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A GENERALISED ALGORITHM FOR THE DEMAND PREDICTION OF A SHORT LIFE CYLCLE PRODUCT SUPPLY CHAIN AND ITS IMPLEMENTAION IN A BAKED PRODUCT

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A GENERALISED ALGORITHM FOR THE DEMAND PREDICTION OF A SHORT LIFE CYLCLE PRODUCT SUPPLY CHAIN AND ITS IMPLEMENTAION IN A BAKED PRODUCT Powered By Docstoc
					 International Journal of              Engineering and Technology (IJMET), ISSN 0976
INTERNATIONALMechanical Volume 4, Issue 1, January - February (2013) © IAEME–
                            JOURNAL OF MECHANICAL ENGINEERING
 6340(Print), ISSN 0976 – 6359(Online)
                          AND TECHNOLOGY (IJMET)
ISSN 0976 – 6340 (Print)
ISSN 0976 – 6359 (Online)
Volume 4 Issue 1 January- February (2013), pp. 44-53                          IJMET
© IAEME: www.iaeme.com/ijmet.asp
Journal Impact Factor (2012): 3.8071 (Calculated by GISI)
www.jifactor.com                                                         ©IAEME


   A GENERALISED ALGORITHM FOR THE DEMAND PREDICTION
    OF A SHORT LIFE CYLCLE PRODUCT SUPPLY CHAIN AND ITS
            IMPLEMENTAION IN A BAKED PRODUCT
                               Bijesh Paul1 Dr Jayadas.N.H.2
                             1
                             Research scholar, 2Associate Professor
                  Division of Mechanical Engineering, School of Engineering,
                   Cochin University of Science and Technology (CUSAT),
                               Cochin - 682 022, Kerala, India.
                              E-mail: bijeshpaul@hotmail.com



  ABSTRACT

          This paper deals with optimization of demand prediction of a short life product with
  a minimum shelf life. A generalized algorithm for optimizing the future demand was
  developed by using markov chain. The purpose of this paper is to develop an algorithm for
  optimizing demand which will act as a benchmark for future production and will lead to huge
  annual savings for each product.For very short life cycle products or perishable products such
  as baked products and newspapers with maximum shelf life of one day it’s very difficult to
  predict the future demand. For such products demand forecast erroris found in between 40 –
  100%.This gives an opportunity to identify the problem of demand forecast error in short life
  cycle products with very low shelf life. Moreover little literature is available to predict the
  demand of short life cycle or very perishable product. Hence a generalized algorithm for
  optimal future demand was developed by using Markov chain.This is very relevant in Indian
  scenario where small firms are going to face stiff competition from multinational and
  indigenous retail chains. Thealgorithm was implemented for a baking product and the optimal
  demand forecast wasdetermined. This paper offers a novel optimization technique for
  optimizing demand forecast of short life cycle supply chain products by using Markov
  chains.The algorithm can be also implemented for novel products as it requires only demand
  data of any two consecutive time periods.

  Keywords: Demand, Markov Chain, Algorithm, Optimization, Short life cycle, Supply chain.



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International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 –
6340(Print), ISSN 0976 – 6359(Online) Volume 4, Issue 1, January - February (2013) © IAEME

