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Unformatted text preview: Single Period
“Newsvendor” Problem IE 251 1 Probabilistic Demand: Single Period Problem
Consider a single ordering/production decision, Q, to be made
at the beginning of a period
Demand, which is probabilistic, occurs during the period
Relevant costs are incurred (inventory holding, variable
purchasing or production costs, fixed costs (when exists),
backordering or lost sales costs) at the end of the period
Single period model is a building block for more complicated
models under uncertainty
Direct applications: style goods and perishable items (dated
publications, garment manufacturing, fresh produce, blood,
medical supplies)
2 Probabilistic Demand: Single Period Problem inventory Q
inventory
End of period 0 Backorder or
lost sales
Total demand within the period
3 Model Environment
Relatively short selling season (weeks, 2 months,…)
with a welldefined beginning and end
When the total demand in the period exceeds the
stock available, there is an associated underage cost, cu
– cost of lost profit and/or goodwill
– expediting or acquiring the material at a higher unit
value When the total demand is less than the stock available,
overage cost occurs, co
4 Simple Case: The Newsvendor Problem
Zero initial inventory at the beginning of the
period
Only overage and underage costs
We are interested in the “expected cost”
Expected Underage Cost:
cu {1× P( X = Q + 1)
+2 × P( X = Q + 2) + ...}
= cu ∞ ∑ (i − Q )P( X = i) i =Q +1 5 Simple Case: The Newsvendor Problem
Expected Overage Cost
co {Q × P( X = 0) + (Q − 1) × P( X = 1) + L
+1× P( X = Q − 1)}
Q = co ∑ (Q − i)P( X = i)
i =0 Total Expected Cost:
Q C(Q ) = co ∑ (Q − i)P( X = i) + cu
i =0 ∞ ∑ (i − Q )P( X = i) i =Q +1 6 Marginal Analysis
From the graph we see that q* is
the smallest value of q for which
C (q*+1) – C (q*) ≥ 0.
If C(q) is a convex function of q
we can minimize expected cost
by finding the smallest value of
q that satisfies the inequality.
Determine q* by repeatedly
computing the effect of adding a
marginal unit to the value of q.
Method is called marginal
marginal C(q) q*–1 q* q
q*+1 analysis. 7 The Concept of Marginal Analysis
Marginal analysis: finding the expected profit of
ordering one more unit.
one Probability of not selling
Your Last item in stock
P(X < Q) Probability of
Selling everything
P(X ≥ Q) μ Q 8 Single Period Inventory Model
Marginal Analysis
Goal in finding the proper Q:
E (revenue on last sale) = E (loss if last paper not sold) P(X ≥ Q)Cu = P(X < Q)Co
⎡1 − P(X < Q)⎤ Cu = P(X < Q)Co
⎣
⎦
Cu
P(X < Q) =
Cu + Co Critical Ratio where:
Cu = unit contribution from newspaper sale ( opportunity cost of
underestimating demand)…underage cost is avoided by Qth newspaper
Co = unit loss from not selling newspaper (cost of overestimating
demand)… adding Qth newspaper adds “exposure” to overage cost
X = demand
Q = newspaper stocked 9 Critical fractile for the newsvendor problem
When the demand is a discrete random variable, the
cu
condition
PX < (Q ) =
cu + c o
may not be satisfied at equality (jumps, due to
discreteness).
The appropriate condition to use:
Q * = min{Q : PX ≤ (Q ) ≥ Q cu
}
cu + co C(Q+1)C(Q) ≥0
thus…
cuPX≥(Q+1) ≤ coPX<(Q+1) 10 SinglePeriod & Discrete Demand: Lively Lobsters
Lively
Lively Lobsters (L.L.) receives a supply of fresh, live
lobsters from Maine every day. Lively earns a profit of
$7.50 for every lobster sold, but a dayold lobster is
worth only $8.50. Each lobster costs L.L. $14.50.
