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Inventory Control
Dr. T. Anitha
Assistant Professor of Mathematics (SF)
V.V.Vanniaperumal College for Women
Virudhunagar
1. Deterministic Inventory Problem with no
shortages
Case 1: The Fundamental Problem of EOQ
Case 2: Problem of EOQ with Several Production
Runs of Unequal Length
Case 3: Problem of EOQ with Finite Production
2. Deterministic Inventory Problem with
shortages
Case 1: Problem of EOQ with Instantaneous
Production and Variable order Cycle Time
Case 2: Problem of EOQ with Instantaneous
Production and Fixed order Cycle
Case 3: Problem of EOQ with Finite Production
1. Deterministic Inventory Problem with no
shortages
Demand is assumed to be fixed and completely pre-determined.
Case 1: The Fundamental Problem of EOQ
Objective: To determine an optimum order quantity(EOQ) such that the total
inventory cost is minimized.
Assumptions:
 Demand is known and uniform
Shortages are not permitted.
Lead time is zero.
Production or supply of commodity is instantaneous.
Notation
D - Total number of units purchased/produced per time
Q - Lot size in each production run.
𝑪𝒐(or A) - Set-up cost (or ordering cost)
𝑪𝟏- Holding cost per unit.
𝑪- Unit cost
I- Inventory carrying cost
Assume that after each time t, the quantity Q is produced throughout the entire
period, say one year.
If 𝑛 is the total number of runs of the quantity produced during the year.
Then
1 = 𝑛𝑡 and 𝐷 = 𝑛𝑄 --------- (1)
Total inventory over the time period t days is the area of the first
triangle =
1
2
Qt.
The average inventory over the time period t days=
1
2
Qt
𝑡
=
1
2
Q
Annual Inventory holding cost 𝑓 𝑄 =
1
2
QC1
Annual costs associated with runs of size Q is 𝑔 𝑄 = 𝑛𝐶0 =
𝐷
𝑄
𝐶0 (by (1))
Since the minimum total cost occurs at the point where
the ordering cost equal to the total inventory carrying cost.
𝑓 𝑄 = 𝑔(𝑄)
1
2
𝑄𝐶1 =
𝐷
𝑄
𝐶0
𝑸𝒐
= 𝟐𝑫𝑪𝟎/𝑪𝟏
The above EOQ formula can also be expressed in terms of the
economic order value as
𝑸𝒐 = 𝟐𝑫𝑨/𝑪𝑰
Characteristics of Case 1.
1. Optimum number of order placed per year
𝑛0 =
𝐷
𝑄0 =
𝐷𝐶1
2𝐶0
2. Optimum length of time between order (or economic review period ERP)
𝑡0
=
𝑇
𝑛0 = 𝑇
2𝐶0
𝐷𝐶1
𝑇𝐸𝐶0 = Annual inventory carrying cost + Annual ordering cost
=
1
2
𝑄0𝐶1 +
𝐷
𝑄0
𝐶0
Corollary 1
In the EOQ problem, if the set-up (ordering)
cost is 𝐶0 + 𝑏𝑄 instead of being fixed then there is no
change in the optimum order quantity produced due to
change in the set-up cost.
Proof:
The annual total cost is 𝑇𝐶 =
1
2
𝑄𝐶1 +
𝐷
𝑄
(𝐶0 + 𝑏𝑄)0
The optimum value of Q occur at
𝑑
𝑑𝑄
(𝑇𝐶) = 0
⇒ 𝑄 =
2𝐷𝐶0
𝐶1
and
𝑑2
𝑑𝑄2 𝑇𝐶 > 0 for 𝑄 > 0.
Hence 𝑄𝑜
= 2𝐷𝐶0/𝐶1
Case 2 : Problem of EOQ with Instantaneous Production and
Fixed order Cycle
All the assumption are same as in Case 1 except that the demand
is uniform and the production run differ in units.
Let 𝑡1, 𝑡2, … , 𝑡𝑛 denote the times of successive production runs
such that 𝑡1 + 𝑡2 + ⋯ + 𝑡𝑛 = 1 year.