1.     INTRODUCTION

        Short Life cycle product is a kind of product with a comparatively short and fixed
selling time, such as baked products, fashion clothes, books, magazines, electronics
merchandise, festival adornment etc [1]. In this product supply chain the indetermination
degree of downstream demand is very high, and the indetermination degree of the upper
stream will enlarge further. The seller’s average out of stock rate is even up to 10-40%. The
demand forecast error of supplier or manufacture is generally between 40 – 100% [2].
Actually, with the quick development of science and technology and with continuous rise of
peoples demand more and more products will have the characteristic of short life cycle. This
phenomenon will play more in the commodity market of fierce competition. The main
reasons are 1. Speed of technology refreshing is more and more 2. The consumption is more
and more of short life characteristic [3] Many researchers in the field of logistics analyzed the
returning goods problem of short life cycle product. Most investigated impact of supply
chain management on logistical performance indicators in food supply chains especially in
the case of baked products. [4]. Lee found that product that has not been sold at the end of the
season may be either returned to the manufacturer or processed at the discount shop [5].
It’s important for a retailer of a short life product with minimum shelf life to predict the
optimal demand as under stocking will result in the switching of loyalty of the customers and
loss of possible profit and overstocking will result in obsolete products which will enhance
the financial burden. For small retailers it’s expensive to use software tools to predict demand
in advance. Hence an attempt is made to develop an algorithm by using Markov chain with
past sales data of any two successive months. This technique is to be applied when demand is
treated as a random variable.ie trend;seasonality and cycleness associated with demand data
are negligible.
Agrawal and Smith used negative binomial distribution (NBD) for the demand model and
suggested that NBD model provides a better fit than the normal or Poisson distributed
data.[6]Cachon used the negative binomial distribution model to analyze the demand of the
fashion goods where it is assumed that the demand process follows the Poisson distribution
and demand rate varies according to a gamma distributed model [7].Hammond studied the
Quick Response policy with ski apparel (ski suits, ski pants, parkas, etc), and showed that
forecast accuracy can be substantially improved by adopting QR policy [11].A Markov chain
model is a stochastic process, with discrete states and continuous time in which modeling is
done on observable parameters. This model can be utilized to evaluate the probability of
different states with respect to time. The earlier states are irrelevant for predicting the
following states, since the current state is known [19]. In 2001, Zhang and He have developed
a Grey–Markov forecasting model for forecasting the total power requirement of agricultural
machinery in Shangxi Province [21]. In 2007, Akay and Atak have formulated a Grey
prediction model with rolling mechanism for electricity demand forecasting of Turkey [22].
A Grey–Markov forecasting model has been developed by Huang, He and Cen in 2007. The
Markov-chain forecasting model is applicable to problems with random variation, which
could improve the GM forecasting model [23].Markov models are used in many disciplines
for many different applications, from thermodynamic modeling in physics to the population
modeling in biology.Theycan be used to model almost any dynamical system whose
evolution over time involves uncertainty" [24]. Due to the uncertainty and randomness of the
data it’s appropriateto use a Markov chain to predict the demand.



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International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 –
6340(Print), ISSN 0976 – 6359(Online) Volume 4, Issue 1, January - February (2013) © IAEME

Markov chains are dynamic systems that describe the evolution of a probability distribution.
Since this analysis is concerned with demand prediction based on a finite time interval,
discrete time stationary Markov chains with a fixed number of states are used. A state of a
system is where the system is at a point of time. Transition probability is the probability of
transforming from one state to another in a specific time period.AMarkov model is described
in terms of its transition probabilities, pij , which can berepresented in a transition probability
matrix
Pij = P11 P12 P13 ………P1n
P21 P22 P23……….P2n
          …………………………….
          ……………………………..
Pn1 Pn2Pn3....…..Pnn
The columns of P are stochastic, meaning the entries are non-negative and sum toone. At
each time step, k, the state of the chain, xk, is determined by the previousstate and the
transition probabilities associated with that state. The assumptions of markov chain analysis
are that
     1. Theprobabilities of travelling from one state to all other sates add to one.
     2. The states are independent of time.
The evolution ofthe system is determined by multiplying the transition matrix by the previous
statevector, which is a stochastic vector representing the probabilities of the system beingin
any one of the given states. The stationary characteristics of the Markov chain reveal that
same output will be produced irrespective of the input. This Property is utilized for
generating the optimal demand forecast for a short life cycle product with minimum shelf life
because options of stocking beyond one day is not possible and the minimization of financial
burden that overstocked products brings to the firm is crucial in these days of intense
competition. Moreover as time passes the probability of a particular state increases and
reaches a steady state probability and the demand corresponding to this state is taken as the
optimal demand forecast. An Algorithm is developed by incorporating the above mentioned
features of a Markov chain to predict the demand forecast.

