1 Analysis of Inventory Models Special topic # 1 ~ Multiple products EPQ analysis ~

2 §. Inventory Management for Finite Production Rate Inventory Levels for Finite Production Rate Model Slope=P- λ Slope= - λ

3 P : production rate (per unit time) λ: demand rate (per unit time) P > λ

4 ………... [Eq.1] §. Finite Production Rate Model

5 ~ EPQ (Finite Production Rate ) Models for Production Planning ~ ◆ Producing n products on a single machine (or line) ◆ Goals ： (1) To determine the optimal procedure for producing n products on the machine to minimize the cost of holding & setups. (2) To guarantee that no stock-outs occur during the production cycle. Demand rate for product j Production rate for product j Holding cost per unit per unit time for product j Cost of setting up the production facility to produce product j ‧

6 ◆ Assumption (required) ◇ To ensure that the facility has sufficient capacity to satisfy the demand for all products. ~ Multiple items production planning ~ ………... [Eq.2]

7 ◆ Rotation cycle policy assumption : means that in each cycle there is exactly one setup for each product, and products are produced in the same sequence in each production cycle. ◇ Recall “the finite production rate “ solution where H T ~ Multiple items production planning ~

8 ◆ Let T be the cycle time ( new, for the rotation cycle policy) During time T, we assume that exactly one lot of each product is produced. ◇ In order that the lot for product j be large enough to meet the demand occurring during time T, it follows that the lot size should be ◇ The average annual cost ( of setup & holding costs only ) (e.g. if T=0.5 year, =12,000 then =6,000) ………... [Eq.3]

9 ◇ The average annual cost for all products ◇ The avg. annual cost for n products in terms of the cycle time T ◇ To find T, minimize G(T) ; (check G”(T)>0 for minimal) ………... [Eq.4]

10 where ………... [Eq.5]

11 ◆ If setup times are a factor, we must check that if there is enough time each cycle to account for both setup times and production time of the n products. Let be the setup time for product j T1 year 0 Idle time fits into idle time ~ Multiple items production planning ~ ………... [Eq.6]

12 then [Eq.6] become (or )  ◆ The optimal solution is to choose the cycle time T ( Let ) T = max { T*, } if considering   ………... [Eq.7]

13 Example 4.7 Bali produces several styles of men’s and women’s shoes at a single facility near Bergamo, Italy. The leather for both the uppers and the soles of the shoes is cut on a single machine, This Bergamo plant is responsible for seven styles and several colors in each style. (The colors are not considered different products for our purposes, because no setup is required when switching colors.) Bali would like to schedule cutting for the shoes using a rotation policy that meets all demand and minimizes setup and holding costs. Setup costs are proportional to setup times. The firm estimates that setup costs amount to an average of $110 per hour, based on the cost of worker time and the cost of forced machine idle time during setups. Holding costs are based on a 22% annual interest charge. The relevant data for this problem appear in Table 4-1.

14 ◆ Example 4.7 # Style Production Rate Setup Time (hr) Variable Cost Assuming $110 per hr Assuming ί =22% 1WP4,52035, ,759 2WL6,60062, ,792 3WB2,34041, ,249 4WS2,60071, ,906 5MW8,80046, ,752 6ML6,20071, ,844 7MO5,20056, ,188 ∵ $110 * (Setup Time) Holding Cost = ί*c =(22%)*c Σ= [Eq.2]

15 ◆ Assuming 250 working days a year ( year ) * 250 = 38.2 days #style 1WP WL WB WS MW ML MO ◆ Example 4.7

Original – by EPQ single item  Not feasible ◆ Example 4.7 in SPT ?

17 #style 1WP WL WB WS MW ML MO Σ=26.1 Σ=26.5 ◆ 1 year = 250 days ◆ Example 4.7 Original by Rotation cycle policy

Original ◆ Example 4.7 in SPT

19 ◆ Example 4.7 by Rotation cycle policy T* = 38.2 in SPT # 1 # 2 # 3 # 4 # 5 # 6 # 7

20 ◆ The optimal solution is to choose the cycle time T ( Let ) T = max { T*, } if considering   ………... [Eq.7] If including setup time s j : T = max { , } = Setup Time (hr)

21 §. I.eg.4.7.1: Class Exercise “IF there are only 2 products” Preparation Time : 15 ~ 20 minutes Discussion : 10 ~ 15 minutes Setup Time #2WL6,60062, *4* WS2,60071,

22 Setup Time #2WL6,60062, $ *4* WS2,60071, year = 250 days ◆ Example “IF there are only 2 products” Original by Rotation cycle policy #2WL = , , *4* WS = , * Data being modified

23 By rotation cycle policy T* Costs - when using T* = ( ) + ( ) = (4529) + (3129) = $7,658 ◆ Example “IF there are only 2 products”

24 #2WL = , , *4* WS = , To Think about - when producing 2 products ◆ To Think about - when producing 2 products Item “WL” has T 1 = 31.9 days Item “WS” has T 2 = 70.6 days Why using R-C-P ? To allow producing 1 run per item per period ?