The annual inventory holding cost is
𝑓 𝑄 =
1
2
𝑄𝑡1 𝐶1 +
1
2
𝑄𝑡2 𝐶1 + ⋯ +
1
2
𝑄𝑡𝑛 𝐶1
=
1
2
𝑄 𝑡1 + 𝑡2 + ⋯ + 𝑡𝑛 𝐶1
=
1
2
𝑄𝐶1
Set-up cost 𝑔 𝑄 =
𝐷
𝑄
𝐶0 Since 𝑛𝑄 = 𝐷
Total annual cost 𝑇𝐶 = 𝑓 𝑄 + 𝑔 𝑄 =
1
2
𝑄𝐶1 +
𝐷
𝑄
𝐶0
These cost are same as was obtained in Case 1, so the optimum
quantities are
𝑸𝒐 = 𝟐𝑫𝑪𝟎/𝑪𝟏
𝐓𝐂𝟎
= √𝟐𝑫𝑪𝟏𝑪𝟎.
Remark:
If the period is T instead of one year, then
the optimum order quantity becomes
𝑸𝒐 = 𝟐𝑫𝑪𝟎/𝑪𝟏𝑻
and the minimum total cost
𝐓𝐂𝟎
= 𝟐𝑫𝑪𝟏𝑪𝟎/T.
Thus the uniform rate of demand is replaced by
average rate of demand. ie., D is replaced by D/T.
Case 3: Problem of EOQ with Finite Production
All the assumption are same as in Case 1, except that of
instantaneous replenishment.
Assume that each production run of length t consists of two
parts, say 𝑡1 and 𝑡2 such that
(i) the inventory is building up at a constant rate or (k-r)
units, per unit of time during𝑡1, 𝑘 > 𝑟
(ii) there is no replenishment (or production) during time
𝑡2 and the inventory is decreasing at the
rate of r per unit of time.
The graphical representation of the situation is shown in Fig.
Here, the total (order) quantity Q is produced over a period,
𝑡1, which is defined by the production rate 𝑘. Since the
inventory does not pile up in one shot but rather continuously
over a time period and is also consumed simultaneously, the
average inventory level would be determined not only by the
lot size Q, but also be affected by the production rate k and
depletion (demand) rate r.
To determine the average inventory:
Since 𝑡1 is the time required to produce Q at a rate k, we shall
have
𝑄 = 𝑘𝑡 or 𝑡1 =
𝑄
𝑘
During production period 𝑡1, inventory is increasing at the rate of k and
simultaneously decreasing the rate of r. Thus, inventory accumulates at the
rate of (k-r) units.
Therefore, the maximum inventory level shall be equal to
𝑡1(𝑘 − 𝑟).
Average inventory =
1
2
𝑡1 𝑘 − 𝑟 =
1
2
𝑄(1 −
𝑟
𝑘
). Since 𝑡1 =
𝑄
𝑘
The total annual holding cost is
𝑓 𝑄 = 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑖𝑛𝑣𝑒𝑛𝑡𝑜𝑟𝑦 × 𝐶1 =
1
2
𝑄𝐶1(1 −
𝑟
𝑘
)
Annual ordering cost
𝑔 𝑄 = 𝐶0 ×
𝐷
𝑄
where D is the total demand in a year.
Since the minimum total cost occurs at the point
where annual ordering cost and annual holding
cost are equal. So we must have
𝑓 𝑄 = 𝑔(𝑄)
This implies that
1
2
𝑄𝐶1 1 −
𝑟
𝑘
= 𝐶0 ×
𝐷
𝑄
Hence, the optimum value of Q is
𝑸𝟎 =
𝟐𝑫𝑪𝟎
𝑪𝟏(𝟏 −
𝒓
𝒌
)
=
𝟐𝑫𝑪𝟎
𝑪𝟏
𝒌
𝒌 − 𝒓
Characteristics of Case 3
1. Optimum number of production runs per year
𝒏𝟎 =
𝑫
𝑸𝟎
=
𝑫𝑪𝟏
𝟐𝑪𝟎
𝟏 −
𝒓
𝒌
2. Optimum length of each lot size production run
𝒕𝟏
𝟎
=
𝑸𝟎
𝒌
=
𝟐𝑫𝑪𝟎
𝑪𝟏𝒌(𝒌 − 𝒓)
3. Total minimum production inventory cost
𝑻𝑪𝟎 =
𝐶0𝐷
𝑄0
+
1
2
𝑄0 1 −
𝑟
𝑘
𝐶1
= 𝟐𝑫𝑪𝟎𝑪𝟏(𝟏 −
𝒓
𝒌
)
Thank you

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Inventory Control File II.pptx

  • 1. Inventory Control Dr. T. Anitha Assistant Professor of Mathematics (SF) V.V.Vanniaperumal College for Women Virudhunagar
  • 2. 1. Deterministic Inventory Problem with no shortages Case 1: The Fundamental Problem of EOQ Case 2: Problem of EOQ with Several Production Runs of Unequal Length Case 3: Problem of EOQ with Finite Production
  • 3. 2. Deterministic Inventory Problem with shortages Case 1: Problem of EOQ with Instantaneous Production and Variable order Cycle Time Case 2: Problem of EOQ with Instantaneous Production and Fixed order Cycle Case 3: Problem of EOQ with Finite Production
  • 4. 1. Deterministic Inventory Problem with no shortages Demand is assumed to be fixed and completely pre-determined. Case 1: The Fundamental Problem of EOQ Objective: To determine an optimum order quantity(EOQ) such that the total inventory cost is minimized. Assumptions:  Demand is known and uniform Shortages are not permitted. Lead time is zero. Production or supply of commodity is instantaneous.
