pi [i( p c) (10 i)(c s)] [1 P 10)( p c) 10 ](
i 4 10
17
Expected profit if order k
pi [i( p c) (k i)(c s)] [1 Pk ](k )( p c)

High availability =>


larger inventory higher costs risk of obsolescence
12
Newsboy Model Newsvendor

single period model (one selling season)
(one-time order, e.g. for quantity discount)
L.L. Bean Example – Demand Distribution
Table 13-1 Demand Di (in hundreds) 4 5 Probability pi 0.01 0.02 Cumulative Probability of Demand Being Di or Less (Pi) 0.01 0.03 Probability of Demand Being Greater than Di 0.99 0.97
1 more unsold
Additional contribution
0
order k+1 instead of k
Co
Cu
1 fewer lost sale
Order k+1 instead of k if Pr(D>k) Cu Pr(D<k) (Co) > 0 or Pr(D< k ) (Co) + [1-Pr(D<k)] Cu > 0
Copyright ©2013 Pearson Education. 13-15
Example: Selling parkas at LL Bean
Cost per parka = c = $45 Sale price per parka = p = $100 Inventory holding (until season end) and transportation cost (to outlet store) per parka = $10 Discount price per parka (season end sales) = $50 Salvage value per parka = $50 -$10 = $40 = s
Demand d(i) 4 5 6 7 8 9 10 11 12 13 14 15 16 Probability p(i) 0.01 0.02 0.04 0.08 0.09 0.11 0.16 0.20 0.11 0.10 0.04 0.02 0.01 Cumulative Prob. P(i) = Pr( D < d(i) ) 0.01 0.03 0.07 0.15 0.24 0.35 0.51 0.71 0.82 0.92 0.96 0.98 0.99
n r 0
售出n, 赚(a b)n
G(n) [( a b)r (b c)( n r )] f (r )
求n使G(n)最大
r n 1
(a b)nf (r )

求 解
将r视为连续变量
n
f (r ) p(r ) (概率密度)

G(n) 0 [( a b)r (b c)( n r )] p(r )dr n (a b)np(r )dr
i 4 k
18
LLBean: Expected profit
Demand d(i) 4 5 6 7 8 9 10 11 12 13 14 15 16 Probability p(i) 0.01 0.02 0.04 0.08 0.09 0.11 0.16 0.20 0.11 0.10 0.04 0.02 0.01 Sum(d(i)xp(i)) Cumulative Prob. Prob. demand greater Expected profit P(i) = Pr( D < d(i) ) 1 - P(i) = Pr( D > d(i) ) if stock d(i) 0.04 0.01 0.99 220.00 0.14 0.03 0.97 274.40 0.38 0.07 0.93 327.60 0.94 0.15 0.85 378.40 1.66 0.24 0.76 424.40 2.65 0.35 0.65 465.00 4.25 0.51 0.49 499.00 6.45 0.71 0.29 523.40 7.77 0.82 0.18 535.80 9.07 0.92 0.08 541.60 9.63 0.96 0.04 541.40 9.93 0.98 0.02 538.80 10.09 0.99 0.01 535.00
3、模型的建立与求解
准 备 建 模
调查需求量的随机规律——每天需求量为 r 的 概率 f(r), r=0,1,2…
• 设每天购进 n 份，日平均收入为 G(n)
• 已知售出一份赚 a-b；退回一份赔 b-c
rn
售出r , 退回n r
赚(a b)r, 赔(b c)(n r )
r n
Cost of overstocking = Co = $45 + $10 - $50 = c - s = $ 5 Marginal profit from selling parka = cost of understocking = Cu = $100 - $45 = p - c = $55
16
6
7 8 9 10 11 12 13 14
0.04
0.08 0.09 0.11 0.16 0.20 0.11 0.10 0.04
0.07
0.15 0.24 0.35 0.51 0.71 0.82 0.92 0.96
0.93
0.85 0.76 0.65 0.49 0.29 0.18 0.08 0.04


demand uncertainty order placed (and delivered) before demand is known unmet demand is lost unsold inventory at the end of the period is discard (or salvaged at lower value)

Understocking cost Cu


Customer/Cycle service level CSL

Determining Optimal Level of Product Availability
• Single period • Possible scenarios
– Seasonal items with a single order in a season – One-time orders in the presence of quantity discounts – Continuously stocked items – Demand during stockout is backlogged – Demand during stockout is lost
n dG (a b)np(n) 0 (b c) p(r )dr (a b)np(n) dn
n (a b) p(r )dr (b c) 0 p(r )dr (a b) n p(r )dr
n

dG 0 dn
p (r )dr a b p (r )dr b c
报童问题的推广与应用：
多产品报童问题； 考虑风险偏好的报童问题； 基于需求预测的报童问题； 考虑采购提前期的报童问题；
Product Availability: Tradeoffs

High availability =>

responsive to customers attract increased sales higher revenue
19
Newsvendor : Marginal Analysis
Stock one unit if … Stock 2 units (instead of 1 unit) if ...
Stock 1
D=0 D=1
Stock 2
Stock 3
D=2
D=3
20
keep order size at k
How much to order?
13
Factors affecting availability


Demand uncertainty Overstocking cost C0

= loss incurred when a unit unsold at end of selling season = profit margin lost due to lost sale (because no inventory on hand) =level of product availability = Prob(Demand < stock level) 14
报童问题模型
1、报童问题的提出
2、报童问题所属范畴
3、报童模型的建立与求解
的提出
在日常生活中，经常会碰到一些季节性强、更新快、不易保 存等特点的物品，如海产、山货、时装、生鲜食品和报纸等，当 商店购进这些商品时，买的数量越多，价格越便宜获利越大。但 买得太多也可能卖不出去，需要削价处理，人力物力都受损；如 果进货太少，又可能发生缺货现象，失去销售机会而减少利润。