eoq model presentaion

8/8/2019 EOQ model presentaion

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Group D

Asim Ali

Amir QasimAsim Javed

Faisal Mehboob

Nasir Aziz

Syed Muhammad Hilal



Contents Of EOQ

Definition of EOQ

How to use the EOQ model in abusiness organization

Total Cost Book Case



EOQ MODEL

Economic Order Q uantity

Definition

EOQ , or Economic Order Q uantity, is defined as the optimal quantity of

orders that minimizes total variable

costs required to order and holdinventory.



How to use EOQ in your organization

How much inventory

should we order each

month?

The EOQ tool can be usedto model the amount of

inventory that we should

order each month or year.



EOQ Model Basics

Total Cost = Purchase Cost +Order Cost + Holding Cost

Purchase cost: cost of the inventoryitems.

Ordering cost: cost of purchasing an

receiving an order

Holding Cost :includes costs such asstorage cost



Total Cost Book Equation

TC=f(q)=D C0/Q +

qPCh /2+pD

D=Annual Demand Unit

C0=Ordering Cost per Unit

Ch = Carrying Cost

P=Purchase Price Per Unit

Q =Order Quantity



Example In Our Book

Question

1: For a given inventory items D=5000, Coor S =125,p=100,Ch orH=0.20.detrmine

the value of q which minimizes thetotal inventory costs .what are theminimum annual inventory costs? How

many orders must be placed eachyear? What are annual ordering costs?Annual carrying costs ?



Continued

Answer

we have found all the values of these values

by this formula:Tc=f(q)=d/qC0+q/2PCh+PD

Q that minimizes the inventory cost i.e. Q = 250

Tc total minimum cost i.e. Tc = 505000

Number of order per year i.e. D/Q = 20 Annual ordering cost i.e. D/Q.C0 = 2500

And annual carrying costs i.e. Q/2PCh= 2500



Continued

Question (2)

The order quantity which minimizes theannual inventory cost is termed as EOQ.Determine the general expression for qthat minimizes the annual inventory costs.

Solution

We have found the solution by taking the

derivative of the formula, we got

Tc=f(q)=d/qC0+q/2PCh+PD

f(Q) = -DC/Q 2+1/2PCh



Continued

Question (3)

Prove that the critical value for Q doesresults in the relative minimum on the

cost functionSolution

We can prove this by substituting the valueof critical point i.e.250 in the secondderivative.

f(Q) = .008 which is > 0

So f(250)= .008 which is also >0

Hence proved.



Continued

Question (4)

Using the expression for Q in part 2, show thatannual ordering costs equals annual carryingcosts when operating at EOQ level?

Solution

Tc=f(q)=d/qC0+q/2PCh+PD

f(Q)=-DCo/Q 2+1/2 PCh=0

DCo/Q 2=1/2 PCh

Putting the values from part 1 we get10=10

Hence proved



Continued

Question (5)

The annual inventory cost can be expressed in terms of the

annual number of orders placed per year,N,recognizing

that N=D/Q. rewrite the generalized cost function in

terms of N instead of Q.Detrmine the general expressionfor the values of N which minimizes annual inventory

cost. Confirm that the critical value for N does result in

the minimum value of the cost function.

Solution

f(N)=NCo+D/2N.PCh+PD

f(N)= 12.5 which is > 0

Hence proved that N results in minimum values.



Q uestion no 10:

Solution

C(X) = 40,000,000 ( 36,000,000 10000x ) » x

= 4000,000 » x + 10,000

O(x) = 500 + 0.40x

A(X) = C(x) + O(x)

A(x) = 4000,000 » x + 10500 + 0.40x

Taking Derivative we get

X=3162.27 hours

Taking 2nd Derivative we get

A ( 3162.27 ) = 8000,000 » (3262.27)³ > 0

(B) A(3162.27) = 4000,000 » 3162.27 + 0.40(3162.27)

+ 10500

= 13029.822 $ (Minimum Cost Per Hour)

(C) S(3162.27) = 36000,000 - 10,000 ( 3162.27 )

=4377300 (Salvage Value)



Q uestion no 11:Solution

(A) p = 100 x (Price per Person)(B) 0 x 100 (Yes there are restriction as described)

(C) R = h(x) = ( price for trip ) (number signed up )

= 5000 50x x²

(D) The value result is

R = 50 2x = 0 X = 25 travel above 50(E) Let Y =no of people who signup for trip

The number of people who should sign up for the trip is

50 + x or 50 + 25 = 75

(F) R = h(25) = 5,000 + 50(25) (25)²

= $ 5625 (maximum value of R)(G) Price per ticket in maximum revenue is

P = f (25) = 100 25 = 75

(H)The club will be generate fewer person as

No: 50 * 100 = 5000 (which is less then 5625)




(A) Net profit = gross profit advertising costsGross profit = profit margin * R * population

= (2) ( 1 e-0.02x ) ( 2,000,000 )

= 4,000,000 4, 000,000e-0.02x

Cost of campaign = 10,000$ per day

Net profit = G.P-Cost of campaign

Net profit = P = 4,000,000 4,000,000e-0.02x - 10000xP = 80,000 e0.02x 10,000 = 0

80,000 e-0.02x 10,000

e 0.02x = 10,000

e 0.02x = 0.125

So X = 105

Taking 2

nd

Derivative we getP = - 195-93 < 0 So Relative Maxima

(B) Put Value in the function we get

P = 3531174.287

R = 87.75 %




Q =F(p)=12000-10p²P=10$ ,p 20$ , p=30$

N=p/q.f(p)

F(p)=-20p

N=p/q-20p

N=-20p²/q -2p²/1200-p²

At p=10$=-2(10) ²/1200-(10) ² -0.1818<1 inelastic

At -p=20$

= -2(20) ²/1200-(20) ² -1<1 inelastic

At P=$30

=-2(30) ²/1200-(30) ² -6<1 inelastic

eoq model presentaion

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