  • 5. Notation D - Total number of units purchased/produced per time Q - Lot size in each production run. 𝑪𝒐(or A) - Set-up cost (or ordering cost) 𝑪𝟏- Holding cost per unit. 𝑪- Unit cost I- Inventory carrying cost
  • 6. Assume that after each time t, the quantity Q is produced throughout the entire period, say one year. If 𝑛 is the total number of runs of the quantity produced during the year. Then 1 = 𝑛𝑡 and 𝐷 = 𝑛𝑄 --------- (1)
  • 7. Total inventory over the time period t days is the area of the first triangle = 1 2 Qt. The average inventory over the time period t days= 1 2 Qt 𝑡 = 1 2 Q Annual Inventory holding cost 𝑓 𝑄 = 1 2 QC1 Annual costs associated with runs of size Q is 𝑔 𝑄 = 𝑛𝐶0 = 𝐷 𝑄 𝐶0 (by (1))
  • 8. Since the minimum total cost occurs at the point where the ordering cost equal to the total inventory carrying cost. 𝑓 𝑄 = 𝑔(𝑄) 1 2 𝑄𝐶1 = 𝐷 𝑄 𝐶0 𝑸𝒐 = 𝟐𝑫𝑪𝟎/𝑪𝟏 The above EOQ formula can also be expressed in terms of the economic order value as 𝑸𝒐 = 𝟐𝑫𝑨/𝑪𝑰
  • 9. Characteristics of Case 1. 1. Optimum number of order placed per year 𝑛0 = 𝐷 𝑄0 = 𝐷𝐶1 2𝐶0 2. Optimum length of time between order (or economic review period ERP) 𝑡0 = 𝑇 𝑛0 = 𝑇 2𝐶0 𝐷𝐶1 𝑇𝐸𝐶0 = Annual inventory carrying cost + Annual ordering cost = 1 2 𝑄0𝐶1 + 𝐷 𝑄0 𝐶0
  • 10.
  • 11. Corollary 1 In the EOQ problem, if the set-up (ordering) cost is 𝐶0 + 𝑏𝑄 instead of being fixed then there is no change in the optimum order quantity produced due to change in the set-up cost. Proof: The annual total cost is 𝑇𝐶 = 1 2 𝑄𝐶1 + 𝐷 𝑄 (𝐶0 + 𝑏𝑄)0 The optimum value of Q occur at 𝑑 𝑑𝑄 (𝑇𝐶) = 0
  • 12. ⇒ 𝑄 = 2𝐷𝐶0 𝐶1 and 𝑑2 𝑑𝑄2 𝑇𝐶 > 0 for 𝑄 > 0. Hence 𝑄𝑜 = 2𝐷𝐶0/𝐶1
  • 13. Case 2 : Problem of EOQ with Instantaneous Production and Fixed order Cycle All the assumption are same as in Case 1 except that the demand is uniform and the production run differ in units. Let 𝑡1, 𝑡2, … , 𝑡𝑛 denote the times of successive production runs such that 𝑡1 + 𝑡2 + ⋯ + 𝑡𝑛 = 1 year.
