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(Problem after bottleneck)
P. 276 in the book

Problem:
D= OT / CT
CT opt = 1.6
CTmin= 1.2
CTmax= t = 4.4

3. Number of stations:
Nmax= t / CTmin = 4.2 / 1.2 = 3.5 = 4
Nmin= 4.2 / 4.2 = 1
Nopt = 4.2 / 1.6 = 2.62 = 3
1.6                       1.6                    1.6
A 0.3                 d. 1.2               F 0.6
C 0.4                                      H 0.5
e. 0.2
g 0.1
b 0.6
Idle = 0                  idle = 0.4             idle = 0.2
Efficiency = Available time / Actual time = t / N.CT = 4.2 / 4.8 = 87.5%

If N= 1 Efficiency = 100%

If 4 Stations:
1.2                               1.2                     1.2                   1.2
A 0.3                     D 1.2                      C 0.4                  G 0.1
B 0.6                                                E 0.2                  H 0.5
                                                     F 0.6                  I 0.3

Idle = 0.3                Idle = 0                   Idle = 0               Idle = 0.3

Efficiency: t / N.CT = 4.2 / 4x1.2 = 4.2 / 4.8 = 87.5%

6. Comment:
Compare the three solutions and choose the best number of stations.

D = OT / CT = 440 / 1.2 = 307 units
Therefore, the 3 stations is better because it produces enough units for the demanded quantity, the
4 stations will produce more (307) but we don’t need them

+++++++++++++++++++++++++++++++++++++
Assignment:

Hazel Case P. 36
Highline Fin Services P. 126
P. 2 & P.8 P. 118 & 119
P. 1, 11, 19 P. 229, 232, 233
P. 6 & 10 P. 314
Son Ltd p. 36
Prob. 2 & 3 P. 401
CHAPTER 6: Inventory
                 Management
Management = POLC

P= Plan = 5Ws: Why keep inventory?, what, where, when,
O= Organise = handling, logistics, shelving, shelf life,
Leading / directing of staff, suppliers,
C= Control inventory: this class

Inventory: A stock of goods
Independent demand: demand on my final products (from customers)
Dependent demand: demand on parts of products (from production department)

Types of inventory:
- Raw Materials and Purchased parts
- Work in progress
- Replacement parts, tools & supplies
- Good in transit
- Finished goods inventories

Function of Inventory:
Main purpose of inventory is to keep the production running  Continuity

Inventory Counting System:
Periodic System: Physical count each fiscal year
Perpetual System:
Universal bar code:

Lead time= time from order till receiving the order
Holding cost: cost u pay per unit annually to keep the product in house
Ordering / set up cost: cost of the purchasing process (from decision to buy till goods are received)
Shortage cost

ABC Classification System:

EOQ Model: Economic Optimal Quantity
Advised to manage critical inventory only.
Only 1 product, annual demand known, demand even each year, ……

Fig 11.2:
Reorder point depends on the lead time, the shorter the lead time the less the reorder point.

Total cost: Annual carrying cost + Annual ordering cost
TC = Q/2 H + D/Q S
S= set up cost (per cycle)
D= annual demand
H: avg annual total cost per unit
Q: Quantity produced each cycle
D/Q= number of cycles

D: from production
H: from inventory mgr (last year, total spending on inventory / Avg quantity of inventory available)
S: from purchasing
Q: we calculate it

Min TC: Q/2H = D/Q S
QH.Q = 2DS
Q2= 2DS/H
Q= sqrt 2DS/H

Let D= 1000units
S= LE40 / order
H= LE 2/unit
Q= sqrt 2x1000x40/2 = 200 units

Min TC: Q/2H = D/Q S
200/2(2)       =     1000/200x40
200            =     200
TC= LE400 / year

ROP= 10 days = 40 units
OR accountants can simply calculate it as: H= 5%-10%


Economic Production Quantity (EPQ):
This model assumes we r producing what we consume! (Eg. Mardini (curtains) and its textile
factory).

Problem: P. 588

OT = 220 days / year
Usage = 50 couplings / day
Production = 200 couplings / day
Holding cost = $2 / coupling
Setup cost = $70 / order

Qo = sqrt 2DS/H x sqrt P/P-U

D = usage x operating time
= 50 x 220 = 11000.

Qo      = sqrt 2DS/H x sqrt P/P-U
        = sqrt 2x11000x20 / 2 x sqrt 200/150
        = 1013
Number of runs per year: 11000/1013 = 10.858 = 11 runs


Qo / u = 1013/50 = 20 USE
Qo / p = 1013/200 = 5 PRODUCE Only
Qo = 20-5 = 15 days   Use(?)

