is still necessary while Condition 2 simplifies to b ≥ 2.
4
Thus, under these conditions, the optimal z can be found via dΠM (z, p(z))/dz = 0.
Managerially, the curious issue is pA (z) ≤ p0,A while pM (z) ≥ p0,M . Additive or multiplicative randomness modelling yields seemingly conflicting results: in the former (latter) case the optimal price for a given
z is always smaller (larger) than or equal to the price under perfect information. A possible explanation
requires first observing that the coefficient of variation of the demand is increasing (nonincreasing) in p in
the additive (multiplicative) case. When switching from perfect information and p0 to a stocahstic model,
the price is used to control the variability in the demand. If the variability is measured by the coefficient
of variation, then price should decrease (increase) in the additive (multiplicative) case with respect to their
full information price.
2
<²αδ Chapter 1 of Graves et al
3
Infinte Horizon Constant Demand: EOQ Model
3.1
Shelf-Inventory Dependent Demand
A store experiences shelf-inventory I(t) dependent but predictable demand rate D(t) for a product as follows,
½
¾
D1 if I(t) ≤ J
D(t) =
D2 if I(t) ≥ J
where J is a given number of units and t denotes time. The store incurs usual ordering cost S (as oppposed
to usual notation in K) and holding costs H.
a) Do you expect D1 > D2 or D1 < D2 if the product is bread, why?
Fresh food items have larger demand. High inventory indicates freshness so D2 > D1 .
b) Do you expect D1 > D2 or D1 < D2 if the product is a stylish necklace, why?
Stylish products are more valuable when they are rare so D2 < D1 .
c) If the store orders Q units and Q ≤ J, what is the optimal Q that minimizes the sum of ordering and
holding costs?
p
Since Q ≤ J, D = D1 always. This is the setting for the EOQ model so Q∗ = min{ 2D1 S/H, J}.
d) Suppose that Q > J for the remaining parts except for part i). Draw the inventory level I (y-axis) as a
function of time (x-axis). Clearly indicate Q, J, D1 and D2 on your graph.
You could start the inventory level at Q at time zero, it drops with a slope of D2 until the level reaches
J. Then the inventory drops with a slope of D1 until it reaches 0. As soon as the inventory reaches 0, it is
brought back to Q. And the cycles continue in this manner.
e) What is the length of an inventory cycle for a given Q ?
5
From the graph, it takes (Q − J)/D2 to reach inventory level J and an additional J/D1 to reach 0. Then
the cycle time is (Q − J)/D2 + J/D1 .
f) Write down the per time ordering cost? Check that your expression reduces to the traditional EOQ cost
when D1 = D2 or J = 0.
Per cycle ordering cost is S, per time ordering cost is S/((Q − J)/D2 ) + J/D1 ). This expression reduces
to the traditionl cost SD2 /Q when D1 = D2 or J = 0.
g) A friend of yours claims that the inventory holding cost is
¶
µ
J2
(Q + J)(Q − J)
+
.
H
2D2
2D1
Determine if this per cycle or per time cost. If you say per cycle cost, then convert it to per time cost.
This is per cycle cost, the two terms inside the parantheses are total inventory held during [0, (Q−J)/D2 ]
and [(Q − J)/D2 , J/D1 ]. To convert it to per time cost, divide by the cycle length:
´
³
(Q+J)(Q−J)
J2
+
2D2
2D1
H
.
Q−J
J
D2 + D1
h) Write down the total cost T C in terms of Q. Take the derivative to establish that optimal Q solves
2SD1
= Q2 α + 2QJ(1 − α) − J 2 (1 − α)
H
where α := D1 /D2 . Specialize this to the case of α = 1/2 to establish that Q =
T C(Q) =
(2)
p
2J 2 + 4SD1 /H − J.
(Q + J)(Q − J)/(2D2 ) + J 2 /(2D1 )
S
+H
(Q − J)/D2 + J/D1
(Q − J)/D2 + J/D1
SD1
H Q2 α + J 2 (1 − α)
=
+
Qα + J(1 − α)
2 Qα + J(1 − α)
Taking the derivative:
dT C(Q)
SD1
H 2αQ(Qα + J(1 − α)) − α(Q2 α + J 2 (1 − α))
= −α
+
dQ
(Qα + J(1 − α))2
2
(Qα + J(1 − α))2
Rearranging we obtain (2):
2SD1
= 2Q(Qα + J(1 − α)) − (Q2 α + J 2 (1 − α)) = Q2 α + 2QJ(1 − α) − J 2 (1 − α).
H
When α = 1/2
4SD1
= Q2 + 2QJ − J 2 = (Q + J)2 − 2J 2 .
H
p
p
p
This yields Q = 2J 2 + 4SD1 /H − J. Note that when J = 0, Q = 4SD1 /H = 2SD2 /H as expected.
i) Find the optimal lot size when J = 4, D1 = 1, D2 = 2, H = 1, S = 16.
6
For Q ≤ J = 4, Q∗ =
p
2 · 16 · 1/1 > 4. We use Q = J = 4 in this region, total cost is
D1
Q
+ H = 16/4 + 2 = 6
Q
2
p
√
For Q ≥ J = 4, (2) yields Q = 2 · 42 + 4 · 16 · 1/1 = 4( 6 − 1) = 5.8. Then the cycle time is (Q − J)/D2 +
J/D1 = (5.8 − 4)/2 + 4/1 = 4.9 and the total cost is
S
T C(Q = 5.8) =
16
(5.8 + 4)(5.8 − 4)/(2 · 2) + 42 /2
+
= 3.27 + 0.9 + 1.63 = 5.8
4.9
4.9
Thus, the overall optimal lot size is Q∗ = 5.8.
