Unit III - Inventory Problems

Syllabus
• Introduction
• Types of Inventory
• Costs Involved in Inventory Problems
• Notations
• Economic Order Quantity (EOQ) Model with Constant Rate of Demand
• Limitations of the EOQ Formula
• EOQ Model with Finite Replenishment Rate
• EOQ Model with Shortages
• Order – Level Lot - Size System
• Order – Level Lot – Size System with Finite Replenishment Rate
• EOQ Model with Quantity Discounts
• EOQ with One-Price Break
• EOQ with Two-Price Breaks

Economic Order Quantity (EOQ) Model with
Constant Rate of Demand
• Economic order quantity is one of the oldest and most commonly
known technique.
• Model developed  Ford Harris and R. Wilson in 1915.
• Objective  Determine economic order quantity Q, which minimizes
the total cost of an inventory system when the demand occurs at a
constant rate.

Following assumption
• The system deals with single item.
• The demand rate of R-units per time unit is known and constant.
• Shortages aren’t allowed.
• Lead-time is zero.
• Replenishment rate is infinite.
• Replenishment size, Q, is the decision variable.
• T is cycle time.
• The inventory holding cost, h unit per time unit and ordering cost, A per
order is known and constant during the period under review.

• At each replenishment, Q-units are ordered and stocked in the
system.
• Demand is occurring at the rate of R-units per time during cycle time
T.
Q = RT (analyze one cycle)
Q
0 T Time
- R
Inventory level
Figure 1

• Let Q(t) denote on-hand inventory at time t of a cycle.
• Using differential equation that describes the instantaneous states of
Q(t), 0 ≤ 𝑡 ≤ 𝑇
• 𝑑𝑄 𝑡 = −Rdt
•
𝑡
0
𝑑𝑄 𝑡 =
𝑡
0
− Rdt
•
𝑑𝑄(𝑡)
𝑑𝑡
= −𝑅, 0 ≤ 𝑡 ≤ 𝑇 _____________ (1)
• With initial condition Q(0) = Q = RT. Then solution of Equation (1) is
given by
• Q(t) = −Rt
• Q(t) = R(T – t), 0 ≤ 𝑡 ≤ 𝑇 _____________ (2)

• Average inventory, I1(Q), in the system per time unit is
• I1(Q) =
1
𝑇
𝑇
0
𝑄 𝑡 𝑑𝑡 = 𝑅𝑇 =
𝑄
2
_____________________(3)
• Total cost, TC(Q), of an inventory system per time unit is
• TC(Q) = IHC + OC =
ℎ𝑄
2
+
𝐴𝑅
𝑄
_____________________(4)
• Using the classical optimization technique, the optimum value of
• Q = Q0 can be obtained by setting
𝑑𝑇𝐶(𝑄)
2
= 0.
• Q0 =
2𝐴𝑅
ℎ
(Willison) _____________(5)
• Total cost TC(Q0) = 2𝐴ℎ𝑅 _____________(6)

•
𝑑2
𝑇𝐶(𝑄)
𝑑Q2
0
=
2𝐴𝑅
Q3
0
> 0 for all Q0 and total cost TC(Q0), obtained in equation
(6) is minimum.
• The optimum cycle time T0 =
Q0
𝑅
=
2𝐴
ℎ𝑅
_____________ (7)
• Graphically, cost equation TC(Q) =
ℎ𝑄
2
+
𝐴𝑅
𝑄
= TC1(Q) + TC2(Q) can be
represented from follows figure 2.

Note 1
• If the unit cost is taken into account then
• TC(Q) = CR +
ℎ𝑄
2
+
𝐴𝑅
𝑄
___________ (8)
• This gives Q0 =
2𝐴𝑅
ℎ
and TC(Q0) = CR + 2𝐴ℎ𝑅

Note 2
• Let P be the selling price per unit.
• Then the gross revenue per time unit is GR = (P – C)R.
• Hence, the net profit per time unit is NP(Q) = GR – TC(Q).
• Then
𝑑𝑁𝑃(𝑄)
𝑑𝑄
= 0 gives Q0 =
2𝐴𝑅
ℎ
• The maximum net profit per time unit is NP(Q0 ) = (P – C)R - 2𝐴ℎ𝑅

Figure 2 : Order – quantity and cost representation
0 Q
Cost
Q0
Q0, TC(Q0)
TC1(Q)
TC2(Q)

• When the lot-size Q is restricted to take discrete values.
• It can’t determine optimum lot-size Q = Q0 by using the differential
equation.
• In this case, the difference equation approach.
• Let the lot-size Q be restricted to take the values u, 2u, 3u, …. Etc.
• Then the necessary condition for Q0 to be optimum lot-size is
• TC(Q0) ≤ TC(Q0 + u) ________________ (9)
• TC(Q0) ≤ TC(Q0 − u) ________________ (10)
• Equation (9) gives
ℎ𝑄0
2
+
𝐴𝑅
𝑄0
≤
ℎ(𝑄0
+ 𝑢)
2
+
𝐴𝑅
Q0 + u
• On simplication is 𝑄0(Q0 + u) ≥
2𝐴𝑅
ℎ
________________ (11)

• From equation (10) and (11), the lot-size Q = Q0 is optimum if
• 𝑄0(Q0 + u) ≤
2𝐴𝑅
ℎ
≤ 𝑄0(Q0 + u)

Sensitivity of lot-size model
• For the lot-size model, we have total cost of an inventory system per time
unit as
• TC(Q) =
ℎ𝑄
2
+
𝐴𝑅
𝑄
and Q0 =
2𝐴𝑅
ℎ
• Suppose that instead of ordering for Q0-units (given above), we replenish
another lot-size Q1.
• Q1 = b Q0, b> 0 and let TC1(Q1) be the corresponding total cost of an
inventory system.
• The ratio
TC1(Q1)
TC(Q0)
=
1+𝑏2
2𝑏
is known as the measure of sensitivity of the lot-
size model.

