EOQ model presentaion
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Transcript of 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
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Contents Of EOQ
Definition of EOQ
How to use the EOQ model in abusiness organization
Total Cost Book Case
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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.
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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.
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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
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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
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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 ?
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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
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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
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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.
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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
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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.
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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)
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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)
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Q uestion no 12:Solution
(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 %
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Q uestion no 23:Solution
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
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