– unit cost of a L.L. stockout
Cu = 7.50 = lost profit – unit cost of having a leftover lobster
Co = 14.50  8.50 = cost – salvage value = 6. – target L.L. service level
CR = Cu/(Cu + Co) = 7.5 / (7.5 + 6) = .56
11 Lively Lobsters
Demand
follows a
discrete
(relative
frequency)
distribution:
Result: order 25
Lobsters,
because
that is the
smallest
amount
that will
serve at
least 56% of
the
demand on
a given
night. Cumulative
Relative
Relative
Frequency Frequency
(cdf)
(pmf)
D emand
19
0.05
0.05
20
0.05
0.10
21
0.08
0.18
22
0.08
0.26
23
0.13
0.39
24
0.14
0.53
25
0.10
0.63
26
0.12
0.75
27
0.10
…
28
0.10
…
29
0.05
…
* pmf = prob. mass function Probability that demand
will be less than or equal to x P(D < 1 9 )
P(D < 2 0 )
P(D < 2 1 )
P(D < 2 2 )
P(D < 2 3 )
P(D < 2 4 )
P(D < 2 5 )
P(D < 2 6 )
P(D < 2 7 )
P(D < 2 8 )
P(D < 2 9 ) 12 Notation: Profit Maximization Q = order quantity
v = unit cost
A = ordering cost
p = selling price pv = profit per item sold g = salvage value (g < v) if g < 0, then storage, holding, or removal cost
v  g = loss per unsold item B = goodwill cost (lost sales penalty)
X = a random variable, the demand for the item
p(x) = probability function for X. 13 Model formulation: Profit Maximization For x ≤ Q:
Profit = px + g(Q  x)  A  vQ For x > Q:
Profit = pQ  B(x  Q)  A  vQ Q E ⎡ profit ⎤ = ∑ ⎡ px + g(Q − x) − A − vQ ⎤ p(x)
⎣
⎦
⎣
⎦
x =0 + ∞ ∑ x =Q +1
Q ⎡ pQ − B(x − Q ) − A − vQ ⎤ p(x)
⎣
⎦ = ∑ ⎡(p − g) x + gQ ⎤ p(x) +
⎣
⎦
x =0 ∞ ∑ x = Q +1 ⎡(p + B) Q − Bx ⎤ p(x) − A − vQ
⎣
⎦
14 Newsvendor Problem
Interpreting as Simple Newsvendor Problem:
– Cost of overage:
» Purchase cost  Salvage Value = v  g
– Cost of underage:
» Lost profit: Selling price – Purchase cost = p  v
» Or, p – v + (any additional penalty, B)
– Critical ratio is then:
» cu / (cu+co) = (pv+B)/(pg+B)
»For the discrete demand case take the smallest
value of Q* that satisfies the condition
P(D ≤ Q*) ≥ (p  v + B) / (p  g + B)
15 Single Period Inventory Model
Newsvendor Problem Example D = newspapers demanded
p(D) = probability of demand
Q = newspapers stocked
P = selling price of newspaper, $10
v = cost of newspaper, $4
g = salvage value of newspaper, $2
Cu = unit contribution: pv = $6
Co = unit loss: vg = $2
16 Single Period Inventory Model
Expected Value Analysis
p(D) D 6 7 Stock Q
8 9 10 .028
.055
.083
.111
.139
.167
.139
.111
.083
.055
.028 2
3
4
5
6
7
8
9
10
11
12 4
12
20
28
36
36
36
36
36
36
36 2
10
18
26
34
42
42
42
42
42
42 0
8
16
24
32
40
48
48
48
48
48 2
6
14
22
30
38
46
54
54
54
54 4
4
12
20
28
36
44
52
60
60
60 $34.43 $35.77 $35.99 Expected Profit $31.54 $35.33 17 THE SENTINEL NEWSPAPER
Management at Sentinel wishes to know how
many newspapers to put in its new vending
machine location.
Data
–
–
–
–
– Unit selling price is $0.30
Unit production cost is $0.38.
Advertising revenue is $0.18 per newspaper.