Limitations of the EOQ Formula
• It is derived under several rigid assumptions which give rise to
limitations on its applicability.
• In practice, the demand is neither known with certainty nor is
uniform over the time period.
• If the fluctuations are mild, the formula is practically valid; but when
fluctuations are wild, the formula loses its validity.
• It is not easy t measure the inventory holding cost and the ordering
cost accurately.
• The ordering cost may not be fixed but will depend on the order
quantity Q.

• The assumptions of zero lead-time and that the inventory level will
reach to zero at the time of the next replenishment is not possible.
• The stock depletion is rarely uniform and gradual.
• One may have to take into account the constraints of floor-space,
capital investment, etc, in stocking the items in the inventory system.

Example 1
• Using the formula information, obtain the EOQ and the total variable
cost associated with the policy of ordering quantities of that size.
Annual demand = 20,00o units, ordering cost = Rs.150 per order, and
inventory carrying cost is 24% of average inventory value.
• O/R = 20,000 units, A = Rs.150/order, h = Rs. ? 24% = 0.24 unit/annum.
• Q0 =
2𝐴𝑅
ℎ
=
2 𝑥 150 𝑥 20,000
0.24
=
6,000,000
0.24
= 25,000,000 = 5,000 units
• TC(Q0) = 2𝐴ℎ𝑅 = 2 𝑥 150 𝑥 0.24 𝑥 20,000 = 1,440,000
• = 1440000 = Rs. 1,200

Example 2
• An oil engine manufacturer purchases 42 lubricants pieces from a vendor.
The requirement of these lubricants is 1,800 per year. What should be the
order quantity per order, if the cost per placement of an order is Rs.16
and inventory carrying charge per rupee per year os 20 paise.
• R = 1,800 x 42 = 75,600 units, A = Rs.16/order and h = 0.20 per unit/year
• Q0 =
2𝐴𝑅
ℎ
=
2 𝑥 16 𝑥 75,000
0.20
=
2,419,200
0.20
= 12,096,000 = 34,776 units
• The optimum inventory quantity of lubricant at the rate of Rs.42 =
𝑄0
42
• =
34,776
42
= 83 lubricants.

Example 3
• A company uses rivets at a rate of 5,000 kg per year, rivets costing
Rs.2 per kg. It costs Rs.20 to place an order and the carrying cost of
inventory is 10% per annum. How frequently should order for rivets
be placed and how much?
• R = 5,000 kg/year, C = Rs.2/kg, A = Rs.20/order and
h = Ci = 2 x 10% per unit/year
• Q0 =
2𝐴𝑅
𝐶𝑖
=
2 𝑥 20 𝑥 5,000
0.2
=
200,040
0.2
= 1,000,200 = 1000 kg
• T0 =
𝑄0
𝑅
=
1,000
5,000
=
1
5
years = 2.4 months

Example 4
• A supplier ships 100-units of a product every Monday. The purpose
cost of the product is Rs.60 per unit. The cost of ordering and
transportation from the supplier is Rs.150 per order. The cost of
carrying inventory is estimated at 15% per year of the purchase cost.
Find the lot-size that will minimize the cost of the system. Also,
determine the optimum cost.
• R = 100 units per week , A = Rs.150/order, h = 15% of 60
= 1.5 x 60 = Rs.9 per unit/year = Rs.9/52 per unit/week.
• Q0 =
2𝐴𝑅
ℎ
=
2 𝑥 150 𝑥 100
0.17
=
30,000
0.17
= 176470.58 = 416 or 420 units
• Optimum cost = CD + TC(Q0) = CR + 2𝐴ℎ𝑅 = Rs.6,072

Example 5:
• A company plans to consume 760 pieces of a particular component.
Past records indicate that the purchasing department sent Rs.12,555
for placing 15,500 purchase orders. The average inventory was valued
at Rs.45,000 and the total storage cost was Rs.7,650 which included
wages, taxes, rent, insurance, etc. related to the store department.
The company borrows capital at the rate of 10% per year. If the price
of a component is Rs.12 and the lot-size is 10, find the following:
• Purchase price per year
• Purchase expenses per year
• Storage expenses per year
• Capital cost per year
• Total cost per year

• R = 760 pieces
• Volume = 15,500 purchase order
• Storage cost = Rs.7,650
• Ordering cost, A = Rs.12,555/year cost per order
= 12,555/15,500 = Rs.0.81
• Average inventory = 45,000 units.
• i = Inventory carrying charge fraction =
𝑆𝑡𝑜𝑟𝑎𝑔𝑒 𝑐𝑜𝑠𝑡
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑖𝑛𝑣𝑒𝑛𝑡𝑜𝑟𝑦
=
7,650
45,000
= 0.17

• Purchase price per year = 12 x 760 = Rs.9,120
• Purchase expenses per year = 0.81 x 760 = Rs.615.60
• Storage expenses per year = (1/2)QCi = Rs.10.20
• Capital cost per year = (1/2) x Q x C x 10% = Rs.6
• Total cost per year = (1) + (2) + (3) + (4) = Rs.9,751.80

Example 6
• If in the model developed in the set-up cost, A, is replaced by A + bQ
where b is the set-up cost per unit item produced, then show the
optimum order quantity produced due to this change if the set-up
cost remains unaffected.
• With the new set-up cost equation (4) in the form of T becomes
• TC(T) =
ℎ
2
RT +
(𝐴+𝑏𝑄)
𝑇
=>
ℎ
2
RT +
(𝐴+𝑏𝑅𝑇)
𝑇
=>
ℎ
2
RT +
𝐴
𝑇
+ bR
• For the optimum cost,
𝑑𝑇𝐶(𝑇)
𝑑𝑇
=
1
2
hR -
𝐴
𝑇2 = 0 which gives T0 =
2𝐴
ℎ𝑅
and
Q0 =
2𝐴𝑅
𝑅
• This is same as equation (5) and there will be no change in the
optimal order quantity due to change in the set-up cost.

