Upload
devika-arumugam
View
221
Download
0
Embed Size (px)
Citation preview
7/30/2019 Or Sol7 Inventory Control
1/35
TTN-DoMS, IIT-MADRAS
T.T. NarendranDepartment of Management Studies
Indian Institute of Technology Madras
7/30/2019 Or Sol7 Inventory Control
2/35
TTN-DoMS, IIT-MADRAS
Problem 1Let Q* be the optimal order quantity in the sample EOQ model
(constant rate of demand, instantaneous supply, no shortages)
Let the actual order quantity Q be given by
Q=K.Q*, K>0
a) Derive an expression for the ratio of the actual total
cost/unit time to the optimal cost.
b) If the holding cost/unit/unit time is overestimated by 20
percent, what is the percentage increase in the total cost
over the optimal cost?
7/30/2019 Or Sol7 Inventory Control
3/35
TTN-DoMS, IIT-MADRAS
a) Q*=(2DC0/CC)
Q = k.Q*
TC = DC0/kQ*+kQ*CC/2
TC = DC0CC/ k(2DCo) + k(2DCo) CC/2 CC
= (DC0CC)/(2).k + k(DC0CC)/2
TC = (DC0CC/2) (k+1/k)
TC* = (DC0CC/2)
TC/TC* = (k+1/k)
b) If CC = 1.2 CC*
TC = 2DC(1.2) CC
TC/TC* = 1.2
7/30/2019 Or Sol7 Inventory Control
4/35
TTN-DoMS, IIT-MADRAS
M/s Gujjubhai Shah and company is an old firm where scientific
inventory management is totally unknown. For a certain raw
material that they buy, it is estimated that each purchase order
would cost Rs.200/-. The cost of the raw material is Rs.30 per
kg; the annual inventory carrying cost is 10 per cent of the cost
per kg; the monthly requirement is 100 kg.
The ordering quantity has been arbitrarily fixed as 440kg. If
this ordering quantity is to be optimum by fluke, what should be
the value of the shortage cost, assuming that shortages can be
allowed?
Problem 2
7/30/2019 Or Sol7 Inventory Control
5/35
TTN-DoMS, IIT-MADRAS
Co = Rs.200 ; C = Rs.30
I = 14% per year
D = 100 kg/Month = 1200 Kg/year
Q = 440 ; CS = ?
Q = (2DCo/Cc) (1 + Cc + Cs)
440 = (2*1200*200/4.2)(1 + 4.2/Cs)
Cs = 6.052
7/30/2019 Or Sol7 Inventory Control
6/35
TTN-DoMS, IIT-MADRAS
Marchand and Wadichand Fabrics run a huge show room forReadymade Garments. The Show room is famous for garments madeof a certain variety of cotton. The firm has a quarterly requirementsof 2000 metres length of this variety of cloth, which costs Rs.38/-per metre.
The proprietors have made a precise estimate of their ordering cost -
thirty two rupees, thirty paise per order. The carrying cost per year isestimated to be 8.5 per cent of the cost per metrelength. Marchands son Lalit Chand, who has recently joined thebusiness after completing his MBA, insists that they would procurethe cloth in economic order quantities. However, their supplier, NaiduGaru Textile Mills, Coimbatore, is not willing to sell less than 2000metres of the cloth per order.
What is the discount on the unit cost that Lalit Chand should demand
in order to give up his insistence on EOQ?
Problem 3
7/30/2019 Or Sol7 Inventory Control
7/35
TTN-DoMS, IIT-MADRAS
Demand = 2000 m/quarter
D = 8000 per year
C = Rs.38
Co = Rs.32.30
i = 8.5%
Cc= 38*8.5/100 = 3.23
Q* = (2DCo/Cc)
= 2*8000*32.30/3.23 = 400
TC =(2DCoCc) + 8000*38 = 305292
Let discount be d%
Cc = 38*8.5(1-d)/100
= Rs.3.23(1-d)
7/30/2019 Or Sol7 Inventory Control
8/35
TTN-DoMS, IIT-MADRAS
TC (With discount) = 8000*32.3/2000 + 8000*38*(1-d)
+ 2000*38*8.5*(1-d)/2*100
=129.2+3230(1-d)+304000(1-d)
307230(1-d) = 305292 - 129.2
1-d = 0.9937
d = 0.0063
i.e. 0.63% discount
7/30/2019 Or Sol7 Inventory Control
9/35
TTN-DoMS, IIT-MADRAS
The Central Supplied Unit, IIT, Madras is consideringquotations from three vendors to whom it has to place ordersfor rice.
