Project and Production Management Module 8 Production Planning over the Short Term Horizon Prof Arun Kanda & Prof S.G. Deshmukh, Department of Mechanical

Project and Production Management

Module 8

Production Planning over the Short Term Horizon

Prof Arun Kanda & Prof S.G. Deshmukh, Department of Mechanical Engineering,Indian Institute of Technology, Delhi

module 8: Production Planning

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MODULE 8: Production Planning over the Short

Term Horizon1. Forecasting

2. The Analysis of Time Series

3. Aggregate Production Planning: Basic Concepts

4. Aggregate Production Planning: Modelling approaches

5. Illustrative Examples

6. Self Evaluation Quiz

7. Problems for Practice

8. Further exploration


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1.Forecasting


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FORECASTING

Forecasting is essential for a number of planning decisions

LONG TERM DECISIONS New Product Introduction Plant Expansion



MEDIUM TERM DECISIONS Aggregate Production Planning Manpower Planning Inventory Policy

SHORT TERM DECISIONS Production planning Scheduling of job orders



PLANNING PROCESSA Forecast of Demand is an essential Input for

PlanningSystem

Objectives

System to beManaged

Constraints- Budget / Space

Resources - Men

- Equipment

DemandForecast

Plan ofAction



METHODS OF FORECASTING

(a) Subjective or intuitive methods Opinion polls, interviews DELPHI

(b) Methods based on averaging of past data Moving averages Exponential Smoothing



(c) Regression models on historical data Trend extrapolation

(d) Causal or econometric models

(e) Time - series analysis using stochastic models

Box Jenkins model



FORECASTING Objective Scientific Free from ‘BIAS’ Reproducible Error Analysis Possible

PREDICTION Subjective Intuitive Individual BIAS Non - Reproducible Error Analysis Limited





COMMONLY OBSERVED “NORMAL” DEMAND

PATTERNS

D

t

Constant

D

t

LinearTrend



D

t

CyclicD

t

SeasonalPattern with

Growth



ABNORMAL DEMAND PATTERNS

TransientImpulse

SuddenRise

SuddenFall



OPINION POLLSPersonal interviews

e.g. aggregation of opinion of sales representatives to obtain sales forecast of a region

Knowledge base (experience) Subjective bias

Questionnaire method questionnaire design choice of respondents obtaining respondents analysis and presentation of results (forecasting)

Telephonic conversation Fast

DELPHI



DELPHIA structured method of obtaining responses from

experts.

Utilizes the vast knowledge base of experts

Eliminates subjective bias and ‘influencing’ by members through anonymity

Iterative in character with statistical summary at end of each round (Generally 3 rounds)

Consensus (or Divergent Viewpoints)

usually emerge at the end of the exercise.



Coordinator

Expert 1

Expert 2

Expert n

1990 1995 2000 2005 20101990

year

• Mean•Median•Std. deviation

A Statisticalsummary

can be givenat end of

each round



DELPHI (Contd.)

Round1

Round2

Round3

MovingTowards

Consensus



DELPHI (Contd.)

Round1

Round2

Round3

MovingTowardsDivergent

View Points



MOVING AVERAGESMonth Demand 3 Month MA 6 Month MA

Jan 199 Feb 202 Mar 199 200.00 Apr 208 203.00 May 212 206.33 Jun 194 203.66 202.33 Jul 214 205.66 207.83 Aug 220 208.33 210.83 Sep 219 216.66 213.13

Oct 234 223.33 217.46 Nov 219 223.00 218.63 Dec 233 227.66 225.13



K PERIOD MA = AVERAGE OF K MOST RECENT OBSERVATIONS

For instance : 3 PERIOD MA FOR MAY

= Demands of Mar, Apr, May / 3= (199 + 208 + 121) / 3 = 206.33



CHARACTERISTICS OF MOVING AVERAGES

Dt

t

Dt

t

(1) MOVING AVERAGES LAG A TREND



Dt

t

(2) MOVING AVERAGES ARE OUT OF PHASE

FOR CYCLIC DEMAND



Dt

t

(3) MOVING AVERAGES FLATTEN PEAKS



EXPONENTIAL SMOOTHINGFt = one period ahead forecast made at

time time t

Dt = actual demand for period t

= Smoothing constant (between 0 & 1)

(generally chosen values tie between 0.01 and 0.3)

Ft = Ft-1 + (Dt - Ft-1)



Ft = Dt +(1 - ) Ft-1

= Dt +(1 - ) [ Dt-1 +(1 - )2 Ft-2 ]

=……..

