Aggregate Planning

Production Planning and Control

Anadolu Üniversitesi Mühendislik Mimarlık FakültesiEndüstri Mühendisliğ Bölümü

Example 3.1

Model Number

Number of Worker-Hours

Required to Produce

Selling Price

A 5532 4.2 285

K 4242 4.9 345

L 9898 5.1 395

L 3800 5.2 425

M 2624 5.4 525

M 3880 5.8 725

Percentages of total

number of sales

32

21

17

14

10

6

Washing Machine Plant

MFICTIOUS = (4.2)(.32)+ (4.9)(.21)+ (5.1)(.17)+ (5.2)(.14)+ (5.4)(.10)+ (5.8)(.06) = 4.856

The ratio of selling price

divided by the worker-hours

67.86

70.41

77.45

81.73

97.22

125.00

Example 3.2

• Densepack is to plan workforce and production levels for six-month period January to June. The firm produces a line of disk drives for mainframe computers that are plug compatible with several computers produced by major manufacturers. Forecast demand over the next six months for a particular line of drives product in Milpitas, California, plant are 1280, 640, 900, 1200, 2000 and 1400. There are currently (end of December) 300 workers employed in the Milpitas plant. Ending inventory in December is expected to by 500 units, and the firm would like to have 600 units on hand at the end of June.

Problem parametersPlanning horizon January-June, six months

{January, February, March, April, May, June}{1,2,3,4,5,6}

Current number of workers 300

Starting inventory I0 =500 (Ending inventory in December)

Ending Inventory I6=600 (Inventory level end of June)

Cost of hiring one worker CH=$500

Cost of firing one worker CF=$1000

Cost of holding one unit of inventory for one month CI=$80

Number of aggregate units produced by one worker in one day

It is known that in the past for 22 working days with workforce level 76 workers, the firm produced 245 disk drivers.

245/22 = 11.1364 drivers per day when 76

workers employed

K = 11.1364 / 76 = 0.14653 drive per worker

Problem Parameters

Month

Net Predicted

Demand

Net Cumulative

Demand

January 780 (1280-500) 780

February 640 1420

March 900 2320

April 1200 3520

May 2000 5520

June 2000 (1400+600) 7520

Evaluation of chase strategy(Zero Inventory Plan )

A B C D E

MonthNumber of working

Days

Number of Unitst

Produced per worker

(Bx0.14653)

Forecast

Demand

Minimum Number of

Workers Required

(D/C rounded up)

January 20 2.931 780 267

February 24 3.517 640 182

March 18 2.638 900 342

April 26 3.810 1200 315

May 22 3.224 2000 621

June 15 2.198 2000 910

Evaluation of chase strategy(Zero Inventory Plan )

• Total cost of hiring, firing and holding is:(755)(500)+(145)(1000)+(30)(80) = $524900

+ the cost of holding for ending inventory$524900 + (600)(80) = $572900

A B C D E F G H I

MonthNumber of

workersNumber of

hiredNumber of

fired

Number of units per worker

Number of units

produced (BxE)

Cumulative

Production

Cumulative

Demand

EndingInventory (G-H)

January 267 33 2.9306 783 783 780 3February 182 85 3.51672 640 1423 1420 3March 342 160 2.63754 902 2325 2320 5April 315 27 3.80978 1200 3525 3520 5May 621 306 3.22366 2002 5527 5520 7June 910 289 2.19795 2000 7527 7520 7

755 145 30

continue

Evaluation of the Constant Workforce Plan

A B C D

Month Cumulative DemandCumulative Number of

units Produced per worker

Ratio B/C (round up)

January 780 2.931 267February 1420 6.448 221

March 2320 9.086 256April 3520 12.896 273May 5520 16.120 343June 7520 18.318 411

Evaluation of the Constant Workforce Plan

• Total cost of hiring and holding is:(111)(500)+(5962+600)(80) = $580460

A B C D E F

Month Number of units

produced per worker

Monthly Production

(B x 411)

Cumulative Production

Cumulative net Demand

Ending Inventory (D-E)

January 2.931 1205 1205 780 425February 3.517 1445 2650 1420 1230March 2.638 1084 3734 2320 1414April 3.81 1566 5300 3520 1780May 3.224 1325 6625 5520 1105June 2.198 903 7528 7520 8Total 5962

Mixed Strategies and Additional Constraints

• The graphical method can also be used when additional constraints are present.– Suppose that capacity of the plant is only

1800 units per month.

– Maximum change from one month to the nest be no more than 750 units.