  • 14. The annual inventory holding cost is 𝑓 𝑄 = 1 2 𝑄𝑡1 𝐶1 + 1 2 𝑄𝑡2 𝐶1 + ⋯ + 1 2 𝑄𝑡𝑛 𝐶1 = 1 2 𝑄 𝑡1 + 𝑡2 + ⋯ + 𝑡𝑛 𝐶1 = 1 2 𝑄𝐶1 Set-up cost 𝑔 𝑄 = 𝐷 𝑄 𝐶0 Since 𝑛𝑄 = 𝐷 Total annual cost 𝑇𝐶 = 𝑓 𝑄 + 𝑔 𝑄 = 1 2 𝑄𝐶1 + 𝐷 𝑄 𝐶0 These cost are same as was obtained in Case 1, so the optimum quantities are 𝑸𝒐 = 𝟐𝑫𝑪𝟎/𝑪𝟏 𝐓𝐂𝟎 = √𝟐𝑫𝑪𝟏𝑪𝟎.
  • 15. Remark: If the period is T instead of one year, then the optimum order quantity becomes 𝑸𝒐 = 𝟐𝑫𝑪𝟎/𝑪𝟏𝑻 and the minimum total cost 𝐓𝐂𝟎 = 𝟐𝑫𝑪𝟏𝑪𝟎/T. Thus the uniform rate of demand is replaced by average rate of demand. ie., D is replaced by D/T.
  • 16. Case 3: Problem of EOQ with Finite Production All the assumption are same as in Case 1, except that of instantaneous replenishment. Assume that each production run of length t consists of two parts, say 𝑡1 and 𝑡2 such that (i) the inventory is building up at a constant rate or (k-r) units, per unit of time during𝑡1, 𝑘 > 𝑟 (ii) there is no replenishment (or production) during time 𝑡2 and the inventory is decreasing at the rate of r per unit of time. The graphical representation of the situation is shown in Fig.
  • 17.
  • 18. Here, the total (order) quantity Q is produced over a period, 𝑡1, which is defined by the production rate 𝑘. Since the inventory does not pile up in one shot but rather continuously over a time period and is also consumed simultaneously, the average inventory level would be determined not only by the lot size Q, but also be affected by the production rate k and depletion (demand) rate r. To determine the average inventory: Since 𝑡1 is the time required to produce Q at a rate k, we shall have 𝑄 = 𝑘𝑡 or 𝑡1 = 𝑄 𝑘
  • 19. During production period 𝑡1, inventory is increasing at the rate of k and simultaneously decreasing the rate of r. Thus, inventory accumulates at the rate of (k-r) units. Therefore, the maximum inventory level shall be equal to 𝑡1(𝑘 − 𝑟). Average inventory = 1 2 𝑡1 𝑘 − 𝑟 = 1 2 𝑄(1 − 𝑟 𝑘 ). Since 𝑡1 = 𝑄 𝑘 The total annual holding cost is 𝑓 𝑄 = 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑖𝑛𝑣𝑒𝑛𝑡𝑜𝑟𝑦 × 𝐶1 = 1 2 𝑄𝐶1(1 − 𝑟 𝑘 ) Annual ordering cost 𝑔 𝑄 = 𝐶0 × 𝐷 𝑄 where D is the total demand in a year.
  • 20. Since the minimum total cost occurs at the point where annual ordering cost and annual holding cost are equal. So we must have 𝑓 𝑄 = 𝑔(𝑄) This implies that 1 2 𝑄𝐶1 1 − 𝑟 𝑘 = 𝐶0 × 𝐷 𝑄 Hence, the optimum value of Q is 𝑸𝟎 = 𝟐𝑫𝑪𝟎 𝑪𝟏(𝟏 − 𝒓 𝒌 ) = 𝟐𝑫𝑪𝟎 𝑪𝟏 𝒌 𝒌 − 𝒓
  • 21. Characteristics of Case 3 1. Optimum number of production runs per year 𝒏𝟎 = 𝑫 𝑸𝟎 = 𝑫𝑪𝟏 𝟐𝑪𝟎 𝟏 − 𝒓 𝒌 2. Optimum length of each lot size production run 𝒕𝟏 𝟎 = 𝑸𝟎 𝒌 = 𝟐𝑫𝑪𝟎 𝑪𝟏𝒌(𝒌 − 𝒓) 3. Total minimum production inventory cost 𝑻𝑪𝟎 = 𝐶0𝐷 𝑄0 + 1 2 𝑄0 1 − 𝑟 𝑘 𝐶1 = 𝟐𝑫𝑪𝟎𝑪𝟏(𝟏 − 𝒓 𝒌 )