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Inventory Management and Control Lect 5

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Inventory Management and Control
Israa Mohamed
October 2019
[email protected]
EOQ Models for Production Planning
• This model is an extension of the finite production planning.
• The problem here is to find an optimal procedure for
producing n products on a single machine to guarantee that:
– Holding and setup cost are minimized.
– No stock outs are occurred.
• We need to assume that
feasibility.
to ensure system
EOQ Models for Production Planning
• It is further assumed that we follow a rotation cycle policy.
• We need to find the optimal producgion quantity for each
item.
• If we used
for each item, we might
be faced by stock-out in some items.
• Sooooo..............................
Cycle Time
T
EOQ Models for Production Planning
•
•
•
•
To avoid stock out we know that
The average annual cost for this Qj =
This cost for all items =
Substituting for T giving,
• Optimizing the above function (first derivative = 0) will give
optimal
EOQ Models for Production Planning
• We still need to check that T* is above the minimum T.
• If setup times are considered, then minimum T must be
greater than setup and production times for all items.
• Qs for all items are then computed based on the optimal T
EOQ Models for Production Planning - Example
Bali produces several styles of men’s and women’s shoes
at a single facility. The leather for both the uppers and the
soles of the shoes is cut on a single machine. This 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. Setup costs are averaged to
$110 per hour, holding costs are based on a 22 percent
annual interest charge.
EOQ Models for Production Planning - Example
Setup cost = 3.2 * 110 = 352
h' = 0.22 * 40 * (1- (4520 / 3580) = 7.69
( 4520 / 35800) + ( 6600 / 62600) + ( 2340 / 41000) + ( 2600 / 71000)
+........ = 0.693
EOQ Models for Production Planning - Example
34758.8
33792
25248.6
9906
59752
34844
32188

2 * 2695
 0.15 year  38days
230458.4
EOQ Models for Production Planning - Example
• We must check if T* is larger than Tmin .
• Tmin =
24.5 /(8 * 250)
 0.03999  0.04 year  10days
=
1  0.69355
• It should be noted that we assumed a 250 working days
per year and a day of 8 working hours.
• Then T* is the optimal cycle time.
• Next step is to find optimal production size for each item
and the associated average cost.
EOQ Models for Production Planning - Example
• We use the following equation for estimating
optimal Qs
EOQ Models for Production Planning
Interesting notes
• This policy guarantees minimum cost BUT doesn't guarantee
optimal time management.
n Q
j
( )
• Uptime for the previous example j 1 Pj will be 70% of the
time, i.e. the machine will be idel 30% of the time.
• Another interesting point is that if the setup costs were
sequence dependent, we would change the solution
dramatically.
• Also, we assumed a rotation cycle which mightnot be suitable
in all cases, for example when demand rates and setup costs
differ widely, it might be advantageous to do two or more
production runs of a product in a cycle.
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