2.      METHODOLOGY
     1. Observed demand data for a short life cycle product is collected for any two
        successive months.
     2. Implement the generalized algorithm for the selected demand data.
     3. Deduce the initial probability matrix and the Transition probability matrix for the
        different states of demand.
     4. By utilizing the above two matrices the probability of different states of demand for
        any future period can be determined.The evolution ofthe system is determined by
        multiplying the transition matrix by the previous statevector(probability matrix),
        which is a stochastic vector representing the probabilities of the system beingin any
        one of the given states
     5. Choose the state with maximum probability from the obtained current probability
        vector (initial probability matrix).
     6. Determine the annual savings by adopting the demand of the state with maximum
        probability.

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International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 –
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3.      ALGORITHM

1) Collect the observed data for sales of a particular product with minimum shelf life for any two
consecutive or successive months, say t and t+1
2) Determine the upper limit and lower limit of the collected sales data for the tth month.
Determine the range or band width of the collected data as the difference between upper limit and
lower limit for the tth month
3) Discretise the obtained range into states or class intervals with minimum possible no of sample
size. Let us denote these states as X1, X2, X3………..Xn.
4) Determine the initial probability vector P0 for the month t. This matrix gives the initial
probability of all states say X1, X2, X3………..Xnin month t.
4, a) List out all the days (m) in a month in the month t as the first column, in the ascending order
of the table
4, b) In second column enter the state of the observed sales data for all the days of tthmonth listed
in the first column
4, c) Count the no of occurrence of each state in tth month. (For eg say state Xi is occurring j times
in the month t of m days, then initial probability of Xi= j/m)
4, d) Determine the initial probability of all states by using the formulae Xi= J/M where J is the
occurrence of ith state in tthmonth of M days and i= 1, 2,3…….n.
4, e) Represent the initialprobabilities obtained from step 8 as a row vector (1*n) with n no of
entries and is called as initial probability vector denoted by P0.
5) Construct state occurrence table for tth month and t+1th month.
5, a) List out all the days of tthand t+1th month in the ascending order as the first column of the
table. Assume the number of working days in both months as same
5, b) In the second column of the table enter the state corresponding to sales data for all the days
listed in tthmonth.
5, c) In column three enter the state corresponding to sales data for all days listed in the t+1th
month.
6) Deduce transition probability matrix from the event occurrence table.
6,a ) Any current state Xi in a particular day of tth month can transform into states X1, X2,
X3………..Xn during the same day of t+1th month. Hence there exits n probabilities which results
from the probable transformation of current state Xi to other possible states X1, X2, X3………..Xn.
Represent these probabilities as P11,P12………..P1n
6, b) Form the Transition probability matrix by representing all the current states as rows and next
states as columns. Now enter the probabilities as P11,P12………..P1n in 1st row and repeat the same
procedure for other rows.
Any entry say Pij= No of transformations of current state i of tth month in a particular day to next
state j of t+1 th month in the same day/ Total no of occurrence of current state i in the t th month
7) Deduce the current probability vector for the succeeding months t+2, t+3 as
P1=P0* TPM
P2=P1* TPM
………
………
Pm= Pm-1 *TPM
8) Choose the state with maximum probability from the obtained current probability vector for
say the mthmonth which is a row matrix with probability of each state during say mthmonth.
9) Determine the possible profit to firm by the adoption of this state of production as indicated by
the step 8



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International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 –
6340(Print), ISSN 0976 – 6359(Online) Volume 4, Issue 1, January - February (2013) © IAEME

4.     IMPLEMENTATION OF THE ALGORITHM FOR A BAKED PRODUCT

        The data collected from a reputed baking firm is furnished below.The firm has been
producing 1300 items per day and selling this item @ Rs 12. Any leftover item is sold at a
rate of Rs 3 and there by occurring a possible loss of profit of Rs 9 per product for left over
item.Cost of each item is Rs7/-.

Step1
The table below shows the demand data gathered for two successive months of a baked
product with a short life cycle of one day.

