Chapter 7 PPS

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ACCA-F5Performance Management 1

ACCA ± F5Limiting Factors

14th and 15th

July2011




Limiting FactorsChapter 7

ACCA ± F5Performance Management




�Identify limiting factors in a scarce resource situation & selectan appropriate technique

Determine optimal production plan under a single limitingfactor situation, including within the context of ³make-or-buy´ decision

� Formulate & solve multiple limiting factors problem using:� Graph, &� Simultaneous equations

Explain & calculate shadow prices, & discuss their implications on decision-making & performance management

Calculate slack & explain their implications on decision-making & performance management

Limiting FactorsObjectives:




Companies are always limited in their production capacity ± products or services

There is always restrictions in theavailability/supply of one or moreresources

Decisions are needed to optimiseproduction plan

Best option selected will maximisecontribution and profit

Limiting FactorsIntroduction




Limiting FactorsWhat are Limiting Factors?

A limiting factor is:

any factor that is in scarce supply, &that stops the organization fromexpanding its activities further,that it limits the organization¶s activities

It limit the ability of the organization toundertake all available alternatives




Limiting FactorsWhat are Limiting Factors?

Limiting factor limits the organization¶sactivitiesExamples:

Sales demandsProduction constraints

Labours supply

Materials availableManufacturing capacity




Limiting FactorsConcept

Contributionprofit is maximised when contribution is maximised

Fixed costs remains fixed

Contribution is maximised by maximisingcontribution per unit of limiting factor Decision: based on contribution earned per unit of limiting factor by each productThis will ensure that the company uses all itsresources effectively & efficientlyIf there is constraints in sales demand (limited salesdemand) ± profit is maximised by producing thehighest-ranked product to the sales demand limit




Limiting FactorsConcept

V ariable cost is the only relevant cost

Companies may have more than onelimiting factors .Management must decide on Appropriatetechniques:

Single limiting factor : Ranking methodMultiple limiting factors: Linear Programmingmethod




S ingle LimitingFactor





Limiting FactorsSingle Limiting Factor

Example 10 (pg 78):Company makes and sells 2 products: Beauty &Cutie

Capacity/day: Beauty: 50; Cutie: 40Contribution/unit: Beauty: $18; Cutie: $28Direct labour/unit needed: Beauty: 1hr; Cutie:2hrs

Limiting factor ± Direct labour hrs/day ±100 hrsOptimum production target?





Example 1.0 (pg 78)If no limiting factor

Beauty Cutie Total

Capacity/day (u) 50 40

D Labour/unit (hr) 1 2

Contribution/unit ($) 18 28

Total contribution ($) 900 1120 2020





Example 1.0 (pg 78)Limiting factor: 100 D Labour hrs/day

Beauty Cutie TotalCapacity/day (u) 50 40D Labour/unit (hr)/ Total needed 1/50 2/80 130 (available 100 )

Contribution/unit ($) 18 28

Contribution/D Labour ($) 18 14

Ranking: 1 2D Labour allocated/ No. of units 50/ 50 50/ 25 100hrs

Total contribution ($) 900 700 1600Ttl contribution If 40 units Cutie &balance to Beauty (20 units) ($)

360 1120 1400







Example 1.0 (pg 78)Step 2 : calculate contribution per unit limitingfactor ($ per D Labour hr)

Step 3 : Ranking

Beauty Cutie TotalContribution per unit ($) 18 28


Contribution per unit limiting factor($) 18

(18/1)

14

(28/2)

Ranking (allocate limited resource) 1 2





Example 1.0 (pg 78)Step 4 : Determine production plan

Ranking

Beauty Cutie TotalD Labour hr allocated by Ranking (hr) 50

(50x1)50

(balanceof 100-50)

100

Production plan (units) 5 0

(50/1)

25

(50/2)Total Contribution based onproduction plan ($)

900(50x18)

700(25x28)

1600




M ultiple LimitingFactors





Limiting FactorsMultiple Limiting Factors

Example 3.0 (pg 80)If there is excess D Labour & Machine hours

Silver Gold Total

S Price ($/u) 110 140

D Material ($/u) 35 (5kg/u) 33 (6kg/u)

D Labour ($/u) (@$5/H) 25 (5H/u) 40 (8H/u)

V OH ($/u) (DLH) 19 27

FOH ($/u) (McH) 5 25

Contribution/u ($) 31 40

Ranking 2 1E xcess resources Toproduce

less more





Example 4.0 (pg 80)If single limiting factor: D Labour hours (unlimited Machine hours)

Silver Gold TotalContribution/u ($) 31 40

D Material ($/u) 35 (5kg/u) 33 (6kg/u)

D Labour ($/u) (@$5/H) 25 (5H/u) 40 (8H/u)

Contribution/D Labour hours ($)

6.2(31/5)

5(40/8)

Ranking 1 2To produce more less





Example 5.0 (pg 81)If multiple limiting factors: D Labour hours & Machine hours

Silver Gold TotalContribution/u ($) 31 40

D Material ($/u) 35 (5kg/u) 33 (6kg/u)D Labour ($/u) (@$5/H) 25 (5H/u) 40 (8H/u)

Contribution/D Labour hours ($) 6 .2(31/5)

5(40/8)

Ranking 1 2

Contribution/D Labour hours ($) 6 .2(31/5)

6 .6 7(40/ 6 )

Ranking 2 1Decision ? ?