= [Dt +(1 - ) Dt-1 +(1 - )2 Dt-2 + …..

+ (1 - )t-1 D1 + (1 - )t F0]

(1- ) (1- )2

tt-1t-2

Weightages given to past data decline exponentially.



MOVING AVERAGESAND

EXPONENTIAL SMOOTHING

(Equivalence between & N :) = 2 / (N+1)



J F M A M J J A S O N D J190

200

210

220

230

Dem

and

Month

3 Month MA

6 Month MA

= 0.3

= 0.1



Common Regression Functions

dt

t

dt’ forecast

dt actual demand (for time period t)

dt’ = a + bt (parameters a, b)

Linear



dt

t

dt’ = a + u Cos (2/n)t + v Sin (2/n)t (parameters a, u, v)

Cyclic



dt

t

dt’ = a +bt + u Cos (2/n)t + v Sin (2/n)t (parameters a,b,u, v)

Cyclic withGrowth



dt

tdt’ = a +bt + ct2 (parameters a,b,c)

Parameters Determined by Minimizing the Sum of Squares of errors,

Quadratic



REGRESSION


200

210

220

230

Dem

and

Month (t)

Actual Data 218

246

Ft = 193 + 3t(Regression Line)

Forecast for next

JAN

1 2 3 4 5 6 7 8 9 10 11 12 13

232



Standard error of estimate = (Dt - Ft)2

Where

Dt = actual demand for period t = 7.32

Ft = forecast for period t

n = no. of data points

f = degrees of freedom lost (2 in this case)

95 % confidence limits for forecast of next JAN ~ 232 14 (* 2 sigma)

n

t = 1 n - f



CAUSAL MODELSHere demand is related to

Causal variables GNP Per Capita income Consumer Price index …………

Demand for tyres

= f (Production of new automobiles, Replacements by existing autos,

Govt policy on automobiles, …..)



Dt = Pt + Pt-5 +

could be a simplified causal model

(Here parameters , ,, are estimated by regression from data)

For a Causal Model to be Useful

The causal variables should be

Leading

Highly correlated with the variable of interest



TIME SERIES ANALYSISTime series decomposed into

Trend Seasonality Cycle Randomness

And Forecast generated from these components



Stochastic modelling ( Box and Jenkins)

Various processes eg.Autoregressive (AR) order pMoving average (MA) order qARMA order (p,q)ARIMA order (p,d,q)

are used to fit the most appropriate model.These models are accurate (for short term demand forecasting) but highly cumbersome to develop.



Past Data

ForecastGeneration

ManagerialJudgement

&Experience

ForecastControl

Current Data

ModifiedForecast

Forecasting System



Moving Range Chart to Control Forecasts

MR = | (Ft -Dt) - (Ft-1 - Dt-1) |

(Moving Range)

MR = MR / (n – 1) ( There are n-1 moving ranges for n period)

Upper Control Limit (UCL) = + 2.66 MR

Lower Control Limit ( LML) = - 2.66 MR



VARIABLE TO BE PLOTTED = (Ft - Dt)

01020

30

-10

20

-20

30

-30 Month

(Control Chart for Example)



SUMMARY

Importance of forecasting in planning

Various Methods of forecasting Subjective methods like opinion polls & Delphi Moving Averages & Exponential Smoothing Trend extrapolation by regression Causal models Time series decomposition

Forecast Control



2. The Analysis of Time Series



CORRELATION



CORRELATION vs REGRESSION?

Correlation examines if there is an association between two variables,

and if so to what extent.