Example 7
• Data relevant to component A used by Engineering India Private
Limited in 20 different assemblies includes: purchase prices = Rs.15
per 100, annual usage = 1,00,000 units, cost of buying office (fixed)
Rs.15,575 per annum, set-up cost = Rs.12 per order, rent of
component = Rs.3,000 per annum, interest = 25% per annum,
insurance = 0.05% per annum based on total purchase, depreciation
as 1% per annum of all items purchase. Calculate
• EOQ for component A
• The percentage changes in total amount costs relating to component
A if the annual usage was (a) = 1,25,000-units and (b) 75,000-units

• R = 1,00,000 units, A = Rs.12/order, h =
15
100
0.25 + 0.0005 + 0.01
• = Rs.0.039075 per unit per annum
• Q0 = 7,837-units and TC(Q0) = Rs.306.25
• 1. Ra = 1,25,000-units, Qa0 = 8,762-units and TCa(Qa0) = Rs.342.31
• 2. Rb = 5,000-units, Qb0 = 6,787-units and TCb(Qb0) = Rs.265.20
• The total annual cost increases by 12% when the annual demand is
1,25,000 units, whereas it decreases by 13% when the annual
demand is 75,000 units.

EOQ Model with Finite Replenishment Rate
• The replenishment rate is infinite.
• A system in which the replenishment rate, P is finite.
• Obviously, the replenishment rate P is larger than the demand rate R.
• The mathematical model is derived under the same assumptions.
• The replenishment rate is finite at P units/time with P > R.
• Describe and analyze one cycle as follows

• Figure 3 – Representation of time – Inventory level
Inventory level
0
Q
t1 T Time
- R
P - R

• We start with zero inventory level.
• The production starts at this point of time at the rate of P-units per
time unit and the demand occurring at the rate of R ( < P)-units, are
satisfied from the production.
• The production continues till Q = RT-units are produced.
• The demand is satisfied from accumulated inventory at the rate of R-
units,
• The production again starts when on-hand inventory level reaches
zero.
• Figure 3 t1 is time at which production stops and T is the cycle time at
which the inventory level reaches zero.

• Let Q(t) denote the on-hand inventory at any instant of time t, 0 ≤
𝑡 ≤ 𝑇.
• It is assumed that production of units continues in (0, t1) and during
(t1, T) demand is satisfied from accumulated memory.
• Differential equation governing instantaneous state of inventory is
given by
•
𝑑𝑄(𝑡)
𝑑𝑡
= −𝑅,
𝑃 −𝑅, 0 ≤ 𝑡 ≤ 𝑡1
𝑡1 ≤ 𝑡 ≤ 𝑇
_______________ (12)
• With initial condition Q(0) = 0 and boundary conditions Q(t1) = Q =RT
and Q(T) = 0.
• The solution of equation (12) is
• Q(t) = 𝑅(𝑇 −𝑡),
𝑃 −𝑅 𝑡, 0 ≤ 𝑡 ≤ 𝑡1
𝑡1 ≤ 𝑡 ≤ 𝑇
_______________ (13)

Note
• Q(t) is considered to be continuous function of time t.
• Q(t1) = (P – R) t1 = R(T – t1), which gives t1 = QIP.
• The average inventory in the system per time unit is
• I1(Q) =
1
𝑇 0
𝑇
𝑄 𝑡 𝑑𝑡 ⇒
1
𝑇
[ 0
𝑡1
𝑄 𝑡 𝑑𝑡 + 𝑡1
𝑇
𝑄 𝑡 𝑑𝑡] ⇒
1
2
Q 1 −
𝑅
𝑃
• The total cost of the system per time unit is
• TC(Q) = IHC + OC = hI1(Q) +
𝐴𝑅
𝑄
•
1
2
hQ 1 −
𝑅
𝑃
+
𝐴𝑅
𝑄
______________ (14)

• Setting
𝑑𝑇𝐶(𝑄)
𝑑𝑄
0 gives optimum value of lot-size,
• Q = Q0 =
2𝐴𝑅
ℎ[1 −
𝑅
𝑃
]
________________ (15)
• The minimum total cost per time unit is
• TC(Q0) = 2𝐴𝑅(1 −
𝑅
𝑃
) _________________ (16)
• Because
𝑑2
𝑇𝐶(𝑄)
𝑑𝑄2 =
2𝐴𝑅
𝑄3
0
> 0 for all Q0
• Optimum cycle time T = T0 =
2𝐴
ℎ𝑅[1 −
𝑅
𝑃
]
_____________ (17)

Note 1
• If P = R, i.e. if the rate of replenishment is equal to the demand rate
then replenishment will have to take place continuously.
• In this case, there will be neither carrying cost nor replenishment
cost.

Note 2
• If P  ∞, 𝑖. 𝑒. replenishment rate is infinite,
all equations derived here will be same as those derived.

Example 8
• A product is to be manufactured on a machine. The cost, production,
demand, etc.
• Ordering cost per order = Rs.30
• Purchase cost per unit = Rs.0.10
• Inventory holding cost per unit per annum = Rs.0.05
• Production rate = 1,00,000 units per year
• Demand rate = 10,000 units/year
• Determine the economic manufacturing quantity.

• A = Rs.30 per order, C = Rs.0.10 per unit, h = 0.05/unit/annum,
• P = 1,00,000 units per year and R = 10,000 units per year
• Q0 =
2𝐴𝑅
ℎ[1 −
𝑅
𝑃
]
=
2 𝑥 30 𝑥 10,000
0.05[1 −
10,000
1,00,000
]
=
600,000
0.05[1 − 0.1 ]
=
600,000
0.05[0.9]
=
600,000
0.045
= 13333333 = 3651 units

Example 9
• A contractor has to supply 10,000 bearing per day to an automobile
manufacturer. He finds that when he starts a production run, he can
produce 25,000 bearings per day. The cost of holding a bearing in
stock for one year is Rs.2 and the set-up cost of a production run is
Rs.1,800. How frequently should production run be made?
• A = Rs.1800/production run, R = 10,000 bearings/day, h =
Rs.2.00/bearing/year = Rs.0.0055/bearing/day, P = 25,000
bearings/day
• Q =
2𝐴𝑅
ℎ[1 −
𝑅
𝑃
]
=
2 𝑥 1800 𝑥 10,000
0.0055[1 −
10,000
25,000
]
=
36,000,000
0.0055[1 − 0.4 ]
=
36,000,000
0.0055[0.6]

• =
36,000,000
0.0033
= 10,909 = 1,04,447 bearings
• T =
𝑄
𝑅
=
1,04,447
10,000
= 10.5 days
• Lengths of production cycle = t1 =
𝑄
𝑃
=
1,04,447
25,000
= 4 days
• The production cycle starts at an interval of 10.5 days
• The production continues for 4 days.