Where there are multiple limiting factorsthe contribution per limiting factor for differentscarce resources may show conflicting priorities

Use Linear Programming to solve the problem:to identify the optimal production for profitmaximisation





Linear Programming:Is a mathematical technique used to identifyoptimal decision variables

when there is more than one limiting factor Decision variables

refer to the different activities which are competing for the allocation of the limiting (scarce) resources

2 ways to solve the problems:Simultaneous equations methodGraphical method





Simultaneous equations method

Step 1 Identify objective variablesStep 2 Determine objective functionStep 3 Construct constraint equationsStep 4 Substitute variables into the

equations





Example 1.0 (pg 78)limiting factor: 100 D Labour hrs

Beauty (X) Cutie (Y) Total

Capacity/day (u) 50 40


Contribution/unit ($) 18 28Total contribution ($) 900 1120 2020





Simultaneous equations methodExample 1.0 (pg 78)

Step 1 Identify objective variables

The objective variables refer to the quantity of acourse of action which is required to achieve theoptimal solution

Let X = the number of units of Beauty manufacturedLet Y = the number of units of Cutie manufactured






Step 2 Determine objective function

The objective function refers to the objective whichthe company wants to achieveIt refers to the maximisation of profitability ORminimisation of costsThe function is expressed in algebra symbols

Z = 18X + 28YZ = the maximised contribution






Step 3 Construct constraint equations

Express the followings as linear equations:The multiple limiting factorsThe defined decision variables

Beauty production = X; Cutie production =Y

X, Y � 0X 50 (Beauty capacity)Y 40 (Cutie capacity)1X + 2Y 100 (Direct labour hours)






Step 4 Substitute variables into the

equationsBeauty production = X; Cutie production =Y

X = 50 (Beauty capacity)Y = 40 (Cutie capacity)1X + 2Y =100 (Direct labour hours)Z = 18X + 28Y






Step 4 Substitute variables into the

equations

1X + 2Y = 100 Z = 18X + 28Y

X = 50 ; Y = 25; Z = 900 + 700 = 16 00R efer to graph ± highest point of iso-contribution line (objectiveline) can touch in feasible area

If Y = 40; X = 20; Z = 360 + 1120 = 1480





Graphical method:Example 1.0 (pg78)

(Step 1)

Show the constraintobjective andfeasible regionon a graph:Capacity of X(i) X = 50

50 units

X = 50





Graphical method:Example 1.0(pg78)

(Step 2)

Show the constraintobjective andfeasible regionon a graph:Capacity of Y(ii) Y = 40

40 units

50 units

X = 50

Y = 40






(Step 3)Show the constraint

objective andfeasible region ona graph: DirectLabour Hr

(iii) 1X+2Y = 100X=0; Y=50Y=0; X=100

50 units

X = 50

DL: 1X+2Y=100

100 units

B: X=50;Y=25

C: X=20; Y=40 y = 40






(Step 3)Show the constraint

objective andfeasible region ona graph: DirectLabour Hr (iii) 1X+2Y = 100

X=0; Y=50Y=0; X=100

Opt point : B

50 units

28 units

X = 50

DL: 1X+2Y=100

100 units

18 units

Iso-profit line (iso=equal)18x+28y=504

B: X=50;Y=25

C: X=20; Y=40 y = 40




M ultiple LimitingFactors:

S lack&

S urplus





S lack refers to a situation when maximumavailability of a resource is not fully utilised

At optimum production, IF:

a resource available is fully utilisedConstraint is ³ binding ´

There is no ³slack´

a resource available is not fully utilisedConstraint not bindingThere is ³ slack ´

Limiting FactorsSlack




Limiting FactorsSlack: Example

Products X Y Max

Machine H/u 3 10 330H

D M kg/u 16 4 400kgD L H/u 6 6 240H

Contribution$/u

5 7

Max demand any 12





The constraint equations are:

Product X & Y

X, Y 0, Y 12Mc H : 3X + 10Y 330D M: 16X + 4Y 400DL: 6X + 6Y 240

Z = 5X + 7Y (maximise profit)

Substitute into equations and present in graph





Graphical method:Show the constraint

objective andfeasible region ona graph:

