Material Management Class Note # 3-B ~ Multiple product EPQ analysis ~

Material Management Class Note #3-B ~ Multiple product EPQ analysis ~ Prof. Yuan-Shyi Peter Chiu Feb. 2011

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

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

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

~ EPQ (Finite Production Rate ) Models for Production Planning ~ ◆ Producing n products on a single machine (or line) 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 ‧ ◆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.

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

~ Multiple items production planning ~ ◆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 H where T

◆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 (e.g. if T=0.5 year , =12,000 then =6,000) ◇The average annual cost ( of setup & holding costs only ) ………... [Eq.3]

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

where ………... [Eq.5]

~ Multiple items production planning ~ ◆ 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 ………... [Eq.6] 4 fits into idle time 2 1 3 0 T 1 year Idle time

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

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.

◆Example 4.7 [Eq.2]

◆Example 4.7 ◆ Assuming 250 working days a year (0.1529 year ) * 250 = 38.2 days

◆Example 4.7 Original – by EPQ single item Not feasible 31.9 34.3 35 35.6 43.4 49.1 50 3.4 6.5 3.1 4.5 4.0 2.8 1.8 in SPT 7 6 5 4 ? 3 2 2 1 72.0 0 10 20 30 40 50 60 70 80 24.3 31.9 44.9 9.9 17.5 60.9 70.6 74.3 48.6 3.4 13 21.5 26.1 37.7

◆Example 4.7 ◆ 1 year = 250 days Original by Rotation cycle policy

◆Example 4.7 Original 31.9 34.3 35 35.6 43.4 49.1 50 3.4 6.5 3.1 4.5 4.0 2.8 1.8 in SPT 7 6 5 4 3 2 1 0 10 20 30 40 50 60 70 80 24.3 31.9 44.9 9.9 17.5 60.9 74.3 48.6 3.4 13 21.5 26.1 37.7

◆Example 4.7 by Rotation cycle policy in SPT # 4 # 7 # 5 # 6 # 3 # 2 # 1 0 5 10 15 20 25 30 40 50 60 22.9 47 50.6 61.1 38.2 8.8 12.4 43 49.2 57.8 4.8 11 19.6 26.5 T* = 38.2

If including setup time sj :  ( Let )  ◆The optimal solution is to choose the cycle time T T = max { T* , } if considering ………... [Eq.7] T = max { 0.1529 , 0.03995 }= 0.1529

Class Note# 3b §.I.4.9: Home Work Problems Chapter 4 : ( # 29, 30) pp.219-220 (a) using Rotation Cycle ( # 45 ) pp.227 Preparation Time : 15 ~ 20 minutes Discussion : 10 ~ 15 minutes ■

§.I.eg.4.7.1: Class Exercise “IF there are only 2 products” Holding costs are based on a 22% annual interest charge Preparation Time : 15 ~ 20 minutes Discussion : 10 ~ 15 minutes

◆Example “IF there are only 2 products” 1 year = 250 days * Data being modified Original by Rotation cycle policy

◆Example “IF there are only 2 products” • By rotation cycle policy T* • Costs - when using T* = (1570+2959) + (2258+871) = (4529) + (3129) = $7,658

◆To Think about - when producing 2 products Item “WL” has T1 = 31.9 days Item “WS” has T2 = 70.6 days Why using R-C-P ? To allow producing 1 run per item per period ?

Class Note# 3b §.I.eg.4.7.2.A: Class Exercise From eg.4.7.1 If we producing 2 runs of item “WL” and 1 run of item “WS” in a period of 70.6 days (or 0.2824 years) (a) What will the annual costs be? (b) Draw a chart for production schedule. Preparation Time : 15 ~ 20 minutes Discussion : 10 ~ 15 minutes

Solution to eg.4.7.2.A : When producing 2 runs of item “WL” and 1 run of item “WS” in a period of 70.6 days (or 0.2824 years) (a) ∴ T1 = 0.2824 / 2 = 0.1412 years = 35.3 days ∴ Q1 = (λ1) ( T1) = (6600) (0.1412) ≒ 932 ∴ T11 = Q1 / P1 = 932 / 62,600 ≒ 0.01488 years ≒ 3.72 days[ up-time] = (1947.42) + (2385.92) = $4,333 = $2,805

Solution to eg.4.7.2.A : = $4,333 + $2,805 = $7,138 Comparing to R-C-P: $7658 There are ($7658 - $7138) / $7658 = 6.8 % cost saved.

Solution to eg.4.7.2.A : (b) #WL-2 #WL-1 # WS 0 6.31 70.6 74.32 35.3 3.72 39.02 T* = 70.6 days

Class Note# 3b §.I.eg.4.7.2.B:Class Exercise From eg.4.7.1 If we producing 2 runs of item “WL” and 1 run of item “WS” in a period of 63.8 days (or 2*0.1276 = 0.2552 years) (a) What will the annual costs be? (b) Draw a chart for production schedule. Preparation Time : 15 ~ 20 minutes Discussion : 10 ~ 15 minutes ◆G-s-32

Class Note# 3b §.I.eg.4.7.2.C: Class Exercise From eg.4.7.1 If we producing 2 runs of item “WL” and 1 run of item “WS” in a period of 66.4 days (or 0.2656 years)– i.e. the optimal length for period What will the annual costs be? ◆G-s-35

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Solution to eg.4.7.2.B : When producing 2 runs of item “WL” and 1 run of item “WS” in a period of 63.8 days (or 0.2552 years) (a) ∴T2 = 0.2552 years = 63.8 days ∴ Q2 = (λ2) ( T2) = (2600) (0.2552) ≒ 664 ∴ T21 = Q2 / P2 = 664 / 71,000 ≒ 0.00935 years ≒ 2.34 days [ up-time] = (1550.6) + (1268.24) =$2,819 = $4,311

Solution to eg.4.7.2.B : = $4,311 + $2,819 = $7,130 Comparing to R-C-P: $7658 There are ($7658 - $7130) / $7658 =6.9 %cost saved.

Solution to eg.4.7.2.B : (b) #WL-2 #WL-1 # WS 0 31.9 63.8 67.2 5.74 3.4 35.3 ◆G-b-29 T* = 63.8 days

Solution to eg.4.7.2.C : When producing 2 runs of item “WL” and 1 run of item “WS” in a period of 66.4 days (or 0.2656 years)– the optimal period length ∴ for item “WL”, each runT1 = 0.2656 / 2 = 0.1328 years = 33.2 days ∴ for item “WS”, T2 = 0.2656 years = 66.4 days ∴ Q1 = (λ1) ( T1) = (6600) (0.1328) ≒ 876 ∴ Q2 = (λ2) ( T2) = (2600) (0.2656) ≒ 691 ∴ T11 = Q1 / P1 = 876 / 62,600 ≒ 0.01399 years ≒ 3.50 days [ up-time] ∴ T21 = Q2 / P2 = 691 / 71,000 ≒ 0.00973 years ≒ 2.43 days [ up-time] = (2072) + (2243) =$4,315

Solution to eg.4.7.2.C : = (1490) + (1320) =$2,810 = $4,315+ $2,810= $7,125 Comparing to R-C-P: $7658 There are ($7658 - $7125) / $7658 = 7.5 % cost saved.

Solution to eg.4.7.2.C : (b) # WS #WL-2 #WL-1 0 33.2 66.4 69.9 5.93 3.5 36.7 T* = 66.4 days

§.I.eg.4.7.1: 2 items production planning comparison

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Material Management Class Note # 3-B ~ Multiple product EPQ analysis ~