25 Class Note# 3b Class Note # 3b §. I.eg A: Class Exercise From eg Preparation Time : 15 ~ 20 minutes Discussion : 10 ~ 15 minutes Setup Time #2WL6,60062, *4* WS2,60071, If we producing 2 runs of item “WL” and 1 run of item “WS” in a period of 70.6 days (or years) (a) What will the annual costs be? (b) Draw a chart for production schedule.

26 A Solution to eg A : When producing 2 runs of item “WL” and 1 run of item “WS” in a period of 70.6 days (or years)(a) ∴ T 1 = / 2 = years = 35.3 days ∴ Q 1 = (λ 1 ) ( T 1 ) = (6600) (0.1412) ≒ 932 ∴ T 11 = Q 1 / P 1 = 932 / 62,600 ≒ years ≒ 3.72 days [ up-time] = ( ) + ( ) = $4,333 = $2,805

27 = $4,333 + $2,805 = $7,138 A Solution to eg A : Comparing to R-C-P: $7658 There are ($ $7138) / $7658 = 6.8 % cost saved.

T* = 70.6 days #WL-1 # WS A Solution to eg A :(b) 6.31 #WL

29 Class Note# 3b Class Note # 3b §. I.eg B: Class Exercise From eg Preparation Time : 15 ~ 20 minutes Discussion : 10 ~ 15 minutes Setup Time #2WL6,60062, *4* WS2,60071, If we producing 2 runs of item “WL” and 1 run of item “WS” in a period of 63.8 days (or 2* = years) (a) What will the annual costs be? (b) Draw a chart for production schedule. ◆ G-s-32 G-s-32

30 Class Note# 3b Class Note # 3b §. I.eg C: Class Exercise From eg Setup Time #2WL6,60062, *4* WS2,60071, – i.e. the optimal length for period If we producing 2 runs of item “WL” and 1 run of item “WS” in a period of 66.4 days (or years) – i.e. the optimal length for period What will the annual costs be? ◆ G-s-35

31 §. I.4.9: Problems ( # N29, N30 ) (a) using Rotation Cycle (a) using Rotation Cycle ( # N45 ) Preparation Time : 15 ~ 20 minutes Discussion : 10 ~ 15 minutes

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33 B Solution to eg B : When producing 2 runs of item “WL” and 1 run of item “WS” in a period of 63.8 days (or years)(a) ∴ T 2 = years = 63.8 days ∴ Q 2 = (λ 2 ) ( T 2 ) = (2600) (0.2552) ≒ 664 ∴ T 21 = Q 2 / P 2 = 664 / 71,000 ≒ years ≒ 2.34 days [ up-time] = (1550.6) + ( ) = $2,819 = $4,311

34 = $4,311 + $2,819 = $7,130 B Solution to eg B : Comparing to R-C-P: $7658 There are ($ $7130) / $7658 = 6.9 % cost saved.

T* = 63.8 days #WL-1 # WS B Solution to eg B :(b) 5.74 #WL-2 ◆ G-b

36 Solution to eg C Solution to eg C : – the optimal period length When producing 2 runs of item “WL” and 1 run of item “WS” in a period of 66.4 days (or years) – the optimal period length ∴ for item “WL”, each run T 1 = / 2 = years = 33.2 days ∴ for item “WS”, T 2 = years = 66.4 days ∴ Q 1 = (λ 1 ) ( T 1 ) = (6600) (0.1328) ≒ 876 ∴ Q 2 = (λ 2 ) ( T 2 ) = (2600) (0.2656) ≒ 691 ∴ T 11 = Q 1 / P 1 = 876 / 62,600 ≒ years ≒ 3.50 days [ up-time] ∴ T 21 = Q 2 / P 2 = 691 / 71,000 ≒ years ≒ 2.43 days [ up-time] = (2072) + (2243) = $4,315

37 = $7,125 = $4,315 + $2,810 = $7,125 Solution to eg C : Comparing to R-C-P: $7658 There are ($ $7125) / $7658 = 7.5 % cost saved. = (1490) + (1320) = $2,810

T* = 66.4 days #WL-1 # WS Solution to eg C : (b) 5.93 #WL

39 Alternative #Period Length Cost Saved Vs. R-C-P Using Rotation Cycle Policy 44.8 days$7,658— Using original “WL” * days$7, % OptimalLength 66.4 days $7,1257.0% Using original “WS” period70.6 days$7,1386.8% §. I.eg.4.7.1: 2 items production planning comparison comparison

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