                                          Table-1
           Date               Total Demand during each day       Total Demand during each
                                      of April 2012                  day of May 2012
        01-04-2012                         1260                             1260
        02-04-2012                         1267                             1260
        03-04-2012                         1260                             1268
        04-04-2012                         1252                             1255
        05-04-2012                         1248                             1249
        06-04-2012                         1266                             1267
        07-04-2012                         1271                             1267
        08-04-2012                         1260                             1265
        09-04-2012                         1259                             1265
        10-04-2012                         1264                             1268
        11-04-2012                         1256                             1259
        12-04-2012                         1265                             1267
        13-04-2012                         1260                             1264
        14-04-2012                         1271                             1269
        15-04-2012                         1265                             1271
        16-04-2012                         1271                             1265
        17-04-2012                         1259                             1262
        18-04-2012                         1266                             1268
        19-04-2012                         1270                             1269
        20-04-2012                         1260                             1263
        21-04-2012                         1262                             1268
        22-04-2012                         1268                             1265
        23-04-2012                         1265                             1260
        24-04-2012                         1271                             1269
        25-04-2012                         1263                             1265



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International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 –
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                                                     Figure-1

                                         Daily Demand Pattern
                          1275
                          1270
                          1265
           Daily Demand




                          1260
                          1255
                          1250                                                     Series1
                          1245
                          1240
                          1235
                                 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49



The above pattern shows that demand is Random. Hence we treat the demand as random
variable and apply Markov chain based algorithm for predicting demand.

Step 2
The Range = Highest demand – Lowest Demand =1272 -1248 = 24. Also when demand was
highest volume of discount sales was (1300 – 1272) 28and when demand was lowest volume
of discount sales was (1300 – 1248) 52. Range 2 = 52- 28 = 25

Step 3
Discretize the range into class intervals or states with a minimum sample size of 3. I.e. Say 8
states denoted by X1, X2, X3………..X8.

Step 4
Determine the initial probability vector P0 for the month April. This matrix gives the initial
probability of all states say X1, X2, X3………..X8in the month of April.

                                                      Table-2
Class interval of discounted                      State       No of occurrence     Probability
sales
52, 51, 50                                         X1                  1                .04
49, 48, 47                                         X2                  1                .04
46, 45, 44                                         X3                  1                .04
43, 42, 41                                         X4                  2                .08
40, 39, 38                                         X5                  6                .24
37, 36, 35                                         X6                  5                .20
34, 33, 32                                         X7                  4                .16
31, 30, 29,28                                      X8                  5                .20
The Fourth row of the above table gives initial probability vector P0 for the month April. This
matrix gives the initial probability of all states say X1, X2, X3………..X8 in the month of April.
P0 = [.04, .04, .04, .08, .24, .2, .16, .2]

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International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 –
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Step 5 – State transition table

                                            Table-3
               Day                Current state (April)       Subsequent state (May)
                1                          5                           5
                2                          7                           5
                3                          5                           7
                4                          2                           3
                5                          1                           1
                6                          7                           7
                7                          8                           7
                8                          5                           6
                9                          4                           6
               10                          6                           7
               11                          3                           4
               12                          6                           7
               13                          5                           6
               14                          8                           8
               15                          6                           8
               16                          8                           6
               17                          4                           5
               18                          7                           7
               19                          8                           8
               20                          5                           6
               21                          5                           7
               22                          7                           6
               23                          6                           5
               24                          8                           8
               25                          6                           6

Step 6

Deduction of Transition Probability Matrix (T P M)
 a. Any current state Xi in a particular day of tth month can transform into states X1, X2,
    X3………..Xn during the same day of t+1th month. Hence there exits n probabilities
    which results from the probable transformation of current state Xi to other possible states
    X1, X2, X3………..Xn. Represent these probabilities as P11,P12………..P1n
 b. Form the Transition probability matrix by representing all the current states as rows and
    next states as columns. Now enter the probabilities as P11, P12………..P1n in 1st row and
    repeat the same procedure for other rows.
 c. Any entry say Pij= No of transformations of current state i of t th month in a particular
    day to next state j of t+1th month in the same day/ Total no of occurrence of current state
    i in the tth month
 d. Transition Probability Matrix,