3.2
Renewal Reward Theorem
Let {Xi }i≥1 be times between events i − 1 and i. Suppose that {Xi }i≥1 are iid. Let {Ri }i≥1 reward earned
at the end (begining or during) event i. Suppose that {Ri }i≥1 are iid. We allow for Ri to depend on Xi but
it should be independent of {Xj : j ≥ 1, j 6= i}. Define the cumulative number N (t) of events by time t as
N (t) :=
i
X X
1[
Xj ≤ t]
j=1
i≥0
where 1[.] is the indicator function taking events as arguments. Associated with N (t), we can define cumulative reward earned by t as
N (t)
X
C(t) :=
Ri
i=1
C(t) does not have to be monotone; it is nondecreasing (nonincreasing) if Ri denotes revenue (cost) or
zigzagging if Ri is profit. C(t) can have discontinuities when rewards happen only at discrete time periods.
A quantity of interest is the expected cumulative reward by time t, i.e. c(t) := E(C(t)).
Theorem 1. i) Suppose E(X1 ) < ∞, then
N (t)
1
=
t→∞
t
E(X1 )
lim
ii) Renewal reward theorem: In addition, if E|Ri | < ∞, then the limit below exists almost surely and
C(t)
E(R1 ) 2
=
t→∞ t
E(X1 )
lim
3.3
Supply Uncertainty Example
Consider the EOQ model under two types of supply uncertainty: Random yield and random supply availability. Suppose that K is the ordering cost, h is the holding cost and demand rate is constant at D.
2
Warning: E(R1 )/E(X1 ) 6= E(R1 /X1 )
7
a) In the random yield case, Q · R units are received when Q units are ordered where R is the random
yield and E(R) = 1. Write down the expected total cost T CR (Q) incurred per time if order quantity is Q.
Find the optimal order quantity Q∗R .
We use the renewal reward theorem by dividing E(cost per cycle) to E(cycle length):
r
K + hE(Q2 R2 )/(2D)
KD hQ
2KD
1
p
T CR (Q) =
=
+
E(R2 ) and Q∗R =
E(QR)/D
Q
2
h
E(R2 )
If you make a mistke here and set costs equal to E(cost per cycle/cycle length), you get
¶
µ
KD
K + h(Q2 R2 )/(2D)
1
hQ
=
E
E( ) +
.
QR/D
Q
R
2
Cleraly you get an incorrect expression that differs from the correct costs T CR (Q).
b) Is Q∗R ≤ EOQ or Q∗R ≥ EOQ, prove your answer. Does your answer change when E(R) < 1?
Clearly E(R2 ) ≥ E(R)2 = 1,
1
1
Q∗R /EOQ = p
≤p
=1
2
E(R)
E(R )
where the inequality follows from Jensen’s inequality. On the other hand, when E(R) < 1 a counterexample
can be constructed with R = 1/2 with probability 1 to show that Q∗R = 2EOQ.
c) In the random supply availability case, min{Q, C} units are received when Q units are ordered where
C ≥ 0 is the random supply. Establish that the total cost T CC(Q) is given by
T CC(Q) =
KD
h E[(min{Q, C})2 ]
+
E(min{Q, C}) 2 E(min{Q, C})
Now obtain the equation for optimal lot size, which should solve an equation of the form
KD + ? E[(min{Q, C})2 ] + ? E(min{Q, C})+? = 0
where “?” are to be determined. Also compare the optimal solution to this equation with EOQ lot size. If
you need, let the pdf and cdf of C be f and F .
The total cost expression follows from computing total costs per cycle and cycle length and dividing
them while using Renewal Reward Theorem. Now write
T CC(Q) =
1
h
{KD + E[(min{Q, C})2 ]}
E(min{Q, C})
2
where
Z
Q
E(min{Q, C}) =
cf (c)dc + QF̄ (Q)
0
Z
Q
E[(min{Q, C})2 ] =
0
8
c2 f (c)dc + Q2 F̄ (Q)
Now take the derivative of the cost by noting that
F̄ (Q)
d
1
=−
dQ E(min{Q, C})
[E(min{Q, C})]2
d
E[(min{Q, C})]2 = 2QF̄ (Q)
dQ
µ
¶
d
F̄ (Q)
h
2
T CC(Q) = −
KD − {2QE(min{Q, C}) + E[(min{Q, C}) ]}
dQ
[E(min{Q, C})]2
2
Therefore the critical Q solves
KD −
h
{2QE(min{Q, C}) − E[(min{Q, C})2 ]} = 0
2
Now note that
Z
Q
2QE(min{Q, C}) − E[(min{Q, C})2 ] = 2Q
Z
cf (c)dc + 2Q2 F̄ (Q) −
0
Q
c2 f (c)dc − Q2 F̄ (Q)
0
after some algebra
Z
2QE(min{Q, C}) − E[(min{Q, C})2 ] = Q2 −
Q
(Q − c)2 f (c)dc ≤ Q2
0
Thus we obtain that
KD −
h
{2EOQ E(min{EOQ, C}) − E[(min{EOQ, C})2 ]} ≥ 0
2
because EOQ solves
KD −
(+)
h 2
{Q } = 0
2
RQ
Equality in Inequality (+) can be obtained by increasing Q if we show that Q2 − 0 (Q − c)2 f (c)dc is
RQ
increasing in Q. This can be shown by taking the derivative of Q2 − 0 (Q − c)2 f (c)dc with respect to Q:
Z
2Q −
Z
Q
2(Q − c)f (c)dc = 2Q − 2QF (Q) +
0
Q
2cf (c)dc ≥ 0
0
As a result the critical Q value with random supply is larger than the EOQ.