Example 10
• Find the most economic batch quantity of a product on a machine if the
production rate of that item on the machine is 200 pieces per day and the
demand is uniform at the rate of 100 pieces per day. The ordering cost is
Rs.200 per batch and the cost of holding one item in inventory is Rs.0.81
per day. How will the batch quantity vary if the production rate is infinite?
• A = Rs.200/order, h = Rs.0.81/unit/day, R = 100 units/day, P = 200 units/day.
• Q =
2𝐴𝑅
ℎ[1 −
𝑅
𝑃
]
=
2 𝑥 200 𝑥 100
0.81[1 −
100
200
]
=
40,000
0.81[1 − 0.5 ]
=
40,000
0.81[0.5]
=
40,000
0.405
=
40,000
0.405
= 98765.43 = 314 or 317 units.

• T =
𝑄
𝑅
=
317
100
= 3.17 days
• Lengths of production cycle = t1 =
𝑄
𝑃
=
317
200
= 1.5 days
• The production rate is infinite, i.e. P  ∞ then
• Q =
2𝐴𝑅
ℎ
=
2 𝑥 200 𝑥 100
0.81
=
40,000
0.81
= 49,382.71 = 222 units

EOQ Model with Shortages
• It derived were based on the assumption that shortages are not
allowed.
• Allowing shortages increases the cycle time and reduces carrying
costs.
• Also, back-log of units is advantages when the time value of inventory
is very high.
• The assumption is relaxed here by the allowing shortages for some
period of time and shortage cost is 𝜋 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑠ℎ𝑜𝑟𝑡.
• Shortage cost is directly proportional to the average number of units
short.

• This model is also known as Order Level System.
• The scheduling period Tp is prescribed constant, and so the lot-size
Qp = RTp raises the on-hand inventory at the beginning of each
scheduling period ot the order level S.
• Shortages are made up when the next procurement arrives.
• Order level, S is the decision variable.
• The time-inventory representation is exhibited.

• Figure 4 – EOQ model with shortages
Qp
S
0
- R
t1
Tp
Qp - S
Time

• Let Q(t) denote the on-hand inventory at time t (0 ≤ 𝑡 ≤ 𝑇𝑝) of a
cycle.
• The system carries inventory during (0, t1) and runs with shortages
during (t1, Tp).
• Instantaneous state of Q(t) is given by differential equation
•
𝑑𝑄(𝑡)
𝑑𝑡
= −𝑅,
−𝑅, 0 ≤ 𝑡 ≤ 𝑡1
𝑡1 ≤ 𝑡 ≤ 𝑇𝑝
______________(18)
• With initial condition Q(0) = S, Q(t1) = 0 and Q(Tp) = Qp
• Solution of equation (18) is given by
• 𝑄(𝑡) = 𝑅(𝑇𝑝−𝑡),
𝑆−𝑅𝑡, 0 ≤ 𝑡 ≤ 𝑡1
𝑡1 ≤ 𝑡 ≤ 𝑇𝑝
______________(19)

• Q(t1) = 0 gives t1 = S/R.
• Average inventory per time unit is
• I1S =
1
𝑇𝑝
0
𝑇
𝑄 𝑡 𝑑𝑡 =
𝑆2
2𝑄𝑝
_______________(20)
• Average inventory per time unit is
• I2S =
1
𝑇𝑝
𝑡1
𝑇
−𝑄 𝑡 𝑑𝑡 =
(𝑄𝑝 −𝑆)2
2𝑄𝑝
_______________(21)
• The total cost, TC(S), of an inventory per time unit is
• TC(S) = hI1S + 𝜋 I2S
• =
ℎ𝑆2
2𝑄𝑝
+
𝜋(𝑄𝑝 −𝑆)2
2𝑄𝑝
_______________(22)

• For the optimum value of S = S0, we need to set
𝑑𝑇𝐶(𝑆)
𝑑𝑆
= 0, which
gives S0 =
𝜋𝑄𝑝
ℎ+𝜋
________________(23)
• The corresponding total minimum cost per time unit is
• TC(S0) =
ℎ𝜋𝑄𝑝
2(ℎ+𝜋)
________________(24)
• If S is restricted to take only discrete values in multiple of u, them the
reader can easily check that the condition for optimality at S = S0 is
• 𝑆0 −
1
2
𝑢 ≤
𝜋𝑄𝑝
ℎ+𝜋
≤ 𝑆0 +
1
2
𝑢 _____________(25)
• Suppose instead of optimal order level S0 gives in equation (23), one
uses another order-level S’ such that S’ = aS0, a > 0 is constant
• The corresponding cost be TC(S’).

• TC(S’)2 =
ℎ𝑆′
2𝑄𝑝
+
𝜋(𝑄𝑝 −𝑆′
)2
2𝑄𝑝
=
ℎ𝑎𝑆0
2𝑄𝑝
+
𝜋(𝑄𝑝 − 𝑎𝑆0)2
2𝑄𝑝
• TC(S0) = (1 – a)
𝜋
ℎ
TC(S0), 0 ≤ 𝑎 ≤ 1 +
ℎ
𝜋
• The gives
TC(S’)
TC(S0)
= 1 + (1 – a)
𝜋
ℎ
, 0 ≤ 𝑎 ≤ 1 +
ℎ
𝜋
.
• The ratio
TC(S’)
TC(S0)
depends on a, h and 𝜋.
• The assumption of placing an order at the prescribed cycle time.
• The mathematical model in which both lot-size and order level are
decision variables.

