multi item inventory model

1. Deen Dayal Upadhyaya College
Delhi University
MULTI-ITEM DETERMINISTIC MODEL AND
THEIR ALGORITHM
By Harsh Kansal (Roll number - 20MTS5715)
Abhishek (Roll number – 20MTS5702)
Course – B.Sc Mathematical Science(Sem-2)
Submitted To – Kiran Garg Mam

2. Inventory
A stock or store of goods.

3. Multi Item Deterministic Problem
Problem if the inventory consists of several items with the
some limitations such as availability of total investment or the
limit on the maximum no. of orders that can be placed in a
year (time horizon) no. of delivery which can be accepted etc.
Steps :-
 First we solve the problem ignoring the limitations.
 Considering the affect of limitations , suppose we have n-
items in the inventory with instantaneous production and no
lead time, shortages are not allowed

4. Let
= is the demand for the ith item.
= is the inventory holding cost per unit for the
ith item.
= is the total cost for ith item.
= is ordered quantity.
= +
= +
Di
Hi
TCi
Qi
TCi
2
1 Q
H i
i
Q
A
D
i
i
i
TCi


n
i
i
i
i
Q
A
D
1
2
1 

n
i
i
i Q
H
1

5. Problem of EOQ with warehouse capacity
constraint ( or floor space constraint)
Let fi = is the storage space required for one unit of ith product.
W = Total space available with ware house.
Then the space occupied by all the items must be less then or
equal to total floor area available.
Thus the constraint is :
≤ W
≥ 0 for i = 1,2,3 …….. N


n
i
i
i
Q
f
1
Qi

6. In order to find the optimum value for ( i=1,2,…..n) so as to
minimize (TC), we use the technique of Lagrange Multiplier.
Let λ is the Lagrange Multiplier such that 0 ≤ λ ≤ 1
Then the Lagrange’s function become
L( , λ) = + + λ [ - W ]
The Optimum value of λ & Qi (i=1,2,3…..,n) are obtained by
setting
= 0
- + λ = 0
Hi
2
1
2
1
fi

7. =
And = 0
= 0
i.e = W
λ is indicating addition cost associated with the storage of item.

 f
H
A
D
i
i
i
i

2
2


L
w
Q
f i
i


Q
f i
i


8. Problem of EOQ with inventory level
constraint
The average number of units in the inventory of an item i is Qi/2
Average inventory of n items.
Let P denotes the maximum no. of units of all item that can be
held in the inventory,
≤ P


n
i
i
Q
1
2
1


n
i
i
Q
1
2
1

9. Now the problem is to minimize the total cost
Minimize TC = +
Subject to ≤ k
≥ 0 for i =1,2,3 ……n
In order to find the optimum value for (I =1,2,…..n) so as to
minimize(TC), we use the technique of Lagrange Multiplier.
Let is the Lagrange multiplier such that 0 ≤ ≤1,
Then the Lagrange's function become
L( , ) = + + [


n
i
i
i Q
H
1 2
1
 
 

10. The optimum values of and Q (i=1,2,3,,,n) are obtained by
setting
= 0 and = 0
Thus we get
=
And
= K
Where
≤ 0

d
dL


H
A
D
i
i
i
2
Qi
*

11. Problem of EOQ with investment
constraint
Inventory is ideal resource but require substantial amount of
investment, so decision makes places a limit on the amount
invested inventory.
This limitations incurred in the form of investment
constraint if F is total fund available to be invested on the
inventory then investment constraint would be
≤ F
Where HiQi = is the per unit cost of item
F = be the total funds available to be invested on inventory.
The problem then , becomes to minimize the total cost , subject
to investment constraint
Q
H i
n
i
i

1

12. Thus we have,
Minimize (TC) = +
Subject to , ≤ F
≥ 0 ( i=1,2,3……..n)
To find the optimum values for Qi (i=1,2,3,…..n) so as to
minimize (TC) , we use the Lagrange’s Multiplier method.
Let λ = is the Lagrange multiplier such that 0 ≤ λ≤ 1
The Lagrangian Function become
L( Qi , λ ) =
Q
H i
i
n
i

1 2
1


n
i
i
i
i
Q
A
D
1
Q
H i
n
i
i

1
Qi
F
Q
C
Q
A
D
Q
H i
n
i
i
n
i
i
i
i
i
i
n
i


 

 

 1
1
1
[
2
1


13. The optimum values of λ and Qi are obtained by setting
= 0 and = 0
=
And
= F


L
Qi
L


C
H
A
D
i
i
i
i

2
2

Qi
*
Q
C i
n
i
i

1

14. 1. A machine shop produces 1,2,3...n lots p the shop has a warehouse where
total floor area is 4000 square metres. The relevant data for the three item is
given below:
Item 1 2 3
Annual demand 500 400 600
Cost per unit 30 20 70
Set up cost 800 600
1000
Floor area 5 4 10
The inventory carrying charges for the show or 20% of the average inventory
valuation per annum for each item if number of stocks outs are allowed and at
now time can the warehouse capacity be exceeded and determine the
optimum lot size for each item.
Sol; Optimum ordered quantity for each item is given by:= 2Di Ai/(
Hi + 2λFi) ^ ½
Let us compute Qi* for λ= 0 for item 1
Di=500, Hi= 30, Ai= 800, Fi = 5
Qi* = [2*500*800/(0.20*30+0)]^1/2
=365 per unit
Qi*fi = 1825units