4
Finite Horizon Dynamic and Deterministic Demand: Wagner-Whitin
Model
We will actually study a more generalized model than Wagner-Whitin. But we first need an important
result from optimization. Given a concave objective function f : <N → <, a matrix A of size M × N and
column vector b of size m × 1, consider
min{f (x) : Ax = b, x ≥ 0}.
9
This is a problem of minimizing a concave function over a convex set. It is well known that the optimal
solution can be found at one of the extreme points of the feasible set {Ax = b, x ≥ 0}.
This result can be applied directly to network flow formulations with concave costs. Given a network
with node and arc sets N and A. Suppose we have concave arc costs ci,j (x) : < → < and consider


 X

X
X
ci,j (xi,j ) :
xi,j −
xj,k = bj , xi,j ≥ 0 .
min


(i,j)∈A
(i,j)∈A
(j,k)∈A
We call the extreme points of the network flow polytope as extreme flows. It is easy to see that concave cost
network flow problem will be optimized by an extreme flow solution. A cycle (not necessarily directional) in
the residual network (with respect to flows xi,j ) can be written as a sum of two flows. Thus, extreme flows
cannot contain cycles. For example, eliminating cycles in a network flow problem with linear costs drives
the solution towards optimality and such algorithms are known as cycle cancelling algorithms.
Many production and inventory problems can be formulated as concave cost network flows. In these
formulations, it may be possible to generate a super supply node with the only negative bj value. Then all
the flows initially come out of the super supply node. If there is a node with incoming flows on two arcs,
then the associated solution is not an extreme flow; because the two paths cominig from the super supply
node to this node constitute a cycle. This fact is apparent in Wagner-Whitin network but we will illustrate
it on a problem with N serial facilities which reduces to the Wagner-Whitin model when N = 1.
Consider a production operation that requires N serial steps each of which is performed at a single
facility. Also let {dt }Tt=1 be given deterministic demands for the final product. We use i and t for production
step and time indices. Concave production cost function cit is for facility i and time t. It is possible to hold
work in process inventories after finishing step i in period t, let the associated concave cost function be hit .
These data is sufficient to set up the problem. Figure 1 contains an example with N = 4 and T = 5 and
d1 = 5, d2 = 1,d3 = 3, d4 = 2 and d5 = 4. It also illustrates an extreme flow solution.
Now let xit be the production at step i in period t, similarly define inventory yti held at that step. It is
possible to express inventory yti in terms of production. For that collapse all the nodes {(τ, i) : 1 ≤ τ ≤ t}
into a single node and write conservation of flow for that node:
inflow = outflow ⇒
t
X
τ =1
xiτ
=
t
X
xi+1
+ yti
τ
τ =1
We assume that ending inventories should be zero yTi = 0. Inventory shortages are not allowed, i.e. yti ≥ 0.
Production amount must be nonnegative,i.e. xit ≥ 0. Putting all these four types of constraints along with
the objective objective function below, we formalize the model:
min
N
T X
X
cit (xit ) + hit (yti )
t=1 i=1
i xi = 0: If we bring in inventory (product that
Now the extreme flow idea can be reiterated by requiring yt−1
t
has gone through step i) to step i from period t − 1, we should not perform any step i operation in period t.
Or if we perform this operation, the incoming inventory must be zero. Once the superscript i is dropped, we
10
Super Supply Node
Facilities:
1
2
3
4
Demands: 5
1
2
3
4
Time periods: 1
2
3
4
5
Figure 1: A numerical example. Vertical (horizontal) arcs are for production steps (inventory). Only flows
on the thick arcs are positive.
actually obtain the familiar Wagner-Whitin valid (for optimal solution) complementarity equality yt−1 xt = 0.
Let us suppose that costs are stationary in time and further suppose that hit is nondecreasing in i. Under
this conditions extreme flows have the following nestedness property:
Nestedness property: A feasible flow is nested if xit = 0 implies xjt = 0 for all j < i in all time periods t.
It is possible to prove under the assumptions of stationary costs and nondecreasing hit in i that a nested flow
can give the optimal solution to the serial production system with concave cost formulated as a network
flow problem. The flow in Figure 1 is also nested in addition to being extreme. Note that with the valid
complementarity constraint we have restricted the set of solutions to search for optimality. With nestedness
property, we are imposing yet another restriction. However, we luckily do not have to search for the optimal
solution, instead we use an efficient dynamic program.
f
Let Cq,s
be the optimal cost of producing at stage f in period q and using that production to meet
demands of periods from q to s − 1. Observe that how this cost definition uses extreme flow and nestedness
properties. Because of the extreme flow property, we know that if the production at step f in time q is
used to meet demand ds−1 , it will also be used to meet all the demands between q and s − 1. Without this
property we need to define a cost for each subset of time periods, clearly a nonpolynomial affair. Because
of nestedness property, we know that the production at step f in time q must be used to meet dq ; otherwise
f
production can be delayed until the next period without increasing the costs. The cost Cq,s
is actually for
a subsytem with f . . . N steps and q . . . s − 1 time periods; see Figure 2.
