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Bottino Exercises First Lesson
- A toy shop sells fluffy bears at a very regular d=10 pcs/day (considering 310 days in a year). The shop pays for each bear Cu=5€ pcs/day. The inventory holding cost is h=20% of the unitary purchasing cost. -Evaluate the lot size Q which guarantee the planning of 20 order per year in these conditions; -If the shop orders 100 units size lot (considering EOQ), what is teh total order fixed cost due to order plans. -If A=10€ per order evaluate the EOQ Exercise Done Data: d=10 pcs/day working days= 310 days/year Cu=5€ pcs/day h%=20% Cu/year -So h =
( pcs ∗ year )
year ( 310 days )
( pcs ∗ day ) Manual Orders=
D
Q
( d daily 310 days year
Q
, daily demand per working day 2 D orders / year = ( 10 pcs / day ∗ 310 days / year ) Q
So we have Q=155 pcs/order -Q'=100, which is EOQ. (Q'=Q*) A=total order fixed cost Q' =
AD
h so A =
( Q'
2 ∗ h ) ( 2 D )
2 ∗ pcs 2 ∗0,2∗ 5 ∗ € /( pcs ∗ year ) ( 2 ∗ 310 ∗ 10 pcs / year )
10000 ∗ pcs 2 ∗0,0032∗ € /( pcs ∗ year ) ( 20 pcs / year )
So the number of annual orders is ( n ° annual ) order
D
( Q ' )
( 10 pcs / day ∗ 310 days / year ) ( 100 pcs / order ) = 31 orders / year So A annual = 31 order / year ∗1,61 € / order = 50 € / year -If A=10€, we have Q' =
Ad h
√
( 2 ∗ 10 € ∗ 10 pcs / day ) (0.0032 € / pcs ∗ day ) )= 249 pcs So in this case we prefer a low A (1,61), so less orders (100)
2)Product A demands rate D= 416 pcs/year. The unitary purchase cost is 20 €/pcs, each order displacement cost(A) = 150€. The unitary annual storage (h) is 350 € per product per year. -Evaluate the Economic Order Quantity -If a discount of 15% is applied whenever the order exceed the 30 units, how does the EOQ change? -What is the EOQ if we consider a bounded speed of filling of P=60 pcs/month? Data: D= 416 pcs/year Cu= 20 €/pcs A=150€/order h=350 €/pcs year -So we have Q' =
AD
h
( 2 ∗ 150 € / order ∗ 416 pcs / day ) ( 350 € / pcs ∗ day ) )=18,9 pcs -Cu' = Cu(1-15%) if Q>30, so Cu' = Cu(1-0,15)*20=17€/pcs To evaluate discount we need to evaluate the cost, so
C 1,2 ( Q' )=√( 2 ADh )=√( 2 ∗ 150 € ∗ 416 pcs / year ∗ 350 € /( pcs ∗ year ))= 6609 € / year
[This formula for C can be used only for Q' ] But C1,2 is not the total cost, we are missing C 3. So C 3 ( Q ' )= CuD = 20 € / pcs ∗ 416 pcs / year = 8320 € / year and Ctot = 6609 + 8320 = 14929 €/year If Q>30 pcs, then C 1,2 ( Q = 30 )=
AD
Q
hQ 2
( 150 € ∗ 416 pcs / year ) ( 30 pcs )
( 350 € /( pcs year )∗ 30 pcs ) 2 = 7330 € / year C 3 ( Q = 30 )= Cu ' D = 17 € / pcs ∗ 416 pcs / year = 7072 € / year So Ctot = 7330 + 7072 = 14402 €/year So Q=30 is preferable to Q', we have a lower cost--> EOQ= 30 Now P= 30 pcs/month = 30 pcs/month * 12 month/year = 720 pcs/year > 416 pcs/year [being P>D, has sense to continue the exercise] Q' ' =
AD
( h ( 1 −
D
P
( 2 ∗ 150 € / order ∗ 416 pcs / day ) ( 350 € / pcs ∗ day ∗( 1 − ( 416 pcs / year ) ( 720 pcs / year )
)=29,06 pcs with P= 720 before Q' = 18,9 pcs wish P= ∞
Second lesson
- An intersection press is used to produce three different products A, B and C in thermoplastic resin. A total production of 600 pcs must be produced with this mix: qA=40% qB=35% qC=25% For each typology of product the standard operation time SOT is assumed to be constant and equal to 4,5 minutes. The machines set-up time is 3 times the SOT. The AWT is 55 hours. -Verify the feasibility of a single lot production (for each product) -Evaluate the smallest sustainable lot size for each product (it's obviously assumed repetitive lot production approach ). DATA SOT= 4,5 minutes/pcs set-up time= 13,5 minutes QPtot = 600 pcs N= UST = ( set − up time ) SOT = 3 pcs HVP = ( 60 mm / hours ) SOT =13,33 pcs / hours a)
