Advanced Inventory Planning PDF

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Logistics Management Prof. Alessandro Perego Politecnico di Milano

Exercises on Advanced Inventory Planning (collection of exercises from past examinations)

1

Exercise 1 In the distribution centre of FOODSTUFFS S.p.A., the service level (intended as item fill rate) for the coffee AROMA in 250 g packages has been 98% in the last two years. Assuming the following: - reorder quantity equal to 18 pallet loads, each of which made by 2,000 sales units; - the replenishment lead time follows a normal distribution with average equal to one working week and standard deviation equal to 2 days; - the value of a pallet load is equal to 4,100 €/pallet load; - the weekly demand follows a normal distribution with average equal to 12,000 sales units and standard deviation equal to 4,000 sales units; - stock out cost equal to 0.15 €/sales unit; - annual storage cost equal to 77.5 €/pallet load; - cost of capital equal to 10%; you are required to assess if it is advantageous to increase the service level to 99%. N.B.: assume a year made by 52 working weeks and a week made by 5 working days. k

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

1.21

1.22

I(k) 0.351 0.307 0.267 0.230 0.198 0.169 0.143 0.120 0.100 0.083 0.069 0.056 0.055 0.054 k

1.25

I(k)

0.05

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2.0

2.1

2.2

2.3

2.4

2.5

0.045 0.037 0.029 0.023 0.018 0.014 0.011 0.008 0.006 0.005 0.004 0.003 0.002 Table 1 – Values for the unit normal loss integral

Exercise 2 The distribution network of a household cleaner manufacturer is made by two levels with a central warehouse (CW) and three regional warehouses (RW1, RW2 and RW3), as per Figure 1. For the item LINDO, the inventory management policy in the central warehouse is a periodic review policy with an order interval of 30 days and a replenishment lead time which follows a normal distribution with average of 10 days and standard deviation equal to 2 days. The regional warehouses work with a periodic review policy with order interval equal to one week and replenishment lead time from the central warehouse equal to 1 day for RW1, 1.5 days for RW2 e 2 days for RW3. Assuming the following: − customer weekly demand following a normal distribution whose average and standard deviation are equal to: RW1=250 units/week, σ1=30 units/week; RW2=350 units/week, σ2=50 units/week; RW3=320 units/week, σ3=60 units/week; − there is no correlation among the demand seen by the regional warehouses; − the service level (intended as the probability to avoid the stock out in the replenishment cycle) equal to 95% for every warehouse; you are requested to assess the amount of inventory (both cycle and safety stocks) for the item LINDO in the distribution network considering a safety stock allocation with a “coupled system”.

Figure 1 – Distribution network structure

k

1.55

1.60

1.65

1.70

1.75

1.8

1.85

1.9

1.95

2.0

2.05

2.1

PSO 0.9394 0.9452 0.9505 0.9554 0.9599 0.9641 0.9678 0.9713 0.9744 0.9772 0.9798 0.9821 Table 2 – Cumulative distribution function of the standard normal distribution

2

Exercise 3 The distribution network of DE.TER.GI S.p.A. is a two level distribution network, made of a central warehouse (CW), that replenishes several important customers as well as two regional warehouses (RW1 e RW2), as reported in Figure 2. For the product family AZ5 (moquette detergents, in 10 dm3 drums) the inventory management policy in the CW is a reorder point policy, with replenishment lead time equal to 10 days. The regional warehouses work with a periodic review policy with an order interval of 2 weeks and replenishment lead time from the CW equal to 4 days. Assuming the following: - weekly demand to the regional warehouses following a Normal distribution: RW1=100 pieces, σ"#$ = σRW2=20*pieces; σCWd=30 pieces; - there is a linear correlation between the demand as seen by the two regional warehouses (Correlation coefficient rRW1,RW2 = 0.7), between the demand as seen by the RW1 and the demand directly shipped from the CW (correlation coefficient rRW1,CWd = 0.6), between the demand as seen by the RW2 and the demand directly shipped from the CW (correlation coefficient rRW2,CWd = 0.8); − the service level, intended as the probability to avoid the stock out in the replenishment cycle time, equal to 95% for every warehouse, you are requested to calculate the amount of safety stock in case of coupled and independent system.