For simplicity let us suppose that production costs have two components a fixed cost and a proportinal
(to production quantity) cost. It is possible to argue that production costs are irrelevant for our objective
function and set them equal to zero without loss of generality. Let Kf denote the remaining fixed costs for
11
Cfqs
Kf
Facilities
f
hf
hf
hf
Cfrs
Cf+1qr
f+1
N
Time q-1
periods
q
q+1
r-1
r
r+1
s-1
s
Figure 2: Justification of DP recursion under extreme flow and nestedness properties.
step f . Also let hf be the linear (still concave) holding costs for work in process inventory at step i.
f
In computing Cq,s
, we know that flow xfq will be split into two. The first part will go into step f + 1 for
production in period q. The second part will be kept in the inventory until period r, p < r ≤ s. Note that
r = p violates the nestedness property. On the other hand, r = s means no inventory is kept as the second
f
part of the flow xfq . Looking at Figure 2, we can decompose Cq,s
as follows
f
Cq,s
= min Kf + (r − q)hf
q<r≤s
s−1
X
f +1
f
dt + Cq,r
+ Cr,s
for 1 ≤ f ≤ N, 1 ≤ q < s ≤ T + 1
t=r
f
f +1
f
where minimization over r ensures that Cq,s
is the optimal cost as long as Cq,r
and Cr,s
are so. We can
finish the DP formulation after specifying the boundary conditions
N +1
f
Cq,s
= 0 for all q, s; Cs,s
= 0 for all s, f.
Now we are in a position to provide an algorithm:
f
• Set q = T , s = T + 1, f = N and Cq,s
= KF (start at the bottom right).
• Repeat
•
f
If s > q + 1, compute Cq,s−1
and set s := s − 1 (move to right).
•
f
else if s = q + 1 and q > 1, compute Cq−1,T
+1 and set q := q − 1 (consider periods q . . . T + 1).
•
f,f −1
else if s = q + 1, q = 1 and f > 1, compute CT,T
+1 and set f := f − 1 (move up).
• until s = T + 1, q = 1 and f = 1 (arrival at the top left).
12
1
• Stop C1,T
+1 is the optimal solution.
f
Finally observe that there are O(N · T 2 ) cost computations for {Cq,s
: 1 ≤ f ≤ N, 1 ≤ q < s ≤ T + 1}.
Each cost computation involves a minimization (actually comparison of T numbers, one for each possible
r) so it takes O(T ). In total the algorithm takes O(N · T 3 ). This DP algorithm isvery efficient and can
be used in practical situations. Its efficiency is inherited from the Wagner-Whitin algorithm which owes its
efficiency to extreme flow property. However, the current algorithm exploits another property, nestedness,
which does not arise in Wagner-Whitin context. The discussion here is mainly based on Love 1972, more
details can be found in there. For a moree recent and general treatment of minimum concave-cost network
flow problems see Guisewite and Pardalos 1990.
5
Infinite Horizon Stationary and Stochastic Demand
Consider a continuous time inventory model. Let t be the continuous time variable. Let N I(t) be the (net)
inventory level at time t. Inventory level is the on-hand (physically availabale at the warehouse) inventory
minus all the backorders at any time. Inventory related costs are based on inventory level. It can be negative with backorders. Also suppose that N I(0) = 0 and Q1 is placed at t = 0. Let L be the lead time for
deliveries; time from when an order is placed until it arrives. L is an uncontrollable deterministic parameter
for most of our discussion. If you are at time t, the orders that will be placed after t cannot affect the costs
in [t, t + L). This is because such orders will all arrive after t + L.
Suppose that cost incurred at time t is computed by function g : < → <+ . It is very commnon to take
g(y) = hy + + by −
where h and b are holding and backorder costs, both measured per unit and per time. Then g(N I(t)) is the
rate at which the inventory costs are incurred at time t. In addition, every time an order is placed a fixed
cost of K is incurred.
Let {D(t), t ≥ 0} denote the demand stochastic process. Demand over an interval (s, t] is denoted by
D(s, t]. We also define demand rate as
ED(s, t]
λ :=
t−s
Identically distributed demand means for any s, t, u ≥ 0, D(t, t + u] and D(s, s + u] have the same distribution. Note that we allow for (t, t + u] ∩ (s, s + u] 6= ∅. For stochastic processes, identically distributed
process and stationary process are usually used synonmously. To define independence, we pick s, t, u ≥ 0
such that (t, t + u] ∩ (s, s + u] = ∅ and require D(t, t + u] and D(s, s + u] be independent.
Let IIT (t) be the inventory in-transit (pipeline) that has been shipped before t but has not arrived yet
at t. However, IIT (t) will definetely arrive by t + L. Some of IIT (t) may actually arrive before t + L.
Inventory position IP (t) is defined as
IP (t) := N I(t) + IIT (t).