Regression establishes an appropriate relationship between the variables

X

Y



r = 0

SCATTER DIAGRAM

*

*

*

** *

** *

***

**

**

* *

**

*

*

** * *

r > 0 r < 0

Positive correlation Negative correlation

No correlation Non-linear association



THE CORRELATION COEFFICIENT

Pearson’s correlation coefficient, r

= (1/n) Sum [(X- X) (Y-Y)]

sigma X sigma Y

(The numerator is the

Co-variance between X and Y)



METHODS OF COMPUTATION

Direct computations using the formulaCumbersome and lengthy computations

Short-cut or the U-V method Involves any conveniently assumed meanSuitable scaling of variables



S. No. X Y x=X-X

y=Y-Y

x2 y2 xy

1 50 700 21 274 441 75,076 5,754

2 50 650 21 274 441 50,176 4,704

3 50 600 21 174 441 30,276 3,654

4 40 500 11 74 121 5,476 814

5 30 450 1 24 1 576 246 20 400 -9 -26 81 676 2347 20 300 -9 -126 81 15,876 1,134

8 15 250 -14 -176 196 30,976 2,464

9 10 210 -19 -216 361 46,656 4,104

10 5 200 -24 -226 576 51,076 5,424

Total 290 4260 0 0 2,740 3,06,840

28,310

Advertisement expenditure (X) vs Sales (Y) figures for 10 years in Lacs of Rupees.



X = 290/10 = 29 : Y = 4260/10 =426 r = Σxy /[ Σx2Σy2]1/2 = 28310/ (2740 * 306840)1/2 = 0.976 Coefficient of Determination = r2 = 0.953



WHAT IS REGRESSION?

Discovering how a dependent variable (Y) is related to one or more independent variables (X)

Y

X

Y = f(X)



CRITERION FOR BEST FIT?

Y = f(X)

Mean error Minimize ? Mean absolute error Sum of Squares of Errors

Least Squares Criterion is the generally preferred criterion

Positiveerror

Negativeerror



FIITING IN A STRAIGHT LINE

• Ft = a + bt is the equation of the line to be fitted• Ft is the fitted function for time t• Dt is the actual demand for period t• Past data is available for n periods• Parameters a & b have to be estimated from the data using least squares criterion



LEAST SQUARES NORMAL EQUATIONS

SSE = Σ(Dt – Ft)2 = Σ (Dt- a – bt)2

To minimize (SSE)

d(SSE)/da = Σ 2(Dt – a –bt)(-1) = 0

d(SSE)/db = Σ 2(Dt – a – bt)(-t) = 0

Or a (n) + b (Σ t) = Σ Dt

a(Σt) + b (Σ t2) = Σ t Dt



LEAST SQUARES NORMAL EQUATIONS (2)

These are two linear simultaneous equations in the two unknown parameters a and b which can be solved by any of the well known methods eg Cramer’s Rule.

These equations are called

Least Squares Normal Equations



a (n) + b (Σ t) = Σ Dt ( Least Squares Normal

a(Σt) + b (Σ t2) = Σ t Dt Equations)

Σ Dt Σ t

a = Σ tDt Σt2 = ΣDt Σt2 – Σt ΣtDt

n Σt n Σt2 – (Σt)2

Σt Σt2

n ΣDt

b = Σt ΣtDt = n ΣtDt - Σt ΣDt

n Σt n Σt2 – (Σt)2

Σt Σt2



ORGANIZING COMPUTATIONS

S. No ti Di tiDi ti2

1 t1 D1 t1D1 t12

2 t2 D2 t2D2 t22

n tn Dn tndn tn2

Totals ∑ti ∑ Di ∑tiDi ∑ti2



COMPUTATIONAL SIMPLIFICATIONS

By choosing an origin and scale of data such that

Σ t = 0

the values of the parameters become

a = Σ Dt / n

b = Σ tDt / Σ t2

(This is useful for equally spaced data with even or odd number of data points)



DEMAND HISTORYMonth Demand

Jan 199 Feb 202 Mar 199 Apr 208 May 212 Jun 194 Jul 214 Aug 220 Sep 219

Oct 234 Nov 219 Dec 233



REGRESSION


200

210

220

230

Dem

and

Month (t)

Actual Data 218

246

Ft = 193 + 3t(Regression Line)

Forecast for next

JAN

1 2 3 4 5 6 7 8 9 10 11 12 13

232



Standard error of estimate = (Dt - Ft)2

= 7.32WhereDt = actual demand for period tFt = forecast for period tn = no. of data pointsf = degrees of freedom lost (2 in this

case)95 % confidence limits for forecast of next

JAN ~ 232 14 (2 sigma limits)

t=1,n

n-f



Common Regression Functions

dt

t

dt’ forecast

dt actual demand (for time period t)

dt’ = a + bt (parameters a, b)