Figure 2 – Distribution network structure

N.B. assume 5 working days a week and 46 working weeks a year. k

1.55

1.60

1.65

1.70

1.75

1.8

1.85

1.9

1.95

2.0

2.05

2.1

PSO 0.9394 0.9452 0.9505 0.9554 0.9599 0.9641 0.9678 0.9713 0.9744 0.9772 0.9798 0.9821 Table 3 – Cumulative distribution function of the standard normal distribution

3

Exercise 1 It is required to assess the inventory costs (related to Safety Stock only) and stockout costs in the two options: A. IFR = 98% B. IFR = 99% Alternative A (IFR = 98%) >

; 𝐿𝑇 ∙ 𝜎4; + 𝐷; ∙ 𝜎67 =

−

𝜎4,67 =

−

𝐼 𝑘 = 1 − 𝐼𝐹𝑅 ∙ U

T

V,WX

>

∙ 4,000; + 12,000; ∙ $L

= 1 − 0.98 ∙

Z.$;

;; >

= 6,248.2*𝑢𝑛𝑖𝑡𝑠 →

;,MMM

= 3.12*𝑃𝐿*

= 0.115 → 𝑘 = 0.9*

−

𝑆𝑆 = 𝑘 ∙ 𝜎4,67 = 0.9 ∙ 3.12 = 2.8*𝑃𝐿*

−

𝐼𝐶 𝑆𝑆 = 2.8 ∙ 4,100 ∙ 0.1 + 77.5 = 1,365€/𝑦𝑒𝑎𝑟* 𝑆𝑡𝑜𝑐𝑘𝑜𝑢𝑡*𝐶𝑜𝑠𝑡 = 12,000 ∙ 52 ∙ 1 − 0.98 ∙ 0.15 = 1,872*€/𝑦𝑒𝑎𝑟 *

−

J,;KL.;

Alternative B (IFR = 99%) 𝐼 𝑘 = 1 − 𝐼𝐹𝑅 ∙

𝑄 𝜎4,67

= 1 − 0.99 ∙

18 = 0.058 → 𝑘 = 1.2 * 3.12

𝑆𝑆 = 𝑘 ∙ 𝜎4,67 = 1.2 ∙ 3.12 = 3.74*𝑃𝐿* 𝐼𝐶 𝑆𝑆 = 3.74 ∙ 4,100 ∙ 0.1 + 77.5 = 1,823*€/𝑦𝑒𝑎𝑟* 𝑆𝑡𝑜𝑐𝑘𝑜𝑢𝑡*𝐶𝑜𝑠𝑡 = 12,000 ∙ 52 ∙ 1 − 0.99 ∙ 0.15 = 936*€/𝑦𝑒𝑎𝑟 * Therefore, the alternative B yields to an overall lower cost. Exercise 2 In a coupled system the total amount of Safety Stock is the sum of Safety Stock stored in each RW: 𝑆𝑆hij = 𝑘k ∙

𝑇li + 𝐿𝑇li + 𝑇hij + 𝐿𝑇hij ∙ 𝜎4j

;

+ 𝐷k; ∙ 𝜎 ;

7mno67 mno7pn o67pn j

*

j

Provided that k is 1.65 for all three RW, the value of SS can be assessed as: 𝑆𝑆hi

𝑆𝑆hi

𝑆𝑆hi

q

r

= 1.65 ∙

= 1.65 ∙

s

= 1.65 ∙

2 30 + 10 + 1 + 1 ∙ 30; + 250; ∙ 5 5 2 30 + 10 + 1.5 + 1 ∙ 50; + 350; ∙ 5 5 2 30 + 10 + 2 + 1 ∙ 60; + 320; ∙ 5 5