Although IP is not used for cost accounting, it has nice properties that will allow us to handle N I. Observe
that IP (t) includes all the orders placed by t. Until after t + L, no orders will arrive except for those already
13
accounted for in IP (t). However, demand of D(t, t + L] will be materialized until then. These observations
let us to obtain the distribution of N I(t + L) given that IP (t) is known
N I(t + L) = IP (t) − D(t, t + L]
(3)
Since the demand can be a general process, we must be careful in writing demands over intervals to avoid
double counting. Our convention is to exclude (include) the left (right) end point of an interval so we write
D(s, t]. By (3), the costs at time t are then incurred after observing the demand at time t.
We can now define two closely related policies for continuous inventory review.
• (R, Q) policy: Whenever IP (t) ≤ R, order Q units.
• (s, S) policy: Whenever IP (t) ≤ s, order S − IP (t) units.
Although (s, S) policies are generally discussed in the context of periodic review and proven to be optimal,
they can be used in continuous review. Note that the diffrenece between (R, Q) and (s, S) policies are very
small indeed. For example, when the demand happens in units of 1 (as opposed to batch demand arrivals)
and s = R, S = Q + s, the sample paths of N I(t) will exactly be the same under both policies.
5.1
(R, Q) Cost formulas
The challenging part in computing the expected cost per time is finding the long run average holding and
backorder costs
Z
1 T
Eg(N I) := lim sup
g(N I(t))dt
T →∞ T t=0
Note that we are defining Eg(N I) from the above equation. It is not always necessary that Eg(N I(t)) will
converge to Eg(N I) as t → ∞. For example, g should not jump around much, e.g. some fancy requirements,
such as uniformly (in t) upper boundable Eg(N I(t)) and Eg(N I(t))2 , are used. Under general conditions,
it can be shown that
lim Eg(N I(t)) = Eg(N I)
t→∞
Thus, the next step should be obtaining the limiting distribution of N I(t) as t → ∞. This can be achieved
with (3), if the limiting distribution of IP (t) is known. At this point we must appreciate why IP is useful,
its limiting distribution is easy to find and that can be used as a stepping stone towards N I(t) limiting
distribution.
To compute IP (t) as t → ∞, first note its uniform (in t) range: IP (t) ∈ (R, R + Q]. For simplicity,
suppose that demand is in units of one so that R and R + Q must be integers. Consider the continuous time
Markov Chain {IP (t), t ≥ 0} which decreases from R + Q to R with demands and jumps back to R + Q. The
important observation is that state transition times and probabilities are totally governed by the demand
process {D(t) : t ≥ 0}. This is because the customers cannot know the inventory position. Because of the
transition probability symmetry, in the long run all the states have the same limiting probability:
lim P (IP (t) = i) = 1/Q for all states i
t→∞
Then we can compute Eg(N I) by conditioning on the inventory position IP as
Eg(N I) = EIP [Eg(N I)|IP ] = EIP [ED g(N I = IP − D(0, L])]
14
For convenience let G(y) := ED (g(y − D)). Then
Z
Eg(N I) = EIP [G(IP )] =
R+Q
G(y)dy/Q.
R
The last expression is the expected holding and backorder cost accimulation rate (per time) in the long run.
If we add to this the ordering cost per time K/(Q/λ), we obtain the total cost per time:
R R+Q
Kλ + R
G(y)dy
C(R, Q) =
(4)
= Kλ/Q + h(Q/2 + R − Lλ) + (h + b)B(R, Q)
Q
where the B(R,Q) is average backorders per time
R R+Q
B(R, Q) =
R
E(D − y)+ dy
.
Q
(5)
When the cumulative demand is a nondecreasing stochastic process with stationary increments and continuous sample paths (see Serfozo and Stidham 1978), the expected per time cost is given by (4).
5.2
(R, Q) optimization
From the discussion in Zipkin 1986, C(R, Q) is jointly convex in their (R, Q). Consider the first order
conditions
∂C(R, Q)
= G(R + Q) − G(R) = 0
∂R
Z R+Q
∂C(R, Q)
2
= −Kλ/Q −
G(y)dy/Q2 + G(R + Q)/Q = 0
∂Q
R
In other words, we need R and Q to respectively solve
G(R + Q) = G(R)
Z
R+Q
Kλ +
G(y)dy = QG(R + Q).
R
These equations have geometric interpretations, see Figure 3. The optimality condition on R says that the
line piece between (R, G(R)) and (R + Q, G(R + Q)) is parallel to 0 − y axis. The condition on Q says that
Kλ is exactly the area between the curve G(y) and the line piece between (R, G(R)) and (R + Q, G(R + Q)).
6
Exercises
1. Establish the identity in (1).
2. If D ∼ N (25, 52 ) and y = 25, use (1) to compute the fill rate.
3. Either prove or find a counterexample for the following inequality: µ ≥ E(D − x)+ for a random
variable D, µ = E(D) and x ≥ 0. If you can construct a counterexample, think of an additional
condition on D to make the inequality valid.
4. Establish that when the randomness is multiplicative the optimal price under perfect information is
bc/(b − 1).
15
G( y)
G( y)
Kλ
∫
R +Q
R
R
G ( R + Q)
G ( y )dy
Q
R+Q
y
Figure 3: A geometric interpretation of optimal R and Q.