Order-Level, Lot-Size System
• In order-level, lot-size system, we need to balance the sum of carrying
cost, shortage cost and ordering cost.
• The assumption here are same as with an addition the lot-size Q is a
decision variable.
• The pictorial representation of order-level lot-size system.

• Figure 5 – Order-level lot-size (OLLS) system
Q S
0
-R
t1
Tp
Q - S
Time
Inventory level
-R

• We analyze one cycle.
• If T denotes cycle time, we obviously have Q = RT.
• Initially, order quantity Q.
• After clearing shortages of (Q – S) units, the initial inventory level is S.
• Let Q(t) denote the on-hand inventory at time t of a cycle, then
clearly Q(0) = S.
• The system carries inventory during (0, t1) and runs with shortages
during (t1, T).
• The differential equation that describes the instantaneous states of
Q(t), 0 ≤ 𝑡 ≤ 𝑇, is given by equation (18) and using Q(0) = S and
Q(t1) = 0, the solution of differential equation (18) is

• 𝑄(𝑡) = 𝑅(𝑡1−𝑡),
𝑆−𝑅𝑡, 0 ≤ 𝑡 ≤ 𝑡1
𝑡1 ≤ 𝑡 ≤ 𝑇
______________(26)
• The condition Q(t1) = 0 gives t1 = S/R.
• Let us calculate average inventory per time unit and average units
shorts per time unit.
• Average inventory,
• I1(S, Q),per time unit =
1
𝑇 0
𝑡1
𝑄 𝑡 𝑑𝑡 =
𝑠2
2𝑄𝑝
____________(27)
• Average shortages,
• I2(S, Q),per time unit =
1
𝑇 𝑡1
𝑇
−𝑄 𝑡 𝑑𝑡 =
(𝑄 −𝑆)2
2𝑄
_________(28)
• The total average cost, TC(Q, S) per time unit is

• TC(Q, S) = hI1(S, Q) + 𝜋I2(S, Q) +
𝐴𝑅
𝑄
• =
ℎ𝑆2
2𝑄
+
𝜋(𝑄 − 𝑆)2
2𝑄
+
𝐴𝑅
𝑄
________________ (29)
• For optimum value of Q = Q0 and S = S0
• We set
𝜕𝑇𝐶(𝑄,𝑆)
𝜕𝑄
= 0 and
𝜕𝑇𝐶(𝑄,𝑆)
𝜕𝑆
= 0.
• It gives S0 =
2𝐴𝑅
ℎ
𝜋
ℎ+ 𝜋
and Q0 =
2𝐴𝑅
ℎ
ℎ+𝜋
𝜋
_________(30)
• The corresponding total minimum cost per time is
• TC(Q0, S0) = 2𝐴ℎ𝑅
𝜋
ℎ+ 𝜋
___________(31)
• Cycle time T0 =
2𝐴
𝑅
ℎ+𝜋
ℎ𝜋
___________(32)
• Optimum storage level (in units) = Q0 – S0 = Q0
ℎ
ℎ+ 𝜋
________(33)

Example 11
• The demand for a certain item is 16 units per period. Unsatisfied
demand causes a shortage cost of Rs.0.75 per unit per short period.
The cost of initiating purchasing action is Rs.15.00 per purchase and
the holding cost is 15% of average inventory valuation per period.
Item cost is Rs.8.00 per unit. Find the minimum cost and purchase
quantity.
• R = 16 units, 𝜋 = Rs.0.75 per unit short, h = Rs.8 x 15% = Rs.1.20 and
A = Rs.15.00/order

• Q =
2𝐴𝑅
ℎ
ℎ+𝜋
𝜋
=
2 𝑥 15 𝑥 16
1.20
1.20+0.75
0.75
=
480
1.20
1.95
0.75
= 400 𝑥 2.6
• = 1,040 = 32 units (appro.)
• TC = 2𝐴ℎ𝑅
𝜋
ℎ+ 𝜋
= 2𝑥 15 𝑥 1.20 𝑥 16
0.75
1.20+0.75
= 576
0.75
1..95
• = 576 𝑥 0.38 = 218.88 = 14.88 (appro.)

Example 12
• A television manufacturing company produces in own speakers,
which are used in the production of its television sets. The television
sets are assembled on a continuous production line at a rate of 8,000
per month. The company is interested in determining when and how
much to procure, given the following information:
• Each time a batch is produced, a set-up cost of Rs.12,000 is incurred.
• The cost of keeping a speaker in stock is Rs.0.30 per month.
• The production cost of a single speaker is Rs.10.00 and can be
assumed to be a unit cost.
• Shortage of a speaker (if there exists) costs Rs.1.10 per month.

• R = 8,000 televisions per month, A = Rs.12,00 per production run,
h = Rs.0.30 per unit per month, 𝜋 = 𝑅𝑠. 1.10 per unit short per month.
• Case(i): When shortages aren’t allowed.
• Q =
2𝐴𝑅
ℎ
=
2 𝑥 12,000 𝑥 8,000
0.30
=
19,20,00,000
0.30
= 64,00,00,000
• = 25,298 speakers
• T =
𝑄
𝑅
=
25,298
8,000
= 3.2 months
• Thus, 25,298 speakers are to be produced every 3.2 months
• Case(ii): When shortages are permitted
• Q =
2𝐴𝑅
ℎ
ℎ+𝜋
𝜋
= 25,298 x
0.30+1.10
1.10
= 25,298 x
1.40
1.10
= 25,298 x 1.27
• = 25,298 x 1.12815 = 28,540 speakers

• T =
𝑄
𝑅
=
28,540
8,000
= 3.6 months.
• When shortages are permitted, 28,540 speakers are produced at
every 3.6 months.
• Number of shortage of speakers =
𝜋𝑄
ℎ+ 𝜋
=
1.10 𝑥 28,540
1.40
=
31,394
1.40
• = 22,424 speakers
• Thus, a shortage of 6,116 (= 28,540 – 22,424) speakers is permitted.