Eskay Provision Stores quotes Rs.120/- per bag irrespective ofthe quantity ordered. Beavy and Company will accept ordersonly for 800 or more bags but quotes a price of Rs.108/- perbag. Bellam Trading Co. will accept orders only for 1000 ormore bags at Rs.100/- per bag.
The total requirement of the twelve hostels, for whom the CSUbuys, is 3000 bags per semester. Inventory carrying costs are
20 per cent of unit cost and ordering cost for the CSU Rs.400/-per order. If you were the Hostel Affairs Secretary, whichvendor should you recommend to the Warden, CSU, for placingthe orders and for what quantity per order?
Problem 4
7/30/2019 Or Sol7 Inventory Control
10/35
TTN-DoMS, IIT-MADRAS
D = 3000 per semester
= 6000 per year
Co = Rs.400
i = 20% = 0.2
Vendor 1 C = Rs.120
EOQ = (2*6000*400/0.2*120)
= 447.21
TC = (2DCoCc) = (2*6000*400*0.2*120) = 10733.126
7/30/2019 Or Sol7 Inventory Control
11/35
TTN-DoMS, IIT-MADRAS
Vendor 2 C = Rs.108
Q = 800 or more
EOQ = (2*6000*400/0.2*108) = 471.40
TC = (6000/800)*400 + (800/2)*0.2*108 = 3000 + 8640
=11640
Vendor 3 C = Rs.100
Q = 1000
EOQ = (2*6000*400/0.2*100) = 489.89
At Q = 1000 , TC = (6000/1000)*400 + 1000*0.2*100/2
=2400 + 10000 = 12400
Hence Vendor1 is chosen with an order quantity of 447.21 bags
7/30/2019 Or Sol7 Inventory Control
12/35
TTN-DoMS, IIT-MADRAS
A company carries an inventory composed of two products A and Bfor which there is a uniform yearly demand of 10,000 and 20,000
units respectively. Product A costs the company Rs.2/unit topurchase, Rs.1/unit/year to store, and Rs.100 in administrativecosts every time it is reordered. Product B costs the companyRs.4/unit to purchase, Rs.2/unit/year to store, and Rs.200 inadministrative costs for each reorder.
a) Set up the total cost equations for each productseparately and solve for their optimum order quantities.
b) Owing to a liquidity problem, the company requires thatthe average yearly investment in inventory for the two productstaken together be less than or equal to Rs.6000/-. Is the answer
found in part (a) consistent with this added constraint?c) Find the optimum solution when the average yearly
investment in inventory must be less than or equal to Rs.5000.Average yearly investment is the rupee value of the averageamount of inventory present.
Problem 5
7/30/2019 Or Sol7 Inventory Control
13/35
TTN-DoMS, IIT-MADRAS
D = 1000 D2 = 20000C1 = Rs.2 C2 = Rs.4CC1 = Rs.1 CC2 = Rs.2C
O1= Rs.100 C
O2= Rs.200
a) Q1* = ( 2*10000*100/1) Q2
* = ( 2*20000*200/2)= 1414.2 = 2000
TC1
= D1
CO1
/Q1
+ Q1*
CC1
/2 TC2
= D2
CO2
/Q2
+ Q2*
CC2
/2
Q1* = (2D1CO1/ CC1) Q2
* = (2D2CO2/ CC2)
b) Amount invested in inventory (average)
= Q1C1/2 + Q2C2/2 = (1414.2*2)/2 + (2000*4)/2
= 1414.2 + 4000 = 5414.2 6000
Answer found in part (a) is consistent with the added constraint
7/30/2019 Or Sol7 Inventory Control
14/35
TTN-DoMS, IIT-MADRAS
c) If inventory 5000, it is binding. We setup the lagrangean function
L = D1CO1/Q1 + Q1CC1/2 + D2CO2/Q2 + Q2 CC2/2
+ (Q1C1/2 + Q2C2/2 6000)
= 1000*100/Q1 + 1*Q1/2 + 2000*200/Q2 + 2*Q2/2+ (2*Q1/2 + 4*Q2/2 - 6000)
L/Q1=0 gives
-1000*100/Q12 + + = 0
1000*100/Q12 = ( + )
L/Q1 = 0
Q1 =
100000/( +
)
L/Q2 = 0 Q2 = 400000/(1 + )
L/ = 0 Q1 + 2Q2 = 6000
7/30/2019 Or Sol7 Inventory Control
15/35
TTN-DoMS, IIT-MADRAS
Let ,
= 0 Q1 = 1414.2 Q2 = 2000 Q1 + 2Q2 = 5414.2
= 1.0 Q1 = 816.5 Q2 = 1414.2 Q1 + 2Q2 = 3644.9
= 0.2 Q1 = 1195.2 Q2 = 1825.7 Q1 + 2Q2 = 4846.7
= 0.15 Q1 = 1240.3 Q2 = 1865.01 Q1 + 2Q2 = 4970.37
= 0.13 Q1 = 1259.88 Q2 = 1881.4 Q1 + 2Q2 = 5022.7
= 0.14 Q1 = 1250 Q2 = 1873.17 Q1 + 2Q2 = 4996.34
The order quantities are 1250 and 1873 respectively.