Unsold newspaper can be recycled and net $0.01.
Unsatisfied demand costs $0.10 per newspaper.
Demand distribution is
Discrete Uniform from
30 through 49 newspapers. 18 SOLUTION
Input to the optimal order quantity formula
p = 0.30
v = 0.20 [0.380.18]
g = 0.01
B = 0.10
p+ B  v
The optimal service level =
p+ B  g
= 0.30 + 0.10  0.20 = 0.513
0.30 + 0.10  0.01
19 Finding the optimal order quantity Q*
1.0 P(D ≤ 39) = 0.50
P(D ≤ 40) = 0.55
0.513
0.55
0.50 30 Q**= 40
Q = 40 39 40 49 20 Example
Demand for a critical component purchased by the Air Force
for use in the B2 bomber is Poisson with an estimated mean
of 12 over the life of the bomber fleet.
The manufacturer, LockheedGrumman, sells the part for
$5,000.
The unit production cost is $3,750.
Some of the parts found in the component may be salvaged
for $900.
The production line for this component will be shut down in
several months in order to make room for a new product
line.
How many units should LockheedGrumman produce in
order to maximize expected profits? Assume A = 0.
21 Example Solution CR = p − v 5000 − 3750
=
= .305
5000 − 900
p−g
X Pr{X≤x} Q* is smallest Q such that:
Q 8
9
10
11 .155
.242
.347
.462 ∑ p( x) ≥ .305
x =0 Q* = 10
22 Expected Profit
∞ E[P ] = ( g − v ) Q + ( p − g ) μ − ( p − g + B ) ∑ ( x − Q )p(x)
x =Q = ( 900 − 3750 )(10 ) + ( 5000 − 900 )(12 )
∞ − ( 5000 − 900 ) ∑ ( x − 10 ) e−12 (12 ) x =10 x x! x
⎡∞
e−12 (12) ⎤
⎥
= −28,500 + 49,200 − ( 4100 ) ⎢ ∑ ( x − 10 )
x! ⎥
⎢ x =10
⎣
⎦
Q
x
−12
⎡ e (12 )
Q e −12 12
⎛
( ) ⎞⎤
⎢μ
⎟⎥
= −28,500 + 49,200 − ( 4100 )
+ ( μ − Q ) ⎜1 − ∑
⎜ x =0
⎟⎥
Q!
x!
⎢
⎝
⎠⎦
⎣
= −28,500 + 49,200 − ( 4100 )( 2.56 ) = $10,204 23 Newsvendor : Continuous Demand Case
If demand is continuous, we’ll have:
Q ∞ 0 Q C(Q ) = co ∫ (Q − x)fX (x)dx + cu ∫ (x − Q )fX (x)dx
– Where fX(x) is the density function for X Check for the properties of the function:
Q
∞
dC(Q )
= co ∫ fX (x)dx − cu ∫ fX (x)dx
dQ
0
Q
= co PX ≤ (Q ) − cu (1 − PX ≤ (Q ))
d 2C(Q )
= (co + cu )fX (Q ) > 0
dQ 2 24 Newsvendor : Continuous Demand Case
Liebnitz’s Rule to take the derivative of an
integral d ⎡b(Q)⎤
d ⎡a(Q)⎤
d
dg(x,Q)
⎣
⎦
⎣
⎦
∫Q) g(Q,x)dx = dQ g(Q,b(Q)) − dQ g(Q,a(Q)) + a(∫Q) dQ dx
dQ a(
b(Q ) b(Q ) Since the second derivative of C(Q) is positive,
C(Q) is convex dC(Q )
= co PX ≤ (Q ) − cu (1 − PX ≤ (Q ))
dQ
= (cu + co )PX ≤ (Q ) − cu ⇒ PX ≤ (Q *) = =0 cu
cu + c o 25 An example:
Demand in the period is uniformly distributed
over [50,150].
cu=$10, co=$7 PX ≤ (x) = x − 50
for 50 ≤ x ≤ 150
100 Q − 50 10
=
100
17
10
⇒ Q * = 50 + 100
≅ 109
17 Solve PX ≤ (Q ) = 26 Example
The Hotel Holiday rents rooms for $100 a night.