Linear



dt

t

dt’ = a + u Cos (2/n)t + v Sin (2/n)t (parameters a, u, v)

Cyclic



dt

t

dt’ = a +bt + u Cos (2/n)t + v Sin (2/n)t (parameters a,b,u, v)

Cyclic withGrowth



dt

tdt’ = a +bt + ct2 (parameters a,b,c)

Parameters Determined by Minimizing the Sum of Squares of errors,

Quadratic



2. Analysis of Time series



COMPONENTS OF A TIME SERIES

Trend, Tt

Seasonality St

Cycle Ct

Randomness Rt



MULTIPLICATIVE MODEL

Xt = Tt * St * Ct * Rt

Time t Identify

Tt

St

Ct

Rt

Obtain Xt



DE-SEASONALIZING THE TIME SERIES

If the time series represents a seasonal pattern of L periods, then by taking a moving average Mt of L periods, we would get the mean value for the year. This would be free of seasonality and contain little randomness (owing to averaging)

Thus Mt = Tt * Ct



DETERMINATIN OF THE TREND

To the de-seasonalized series, a suitable trend line could be fitted using RegressionThe choices could beLinearQuadraticExponentialOther



ESTIMATING THE CYCLE COMPONENT

After the Trend Tt has been estimated

one can use

Ct = Mt / Tt

to estimate the Cycle Component, Ct



DETERMINATION OF SEASONAL INDICES

To isolate seasonality, one could simply divide the original series by the moving average

Xt/ Mt = Tt *St* Ct* Rt /Tt *Ct

= St * Rt

Averaging over same months eliminates randomness and yields seasonality indices



PROCEDURE FOR DECOMPOSTION

1 Decompose the time series into its components Find seasonal component Deseasonalize the demand Find trend component

2 Forecast future values of each component Project trend component into the future Multiply trend component by seasonal component



EXAMPLE 1

Past Sales Average Sales Seasonal Factor (1000/4)

Spring200 250 200/250 = 0.8Summer 350 250 350/250 = 1.4Fall 300 250 300/250 = 1.2Winter150 250 150/250 = 0.6

Total 1000



EXAMPLE 1(contd)

Expected Average Sales Seasonal Factor next year demand (1100/4) Next year forecastSpring 275 * 0.8 = 220Summer 275 * 1.4 = 385Fall 275 * 1.2 = 330Winter 275 * 0.6 = 165

Total 1100



EXAMPLE 2

Quarter Amount

I- 2000 300II- 2000 200III- 2000 220IV- 2000 530

Quarter Amount

I - 2001 520II- 2001 420III- 2001 400IV- 2001 700

Computing Trend & Seasonal Factor

on a 2 year demand history



EXAMPLE 2 (contd 1)

Quarter Demand

2000I 300II 200III 220IV 5302001I 520II 420III 400IV 700

From Trend Equation Ratio of Seasonal Tt = 170+55t Actual / Factor Trend

225 1.33 I 1.25 280 0.71 II 0.78335 0.66 III 0.69390 1.36 IV 1.25

445 1.17500 0.84555 0.72610 1.15



EXAMPLE 2 (contd 2)Forecast for 2002 using Trend and Seasonal factors

I – 2002 [170+ 55*09] 1.25 = 831II –2002 [170+ 55*10] 0.78 = 562III-2002 [170+ 55*11] 0.69 = 535IV- 2002 [170 + 55*12] 1.25 = 1,038

Trend * Seasonal factor = Forecast



EXAMPLE 3

For the given demand history prepare a forecast using decomposition

Period Actual Period Actual

1 300 5 416 2 540 6 7603 885 7 11914 580 8 760



EXAMPLE 3 (Contd 1)Period Actual Period Seasonal Deseasonalized x Y Average Factor Demand

1 300 358 0.527 568.992 540 650 0.957 564.093 885 1038 1.529 578.924 580 670 0.987 587.795 416 0.527 789.016 760 0.957 793.917 1191 1.529 779.088 760 0.987 770.21Total 5432 2716 8.0Average 679 679 1.0



EXAMPLE 3 (Contd 2)