7/30/2019 Or Sol7 Inventory Control
16/35
TTN-DoMS, IIT-MADRAS
A shop sells two products that it orders periodically from one supplier.The demand rates for both products are predictable and constant.
Demand rates and inventory costs are given by the following table:
a) Find the optimum values of the inventory cost per month and thelength of both inventory cycles under the assumption that bothinventory processes are managed separately.b) The shop finds that they could save on the order cost of product B
if both products were ordered together. This course of action can beevaluated by finding the minimum total inventory cost per month,subject to the constraint that the two inventory periods be equal.Write the objective function and the constraint. Show how to solve this byLagrange Multiplier method.
Problem 6
Product Demand rate Order cost Holding cost
Units/month Rs./order Rs./Unit/Month
A 100 50 25
B 300 50 3
7/30/2019 Or Sol7 Inventory Control
17/35
TTN-DoMS, IIT-MADRAS
Product A D1 = 100 per month
CO1 = Rs.50
CC1 = 2.50
Q1* = (2D1CO1/ CC1) = ( 2*100*50/2.5) = 63.24
N = D/Q = 100/63.24 = 1.581
TC1 = (2DCo/Cc) = (2*100*50*2.5) = 158.1139
Product B D2 = 300
CO2 = Rs.50
CC2 = 3.0
Q2* = ( 2*300*50/3) =100
N = 3
TC2 = 300 Time periods are different
7/30/2019 Or Sol7 Inventory Control
18/35
TTN-DoMS, IIT-MADRAS
b) We assume that we can save order cost of B if they are ordered together.The periods have to be equal D1/ Q1 = D2/ Q2
L = (D1/Q1)*50 + Q1* CC1/2 + Q2* CC2/2 + (D2/Q2 - D1/Q1)
L/Q1 = 0 gives -50D1/Q12 + CC1/2 + (D1/Q1
2) = 0
Q1 = (2D1 (-+50)/ CC1) = 80(50-)
L/Q2 = 0 gives CC2/2 -
(D2/Q2
2
) = 0
Q2 = 2D2/CC2 = 200
L/ = 0 gives 100/Q1 = 300/Q2
Q1/Q2 = 3
200 = -720 + 720*50 = 720*50/920 = 39.13
Q1 = 29.49 Q2 = 88.46
P bl 7
7/30/2019 Or Sol7 Inventory Control
19/35
TTN-DoMS, IIT-MADRAS
Messrs. Ghatilkar and Pillkarni Chemicals Private Ltd., own a plant in Chembur
which processes three chemicals in liquid form for consumption within the
factory itself at a uniform rate. The following are the particulars relating to the
three items:
Item Annual consumption in litres Cost per Processing Time for
Litre (Rs.) 100 Litres
1. 36,000 20 2 hours 13 1/3 min.2. 30,000 40 3 hours 20 min.
3. 50,000 20 1 hour 36 min.
Setting up the plant to start production of any of these items takes 8 hours for
every set up. However, during set-up the plant is not productively engaged; the
estimated cost of the idleness of the plant due to set up is Rs.100/- per hour.
The company works on a two-shifts-5 day week basis for 50 weeks, the
remaining 2 weeks being the annual shut down period for maintenance work.
Each shift is of 8 hours duration.
Problem 7
7/30/2019 Or Sol7 Inventory Control
20/35
TTN-DoMS, IIT-MADRAS
The inventory carrying charges per year are 10 per cent of the
cost per litre for each chemical.
The chemicals processed by the plant are collected and stored
in 10-litres cans. On an average how many such cans should be
made available, if the company desires to turn out these
chemicals in economic manufacturing quantities?
Suppose there are only 500 cans available, and assuming that
the average requirement of cans should never exceed this
number, how should the economic manufacturing quantities be
modified to satisfy this restriction?