The Operations Research Institute is holding a
conference at the Hotel Holiday in two months. They
have negotiated a special rate of $70 a night for those
attending the conference.
Empty rooms cost the hotel $25.
The demand for regular rooms is exponential with a
mean of 140 per night.
How many rooms should be kept at the regular rate in
order to maximize profit? All rooms at the lower rate
can be rented.
27 Example Solution CR =
Q Cu
30
=
= .5455
Cu + Co 30 + 25
Q ∫ f(x)dx = ∫
0 e 0 Q
−
140 e − x μ μ dx = 1 − e − Q μ = .5455 = 1 − .5455 Q * = −140ln (.4545 ) = 110.384 → 110
28 Continuous Demand case: Max Profit
For the continuous case: E[ profit] = ∫ Q 0 ⎡(p − g) x + gQ ⎤ f(x)dx
⎣
⎦
∞ + ∫ ⎡(p + B)Q − Bx ⎤ f (x) dx − A − vQ
⎦
Q⎣
Q Q = gQ ∫ f (x)dx + (p − g)∫ xf (x)dx
0 0 ∞ ∞ Q Q +(p + B)Q ∫ f (x)dx − B ∫ xf (x)dx − A − vQ
29 Continuous Demand case: Max Profit
∞ ∞ 0 Q = gQ ∫ f (x)dx − gQ ∫ f (x)dx
∞ ∞ 0 Q +(p − g)∫ xf (x)dx − (p − g)∫ xf (x)dx
∞ ∞ Q Q +(p + B)Q ∫ f (x)dx − B ∫ xf (x)dx − A − vQ
∞ = gQ + (p − g)μ + ( p + B − g)∫ Qf ( x)dx
Q ∞ −(p + B − g)∫ xf (x)dx − A − vQ
Q 30 Continuous Demand case: Max Profit
∞ = gQ + (p − g)μ + (p + B − g)∫ Qf (x)dx
Q ∞ −(p + B − g)∫ xf (x)dx − A − vQ
Q ∞ = (g − v ) Q + (p − g)μ − A − (p + B − g)∫ (x − Q )f (x)dx
Q 31 Continuous Demand case: Max Profit
∞
d {.}
= (g − v ) − (p + B − g)∫ − f (x)dx = 0
Q
dQ
∞
v−g
f (x)dx =
∫Q
Solving:
Critical
p+B−g ratio Loss per unsold item v−g
v−g
=
p + B − g (p − v ) + B + (v − g)
Item profit + goodwill PX ≤ (Q * ) = p−v +B
p−g+B 32 Example
The historical demand for Christmas wreaths sold
during the annual Boy Scout fundraising drive has
the following PDF:
⎧1
⎪
f ( x) = ⎨ 50
⎪0
⎩ 50 ≤ x ≤ 100
elsewhere Wreaths sell for $35 each and cost the Boy Scouts
$20.
A goodwill cost of $2 is assumed as a matter of
policy by the National Boy Scout Fund Raising
Committee.