SOT
∗( QPtot + N ∗ UST )⩽ AWT So
∗( 600 + 3 ∗ 3 )⩽ 55 which is^ 45,67⩽^55 OK b) being x=QPtot y =
N ∗ UST
HVP ∗ AWT
x
=40,5≃ 41 ⇒ LTPmin LAmin = qA ∗ LTPmin =0,4∗40,54=16,2 pcs = 17 pcs LBmin = qB ∗ LTPmin =0,35∗40,54=14,2 pcs = 15 pcs LCmin = qC ∗ LTPmin =0,25∗40,54=10,1 pcs = 11 pcs ∑ Lnmin =^17 +^15 +^11 =^43 pcs LTN = QPtot ∑ Lnmin
2 nd^ verification LTN ∗ SOT 60 ∗(∑ Lnmin + N ∗ UST )⩽ AWT So
∗( 43 + 3 ∗ 3 )⩽ 55 which is 54,6⩽ 55 Total units are LTN ∗∑ Lnmin =^14 ∗^43 =^602 pcs
- Once produced, circuits are packed by a special machine. The time needed for packing operation is 8mm. Following quantities, m 2008, are required to be packed: -100 type A -40 type B -180 type C -80 type D Every time the production changes the machine needs a set-up time equal to 16mm, AWT is between 65 and 70 hours. The packing has to be done in order to have repetitive lots of the smallest sustainable lot size. -Find the lot size -Evaluate the average time needed to pack the single integrated circuit considering the time loss due to the set-up. Calculate also the number of set-ups realized. -Repeat the previous point doubling the repetitive lot size. DATA N= UST = ( set − up time ) SOT
= 2 pcs SOT 60 ∗( QPtot + N ∗ UST )⩽ AWT so^
∗( 400 + 4 ∗ 2 )⩽ 65 / 70 which is^ 54,4⩽^65 /^70 We have then qA =
qB =
qC =
qD =
AWT= 65/70 hours, we consider 65 because the 5 extra hours costs, so LTPmin =
N ∗ UST
HVP ∗ AWT
QPtot
LAmin = qA ∗ LTPmin =0,25∗36,6=9,1 pcs = 10 pcs LBmin = qB ∗ LTPmin =0,1∗36,6=3,6 pcs = 4 pcs LCmin = qC ∗ LTPmin =0,45∗36,6=16,46 pcs = 17 pcs LDmin = qD ∗ LTPmin =0,2∗36,6=7,3 pcs = 8 pcs ∑ Lnmin =^10 +^4 +^17 +^8 =^39 pcs LTN = QPtot ∑ Lnmin
2 nd^ verification LTN ∗ SOT 60 ∗(∑ Lnmin + N ∗ UST )⩽ AWT
Third Lesson
- A company produces bookshelves. The production of each bookshelf requires the following production stages: -Shelves machining: the average time needed is 2 weeks per lot, we need two shelves for bookshelf. -Bookshelf frame production, which average lead time is 1 week, 1 frame needed for each bookshelf. -Assembly and packing LT=1 week. Build a proper MRP from these data t 1 2 3 4 5 6 7 8 MPS - - 40 20 - 70 10 60 Item master file \ OH 0 LT SR Lot sizing rule A 10 1 / L4L B 0 1 10(1), 20(7) FOP= 2 time buckets C 100 2 10(2) FOQ= 60*n Exercise Done Notes: -If I can postpone a SR, then I have to note it with a * and sign it on the last table -FOP the time I must consider for the OR order, (here 2, so 1 “time” is empty) -FOQ says the multiples of “items” I can put in OR -LT is the time I need to receive my “items”, so I have to anticipate A 1 2 3 4 5 6 7 8 GR 40 20 70 10 60 SR OH with OH 0 =
NR 30 20 70 10 60
OP 30 20 70 10 60
OR 30 20 70 10 60
B 1 2 3 4 5 6 7 8
GR 30 20 70 10 60
SR 10* 20
OHwith OH 0 =
NR 20 20 70 10 40
OP 40 80 40
OR 40 80 40
C 1 2 3 4 5 6 7 8
GR 60 40 140 20 120
SR 20*
OHwith OH 0 =
NR 120 60 120
OP 120 60 120
OR 120 60 120
Code Quantity (Qty) Period Remarks A 30 2 OR 20 3 OR 70 5 OR 10 6 OR 60 7 OR B 40 1 OR 80 4 OR 40 6 OR 10 1-->3 SR defer 20 7 SR OK C 120 3 OR 60 4 OR 120 5 OR 20 2-->5 SR defer