Example 13
• A dealer supplies you the following information with regard to a
product dealt in by him: Annual demand = 5,000 units, ordering cost
= Rs.25.00 per order, inventory carrying cost is 30% per unit per year
of purchase cost Rs.100 per unit.
1. The dealer is considering the possibility of allowing some back-
orders to occur for the product. He has estimated that the annual
cost of back-ordering the product will be Rs.10.00 per unit.
2. What should be the optimum number of units of the product he
should buy in one lot?
3. What quantity of the product should he allow to be back-ordered?
4. How much additional cost will he have to incur on inventory if he
doesn’t permit back-ordering?

1. R = 5,000 units, A = Rs.250 per order, h = 100 * 0.30 = Rs.30 per
unit, 𝜋 = Rs.10.00 per unit.
• Q =
2𝐴𝑅
ℎ
ℎ+𝜋
𝜋
=
2 𝑥 250 𝑥 5,000
30
30+10
10
=
2.500,000
30
40
10
=
83,333.33 4
• = 288.678 x 2 = 576 units
• S =
𝜋𝑄
ℎ+ 𝜋
=
10 𝑥 576
40
=
5760
40
= 144 units.
• The units to be back-ordered = Q – S = 576 – 144 = 432 units
2. TC = 2𝐴ℎ𝑅
𝜋
ℎ+ 𝜋
= 2 𝑥 250 𝑥 30 𝑥 5,000
10
30+10

• = 2 𝑥 250 𝑥 30 𝑥 5,000
10
30+10
= 75,000,000
10
40
• = 8660.254 x 0.25 = 8660.254 x 0.5 = Rs.4,330
3. If back-ordered aren’t permitted, then the total cost of an inventory
system per unit is 2𝐴ℎ𝑅 = Rs.8,660
Additional cost when back-orders aren’t permitted = 8,660 – 4,330
= Rs.4,330.

Example 14
• The annual demand for a product is 3,600 units with an average of
12-units per day. The lead-time is 10 days. The ordering cost per order
is Rs.10 and the annual carrying cost is 25% of the value of the
inventory. The price of the product per unit is Rs.3.00.
1. What will be the EOQ?
2. Find the purchase cycle time.
3. Find the total inventory cost per year.
4. If the safety stock of 100-units is considered necessary, what will be
the reorder level and the total annual cost of inventory which will
be relevant to inventory decision?

1. R = 3,600 units, A = Rs.20 per order, h = Rs.3 * 25% per unit per year
• Lead-time = 10 days
• Since the demand is uniform at 12-units per day, the total number of
working days in the year = 3,600/12 = 300.
• Q =
2𝐴𝑅
ℎ
=
2 𝑥 20 𝑥 3,600
0.75
=
1,44,000
0.75
= 1,92,000 = 438 (appro.)
2. Cycle time =
438
12
= 36.5 days
3. TC = CR +
𝐴𝑅
𝑄
+
ℎ𝑄
2
= 36.5 +
20 𝑥 3,600
438
+
0.75 𝑥 438
2
• = 36.5 +
72,000
438
+
328.5
2
= 36.5 + 164.383562 + 164.25 =

4. Re-order level = Safety stock – Lead-time demand
= 100 + 12 x 10 = 220 units.
Average inventory = Safety stock +
𝑄
2
= 220 +
438
2
= 319 units
TC =
𝐴𝑅
𝑄
+
ℎ𝑄
2
= Rs.164.38

Order – Level Lot – Size System with Finite
Replenishment Rate
• It is based on the assumptions which are same as in the model
developed, except the assumption that “replenishment rate is finite”.
• We analyze one cycle.
• Let the maximum inventory level Q1 be reached at the end of time t1.
• Then Q1 = (P – R)t1.
• After time t1, the stock Q1 is depleted during t2.
• Then Q1 = Rt2.
• During time t3, shortages accumulate at the rate of R.

• The maximum shortages Q2 = Rt3.
• After time t3, the production again starts and shortages are getting
cleared at the rate of (P – R)t4.
• Then Q2 = (P – R)t4.
• Average inventory =
1
2
𝑄1(𝑡1+𝑡2)
𝑇
________ (34)
• Average shortages =
1
2
𝑄2
(𝑡3
+𝑡4
)
𝑇
________ (35)

• Figure 6 – Order-level lot-size (OLLS) system with finite
replenishment
- R
0
Q
t2 t3
Time
Inventory level
-R
P - R
t1 t4
- R P - R
T

• The cycle time, T = t1 + t2 + t3 + t4
• =
𝑄1
𝑃 −𝑅
+
𝑄1
𝑅
+
𝑄2
𝑅
+
𝑄2
𝑃 −𝑅
=
𝑃(𝑄1
+ 𝑄2
)
𝑅(𝑃 −𝑅)
_____ (36)
• If Q is the lot-size then Q1 = Q – Q2 – Rt1 – Rt4 =
𝑃 − 𝑅
𝑃
Q – Q2
or
• Q1 + Q2 =
𝑃 − 𝑅
𝑃
Q
• Then, equation (36) becomes
• T =
𝑄
𝑅
_____________ (37)

• The total cost, TC, of an inventory system per time unit is
• TC(Q1, Q2) =
𝐴𝑅
𝑄
+
1
2
𝑄1(𝑡1+𝑡2)
𝑇
h +
1
2
𝑄2(𝑡3+𝑡4)
𝑇
𝜋
• =
𝐴𝑅
𝑄
+
1
2𝑄
𝑃
𝑃 −𝑅
ℎ
𝑃 − 𝑅
𝑃
𝑄 − 𝑄2
2 + 𝑄2
2𝜋 ___(38)
• The optimum values of Q and Q2 can be obtained by setting partial
derivatives of TC(Q1, Q2) with respect to Q and Q2 equal to zero which
gives
• Q =
2𝐴𝑅
ℎ
𝑃
𝑃 − 𝑅
ℎ+ 𝜋
𝜋
_______(39)
• Q2 = Q 1 −
𝑅
𝑃
ℎ
ℎ+ 𝜋
_______(40)