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International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 –
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                                                Table-4
                   1         2         3          4             5         6       7              8
         1         1         0         0          0            0          0       0              0
         2         0         0         1          0            0          0       0              0
         3         0         0         0          1            0          0       0              0
         4         0         0         0          0            .5        .5       0              0
         5       .167        0         0          0            0         .5     .333             0
         6        .2         0         0          0            .2         0      .4              .2
         7        .5         0         0          0           .25       .25       0               0
         8        .6         0         0          0            0         .2      .2               0

Step 7
Deduce the current probability vector for the succeeding months t+2, t+3 ….t+9 as
P1=P0* TPM
P2=P1* TPM
………
Pm= Pm-1 *TPM
P1=P0* TPM = [.3201, 0, .04, .04, .12, .24, .1994, .04]
P2 = [.5121, 0, .0, .04, .118, .138, .144, .048]
P3 = [.6602, 0, .0, .0, .0836, .1246, .1041, .0276]
P4 = [.7676, 0, .0, .0,.0509, .0733,.0832 .024]
P5= [.8473, 0, .0, .0, .0355, .0512, .0513, .0147]
P6= [.8980, 0,.0, .0, .0231, .0335, .0352, .0102]
P7= [.9323, 0, .0, .0, .0155, .0224, .0231, .0067]
P8= [.9549, 0, .0, .0, .0103, .0149, .0155, .0045]
P9= [.97, 0, .0, .0, .0068, .0099, .0103, .0030]
Step 8
The maximum Probability is for state 1 and as time passes we can see that probability of demand for
state 1 is approaching steady state probability. The Probability V/s Time curve for state one is shown
below
                                                Figure-2
                1.2

                  1

                0.8

                0.6
                                                                                       Series1
                0.4

                0.2

                  0
                      0          2          4             6         8          10


Step 9
The firm should produce the mean of the class interval corresponding to state 1 namely 1249
items .Thus the predicted or forecasted value of demand is Df = 1249.


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International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 –
6340(Print), ISSN 0976 – 6359(Online) Volume 4, Issue 1, January - February (2013) © IAEME

5.     RESULT

        The maximum Probability is for state 1 and as time passes we can see that probability
of demand for state 1 is approaching steady state probability. Hence the firm should produce
the mean of the class interval corresponding to state 1 namely 1249 items which will enhance
the profit of the firm.Annual Possible Profit for a single product = {(P-Df) (E-S)*25- ∑ (Dn-
Df) C}*12= {(1300- 1249) (7-3) 25 – 354*5}12 = Rs 41000, where P= Production rate per
day, Df is the predicted value of demand by using algorithm, E is the cost of making unit
quantity, S is the price of the product after discount to be sold after 8pm, ∑ (Dn-Df) represents
the error in forecast for one month and R represents the profit of selling of unit quantity.The
above concept can be extended for many products and huge annual savings can be achieved.

6.     CONCLUSION AND FUTURE SCOPE

        A generalized algorithm for optimizing the demand data was developed by using
Markov chain for a short life cycle supply chain with least possible shelf life. Little literature
exits in the demand forecasting of short life cycle of baked product. The algorithm has been
validated by implementing it in a baking firm and by the huge annual savings Rs 41000/for
one product. The above concept can be extended for many products and huge annual savings
can be achieved. It acts as a tool for recovering the lost possible profit.This is very relevant in
today’s world when these small firms are going to face stiff competition from multinational
and indigenous retail chains.Further validation can be carried out by comparing the results
obtained by implementing this algorithm with that obtained by other statistical methods.It can
be further validated by applying in different type of short life products such as Newspaper

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[10]Ben-Daya, M. and Raouf, A., “Inventory models involving lead time as decision
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Published by IAEME.


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