;

= 223.1*𝑢𝑛𝑖𝑡𝑠 * ;

= 341.6*𝑢𝑛𝑖𝑡𝑠* ;

= 369.8*𝑢𝑛𝑖𝑡𝑠 *

For a total value of SS in the system equal to 934.5 units. As regards Cycle Stock, they are equal to: 250 + 350 + 320 ∙ 30 5 = 2,760*𝑢𝑛𝑖𝑡𝑠* 2 2 𝐷i,hi$ ∙ 𝑇hi$ 250 ∙ 1* = 125*𝑢𝑛𝑖𝑡𝑠 * 𝐶𝑆hi$ = = 2 2 𝐷i,hi; ∙ 𝑇hi; 350 ∙ 1 = 175*𝑢𝑛𝑖𝑡𝑠 * 𝐶𝑆hi; = = 2 2 𝐷i,hiZ ∙ 𝑇hiZ 320 ∙ 1 = 160*𝑢𝑛𝑖𝑡𝑠 * 𝐶𝑆hi$ = = 2 2 For a total value of 3,220 units. 𝐶𝑆li =

Z 𝐷i,hij kt$

∙ 𝑇li

=

Exercise 3 Coupled system In a coupled system, the total amount of Safety Stock is the sum of Safety Stock stored in each RW (plus, in this case, the SS related to the direct shipment from the CW, stored directly in the CW itself):

4

𝑆𝑆k = 𝑘k ∙

𝐿𝑇li + 𝑇hij + 𝐿𝑇hij ∙ 𝜎4j

;

+ 𝐷k; ∙ 𝜎 ;

67mno7pn o67pn j

*

Provided that k is 1.65 for all three RW, the value of SS can be assessed as: 𝑆𝑆hi

4 10 +2+ ∙ 25; + 100; ∙ 0; = 90.4*𝑝𝑖𝑒𝑐𝑒𝑠* 5 5

= 1.65 ∙

q

𝑆𝑆hi

r

= 1.65 ∙

4 10 +2+ ∙ 20; = 72.3*𝑝𝑖𝑒𝑐𝑒𝑠 * 5 5

𝑆𝑆li = 1.65 ∙

10 ∙ 30; = 70.0*𝑝𝑖𝑒𝑐𝑒𝑠 * 5

For a total value of SS in the system equal to 232.7 pieces. Independent system In an independent system, the total amount of Safety Stock is the sum of Safety Stock stored in each RW and in the CW: 𝑆𝑆hij = 𝑘k ∙

(𝑇hij + 𝐿𝑇hij ) ∙ 𝜎4j

𝑆𝑆li = 𝑘k ∙ 𝐿𝑇li ∙ 𝜎4mn;

;

+ 𝐷k; ∙ 𝜎7;pn o*67pn * j

; ; + 𝐷li ∙ 𝜎67 * mn

Where: x

𝜎4; mn

x|$

𝜎4;j

= kt$

x

𝜌kz ∙ 𝜎4j ∙ 𝜎4{

+2∙ kt$ ztko$ ; ;

= 30 + 25 + 20; + 2 ∙ 0,6 ∙ 30 ∙ 25 + 0,8 ∙ 30 ∙ 20 + 0,7 ∙ 25 ∙ 20 = 4,485*𝑝𝑖𝑒𝑐𝑒𝑠 ;* Therefore: 𝑆𝑆hi

q

= 1.65 ∙

𝑆𝑆hi

r

10 + 4 ∙ 25; + 100; ∙ 0; = 69.0*𝑝𝑖𝑒𝑐𝑒𝑠 * 5

= 1.65 ∙

𝑆𝑆li = 1.65 ∙

10 + 4 ∙ 20; = 55.2*𝑝𝑖𝑒𝑐𝑒𝑠* 5 10 ∙ 4,485 = 156.3*𝑝𝑖𝑒𝑐𝑒𝑠* 5

For a total value of SS in the system equal to 280.5 pieces.

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