Презентация на тему Production management

that minimizes total holding and ordering costs for the

year.

minimizes total inventory holding costs and ordering costs. It

is one of the oldest classical production scheduling models.

EOQ applies only when demand for a product is constant over the year and each new order is delivered in full when inventory reaches zero. There is a fixed cost for each order placed, regardless of the number of units ordered. There is also a cost for each unit held in storage, commonly known as holding cost, sometimes expressed as a percentage of the purchase cost of the item.

EOQ search to determine the optimal number of units to order to minimize the total cost associated with the purchase, delivery and storage of the product.

The required parameters to the solution are the total demand for the year, the purchase cost for each item, the fixed cost to place the order and the storage cost for each item per year. Note that the number of times an order is placed will also affect the total cost, though this number can be determined from the other parameters.

demand is known, and spread evenly throughout the year.
The

lead time is fixed.
The purchase price of the item is constant i.e. no discount is available
The replenishment is made instantaneously, the whole batch is delivered at once.
Only one product is involved.

order quantity
Q* : optimal order quantity
D : annual demand

quantity
K : fixed cost per order, setup cost (not per unit, typically cost of ordering and shipping and handling. This is not the cost of goods)
H : annual holding cost per unit, also known as carrying cost or storage cost (capital cost, warehouse space, refrigeration, insurance, etc. usually not related to the unit production cost)

of Q*, we compute…

we have the same holding cost and order cost…
Q*
C(Q*)

ordered
It is often possible to obtain price discounts when

quantities ordered are below certain thresholds

10 € per item when the order quantity is in interval ]0 ; 400[.

9,5 € per item when the order quantity is in interval [400 ; 600[.

9 € per item when the order quantity is in interval [600 ; inf[.

Formally, order are partitioned to intervals [ai, bi] based on the price of the item

Algorithm

Compute Q*
For each interval :
Determine Qopt
Compute R(Qopt)
Keep Qopt giving min(R(Qopt))

to obtain price discounts when quantities ordered are below

certain thresholds

K = 1000 €, if Q ≤ 400 (road transport)

For each intervals [ai, bi], order cost is different…

Algorithm

For each interval :
Compute Q* based on K
Determine Qopt
Compute C(Qopt)
Keep Qopt giving min(C(Qopt))

K = 1500 €, if Q > 400 (rail transport)

D =1000 item / day

H =1 € / item / day

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to obtain price discounts when quantities ordered are below

certain thresholds

10 € per item in interval ]0 ; 400[.

9,5 € per item in interval [400 ; 600[.

9 € per item in interval [600 ; inf[.

Formally, order are partitioned to intervals [ai, bi] based on the price of the item

Algorithm

Compute Ki’ = K
For each interval :
Determine Qi*
Compute Qopt & R(Qopt)
Ki’ = Ki’+ bi(ci+1-ci)
Keep Qopt giving min(R(Qopt))

stock level up to the maximum quantity at fixed

period

to supply the demand. There are two indicators:
α: Probability

to meet demand without breaking
β: Ratio between the number of items supplied and demand

Exemple

A pharmacist sells every day 6-9 boxes of some infant vaccine. The vaccine should be kept refrigerated. Space is limited and replenishment is daily. The pharmacist decides to fix the reorder point 7 boxes.

that the cost of holding CP is equal to

the cost of unsold for an article.
The cost of shortage CR is proportional to the number of missing items.

A newsagent buys every day a fixed amount of a magazine. It pays each copy 2 € and sold 3 €, representing a margin of 1 €.
If demand exceeds its stock, it finds a shortfall of 1 €, so CR = 1. Any unsold magazine was bought 1.5 € per press group which causes a loss of 0.5 €, so CP = 0,5.

probability that the demand X is equal to x

items on period T and S is the desired level of stock.

if x ≤ S there is (S-x) unsold, representing a cost of CP(S-x)

Si x > S, (x-S) items are missing, representing a cost of CR(x-S)

A request of x articles having a probability px of occurring, the average total cost is given by:

Warning : it is for x ≤ S and x > S. If we have x < S et x ≥ S, we obtain:

Eq. 1

Eq. 2

density function F(S), we obtain
Let
S* is the only optimal

solution

(S*-1) and S* are the two optimal solutions

If

If

over 10 weeks the number of newspapers sold.
Every

day, he sold between 20 and 60 magazines

What is the optimal level of inventory?

CP = 3

CR = 2

number of articles
Model
1. If x ≤ S, there is

no out of stock

By approximating the demand with a line, we obtain:

Let CP and CR holding cost and shortage cost for one item for a period.
Let x the observed demand during the period T.

1. if x ≥ S, after a time T1, there is out of short

Holding costs amounted to

Shortage costs amounted to

Thanks to Chasles relation a :

Given a total cost

will be used per unit of time:
With probability….
And, if

we consider instead x < S et x ≥S, we obtain:

Replenishment with CP, CR based on time and number of articles

can be expressed by:
Let
Replenishment with CP, CR based on

time and number of articles

a similar result to the previous problem...
Moreover, L(S) is

strictly increasing, so:

If

If

S* is the only optimal solution

(S*-1) and S* are the two optimal solutions

Replenishment with CP, CR based on time and number of articles

is estimated that the shortage cost for a period

is equivalent to 20 holding cost (over the same period).
Over 15 days, the demand varie from 20 to 27.

Replenishment with CP, CR based on time and number of articles

on time and number of articles
A workpiece is supplied

every 15 days. It is estimated that the shortage cost for a period is equivalent to 20 holding cost (over the same period).
Over 15 days, the demand varie from 20 to 27.

model differs from the previous by the computation of

the shortage cost

When x ≥ S, after a period T1, ther is out of stock

Shortage cost:

In terms of probability, total cost will be:

Let

on number of articles
S* is the only optimal solution
(S*-1)

and S* are the two optimal solutions

25 computers. Replenishment is done every Thursday night. With

competition strong, anyone who can not take his product will buy directly from a competitor. In this case, the distributor estimates its loss to 200 €. On the other hand, the shop buys a computer 2000 € and estimates the annual holding cost to 25% of the purchase price.

Replenishment with CR based on number of articles