• Cycle time T =
𝑄
𝑅
=
2𝐴
ℎ𝑅
𝑃
𝑃 − 𝑅
ℎ+ 𝜋
𝜋
________ (41)
• Optimum inventory level Q1 = Q – Q2 =
2𝐴𝑅
ℎ
1 −
𝑅
𝑃
𝜋
ℎ+ 𝜋
__(42)
• Total minimum cost = 2𝐴ℎ𝑅 1 −
𝑅
𝑃
𝜋
ℎ+ 𝜋
______(43)
• The obtained in equation (43) is minimum since second order partial
derivatives of TC with respect to Q and Q2 are both positive.

• Note 1: If replenishment rate, P  ∞, then the results derived here
are same.
• Note 2: If 𝜋  ∞, then the above derived results are same as the
one.
• Note 3: If P  ∞ and 𝜋  ∞, then we get EOQ model derived.

Example 15
• The demand for an item in a company is 18,000 units per year, and
the company can produce the item at a rate of 3,000 per month. The
cost of one set-up is Rs.500 and the holding cost of one unit per
month is Rs.0.15. The shortage cost of one unit is Rs.20.00 per
month. Determine the optimum manufacturing quantity and the
number of shortages. Also, determine the manufacturing time and
the time between two set-up?

• h = Rs.0.15 per unit per month, 𝜋 = Rs.20.00 per unit short,
• A = Rs.500 per order, P = 3,000 units per month
• R = 18,000 unit per year = 1,500 units per month
• Q =
2𝜋(ℎ+𝜋)
ℎ𝜋
𝑃𝑅
𝑃 − 𝑅
=
2 𝑥 20(0.15+20)
0.15 𝑥 20
3,000 𝑥 18,000
3,000 −18,000
• =
40(20.15)
3
54,000,000
15,000
=
806
3
36,00 = 268.66 𝑥 36,00
• = 268.66 𝑥 36,000 = 4,488 units (appro.)
• Number of shortages =
ℎ
ℎ+𝜋
Q 1 −
𝑅
𝑃
= 17 units (appro.)

• Manufacturing time =
𝑄
𝑃
= 0.1246 years
• The time between two set-ups =
𝑄
𝑅
= 0.2493 years

EOQ Model with Quantity Discounts
• We have developed models by taking unit cost C to be known and
constant during the cycle time.
• However, the retailer offers discount in purchase price to increase his
sale.
• Quantity discounts are offered.
i. All units discounts
ii. Incremental quantity discounts.

• The assumption are same, except the constant unit purchase cost.
• Since the purchase price of an inventory item is variable, it is to be
included while finding the total cost of an inventory system.
• EOQ when price discounts are available by minimizing the total costs
of an inventory system.
• Suppose there are several price breaks say 0, b1, b2, ……, bn and the
order quantity Q lies in the discount interval bk-1 ≤ 𝑄 ≤ 𝑏𝑘.
• The corresponding unit cost for Q-units is Ck where Ck < Ck-1.
• Then the total annual cost in the discounted period is
• TCk = CkR +
𝐴𝑅
𝑄
+
𝐶𝑘
𝐼𝑄
2
, bk-1, ≤ Q ≤ bk

• Figure 7 – Inventory level with price breaks
0
b2
b3
Total cost
Total inventory
cost
b1 Order quantity Q

• Since the total cost curve is not continuous, the calculus method fails.
• We need to move in each price break to find the optimum lot-size,
which minimizes the total inventory cost.
• Discuss about model with one-price break and two-price breaks in
detail.

EOQ with One-Price Break
• The retailer offers quantity discount at b1, i.e.
• Use the following algorithm to determine
optimal purchase quantity
• Step 1: Consider the lowest price C2 and compute Q2 using the basic
EOQ formula. If b1 ≥ 𝑄2 then Q2 is the required lot −
size to be procured and compute optimal cost associated with Q2.
• Step 2: If Q2 < b1, then compute Q1 with price C1 and the
corresponding total cost TC(Q1).
• Compare TC(b1) and TC(Q1).
• If TC(b1) > TC(Q1) then buy Q1-units, otherwise purchase b1-units.
Quantity Price/Unit
0 ≤ 𝑄1 < 𝑏1 C1
b1 ≤ Q2 C2(<C1)

Example 18
• Find the optimum order quantity for a product with price-breaks as
under.
• The monthly demand for the product is 200-units, the holding cost is
2% of the unit cost and the cost of ordering is Rs.350.00
Quantity Price/Unit
0 ≤ 𝑄1 < 500 10.00
500 ≤ Q2 9.25

• A = Rs.350/order, R = 200 units/month, I = Rs.0.20/unit/annum,
• C1 = Rs.10.00/unit and C2 = Rs.9,25/unit
• C2 < C1
• Q2 =
2𝐴𝑅
𝐶2𝐼
= 870 𝑢𝑛𝑖𝑡𝑠 > 500(= 𝑏1)
• i.e. Q2 is within the range of Q2 > 500.
• The optimum purchase quantity is Q2 = 870 units

Example 19
under.
• The yearly demand for the product is 1,600-units per year, the cost of
placing an order is Rs.5.00 and the carrying charge fraction is 10% per
year.
Quantity Price/Unit
0 ≤ 𝑄1 < 800 Rs.1.00
800 ≤ Q2 Rs.0.98

• R = 1,600 units/year, A = Rs.5.00/order, I = Rs.0.10 units/year
• C1 = Rs.1.00/unit, C2 = Rs.0.98/units
• C2 < C1
• Q2 =
2𝐴𝑅
𝐶2
𝐼
= 404 𝑢𝑛𝑖𝑡𝑠 < 800(= 𝑏1), so go to step 2 of algorithm
• Q1 =
2𝐴𝑅
𝐶1𝐼
= 400 units < b1
• So, we compute the total cost TC(Q1) = 2𝐴ℎ𝑅 + 1,600 = Rs.1,640
• TC(b1) = C2R +
𝐴𝑅
𝑏1
+
𝐶2𝐼𝑏1
2
= Rs.1,617.20
• TC(b1 = 800) < TC(400).
• The optimum purchase quantity is 800-units.