5. Sensitivity of EOQ costs to a finite planning horizon.
√ Average inventory costs over an infinite planning
horizon (T = ∞) in EOQ setting is known to be 2KDh. Consider a finite horizon problem T < ∞
during which n orders are made. The begining (t = 0) and ending (t = T ) inventory levels are
constrained to be zero.
a) Argue that optimal order quantites must be equal to each other.
b) With n ∈ Z+ orders, argue that the per time cost over [0, T ] is given by T C(n, T ) = nK/T +
hDT /(2n).
c) Show that optimal n(T ) for given T is found by searching for the minimum value of n that satisfies
n(n + 1) ≥ hDT 2 /(2K). Note that n(T ) is a step-wise increasing function of T .
d) Let C(T ) be the per time minimum cost if the planning horizon is T . Observe that
C(T ) = min T C(n, T ) = n(T )
n∈Z+
hDT
K
+
.
T
2n(T )
q
√
2K
Draw C(T ) and point out how it relates to T C(n, T ) for n ≥ 1. Prove that C(T = k· Dh
) = 2KDh,
i.e. planning horizons that are integer multiples of an optimal EOQ cycle yield optimal EOQ costs.
(Hint: Choose n = k to keep T /n constant and equal to EOQ cycle length.)
e) As we observed in c), n(T ) switches from a k to k + 1 as T grows. Find the largest value of T (k)
where k cycles are still optimal, i.e. at T (k) + ², k + 1 cycles are optimal as discussed in c). Show that
at the value of T (k) where the switch from k to k + 1 cycles happens
Ãr
!
r
1
k
k+1 √
C(T = T (k)) =
+
2KDh
2
k+1
k
f) Note that switching T values are local maxima of C(T ) and argue that if C(T = T (k)) ≤ C0 for
some C0 then C(T ) ≤ C0 for T ≥ T (k). Find the smallest value of k(2.5%) such that
√
C(T = T (k)) ≤ 1.025 2KDh
16
p
Compute the planning horizon T (k = k(2.5%)) value (in terms of EOQ cycle length 2K/(Dh))
corresponding to the k(2.5%) value. We have just proved that by using a planning horizon longer than
T (k = k(2.5%)), the per time optimal cost of finite planning horizon problem can at most be 2.5%
worse than the optimal cost of the infinite horizon problem (EOQ costs).
6. Prove that in order to minimize a concave function over a polytope, it sufficient to consider only
the extreme points of the polytope. Hint: Use proof by contradiction, suppose x∗ is the optimal
solution but can be written as λx1 + (1 − λ)x2 where x1 and x2 are in the polytope, show that
f (x∗ ) ≥ min{f (x1 ), f (x2 )}. Generalize this argument.
7. Prove that in order to minimize a concave function over a closed convex set, it sufficient to consider
only the boundary of the convex set.
8. Refer to serial production system with concave costs. Suppose that a given (x, y) is an extreme flow
i
and yt−1
+ xit > 0, then argue that there exist integers τ1 and τ2 , t ≤ τ1 ≤ τ2 ≤ T such that
i
yt−1
+
xit
=
τ2
X
dτ .
τ =τ1
Express this property in a sentence. How (if) will you modify this equality if it is also known that
(x, y) is a nested flow?
9. Refer to serial production system with concave costs. Motivate the fact that hit is likely to be nondecreasing in i. Suppose that N = 2 and i = 1 is a production activity in the factory while i = 2 is a
packaging activity at a distribution center. You may choose to give an argument for this special case.
10. Refer to serial production system with concave costs. Under the assumptions of stationary costs and
nondecreasing hit in i prove that a nested flow can give the optimal solution. Do you really need
stationary cost assumption? For example if cit is nonincreasing in time (no inflation), can you repeat
the proof?
11. For the example in Figure 1, draw a feasible flow that is nested but not extreme.
12. Refer to serial production system with concave costs. When production costs have a fixed and proportional cost component, prove that proportinal costs can be set to zero without loss of generality.
Does your proof depend on extreme flow or nestedness properties. If not, find a feature of the model
that is necessary for your proof — in the sense that your proof fails without this feature.
References
[1] G.M. Guisewite and P.M. Pardalos (1990). Minimum concave-cost network flow problems: Applications,
complexity, and algorithms. Annals of Operations Research 25: 75-100.
[2] S.F. Love (1972). A facilities in series inventory model with nested schedules. ManSci Vol.18 No.5: 327338.
17
[3] N.C. Petruzzi and M. Dada (1999). Pricing and the newsvendor problem: a review with extensions.
Operations Research Vol.47 No.2: 183-194.
[4] R. Serfozo and S. Stidham (1978). Semi-Stationary Clearing Processes. Stochastic Process Applications,
Vol.6: 165-178.
[5] P. Zipkin (1986). Inventory Service-Level Measures: Convexity and Approximation. Management Science,
Vol.32, No.8: 975-981.
7
HW Solutions
• Establish the identity in (1).
Z
+
E(D − y)
µ
∞
=
(ξ − y)φ
y
Z
ξ−µ
σ
¶
dξ
∞
=
(σx + µ − y)φ(x)dx
(y−µ)/σ
Z
∞
1
σx √ exp(−x2 /2)dx
2π
(y−µ)/σ
Z ∞
1
√ exp(−z)dz
= (µ − y)(1 − Φ((y − µ)/σ)) + σ
2π
((y−µ)/σ)2 /2
1
= (µ − y)(1 − Φ((y − µ)/σ)) − σ √ exp(−z)|∞
((y−µ)/σ)2 /2
2π
¶¶
µ
¶
µ
µ
y−µ
y−µ
+ σφ
.