Period x Deseazonalized demand, y x2 xy 1 568.99 1 569.0 2 564.09 4 1128.23 578.92 9 1736.74 587.79 16 2351.25 789.01 25 3945.0 6 793.91 36 4763.47 779.08 49 5453.68 770.21 64 6161.7Su ms 5432 204 26,108.8Average 679



EXAMPLE 3 (Contd 3)The regression equation for deseasonalized data:

26108 – (8) (4.5) (679) = 39.64 (slope of st. line) 9204) – (8) )(4.5)2

a = Y – bx = 679- 39.64(4.5) = 500.6 (intercept of st.line)

Thus, Y = 500.6+ 39.64x

is the result of the deseasonalized regression line

b =



EXAMPLE 3 (Contd 4)

Forecasts for the next four quarters of the following year

Period Trend Seasonal Final

Forecast Factor Forecast

9 857.4 * 0.527 = 452.0

10 897.0 * 0.957 = 858.7

11 936.7 * 1.529 = 1431.9

12 976.3 * 0.987 = 963.4



APPLICATION OF AUTO-CORRELATION

Random data

Trend

Seasonal

lag

lag

lag

Autocorrelation (range –1 to 1) plotted on the x-axis



CONCLUSIONS

Choice of forecasting technique Problem context Accuracy desired Cost Planning horizon

Importance of Correlation and Regression in analysis of Time Series Least Squares Normal Equations with examples

Time series decomposition Components (Trend, seasonal, cycle & random) Deseasonalization Re-construction of the time series



CONCLUSIONS (Contd)

Three illustrative examples of decomposition

Use of Auto-correlations to identify the kind of time seriesRandomTrend Seasonality



3. Aggregate Production Planning: Basic Concepts



AGGREGATE PRODUCTION PLANNING

Concerned with planning overall production of

all products combined (in tonnes of steel,

litres of paint etc.) Over a planning horizon

(generally next 3 to 6 months) for a given

(forecast) demand schedule.



CapacitiesCost

Commitments

Forecast ofDemand

Aggregate Production

Plan

Work force

M/c TimeAllocation over planning horizon



MANAGEMENT OPTIONS TO MEET FLUCTUATING

DEMANDBuild inventories in slack periods in anticipation of higher demands later in planning horizon.Carry backorders or tolerate lost sales during peak periods.Use over time in peak periods, under time in slack periods to vary output, while holding work force and facilities constant



Vary capacity by changing size of work force through hiring and firing

Vary capacity through changes in plant and equipment (generally long term option)

Each option involves cost (tangible or intangible). Aim in aggregate production planning is to choose best option.



KINDS OF COSTS INVOLVED

Procurement Costs

Production Costs

Inventory holding Costs

Shortage losses associated with backorders and lost sales

Costs of increasing / decreasing work force

Cost of overtime / under time

Cost of changing production rates (Set ups, opportunity losses etc)