7/30/2019 Or Sol7 Inventory Control
21/35
TTN-DoMS, IIT-MADRAS
Three items A,B,C
Item A Item B
D = 36000 D = 30000
CO = Rs.100 per hour *8hrs = Rs800 CO = Rs.800
C = 20 C = 40
i = 10% i = 10%
CC = Rs.2 CC = Rs.4
P = 350*16*60*3*100/400 P = 350*16*60*100/200
= 25200 litres = 168000 litres
Q = {2DCO
/ CC
(1-D/P)} Q = {2*30000*800/2(1-30/168)}
= {2*3600*800/2(1-36/252)} = 3822.132 litres
= 5796.55 litres
No. of cans = 5796.55/20 = 289.83 cans No. of cans = 191.11 cans
7/30/2019 Or Sol7 Inventory Control
22/35
TTN-DoMS, IIT-MADRAS
Item C
D = 50000
CO= Rs.100 per hour *8hrs = Rs.800C = 20
CC = Rs.2
P = 350*16*60*100/96 = 350000 litres
No. of cans = 341.56 cans
Total cans required = 289.83 + 191.11 + 341.56 = 822.5
Only 500 cans are available
7/30/2019 Or Sol7 Inventory Control
23/35
TTN-DoMS, IIT-MADRAS
Lagrangean L = 3600*800/Q1 + (Q1)2*0.857/2 + 30000*800/Q2+ Q2*4*0.821/2 + 50000*800/Q3 + (Q3)2*0.857/2+ (Q1+ Q2 + Q3/20 500)
L/Q1 = 0 -36000*800/Q12 + 0.857 + /20 = 0
36000*800/Q12 = (17.14 + )/20
Q1 = {36000*800*20/(17.14 + )}Q2 = {30000*800*20/(32.84 + )}Q3 = {50000*800*20/(17.14 + )}Q1 + Q2 + Q3 = 10000
Solving we get = 34Q1 = 3356Q2 = 2680Q3 = 3955
7/30/2019 Or Sol7 Inventory Control
24/35
TTN-DoMS, IIT-MADRAS
M/s Kanjivellam Enterprises Pvt. Limited own a medium-sized
factory divided into 3 manufacturing shops. The factory makesthree products A, B and C.
Manufacturing one unit of A requires 2 hours at Shop I, 1 hour
at Shop II and 2 hours at Shop III.
Manufacturing one unit of B requires 3 hours at Shop II and 2
hours at Shop III.Manufacturing one unit of C requires 1 hour at Shop I and 2
hours at Shop III.
The available capacities for shops, I, II, and III are 90 machine-
hrs. 120 machine-hours and 220 machine-hours respectively.The profits per unit of A, B and C are Rs.5/-, Rs.5/- and Rs.3/-
respectively. What is the optimal weekly product-mix that
maximises the total profit of M/s Kanjivellam Enterprises Pvt.
Ltd.,?
Problem 8
7/30/2019 Or Sol7 Inventory Control
25/35
7/30/2019 Or Sol7 Inventory Control
26/35
TTN-DoMS, IIT-MADRAS
Let x1, x2, x3 be quantities of A, B, and C respectively
Maximize 5x1 + 5x2 + 3x32x1 +x3 90
x1 +3x2 1202x1 +2x2 +2x3 220x1, x2 , x3 0P reqd = 2x1 +2x3 = 150/weekQ reqd = 3x2 = 105/week
R reqd = 3x3 = 180/week
Item P Item Q
D =150*50 = 7500 D = 105*50 = 5250
CO = 50 CO= 50i = 20% of Rs.5 = Rs.1 CC = 20% of Rs4.20 = Rs.0.84Q = {2DCO/CC Q =790.57
= {2*7500*50/1} = 866.02QC/2 = 2165.06 QC/2 = 1660.20
Item R
7/30/2019 Or Sol7 Inventory Control
27/35
TTN-DoMS, IIT-MADRAS
Item R
D = 180*50 = 9000
CO = 50
CC = 20% of Rs.3 = Rs.0.60Q = {2DCO/ CC = {2*9000*50/0.6} = 1224.75
QC/2 = 1837.117
Budget = 2165.06 + 1660.20 + 1837.12
= 5662.373
Violates budget restriction
New values of order quantities areItem P = 866.02*5000/5662.373 = 764.71
Item Q = 790.57*5000/5662.373 = 698.09
Item R = 1224.75*5000/5662.373 = 1081.48
7/30/2019 Or Sol7 Inventory Control
28/35
TTN-DoMS, IIT-MADRAS
Messrs. Spoilda Soil Fertilizer Company has undertaken a
contract to supply 50 tons of fertilizer at the end of the firstmonth, 70 tons at the end of the second month and 90 tons at
the end of the third month. The cost of producing X tons of
fertilizer in any month is given by
(4500X + 20X2
) RupeesThe company can produce more amount of fertilizer in any
month and supply it in the next month. However, there is an
inventory carrying cost of Rs.400/- per ton per month.