It cost $1.00 per wreath for disposal. 33 Example Solution 50 1 − F (Q ) = Q 100 100 − Q
20 − (−1
)
21
v −g
=
=
=
50
)
p + B − g 35 + 2 − (−1 38
⎛ 21 ⎞
100 − Q = 50 ⎜ ⎟
⎝ 38 ⎠
⎛ 21 ⎞
Q * = 100 − 50 ⎜ ⎟ = 72.368 ≈ 72
⎝ 38 ⎠ 34 Expected Profit
E ⎡Profit ⎤ = (g − v ) Q + (p − g)μ
⎣
⎦
∞ − A − (p + B − g)∫ (x − Q )f (x)dx
Q = ( −1 − 20 ) 72 + ( 35 + 1) 75 − A − ( 35 + 2 + 1) ∫ 100 72 = −1512 + 2700 − ( 38 ) (100 − 72 ) (x − 72)
dx
50 2 − A = 890.08 − A
100
= −1512 + 2700 − 38(7.84) − A = 890.08 − A
Order Q if A − Bμ < 890.08
Otherwise, order nothing
35 Normally Distributed Demand
The optimality equation can be written as:
Φ( Q −μ σ )= cu
= Pu≤(k)
cu + co When the demand is normally distributed, expected cost
E(Q) can be written as: E(Q ) = co (Q − μ) + (co + cu )σ Gu ( k ) Expected profit equation may be written as: E ⎡Profit⎤ = (g − v) Q + (p − g)μ
⎣
⎦
− A − (p + B − g)σ Gu (k)
36 Normally Distributed Demand
Example: X~N(100, 36),
μ = 100, σ = 6
Let’s find the optimal order quantity and optimal
expected cost for cu=$10 and c0=$2.5.
k is the 10/12.5 percentile of standard normal
From Table B.1 of appendix B, for Pu≥(k)=10.8=0.2,
k is approximately 0.84.
We also read Gu(k) as 0.1120.
Q*=100+(0.84)(6)=105.04
E(Q*)=2.5(5.04)+12.5(6)(0.1120)=$21 37 Example 2
Demand for the UD daily newspaper at the
campus newsstand is normal with a mean of
100 and a standard deviation of 15.
Newspapers sell for 25 cents and costs 10
cents to produce.
Goodwill costs are assumed to be zero and
there is a 2 cents salvage value. 38 Example 2 Solution ∞ ∫ f(x)dx = Q v−g
.10 − .02
=
= .3478
p + B − g .25 − .02
Pu>(k) = .3478
k = .39
100 Q Q* = μ + k σ = 100 + 15(.39) = 105.85 ≈ 106
39 MultiItem Newsvendor Problem
n: number of different items
fi(x)=probability density function for item i
Fi(x)=cdf for item i
cio: overage cost for item i
ciu: underage cost for item i
W: available budget
We want to minimize
C(Q 1,..., Q n ) =
subject to n Qi ∞ i =1 0 Qi ∑ { cio ∫ (Q i − x ) fi ( x )dx + ciu ∫ ( x − Q i ) fi ( x )dx } n ∑vQ
i =1 i i ≤W 40 MultiItem Newsvendor Problem
Very similar to multiitem EOQ under constraints
First check unconstrained optimum.
Qi
∂C
= −ciu + (ciu + cio ) ∫ fi (x)dx = 0
∂Q i
0 ⇒ Fi (Q i ) = ciu
ciu + cio If ∑ v i Q i ≤ W then we are done.
Otherwise, form the Lagrangian function.
41 MultiItem Newsvendor Problem
n Qi ∞ i =1 0 Qi L(Q1,...,Q n, M) = ∑{cio ∫ (Qi − x)fi (x)dx + ciu ∫ (x − Qi )fi(x)dx } + M(∑viQi − W) Taking derivatives with respect to (Q1,…, Qn, M) :
−ciu + (ciu + cio )Fi (Q i ) + Mv i = 0 for i = 1,2,...,n ∑v Q
i i −W = 0 Usually requires numerical methods to solve. 42 MultiItem Newsvendor Problem p
v
g
B
x hat
sigma Item 1
Item 2
Item 3
Item 4
Total
50.3
40
32
6.1
35.1
15
28
4.8
25
12.5
15.1
2
9.7
0
10.3
1.5
900
800
1200
2300
122
200
170
200 M
p(k)
k
Q
G(k)
E[P(Q)]
Qv 0.509306 0.50930629 0.50930629 0.50930629
0.7993
0.3687
0.9986
0.9365
0.8392
0.3353
2.9783
1.5264
798
867
694
1995
0.9514
0.2535
2.9787
1.5540
$ 10,652 $
18,438 $
(2,442) $
2,104 $
$ 27,996 $
13,006 $
19,423 $
9,575 $ W = Budget $ 70,000 Goal Seek to set Qv total to W by changing M. 28,752
70,000 43 MultiItem Newsvendor Problem
M
1.667
1.567
1.467
1.367
1.267
1.167
1.067
0.967
0.867
0.767
0.667
0.567
0.467
0.367
0.267
0.167
0.067
0.033
0.133 Qv
$
70,000
$
33,572
$
40,760
$
41,875
$
42,631
$
43,235
$
43,756
$
44,228
$
44,668
$
45,090
$
45,502
$
56,801
$
62,006
$
77,137
$
82,319
$
85,993
$
89,289
$
92,574
$
96,191
$ 101,116 E[P(Q)]
$
28,752
$
(5,699)
$
7,611
$
9,307
$
10,380
$
11,177
$
11,812
$
12,339
$
12,787
$
13,174
$
13,511
$
21,456
$
24,615
$
32,291
$
34,477
$
35,647
$
36,363
$
36,746
$
36,803
$
36,373 44 Multiple Period Problems
Difficulty: Technically, Newsvendor model is for a
single period.