Period Expected Demand

Cumulative Demand

1 100 100

2 180 280

3 220 500

4 150 650

5 100 950

6 200 950

7 250 1200

8 300 1500



Period Expected Demand

Cumulative Demand

9 260 1760

10 250 2010

11 240 2250

12 210 2460

13 140 2600

EXPECTED SALES FOR A ONE YEAR PLANNING HORIZON BROKEN INTO 13 (4 WEEK) PERIODS



GRAPHICAL PROCEDURE

1 2 3 4 5 6 7 8 9 10 11 12 13

500

1000

1500

2000

2500

3000

0

Plan 1 - - Constant Production: 200/period

Plan 2 - - Varying Production. 150 /period Period 1 - 5

250 /period Period 6 - 11

175 /period Period 12 - 13

Actual Cumulative Demand



Analysis of plan1Period Prod. Inv. Back

Order Capacity Change

Over Time

Sub Contract

1 200 100 0 +20 0 0

2 200 120 0 0 0 0

3 200 100 0 0 0 0

4 200 150 0 0 0 0

5 200 250 0 0 0 0

6 200 250 0 0 0 0

7 200 200 0 0 0 0

8 200 100 0 0 0 0



Analysis of plan1

Period Prod. Inv. Back Order

Capacity Change

Over Time

Sub Contract

9 200 40 0 0 0 0

10 200 0 10 0 0 0

11 200 0 50 0 0 0

12 200 0 60 0 0 0

13 200 0 0 0 0 0



Analysis of plan2Period Prod. Inv. Back

Order Capacity Change

Over Time

Sub Contract

1 150 50 0 -30 0 0

2 150 20 0 0 0 0

3 150 0 50 0 0 0

4 150 0 50 0 0 0

5 150 0 0 0 0 0

6 250 50 0 +50 40 10

7 250 50 0 0 40 10

8 250 100 0 0 40 10



Analysis of plan2

Period Prod. Inv. Back Order

Capacity Change

Over Time

Sub Contract

9 250 0 10 0 40 10

10 250 0 10 0 40 10

11 250 0 0 0 40 10

12 175 0 35 -25 0 0

13 175 0 0 0 0 0



ASSUMPTIONSAll shortages backlogged

Regular Time Capacity = 200 units/period

Max. Overtime = 20% of Regular Time Capacity

Overtime Preferable to Subcontract

Assumed Initial Inventory = 0

Initial Regular Time Prodn. Capacity = 180



NATURE OF COSTS AND SOLUTION PROCEDURES

Production Inventory Cost Linear Cost

Cost

•LP

•Simplified Transportation

•Transportation

Xt It

Ct (

Xt)

0




Production Inventory Cost Convex Cost

Cost

•Transportation

(after piece wise linearization

•HMMS ( LDR)

(quadratic cost)Xt It

Ct (

Xt)

0




Production Inventory CostConcave and Cost

Piecewise Concave Cost

•Dynamic Programming

(Shortest path Wangner / Whitin)Xt It

Ct (

Xt)

0



LINEAR DECISION RULESHolt, Modigliani, Muth & Simon (HMMS) 1955

(Study in large paint manufacturing unit)

Wt = Work force level in period t, t =1,…..T

Xt = Aggregate prodn.level in period t

It = Actual aggregate net inventory at end of period t

It* = Desired (ideal) aggregate net inventory at end of period t



COSTS INCURRED IN PERIOD IN PERIOD t

1. Regular time payroll cost

C1Wt + C13

2. Work force change cost

C2 (Wt - Wt-1 - C11 )2

Wt

Wt - Wt-1



3. Overtime Cost

C3 (Xt - C4 Wt)2 + C5Xt - C6 X Wt + C12XtWt

4. Inventory related costs

C7 (It - It*)2 = C7 (It - C8 - C9Dt)2

LDR W1*

X1 *

Linear fns. of Dt, W0, I0



Other Methods To Handle Aggregate Planning

Linear Programming Formulation(Hanssmann & Hess)

Search Decision Rules(Taubert)

Goal Programming Formulation(Multiple goals)

Parametric production planning(Jones)

Management coefficients model(Bowman)



SUMMARY

Aggregate Production Planning relevant for fluctuation demands for medium term horizons (6 months -1 year)A simple graphical procedure that generates good solutions by examining the demand pattern was presentedVarious ways to meet a fluctuating demand were consideredA summary of solution procedures including the Linear Decision Rules were indicated.



4. Aggregate Production Planning:Modelling Approaches



SOLUTION TECHNIQUES

Linear costsLinear ProgrammingTransportation Model

Piecewise linear and Convex costsHolt, Modigliani, Muth and Simon’s LDRsTransportation Model

Concave and Arbitrary CostsNetwork based ModelNon linear programming



LP : DEFINITION OF VARIABLES

r,v = cost /unit produced during regular time and overtime respectivelyPt, Ot = units produced during regular time and overtime, respectivelyH,f = hiring and layoff costs per unit, respectivelyAt, Rt = number of units increased or decreased, respectively, during consecutive periodsC = inventory costs [per unit per periodDt = sales forecastMt, Yt = Available regular time and overtime capacities respectively



LP: OBJECTIVE FUNCTION

Min C (production, hiring, layoffs, overtime, undertime and inventory)

= r Pt + h At + f Rt + v Ot + c It



LP: CONSTRAINTS

Pt <= Mt, t=1,2, , k

Ot<= Yt, t= 1,2, , k

It=It-1 + Pt +Ot-Dt , t= 1, 2, , k

At >= Pt-Pt-1 t= 1,2, , k

Rt >= Pt-1 – Pt t= 1,2, k

(All variables non-negative)



TRANSPRTATION MODELPeriod (month) 1 2 3 4Demands 100 105 200 95Production capacity(units)