= (µ − y) 1 − Φ
σ
σ
= (µ − y)(1 − Φ((y − µ)/σ)) +
• Either prove or find a counterexample for the following inequality: µ ≥ E(D − x)+ for a random
variable D, µ = E(D) and x ≥ 0. If you can construct a counterexample, think of an additional
condition on D to make the inequality valid.
Clearly with D = −1 wp 1 fails µ ≥ E(D − x)+ . Then we require D ≥ 0 wp 1 so that
Z ∞
Z ∞
µ=
Df (D)dD ≥
(D − x)f (D)dD = E(D − x)+
0
x
Thus, D ≥ 0 is a sufficient condition. However, D ≥ 0 is not necessary. To see this let D = −1 and
D = 11 with equal probabilities and set x = 11, then µ = 5 > 0 = E(D − 11)+ .
• Establish that when the randomness is multiplicative the optimal price under perfect information is
bc/(b − 1).
First recall
ΨM (p) = (p − c)ap−b E(ξ)
So that
∂ΨM (p)
= ap−b−1 (p − (p − c)b) = 0
∂p
18
⇒
p0,M =
bc
.
b−1
• Sensitivity of EOQ costs to a finite planning horizon. Average
inventory costs over an infinite plan√
ning horizon (T = ∞) in EOQ setting is known to be 2KDh. Consider a finite horizon problem
T < ∞ during which n orders are made. The begining (t = 0) and ending (t = T ) inventory levels are
constrained to be zero.
a) Argue that optimal order quantites must be equal to each other.
Suppose that there are two consecutive orders Q1 and Q2 with Q1 6= Q2 . Then the corresponding
cycle times are Q1 /D and Q2 /D respectively, and the total cost in the two cycles are
C(Q1 , Q2 ) = 2K +
hQ21 hQ22
+
.
2D
2D
Now consider an alternative: order Q = (Q1 + Q2 )/2 twice. Then we have to identical cycles with
cycle time (Q1 + Q2 )/(2D) (note the total time of two cycle does not change), and the resulting total
cost in the two cycles are
C(Q, Q) = 2K + 2
h(Q1 + Q2 )2
h[(Q1 + Q2 )/2]2
= 2K +
.
2D
4D
Since Q21 + Q22 − (Q1 + Q2 )2 /2 = (Q1 − Q2 )2 /2 > 0 for Q1 6= Q2 , we deduce that C(Q1 , Q2 ) > C(Q, Q).
Thus the order sizes must be equal.
b) With n ∈ Z+ orders, argue that the per time cost over [0, T ] is given by T C(n, T ) = nK/T +
hDT /(2n).
Since beginning (t = 0) and ending (t = T ) inventories are restricted to be zero, and the order
quantities are the same, T is divided into n identical cycles. That is, the cycle time is T /n. It follows
that the order quantity is T D/n. Then the cost per cycle is K + hDT 2 /(2n2 ). Hence we can derive
the cost per time is
T C(n, T ) =
K + hDT 2 /(2n2 )
= nK/T + hDT /2n.
T /n
(6)
c) Show that optimal n(T ) for given T is found by searching for the minimum value of n that satisfies
n(n + 1) ≥ hDT 2 /(2K). Note that n(T ) is a step-wise increasing function of T .
We take the first order difference with respect to n in (6) and get
∆n T C(n, T ) = T C(n + 1, T ) − T C(n, T )
= K/T + hDT /(2(n + 1)) − hDT /(2n)
K
=
[n(n + 1) − hDT 2 /(2K)].
T n(n + 1)
(7)
It is easily checked that (6) is convex in n. Thus the optimal n(T ) is an integer such that ∆n T C(n, T ) <
0 for n < n(T ) and ∆n T C(n, T ) > 0 for n > n(T ). Then, from (7), n(T ) can be chosen as the smallest
number which satisfy n(n + 1) ≥ hDT 2 /(2K).
19
C(T )
√
2KDh
q
2K
Dh
q
2K
2 Dh
T
Figure 4: The per time cost C(T ) of finite horizon EOQ model as a function of the horizon length T .
d) Let C(T ) be the per time minimum cost if the planning horizon is T . Observe that
C(T ) = min T C(n, T ) = n(T )
n∈Z+
K
hDT
+
.
T
2n(T )
q
√
2K
Draw C(T ) and point out how it relates to T C(n, T ) for n ≥ 1. Prove that C(T = k· Dh
) = 2KDh,
i.e. planning horizons that are integer multiples of an optimal EOQ cycle yield optimal EOQ costs.
(Hint: Choose n = k to keep T /n constant and equal to EOQ cycle length.)
C(T ) p
is drawn in Figure 4. As you may expect C(T ) converges to
T = k (2K)/(Dh) in the condition derived in part c), then
√
2KDh as T → ∞. Now put
n(n + 1) ≥ hDT 2 /(2K) = k 2 .
So n(T ) = k. Then
p
µ
¶
p
hDk (2K)/(Dh) √
K
C T = k (2K)/(Dh) = k p
+
= 2KDh.
2k
k (2K)/(Dh)
Note that horizon lengths T that are integer multiples of an optimal EOQ cycle length yield optimal
EOQ costs in Figure 4.
e) As we observed in c), n(T ) switches from a k to k + 1 as T grows. Find the largest value of T (k)
where k cycles are still optimal, i.e. at T (k) + ², k + 1 cycles are optimal as discussed in c). Show that
at the value of T (k) where the switch from k to k + 1 cycles happens
Ãr
!
r
1
k
k+1 √
C(T = T (k)) =
+
2KDh
2
k+1
k
20
Since T (k) is the largest where k cycles are still optimal,
k(k + 1) = hD[T (k)2 ]/(2K).