• As the constraints become more complex finding good solutions graphically becomes more difficult.

• Most constraint of this nature can be incorporated easily into linear programming formulations.

0

1000

2000

3000

4000

5000

6000

7000

8000

1 2 3 4 5 6 7

• With more flexibility, small modifications can result in dramatically lower cost.

– Aggregate planning problem can be formulated and solved optimally* by linear programming .

Solution of Aggregate Planning Problems by Linear Programming

Cost parameters and given information

The following values are assumed to be known:

cHCost of hiring one worker,

cFCost of firing one worker,

cICost of holding one unit of stock for one period,

cRCost of producing one unit on regular time,

cOIncremental cost of producing one unit on overtime,

cUIdle cost per unit of production

csCost to subcontract one unit of production,

ntNumber of production days in period t,

K Number of aggregate units produced by one worker in one day,

I0Initial inventory on hand at the start of the planning horizon,

W0Initial workforce at the start of the planning horizon

DtForecast of demand in period t.

Solution of Aggregate Planning Problems by Linear Programming

Problem variablesThe following values are assumed to be known:

WtWorkforce level in period t,

PtProduction level in period t,

ItInventory level in period t,

HtNumber of workers hired in period t,

FtNumber of workers fired in period t,

OtOvertime production in units,

UtWorker idle time units (“undertime”)

StNumber of units subcontracted from outside.

Solution of Aggregate Planning Problems by Linear Programming

Problem Constraints1. Conservation of workforce constraints.

Wt = Wt-1 + Ht - Ft , for 1≤t ≤T

2. Conservation of units constraints.

It = It-1 + Pt + St - Dt , for 1≤t ≤T

3. Constraints relating production levels to workforce levels.

Pt = KntWt + Ot - Ut , for 1≤t ≤T

Solution of Aggregate Planning Problems by Linear Programming

Linear Programming FormulationMinimize ∑t (cHHt + cFFt +cIIt + cRPt + cOOt + cUUt + cSSt)

subject toWt = Wt-1 + Ht - Ft , for 1≤t ≤T

Conservation of workforce constraints.

It = It-1 + Pt + St - Dt , for 1≤t ≤TConservation of units constraints.

Pt = KntWt + Ot - Ut , for 1≤t ≤TConstraints relating production levels to workforce levels.

Ht ,Ft , It , Ot , Ut , St ,Wt ,Pt ≥ 0

Rounding the Variables

• In general optimal values of the problem will not integers.

• Fractional values for many variables do not make sense– Size of workforce, the number of workers hired each period, etc.

• All variables assumes integer (Integer linear programming problem)– Solution algorithm considerably more complex.

– Rounding• Closets Integer ? Inconsistency.

• Round up Wt to the next integer.

Extentions

• Minimum buffer inventory Bt each period.

It ≥ Bt , for 1≤t ≤T

• Capacity constraints on the amount of production each period

Pt ≤ Ct , for 1≤t ≤T

• Backlogging

It = It++ It

-

It+≥ 0, It

- ≥ 0

• convex-piecewise linear cost

∑t (cH1H1t +cH2H2t)

Ht = H1t+ H2t

0≤ H1t ≤H*

0≤ H2t

$Cost

Slope = c1H

H*

Slope = c2H

Solving Aggregate Planning Problems By Linear Programming: an example

Minimize ∑t (500Ht + 1000Ft +80It)subject to

Wt = Wt-1 + Ht - Ft , for 1≤t ≤T W1 - W0 - H1 + F1 =0W2 - W1 - H2 + F2 =0W3 - W2 - H3 + F3 =0W4 - W3 - H4 + F4 =0W5 - W4 - H5 + F5 =0W6 - W5 - H6 + F6 =0

Conservation of workforce constraints.

It = It-1 + Pt + St - Dt , for 1≤t ≤TP1 - I1 + I0 = 1280,P2 - I2 + I1 = 640,P3 - I3 + I2 = 900,P4 - I4 + I3 = 1200,P5 - I5 + I4 = 2000,P6 - I6 + I5 =1400,

Conservation of units constraints.Pt = KntWt + Ot - Ut , for 1≤t ≤TP1 - 2.931 W1=0P2 - 3.517 W2=0P3 - 2.638 W3=0P4 - 3.810 W4=0P5 - 3.224W5=0P6 - 2.198 W6=0

Constraints relating production levels to workforce levels.

Ht ,Ft , It ,Wt ,Pt ≥ 0W0 =300I0 =500I6 =600