Regular time 100 80 120 60Overtime 40 40 50 30

Production Costs (Rs) Regular time 16 20 22 18Overtime 24 30 30 26

Holding cost/unit/period (Rs) 2 2 4 5Initial on hand inventory (units) 50 unitsFinal desired inventory (units) 20 units



D1 D2 D3 D4 Ifinal Dummy I in 50

R1 100

O1 40

R2 80

O2 40

R3 120

O3 50

R4 60

O4 30

100 105 200 95 20 50 570

Setting up the Transportation Problem



D1 D2 D3 D4 Ifinal Dummy I in 0 2 4 8 13 0

50

R1 16 18 20 24 29 0 100

O1 24 26 28 32 37 0 40

R2 20 22 26 31 0 80

O2 30 32 36 41 0 40

R3 22 26 31 0 120

O3 30 34 39 0 50

R4 18 23 0 60

O4 26 31 0 30

100 105 200 95 20 50 570

Introducing Unit Costs




50 2 4 8 13 0

50/0

R1 16 50

18 20 24 29 0 100/50

O1 24 26 28 32 37 0 40

R2 20 22 26 31 0 80

O2 30 32 36 41 0 40

R3 22 26 31 0 120

O3 30 34 39 0 50

R4 18 23 0 60

O4 26 31 0 30

100 105 200 95 20 50 570

Satisfying 1st Period Demand




50 2 4 8 13 0

50/0

R1 16 50

18 50

20 24 29 0 100/50/0

O1 24 26 28 32 37 0 40

R2 20 55

22 26 31 0 80/25

O2 30 32 36 41 0 40

R3 22 26 31 0 120

O3 30 34 39 0 50

R4 18 23 0 60

O4 26 31 0 30

100 105 200 95 20 50 570

Satisfying 2nd Period Demand




50 2

4 8 13 0

50/0

R1 16 50

18 50

20 24 29 0 100/50/0

O1 24 26

28 40

32 37 0 40/0

R2 20 55

22 25

26 31 0 80/25/0

O2 30

32 36 41 0 40

R3 22 120

26 31 0 120/0

O3 30 15

34 39 0 50/35

R4 18 23 0 60

O4 26 31 0 30

100 105 200 95 20 50 570

Satisfying 3rd Period Demand




50 2 4 8 13 0

50/0

R1 16 50

18 50

20 24 29 0 100/50/0

O1 24 26 28 40

32 37 0 40/0

R2 20 55

22 25

26 31 0 80/25/0

O2 30 32 36 41 0 40

R3 22 120

26 31 0 120/0

O3 30 15

34 5

39 0 50/35/30

R4 18 60

23 0 60/0

O4 26 30

31 0 30/0

100 105 200 95 20 50 570

Satisfying 4th Period Demand module 8: Production Planning



50 2 4 8 13 0

50/0

R1 16 50

18 50

20 24 29 0 100/50/0

O1 24 26 28 40

32 37 0 40/0

R2 20 55

22 25

26 31 0 80/25/0

O2 30 32 36 41 0 40

R3 22 120

26 31 0 120/0

O3 30 15

34 5

39 20

0 50/35/30/10

R4 18 60

23 0 60/0

O4 26 30

31 0 30/0

100 105 200 95 20 50 570

Satisfying final inventory restrictions module 8: Production Planning



50 2 4 8 13 0

50/0

R1 16 50

18 50

20 24 29 0 100/50/0

O1 24 26 28 40

32 37 0 40/0

R2 20 55

22 25

26 31 0 80/25/0

O2 30 32 36 41 0 40

40/0

R3 22 120

26 31 0 120/0

O3 30 15

34 5

39 20

0 10

50/35/30/10/0

R4 18 60

23 0 60/0

O4 26 30

31 0 30/0

100 105 200 95 20 50 570

Satisfying Dummy restrictions



OPTIMAL SOLUTION

Total Cost of optimal solution = Sum of (unit cost x

quantity) = Rs 10370 for the planning horizon

This includes the costs of production on regular and overtime and the costs of holding inventories



OPTIMAL PRODUCTION PLAN

Period 1

Period 2 Period 3 Period 4

Regular time production

100 80 120 60

Overtime production

40 ----

(-40)

40

(-10)