(8)
p
Solving this yields T (k) = k(k + 1) · 2K/(hD). Thus,
C(T = T (k)) = k p
=
hD
K
p
k(k + 1) · 2K/(hD)
2k
+
k(k + 1) · 2K/(hD)
Ãr
!
r
1
k
k+1 √
+
2KDh.
2
k+1
k
(9)
f) Note that switching T values are local maxima of C(T ) and argue that C(T = T (k)) is nonincreasing
in k. Find the smallest value of k(2.5%) such that
√
C(T = T (k)) ≤ 1.025 2KDh
p
Compute the planning horizon T (k = k(2.5%)) value (in terms of EOQ cycle length 2K/(Dh))
corresponding to the k(2.5%) value. We have just proved that by using a planning horizon longer than
T (k = k(2.5%)), the per time optimal cost of finite planning horizon problem can at most be 2.5%
worse than the optimal cost of the infinite horizon problem (EOQ costs).
We first show that local minima of C(T ) is between the switching T values. Fix k ≥ 1 arbitrarily. We
only examine {T : T (k − 1) ≤ T ≤ T (k)} where n(T ) = k We argue that
T (k − 1) ≤ arg min T C(k, T ) ≤ T (k)
T
Note that arg minT T C(k, T ) satisfies k 2 = hDT 2 /2K while T (k) satisfies k(k+1) = hDT 2 /2K. Thus,
minT T C(k, T ) ≤ T (k). Similiarly argue for T (k − 1) ≤ arg minT T C(k, T ). Since C(T ) is convex and
its local minima are between switching T values, C(T ) is locally maximized at the switching T values.
It is instructive to go over these arguments in view of Figure 4.
From (9),
C(T = T (k + 1)) − C(T = T (k))
Ãr
!
r
r
r
k+1
k+2
k
k+1
1√
2KDh
+
−
−
≤0
=
2
k+2
k+1
k+1
k
√
The last inequality follows from the fact that . is a concave function and k/(k + 1) ≤ (k + 1)(k + 2) ≤
(k + 2)(k + 1) ≤ (k + 1)(k). Consequently, C(T = T (k)) is nonincreasing in k. Combining this with
the fact that local maxima happens only at the switching points, we establish that for T1 ≤ T2
C(T2 ) ≤ C(T (k(T2 ) − 1)) ≤ C(T (k(T1 ) − 1))
In English, cost at any T is smaller than or equal to costs at all switching T values that are smaller
than T . This upper bound on the cost helps us to obtain worst case performance.
µq
¶
q
k
k+1
To get k(2.5%), find the minimum k such that 21
+
≤ 1.025. That gives k = 2. From
k+1
k
p
(8), T (2) = 2 2K/(Dh). As a result, if you have a finite horizon longer than 2 EOQ cycles, per time
costs will be only 2.5% worse than EOQ costs: Per time costs are not sensitive to horizon length T .
21
• Prove that in order to minimize a concave function over a polytope, it sufficient to consider only the
extreme points of the polytope.
Suppose x∗ is the minimizer of a concave function f (·) over a polytope P
P . If x∗ is not an extreme
∗
∗
point,
then x is a convex combination of extreme points of P , i.e., x = i∈I λi xi where λi ≥ 0 and
P
λ
i∈I i = 1, and {xi ∈ P |i ∈ I} is the set of extreme points of P . Then by concavity
f (x∗ ) ≥
X
i∈I
ai f (xi ) ≥
X
ai min f (xi ) = min f (xi )
i∈I
i∈I
i∈I
If f is strictly concave, we have a contradiction. If f is not strictly concave, the minimality of x∗
implies f (x∗ ) = mini∈I f (xi ). That is, there is an extreme point in P which also minimizes f . Hence,
it is sufficient to check the minimizer only at the extreme points.
• Refer to serial production system with concave costs. Suppose that a given (x, y) is an extreme flow
i
and yt−1
+ xit > 0, then argue that there exist integers τ1 and τ2 , t ≤ τ1 ≤ τ2 ≤ T such that
i
yt−1
+ xit =
τ2
X
dτ .
τ =τ1
Express this property in a sentence. How (if) will you modify this equality if it is also known that
(x, y) is a nested flow?
First, observe that there cannot exist any demand dt which is only partially satisfied by the inflow
i
yt−1
+ xit into node (i, t) due to extreme flow property. Then let T be a subset of {1, . . . , T } and we
can always write
X
i
yt−1
+ xit =
dτ .
τ ∈T
But in that case attempt to pick a τ ∈
/ T such that τ1 < τ < τ2 and τ1 , τ2 ∈ T . If no such τ can be
found, the proof is complete. If there is such a τ then the flow from the source to node (N, τ ) and the
flow from (i, t) to (N, τ2 ) collide at a node because the graph is planar. But this collison also violates
the extreme flow property, establishing a contradiction — no such τ can exist.
In English: Inflow to any node is the sum of demands of some adjacent periods starting at or after
the period the node belongs to.
In case of nestedness, τ1 = t. In English: Inflow to any node is the sum of demands of some adjacent
periods starting at the period the node belongs to.
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