Extensions: But Newsvendor model can be applied
to multiple period situations, provided:
– demand during each period is independent and identically
distributed
– amount ordered simply the difference between Q* and the
present backorder or positive inventory level Key: make sure co and cu appropriately represent
overage and shortage cost.
45 Example
Scenario:
–
–
–
– GAP orders a particular clothing item every Friday
mean weekly demand is 100, std dev is 25
wholesale cost is $10, retail is $25
holding cost has been set at $0.5 per item per week (to
reflect obsolescence, damage, etc.) Problem: how should they set order amounts? 46 Example (cont.)
Newsvendor Parameters:
co = $0.5
cu = $15
15
= 0.9677
0.5 + 15
⎛ Q − 100 ⎞
Φ⎜
⎟ = Pu ≤ (k) = 0.9677
⎝ 25 ⎠
Every Friday, they should
orderupto 146, that is, if
Q − 100
= 1.85
there are x on hand, then
25
order 146x.
Q = 100 + 1.85(25) = 146
47 Solution: CR = Newsvendorlike Problem
Problem
Condo Construction Company is bidding on an
important construction job.
The job will cost $2 million to complete.
One other company is bidding for the job.
Condo believes that the opponent’s bid is equally likely
to be an amount between $2 million and $4 million.
If Condo wants to maximize expected profit, what
should its bid be? 48 Newsvendorlike Problem
Problem
Let
B = random variable representing bid of Condo’s
opponent
b = actual bid of Condo’s opponent
q = Condo’s bid
If b > q, Condo outbids the opponent and earns a profit
of q – 2,000,000.
If b < q, Condo is outbid by the opponent and earns
nothing. 49 Newsvendorlike Problem
Problem
Let E(q) be Condo’s expected profit if it bids q. Then E(q) = (0)∫ q 2,000,000 f (b)db +(q − 2,000,000)∫ 4,000,000 q f (b)db Since f(b) = 1/2,000,000 for 2,000,000 ≤ b ≤ 4,000,000,
we obtain E(q) = (q − 2,000,000)(4,000,000 − q)
2,000,000
50 Newsvendorlike Problem
Problem
To find the value of q maximizing E(q), we find
E ′(q) = −(q − 2,000,000) + (4,000,000 − q) 6,000,000 − 2q
=
2,000,000
2,000,000 Hence, E ′(q) =0 for q=3,000,000. We know that E (q)
is a concave function of q, and 3,000,000 does indeed
maximize E (q).
Hence, Condo should bid $3 million. Condo’s expected
profit will be $500,000.