30



NETWORK FLOW PROCEDURE

General cost structure

Linear Non-linear arbitrary Concave with set ups

Nature of Inventory Holding and Shortage Costs

It ItIt

cost



PRODUCTION COST

Qty Qty Qty

Cos

t of

Pro

duct

ion

Linear Piecewise linear Piecewise linear convex concave with set up



EXAMPLE PROBLEM

Four period problem (t = 1, 2, 3, 4)Production in period t involves a set up cost At

and a unit variable cost ct for raw materials, utilities, labour etc

Ct(Xt) = At + ct Xt , if Xt >0 = 0, if Xt =0

Linear holding costs (no shortage)

At

ct

Xt

Ct(

Xt)



PROBLEM DATA

Period 1 Period 2 Period 3 Period 4

Demand

(units)10 20 5 15

Set up

(Rs)100 120 120 140

Variable cost (Rs/unit)

8 9 10 10

Holding cost (Rs/unit)

2 4 5 7



SETTING UP THE NETWORK

0 1 2 3 4

For N periods there will be N+1 nodes and N(N+1)/2 arcs, withN emanating from Node 0, N-1 from Node 1,…,1 from Node N-1



ARC INTERPRETATIONS

Mjk = Cost of producing in period (j+1) for

the requirements of the periods

(j+1), (j+2), … (k). (j<k)

j k



COST COMPUTATIONS I

M01= A1+c1(D1)

M02 =A1 + c1 (D1+D2) +h1(D2)

M03 =A1 + c1 (D1+D2+D3) + h1 (D2+D3)+ h2 (D3)

M04 = A1 +c1 (D1+D2+ D3+D4) + h1 (D2+D3+ D4) +

h2(D3+ D4) + h3( D4)

M12 = A2 +c2(D2)

M13 = A2 +c2(D2+ D3) + h2(D3)

M14= A2 + c2 (D2+D3+D4)+ h2 (D3+D4) +h3 (D4)



COST COMPUTATIONS II

M23 = A3 + c3(D3)

M24 = A3 + c3(D3+D4) + h3 (D4)

M34 = A4 + c4 (D4)



COMPUTATION OF ARC LENGTHS

M01 = 100+ 8(10) = 180

M02 = 100+8(30) +2(20) = 380

M03 = 100+8(35) +2(25) + 4(5) = 450

M04 = 100 + 8(50) +2(40) +4(20) +5(15) = 735

M12= 120 +9(20)= 300

M13 = 120+9(20)+ 4(5) = 365

M14 = 120 + 9(40) +4(20) +5(15) = 635



COMPUTATION OF ARC LENGTHS

M23 = 120+ 10(5) = 170

M24 = 120+10(20) +5(15) = 395

M34 = 140+10(15) = 290



SETTING UP THE NETWORK

0 1 2 3 4M01=180

M02= 380

M03= 450

M04= 735

M12= 300

M13=365

M14 = 635

M23=170

M24= 395

M34=290



DETERMINING THE SHORTEST PATH IN THE

NETWORK

0 1 2 3 4M01=180

M02= 380

M03= 450

M04= 735

M12= 300

M13=365

M14 = 635

M23=170

M24= 395

M34=290

0 180 380 450 735



DETERMINING THE OPTIMAL PRODUCTION PLAN

The Shortest path is M04 which yields the optimal production plan with a total cost of Rs 735.

Period

1 2 3 4

Production

50 -- -- --



ARC INTERPRETATIONS(for the problem with shortages)

Mjk = Minimum Cost of producing for the

requirements of the periods

(j+1), (j+2), … (k). (j<k)

(Notice that production may take place

in any period between (j+1), (j+2),… (k) )

j k



SUMMARY

LP is a versatile tool to handle Aggregate Production Planning Problems with a variety of linear costs and constraintsThe Transportation Model is capable of dealing with problems with piecewise linear convex costs (regular time, overtime, subcontracting) No shortage case handled by a simple greedy procedure Case with shortages treated by regular transportation

problem procedure



SUMMARY II

The case of concave and arbitrary costs can be solved using a shortest path procedure on a network. Both cases with and without shortages

can be handled

Sample problems for both procedures were solved to illustrate the procedures



Documents

Project and Production Management Module 8 Production Planning over the Short Term Horizon Prof Arun Kanda & Prof S.G. Deshmukh, Department of Mechanical