EOQ with Two-Price Break
• The price discount offer is as follows with two-price breaks b1 and b2.
• With C1 > C2 > C3.
Quantity Price/Unit (In
Rs.)
0 ≤ 𝑄1 < 𝑏1 C1
b1 ≤ Q2 ≤ 𝑏2 C2
b2 ≤ Q3 C3

Steps as under to determine the optimal lot-size
Step 1: Consider the lowest price C3.
Compute Q3 using the basic EOQ formula.
If Q3 ≥ 𝑏2 then Q3, TC Q3 is the optimal solution.
If Q3 < b2, then go to step 2.
Step 2: Determine Q2 with purchase cost C2.
If b1 ≤ 𝑄2 ≤ 𝑏2, 𝑡ℎ𝑒𝑛 𝑐𝑜𝑚𝑝𝑎𝑟𝑒 𝑇𝐶 𝑄2 𝑎𝑛𝑑 𝑇𝐶 𝑏2 .
If 𝑇𝐶 𝑄2 𝑎𝑛𝑑 𝑇𝐶 𝑏2 , then b2, TC b2 is the solution.
Otherwise, Q2, TC Q2 is the optimal solution.
If Q2 < b1 as well as b2, then go to step 3.
Step 3: Find Q1 with unit cost C1.
Compare TC(b1), TC(b2) and TC(Q1) to find optimum EOQ.
The quantity with the lowest cost will be the optimum purchase quantity.

Example 20
under.
• The monthly demand for the product is 200-units per year, the
carrying charge function is 20% of the unit cost, and the set-up cost is
Rs.350 per order.
Rs.)
0 ≤ 𝑄1 < 500 10.00
500 ≤ Q2 ≤ 750 9.25
750 ≤ Q3 8.75

• A = Rs.350/order, I = Rs.0.20/unit/annum and R = 200 units/month.
• Q3 =
2𝐴𝑅
𝐶3𝐼
= 894 units > 750 (=b2).
• Q3 is the optimum purchase quantity.

Example 21
under.
Rs.20.00 per order.
Rs.)
0 ≤ 𝑄1 < 50 10.00
50 ≤ Q2 ≤ 100 9.00
100 ≤ Q3 8.00

• R = 200 units/month, I = Rs.0.25/unit/annum
• A = Rs.20.00/order
• Q3 =
2𝐴𝑅
𝐶3
𝐼
= 63 units <b2(= 100).
• So, we compute Q2 =
2𝐴𝑅
𝐶2
𝐼
= 60 units.
• Since Q2 > b (=50), to decide optimum purchase quantity we compare
• TC(Q2) = R x I2+A x
𝑅
𝑄2
+ I2xI +
𝑄2
2
= 200 x 9 + 20 x
200
60
+ 9 x 0.25 x
60
2
= Rs.1,934.16
• TC(b2) = R x I3+A x
𝑅
𝑄3
+ I3xI +
𝑄3
2
= 200 x 8 + 20 x
200
100
+ 8 x 0.25 x
100
2
= Rs.1,740.00.
• Then, TC(Q2) > TC(b2).
• Hence, b2 = 100 units will be the optimum lot-size.

Example 22
under.
Rs.25.00 per order.
Rs.)
0 ≤ 𝑄1 < 100 20
100 ≤ Q2 ≤ 200 18
200 ≤ Q3 16

• R = 400, I = Rs.0.20 and A = Rs.25.00
• Q3 =
2𝐴𝑅
𝐶3
𝐼
= 79 units < b2(= 200).
• Then compute Q2 =
2𝐴𝑅
𝐶2
𝐼
= 75 units < b1(= 100).
• So we need to compute Q1 =
2𝐴𝑅
𝐶1
𝐼
= 70 units < b1.
• So we need to compare costs.
• TC(Q1) = 400 x 20 + 25 x
400
70
+ 20 x 0.20 x
70
2
= Rs.8,283.
• TC(b1) = 400 x 18 + 25 x
400
100
+ 18 x 0.20 x
100
2
= Rs.7,480.
• TC(b2) = 400 x 16 + 25 x
400
200
+ 16 x 0.20 x
200
2
= Rs.6,770.
• Since TC(b2) < TC(b1) < TC(Q1), the optimum purchase quantity is b1 = 100
units.

General
• If there are n-price breaks, then use the following computational procedure
for obtaining optimal purchase quantity:
• Step 1: Compute Qn. If Qn ≥ 𝑏 𝑛 − 1,
the optimum purchase quantity is
decided.
• Step 2: If Qn < 𝑏 𝑛 − 1,
compute Qn-1. If Qn-1 ≥ 𝑏 𝑛 − 1,
proceed as in the case
of two-price break.
• Step 3: If Qn-1 < 𝑏 𝑛 − 2,
compute Qn-2.If Qn-2 ≥ 𝑏 𝑛 − 2,
proceed as in the case
of two-price break.
• Step 4: If Qn-2 < 𝑏 𝑛 − 2,
compute Qn-3.If Qn-2 ≥ 𝑏 𝑛 − 3,
then compare TC(Qn-3)
with TC(𝑏 𝑛 − 3), TC(𝑏 𝑛 − 2) and TC(𝑏 𝑛 − 1).
• Step 4: Continue the steps until Qn-j < 𝑏 𝑛 − (j + 1)
and the compare TC(Qn-j)
with TC(𝑏 𝑛 − 𝑗 − 1), TC(𝑏 𝑛 − 𝑗 − 2) and TC(𝑏 𝑛 − 1).

Unit III - Inventory Problems

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Unit III - Inventory Problems