Formulate the problem of finding the optimal level of productionin each of the three periods and the total cost incurred as a
dynamic programming problem and solve.
Assume initial and final inventory to be equal to Zero.
Problem 9
7/30/2019 Or Sol7 Inventory Control
29/35
TTN-DoMS, IIT-MADRAS
This problem can be solved by forward recursion.
Stage : Each monthStates : Inventory carried from the previous month
Decision Variable: Quantity to be produced each month
Criterion : Minimize cost
X1 X2 X3
Month 1 Month 2 Month 3
Initial Inventory = 0 S1 S2
7/30/2019 Or Sol7 Inventory Control
30/35
TTN-DoMS, IIT-MADRAS
n = 1; 2 more months remaining:f1(0,X1) = 20X1
2 + 4500X1f1*(0) = min {20X1
2 + 4500X1}
= 20(50 + S1)2 + 4500(50 + S1)= 20(2500 + S1
2 + 100S1) + 225000 + 4500S1= 275000 + 20S1
2 + 6500S1 at X1* = 50 + S1
n = 2; 1 more month remaining:f2(S1,X2) = 20X2
2 + 4500X2 + 400S1 + f1*(0)
= 20X22 + 4500X2 + 400S1 + 275000 + 20S1
2 + 6500S1f2*(S1) = min {20X2
2 + 4500X2 + 400S1 + 275000 + 20S12
+ 6500S1}
= 20(70 + S2 S1)2 + 4500(70 + S2 S1) + 400S1
+ 275000 + 20S12 + 6500S1
= 688000 + 40S12 + 20S2
2 - 400S1 + 7300S2 - 40S1S2at X2* = 70 + S2 - S1
7/30/2019 Or Sol7 Inventory Control
31/35
TTN-DoMS, IIT-MADRAS
n = 3; last month:f3(S2,X3) = 20X3
2 + 4500X3 + 400S2 + f2*(S1)
= 20X32 + 4500X3 + 400S2 + 688000 + 40S1
2 + 20S22
- 400S1
+ 7300S2
- 40S1S
2f3*(S2) = min {20X3
2 + 4500X3 + 400S2 + 688000 + 40S12 + 20S2
2
- 400S1 + 7300S2 - 40S1S2}
= 20(90 - S2)2 + 4500(90 S2) + 400S2 + 688000 + 40S1
2
+ 20S22 - 400S1 + 7300S2 - 40S1S2
= 1255000 + 40S12 + 40S22 - 40S1S2 - 400S1 - 400S2at X3* = 90 - S2
Differentiating with S1 and S2 and solving the equations we get,
S1 = 10 and S2 = 10
Hence, X3* = 80 Total cost = 12,51,000
X2* = 70
X1* = 60
7/30/2019 Or Sol7 Inventory Control
32/35
TTN-DoMS, IIT-MADRAS
Consider the problem of determining economic lot sizes for four
different items. Assume that the demand occurs at a constantrate over time. The constant order quantity for the jth item is
denoted by Qj. The total storage space available is A, where
each unit of jth item occupies an area aj.
The objective is to find the Qjs which minimize the total cost
subject to the total area constraint. The total cost is of the form.Uj
TC = j=14
-------- + VjQj , Qj > 0
Qj
Where Uj and Vj are constants.
Formulate the problem as a dynamic programming model.
Assume, if necessary, that Qj is discrete.
Problem 10
7/30/2019 Or Sol7 Inventory Control
33/35
TTN-DoMS, IIT-MADRAS
Stage : Each item
State :Area remaining for the last n items
Decision Variable: How much to orderCriterion : Minimize cost
n = 1; last item:
f1(S1,Q1) = + V1Q1
f1*(S1) = Min + V1Q1
= Min ,
U1Q1 U1
Q10 Q1 S1a1
U1Q1
S1a1
7/30/2019 Or Sol7 Inventory Control
34/35
TTN-DoMS, IIT-MADRAS
n = 2; two more items:
f2(S2,Q2) = + V2Q2 + f1*(S2
Q2a2)
In general, the recursive relation is
fn(Sn,Qn) = + VnQn + fn-1*(Sn Qnan)fn*(Sn) = Min + VnQn + fn-1*(Sn Qnan)
For the final stage (n = 4),
Sn = A [The total storage space available ]
UnQn Un
Qn0 Qn Snan
U2Q2
7/30/2019 Or Sol7 Inventory Control
35/35
Thank You !!!