Imax = Qo (P-U/P) or 1-U/P

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Supply chain mgt lectures

  • 1. (Problem after bottleneck) P. 276 in the book Problem: D= OT / CT CT opt = 1.6 CTmin= 1.2 CTmax= t = 4.4 3. Number of stations: Nmax= t / CTmin = 4.2 / 1.2 = 3.5 = 4 Nmin= 4.2 / 4.2 = 1 Nopt = 4.2 / 1.6 = 2.62 = 3 1.6 1.6 1.6 A 0.3 d. 1.2 F 0.6 C 0.4 H 0.5 e. 0.2 g 0.1 b 0.6 Idle = 0 idle = 0.4 idle = 0.2 Efficiency = Available time / Actual time = t / N.CT = 4.2 / 4.8 = 87.5% If N= 1 Efficiency = 100% If 4 Stations: 1.2 1.2 1.2 1.2 A 0.3 D 1.2 C 0.4 G 0.1 B 0.6 E 0.2 H 0.5 F 0.6 I 0.3 Idle = 0.3 Idle = 0 Idle = 0 Idle = 0.3 Efficiency: t / N.CT = 4.2 / 4x1.2 = 4.2 / 4.8 = 87.5% 6. Comment: Compare the three solutions and choose the best number of stations. D = OT / CT = 440 / 1.2 = 307 units Therefore, the 3 stations is better because it produces enough units for the demanded quantity, the 4 stations will produce more (307) but we don’t need them +++++++++++++++++++++++++++++++++++++ Assignment: Hazel Case P. 36 Highline Fin Services P. 126 P. 2 & P.8 P. 118 & 119 P. 1, 11, 19 P. 229, 232, 233 P. 6 & 10 P. 314 Son Ltd p. 36 Prob. 2 & 3 P. 401
  • 2. CHAPTER 6: Inventory Management Management = POLC P= Plan = 5Ws: Why keep inventory?, what, where, when, O= Organise = handling, logistics, shelving, shelf life, Leading / directing of staff, suppliers, C= Control inventory: this class Inventory: A stock of goods Independent demand: demand on my final products (from customers) Dependent demand: demand on parts of products (from production department) Types of inventory: - Raw Materials and Purchased parts - Work in progress - Replacement parts, tools & supplies - Good in transit - Finished goods inventories Function of Inventory: Main purpose of inventory is to keep the production running  Continuity Inventory Counting System: Periodic System: Physical count each fiscal year Perpetual System: Universal bar code: Lead time= time from order till receiving the order Holding cost: cost u pay per unit annually to keep the product in house Ordering / set up cost: cost of the purchasing process (from decision to buy till goods are received) Shortage cost ABC Classification System: EOQ Model: Economic Optimal Quantity Advised to manage critical inventory only. Only 1 product, annual demand known, demand even each year, …… Fig 11.2: Reorder point depends on the lead time, the shorter the lead time the less the reorder point. Total cost: Annual carrying cost + Annual ordering cost
  • 3. TC = Q/2 H + D/Q S S= set up cost (per cycle) D= annual demand H: avg annual total cost per unit Q: Quantity produced each cycle D/Q= number of cycles D: from production H: from inventory mgr (last year, total spending on inventory / Avg quantity of inventory available) S: from purchasing Q: we calculate it Min TC: Q/2H = D/Q S QH.Q = 2DS Q2= 2DS/H Q= sqrt 2DS/H Let D= 1000units S= LE40 / order H= LE 2/unit Q= sqrt 2x1000x40/2 = 200 units Min TC: Q/2H = D/Q S 200/2(2) = 1000/200x40 200 = 200 TC= LE400 / year ROP= 10 days = 40 units OR accountants can simply calculate it as: H= 5%-10% Economic Production Quantity (EPQ): This model assumes we r producing what we consume! (Eg. Mardini (curtains) and its textile factory). Problem: P. 588 OT = 220 days / year Usage = 50 couplings / day Production = 200 couplings / day Holding cost = $2 / coupling Setup cost = $70 / order Qo = sqrt 2DS/H x sqrt P/P-U D = usage x operating time = 50 x 220 = 11000. Qo = sqrt 2DS/H x sqrt P/P-U = sqrt 2x11000x20 / 2 x sqrt 200/150 = 1013
  • 4. Number of runs per year: 11000/1013 = 10.858 = 11 runs Qo / u = 1013/50 = 20 USE Qo / p = 1013/200 = 5 PRODUCE Only Qo = 20-5 = 15 days Use(?) Imax = Qo (P-U/P) or 1-U/P