15. For item 2 :
D2=400, H2=20 , A2=600, F2=4
Q2*= [2*400*600/(0.2*20+0)]^1/2
= 346 units
Q2*F2 = 1384 sq.
.
For item 3,
D3=400 , H3=70, A3= 1000, F3=10
Now Q3*= [2 *600*1000/(0.2*70+0)]^1/2
= 292 units
Q3*f3= 2920 sq
Σ(j=1 to n) * Fi*Qi = 1825+1384+2920
= 6129sq metre
As Σ(j=1 to n) Qi*Fi> 4000 sq metre
Thus floor space constraint is not satisfied so now we have to computed Qi* for
each item for different values of
λ.

16. For λ=0.5 ,we get
For item1
Qi*= [2*500*800/(0.20*30 + 2*0.5*5)]^1/2
=270
For item 2 :
D2=400, H2=20 , A2=600, F2=4
Q2*= [2*400*600/(0.2*20+4)]^1/2
= 245 units
.
For item 3,
D3=400 , H3=70, A3= 1000, F3=10
Now Q3*= [2 *600*1000/(0.2*70+0.5*20)]^1/2
= 223 units
Σ(j=1 to n) Fi*Qi = 1350+ 980 + 2230
= 4560sq. units

17. As Σ(j=1 to n) Fi*Qi > 4000 sq. metre
Floor space constraint is not satisfied.
Hence try another value for λ=0.8,we get
For item1
Qi*= [2*500*800/(0.20*30 + 2*0.8*5)]^1/2
=239
For item 2 :
D2=400, H2=20 , A2=600, F2=4
Q2*= [ 2*400*600/(0.2*20+ 6.4)]^1/2
= 274 units
For item 3,
D3=400 , H3=70, A3= 1000, F3=10
Now Q3*= [ 2 *600*1000/(0.2*70+0.8*20)]^1/2
= 200 units
Σ(j=1 to n) F3*Q3= 1195+856+2000
= 4051sq units

18. As Qi*fi = 4051 sq. metre which is slightly more than 4000 sq. metre.
Hence the best value of λ should lie between 0.8 to 1.
2. A small shop produces three machine part 1 2 and 3 in lots, the shop
has limited storage space sufficient only for 500 units of all type of items
the relevant data for the three items is given below:
Item 1 2 3
Demand rate 600 1200
1500
Cost per unit 5 10
15
Set up cost per unit 100 50
200
the inventory carrying charges for the show at 20% of the average
inventory valuation per month for each item if stock out are not allowed,
determine the optimum lot size for each item.
Solution. Let us calculate Q* for λ=0
Q1*= [2*600*100/(0.2*5)]^1/2
= 346 units

19. Q2* =[ 2*1200*50/(0.20*10)]^1/2
= 250 units
Q3* = [2*1500*200/(0.20*15)]^1/2
= 447units
The average inventory level will be :
1/2 Σ(i=1 to n) Qi*= 1/2 (346+250+447)= = 526.5units
As Σ(j=1 to n) Qi> 500
We will calculate for λ=0.1
We get
Q1* = [2*600*100/(0.2*5 + 0.1)]^1/2
= 316units
Q2* = [2*1200*50/(0.20*10 + 0.1)]^1/2
= 230 units
Q3* = [2*1500*200/(0.20*15 +0.1)]^1/2
= 435units
The average inventory level will be :
1/2 Σ(i=1 to n) Qi*= 1/2 (316+230+435)
= 491units
This inventory level is very close to available storage space .hence optimum
lot size is 316unit,230units,435unit respectively.

20. 3. Consider a shop which produces three items the items are
produced in Lourdes the demand rate for each item is constant and
can be assumed to be deterministic no back orders are to be allowed
the pertinent data for the items is given in the following table
determine approximately the EOQ when the total value of average
inventory levels of these items does not exceed rupees thousand.
By ignoring the restriction of total value for inventory level,we have for λ=0
Q1*= (2*50*10000/20) ^0.5=224 units
Q2*= (2*40*12000/20) ^0.5=219 units
Q3*= (2*60*7500/20) ^0.5=212 units
corresponding average value of inventory at any time is:
=224/2*6 + 219/2*7 + 212/2*5=1968.5
1968.50> rs1000 therefore constraint is not satisfied.
For λ=1,
Q1*= (2*50*10000/20+ 2*1*6) ^0.5=177 units
Q2*= (2*40*12000/20+ 2*1*7) ^0.5=168 units
Q3*= (2*60*7500/20 + 2*1*5) ^0.5=173units
Inventory over the average inventory is given by
1/2 Σ(i=1 to n) * ki*Qi =1/2(177*6 + 168*7 +173*5)
= 1551rs.

21. Again rs1551>rs1000 we try for another value of λ>0
For λ=4 we get Q1*=121, Q2*=112, Q3*=123
And the cost of avg. inventory is rs 1112.5
This indicates the best value of λ lies between λ=4 and λ=5. The
most suitable value is λ=4.7
we get Q1*=114, Q2*=105, Q3*=116
And the cost of avg. inventory is Rs 999.50.