51 Impact of Supply Chain Contracts
on Profitability: Buyback Contracts
Buyback
Contracts
Specifies the parameters within which a buyer places
orders and a supplier fulfills them
Example parameters: quantity, price, time, quality
Double marginalization:
– buyer and seller make decisions acting independently
instead of acting together
– gap between potential total supply chain profits and
actual supply chain profits results Buyback contracts can be offered that will increase total
supply chain profit
52 Returns Policy:
Buyback Contracts
A manufacturer specifies a wholesale price and a buyback
price at which the retailer can return any unsold items at
the end of the season
Results in an increase in the salvage value for the
retailer
– which induces the retailer to order a larger quantity The manufacturer is willing to take on some of the cost
of overstocking (market risk) because the supply chain
will end up selling more on average
Manufacturer profits and supply chain profits can
increase
53 Returns Policy:
Buyback Contracts
An example with high demand variability
v = manufacturer production cost = $50
c = wholesale price to retailer = $250
p = retail selling price = $300
sR = retail clearance price = $100 Retailer :
Pu ≤ (k) = cu
(300 − 250)
=
= .25
cu + co (300 − 250) + (250 − 100) k = −.6745 μ = 5000 σ = 3000
Q * = 5000 − .6745(3000) = 2977 = Manufacturer's Sales
54 Returns Policy:
Buyback Contracts
Retailer’s profit: E(PR (k)) = ( p − s)μ − (c − s)Q * −( p − s)σ x Gu ( k ) where Q * = μ + kσ E (PR (−.6745)) = (300 − 100)(5000) − (250 − 100)(2977)
−(300 − 100)(3000)(.8236)
= 59,334 Manufacturer’s profit E(PM (Q *) = (c − v )Q * = (250 − 50)(2977) = 595,400 Manufacturer makes significantly more profit, but
faces no market risk
55 Returns Policy:
Buyback Contracts
Independent
decisions by
Manufacturer
and Retailer Inputs
Mean Demand, Mu
Standard Deviation of demand
Retailer's Cost, c =
Retailer's Sale price, p =
Retailer's Salvage value, s=b =
Retailer's Return cost =
Mfg's Cost, v
Mfg's Sale price, c
Mfg's Buyback Price, b
Mfg's Salvage Value, sm $
$
$
$
$
$
$
$ 5000
3000
250
300
100
50
250
 Intermediate Calculations
Cost of Understocking, Cu
Cost of Overstocking, Co $
$ 50
150 Outputs
Order size, O*
overstock
understock
Retailer's Expected Profit
Manufacturer's E(Profit)
Total Supply Chain Profit = 2977
447
2471
$ 59,334
$ 595,306
56
$ 654,640 Returns Policy:
Buyback Contracts
Product Availability maximizing supply chain profit
– solving single period problem from a supply chain
perspective
– shift some overstock risk to manufacturer so that retailer
will order more
– assume that manufacturer can salvage overstock for
$20/unit ( sM ) co = v − s M = 50 − 20 = 30
cu = p − v = 300 − 50 = 250
CR = cu
250
=
= .8929 → k* = 1.2419
cu + co 280 Q * = 5000 + (1.2419)(3000) = 8726 57 Returns Policy:
Buyback Contracts
Define
– b = price manufacturer will pay to buy back remaining
retailer inventory at the end of the selling period
– sM = salvage value of any items the manufacturer buys
back Expected manufacturer’s profit under buyback
= Q * (c − v) − (b − sM )E[Retailer overstock]
Q* E[Retailer overstock] = ∫ (Q * −x)f(x)dx = (Q * −μ)Pu≤(k) + σ fu (k)
0 = σ (k + Gu (k))
Q * = μ + kσ 58 Returns Policy:
Buyback Contracts
Determining buyback price that maximizes supply chain
profit
