ISYE 3232 B, Fall 2014
Week #3, September 1-5, 2014
Stochastic Manufacturing & Service Systems
Xinchang Wang
H. Milton Stewart School of Industrial and Systems Engineering
Georgia Institute of Technology
Newsvendor Model (cont’d)
1
The Discrete Case
Let us turn our attention to the discrete case. Assume that D is a discrete random variable taking
values in {d0 , d1 , d2 , · · · } with p.m.f P[D = di ] = pi .
d
P[D = d]
Then, F (x) = P[D ≤ x] =
such that
P
i pi .
d0
p0
d1
p1
···
···
d2
p2
For discrete D, an optimal order quantity y ∗ is the smallest y
F (y) ≥
cp − cv
.
cp − cs
(1)
Example 1.1. Still, cp = 30, cv = 10, cs = 5.
d
P(D = d)
2
0.15
What is y ∗ — the smallest y such that F (y) ≥
2
5
0.80
30−10
30−5
10
0.05
= 54 ?
Summary of Solutions to the Profit Maximization Problem
1. When D is a continuous random variable, choose y ∗ such that
F (y ∗ ) =
cp − cv
.
cp − cs
2. When D is a discrete random variable, choose y ∗ to be the smallest y such that
F (y) ≥
1
cp − cv
.
cp − cs
3
Examples for Profit Maximization
Example 3.1. David buys fruits and vegetables wholesale and retails them at David’s Produce
on La Vista Road. One of the difficult decisions is the amount of bananas to buy. Let us make
some simplifying assumptions, and assume that David purchases bananas once a week at 20 cents
per pound and retails them at 50 cents per pound during the week. Bananas that are more than a
week old are too ripe to sell and David will pay workers to take them away. It costs 2 cent to get
rid of each pound of unsold bananas. Suppose that the weekly demand for bananas is uniformly
distributed between 500 and 1500 pounds.
(a) How many pound of bananas should David order each week?
(b) What is the optimal expected weekly profit?
(c) Now assume that the demand for bananas is exponentially distributed with mean 1000. How
many pound of bananas should be ordered?
Solution. We have that cp = 50, cv = 20, cs = −2, and that demand D ∼ uniform[500, 1500].
(a) Optimal order quantity y ∗ solves F (y ∗ ) = 50−20
50+2 because the demand is continuous. Since
y ∗ −500
∗
D ∼ uniform[500, 1500], F (y ) = 1500−500 . It follows that
y ∗ − 500
50 − 20
=
,
1500 − 500
50 + 2
which gives that y ∗ = 1076.92 ≈ 1077.
(2)
(b) If order quantity is y ∗ = 1077, then
E[1077 ∧ D] = E[min(1077, D)]
Z ∞
(1077 ∧ u)f (u)du
=
−∞
1077
Z
1
=
u·
du +
1000
500
= 910.5355
Z
1500
1077 ·
1077
1
du
1000
E[(1077 − D)+ ] = E[max(1077 − D, 0)]
Z ∞
=
(1077 − u)+ f (u)du
−∞
1500
Z
1
du
1000
500
Z 1077
Z 1500
1
1
=
(1077 − u) ·
du +
0·
du
1000
1000
500
1077
= 166.4645
=
(1077 − u)+
E[Profit(1077, D)] = (cp − cv )E[1077 ∧ D] − (cv − cs )E[(1077 − D)+ ]
= (50 − 20) · 910.5355 − (20 − (−2)) · 166.4645
= 23653.846
2
(c) By using the formula given in Lecture Note 2, it follows that
1
1 − e− 1000 y
∗
=
50 − 20
.
50 + 2
Solving for x∗ gives optimal order quantity y ∗ = −1000 · ln(1 − 30/52) ≈ 860.
Example 3.2. Dan buys fruits and vegetables wholesale and retails them at the Georgia Tech
Farmer’s Market (Thursdays on Tech Parkway). One of the more difficult decisions is the amount
of peaches to buy. Assume that Dan purchases organic, locally grown Georgia Peaches once a week
at $10 per pound and retails them at $30 per pound at the Farmer’s Market. Peaches he can’t sell
during market ours, he sells to a local grocer for $5 per pound. Suppose the demand for the fresh
peaches follows the following p.m.f.

1/10 if k = 5




3/10
if k = 6



2/5 if k = 7
P(D = k) =
1/10 if k = 8





1/10 if k = 9


0
otherwise
(a) What is Dan’s expected profit per week if he buys 6 pounds of peaches?
(b) What is the optimal amount of peaches Dan should buy to maximize his profits?
(c) What is the expected optimal profit per week Dan will make?
Solution:
(a) cp = 30, cv = 10, cs = 5. From the newsvendor class notes
Profit(6, D) = (cp − cv )(D ∧ 6) − (cv − cs )(6 − D)+ .
Hence
E[Profit(6, D)] = 20E[(D ∧ 6)] − 5E[(6 − D)+ ] = 20 × 5.9 − 5 × 0.1 = 117.5
(b) The optimal amount is the smallest y ∗ such that F (y ∗ ) ≥
y ∗ = 7.
cp −cv
cp −cs
= 20/25 = 4/5, which implies
(c) It follows that
E[(D ∧ 7)] = 5 × 0.1 + 6 × 0.3 + 7 × 0.6 = 6.5,
E[(7 − D)+ ] = 2 × 0.1 + 1 × 0.3 = 0.5.
Thus, the optimal profit
E[Profit(7, D)] = 20E[(D ∧ 7)] − 5E[(7 − D)+ ] = 20 × 6.5 − 5 × 0.5 = 127.5.
3
4
Cost Minimization Problem
We are sometimes only interested in cost minimization in a production/inventory management
system. Let’s re-define the notation or look at it from a different perspective.
y := quantity of order — the decision variable
D := the demand of the period (random variable: we know the distribution)
cf
:= fixed production/ordering cost — shipping cost and taxes
cv
:= variable cost
cu := understock cost, penalty cost for each lost sale
co := overstock cost, the holding cost for each leftover, co = −cs .
We are interested in the following quantities:
(1) Production quantities: y
(2) Understock quantities: (D − y)+
(3) Overstock quantities: (y − D)+
The cost minimization problem aims to solve the following optimization model:
min E[Cost(D, y)] = ordering cost + understock cost + overstock cost
y
= cf + cv y + cu E[(D − y)+ ] + co E[(y − D)+ ].
4.1
(3)
Solution to Cost Minimization Problem
By using a similar method to solve the profit maximization problem, we have the following useful
theorem.
Theorem 1. Let D be the random demand with c.d.f. F (x) = P(D ≤ x) for all x ∈ (−∞, ∞). If D
is a continuous rv, then the optimal production quantity that solves the cost minimization problem
(3) is y ∗ such that
cu − cv
F (y ∗ ) =
.
cu + co
If D is a discrete rv, then the optimal production quantity that solves the cost minimization problem
(3) is the smallest y ∗ such that
cu − cv
F (y ∗ ) ≥
.
cu + co
I would like to give several interesting remarks:
(1) Fixed cost (cf ) again does not affect the optimal production quantity.
(2) If understock cost (cu ) is equal to unit production cost (cv ), which makes cu − cv = 0, then you
will not produce anything.
(3) If unit production cost and overstock cost are negligible compared to understock cost, meaning
cu cv , co , you will prepare as much as you can.
4