– Retailer CR must match Supply Chain CR CR* = .8929 = cu
300 − 250
=
cu + co (300 − 250) + (250 − b) b = 244 b must be greater than sR to provide any incentive for
the retailer to order a larger amount 59 Buyback
Contracts
“Breakeven”
buyback price Inputs
Mean Demand, Mu
Standard Deviation of demand
Retailer's Cost, c =
Retailer's Sale price, p =
Retailer's Salvage value, s=b =
Retailer's Return cost =
Mfg's Cost, v
Mfg's Sale price, c
Mfg's Buyback Price, b
Mfg's Salvage Value, sm $
$
$
$
$
$
$
$ 5000
3000
250
300
100
50
250
 $
$
$
$
$
$
$
$ Intermediate Calculations
Cost of Understocking, Cu
Cost of Overstocking, Co $
$ 50
150 $
$ Outputs
Order size, O*
overstock
understock
Retailer's Expected Profit
Manufacturer's E(Profit)
Total Supply Chain Profit = 2977
447
2471
$ 59,334 $
$ 595,306 $
$ 654,640 $ 5000
3000
250
300
116
50
250
116.00
20.00
50
134 3177
500
2323
66,898
587,480
60
654,378 Buyback
Contracts Inputs
Mean Demand, Mu
Standard Deviation of demand
Retailer's Cost, c =
Retailer's Sale price, p =
Retailer's Salvage value, s=b =
Retailer's Return cost =
Mfg's Cost, v
Mfg's Sale price, c
Mfg's Buyback Price, b
Mfg's Salvage Value, sm Less demand
uncertainty
…less
benefit from
buybacks $
$
$
$
$
$
$
$ $
$ 50
150 $
$ Outputs
Order size, O*
overstock
understock
Retailer's Expected Profit
Manufacturer's E(Profit)
Total Supply Chain Profit = Buyback
Contracts $
$
$
$
$
$
$
$ Intermediate Calculations
Cost of Understocking, Cu
Cost of Overstocking, Co Buyback
price
optimizing
Supply Chain
profit 5000
3000
250
300
100
50
250
 5000
3000
250
300
244
50
250
244.00
20.00 2977
8726
447
3880
2471
154
$ 59,334 $ 219,002
$ 595,306 $ 876,010
61
$ 654,640 $ 1,095,012 50
6 Mean Demand, Mu
Standard Deviation of demand
Retailer's Cost, c =
Retailer's Sale price, p =
Retailer's Salvage value, s=b =
Retailer's Return cost =
Mfg's Cost, v
Mfg's Sale price, c
Mfg's Buyback Price, b
Mfg's Salvage Value, sm $
$
$
$
$
$
$
$ 5000
300
250
300
100
50
250
 Intermediate Calculations
Cost of Understocking, Cu
Cost of Overstocking, Co $
$ 50 $
150 $ Outputs
Order size, O*
overstock
understock
Retailer's Expected Profit
Manufacturer's E(Profit)
Total Supply Chain Profit = 4798
5373
45
388
247
15
$ 230,933 $ 246,900
$ 959,531 $ 987,601
$ 1,190,464 $ 1,234,501 $
$
$
$
$
$
$
$ 5000
300
250
300
244
50
250
244.00
20.00
50
6 62 Buyback
Contracts
Restricted
manufacturer
margin…
buyback
contract can
not be
justified Mean Demand, Mu
Standard Deviation of demand
Retailer's Cost, c =
Retailer's Sale price, p =
Retailer's Salvage value, s=b =
Retailer's Return cost =
Mfg's Cost, v
Mfg's Sale price, c
Mfg's Buyback Price, b
Mfg's Salvage Value, sm $
$
$
$
$
$
$
$ 5000
3000
250
300
100
200
250
 Intermediate Calculations
Cost of Understocking, Cu
Cost of Overstocking, Co $
$ 50 $
150 $ $
$
$ 2977
447
2471
59,334 $
148,827 $
208,161 $ Outputs
Order size, O*
overstock
understock
Retailer's Expected Profit
Manufacturer's E(Profit)
Total Supply Chain Profit = $
$
$
$
$
$
$
$ 5000
3000
250
300
160
200
250
160.00
20.00
50
90 3902
727
1825
93,305
93,305
186,611 63 ...
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This note was uploaded on 10/24/2011 for the course IE 351 taught by Professor Wil during the Spring '08 term at Lehigh University .
 Spring '08
 Wil

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