X = 12Mc H: 3X+10Y = 330DM: 16X+4Y = 400DL: 6X+6Y =240Optimal Pt:X=10, Y=30

DM: 16X+4Y=400 = 50

MC H: 3X+10Y=330

DL: 6X+6Y = 240

X=10; Y=30

Y = 12





Solving the equations:Where optimal production occurs (from graph):Mc H : 3X + 10Y = 330 (x2)

DL: 6X + 6Y = 240

6X+20Y =6606X+6Y =240

14Y =420Y=30; X=10





Substitute X=10, Y=30, into the constraint equations:Mc H : 3X + 10Y = 330

(3x10)+(10x30)=330=availabilityConstraint is binding

DL: 6X + 6Y = 240(6x10)+(6x30)=240=availabilityConstraint is binding

D M: 16X +4Y = 400(16x10)+(4x30)=280<400There is slack of 120kg of DM

More units can be produced if there ismore machine and labour hours availablesince there are slacks in materials




S urplus refers to a situation where at optimalproduction:

the units of a product produced:

is more than the maximum demandIn the example, there is a surplus of 18 units of

Y (30 ± 12) produced over the max of 12 units

Limiting FactorsSurplus




M ultiple LimitingFactors:

S hadow Price(dual price)







S hadow price :is the increase in contribution from the

original

Is the price of a scarce resource that acompany is willing to pay for

over and above the present price ±still obtain a contribution towards its

fixed costsNo shadow price for resources which is

not binding ± in excess

Limiting FactorsShadow (dual) price




Limiting FactorsShadow price: Example (Example 10 pg86)

Product X Y MaxSelling Price/u 200 250

DM ($5/kg) $ 6012kg

408kg 540,000kg

DL ($10/H) $ 505H

808H 400,000H

OH ($) 60 90

TotalV

Cost ($) 170 210Contribution/u $ 30 40

Demand (U) all Max 40,000




Limiting FactorsShadow price: Example (Example 10 pg86)

Identify objective variable and determineobjective function :

X, Y 0, Y 40,000D M: 12X + 8Y 540,000D L: 5X + 8Y 400,000Z = 30X + 40Y (maximise profit)

Substitute into equations and present in graph




Limiting FactorsShadow price: Example (Example 10 pg 86)


objective andfeasible region ona graph:

Y = 40,000

DM: 12X+8Y = 540,000

DL: 5X+8Y =400,000

Z=30X+40Y

Opt point :

Y=40,000

DM 12X+8Y=540,000

DL 5X+8T=400,000

X=20,000; Y=37,500





Solving the equations:Where optimal production occurs (from graph):DM : 12X + 8Y = 540,000

DL: 5X + 8Y = 400,000

7X = 140,000X = 20,000

Y = 37,500

Z = $600,000+$1,500,000=$2,100,000





Substitute X and Y value into the constraintequations:

Where optimal production occurs:

DM : 12X + 8Y = 540,000240,000 + 300,000 = 540,000(binding)

DL: 5X + 8Y = 400,000

100,000 + 300,000 = 400,000(binding)





Shadow price of DM ± if one extra unit of DM isavailable:

DM : 12X + 8Y = 540,00 1

DL: 5X + 8Y = 400,0007X = 140,001X = 20,000.14 x $30 = $ 600,004.2Y =37,499.91 x $40 = $1,499,996.4Revised contribution $2,100,000.6

Original contribution $2,100,000.0Increase in contribution $0.6Shadow price of DM = $0.6 / kg





Shadow price of DL ± if one extra unit of DLhour is available:

DM : 12X + 8Y = 540,000

DL: 5X + 8Y = 400,00 17X = 139,999X = 19,999.86 x $30 = $ 599,995.7Y =37,500.21 x $40 = $1,500,008.4Revised contribution $2,100,004.1

Original contribution $2,100,000.0Increase in contribution $4.1Shadow price of DL = $4.1 / H





Shadow price ± Implication:

Any binding scarce resource in the optimal solutionhave a shadow priceUsed to make decision if an existing resource shouldbe relieved of its effective value to the businessOnly valid for a small range before the constraintbecomes non-binding, or different resources becomecritical

A significant change would call for a revision in thelevel of resources, and a new shadow price will be

createdIt allows better informed decisions on payment of overtime premiums, bonuses, premiums on smallorders of DM




Limiting FactorsOther issues

Other issuesIn the short run, management should find ways toovercome the constraints

Look for substitutes

Work overtimeRent more space

In the long run, management should remove thelimiting factor

However, another limiting factor will surface onceone is removed




Limiting Factors:Practice Questions

(pg 90)





Limiting FactorsQ1 pg 90

Co produce 2 products, X & YUse 2 machinesMachine A: 3X+2y � 18Machine B: X+2Y � 10Z=5X+4Y

Substitute into equations:McA: 3X+2Y=18McB: X+2Y=10



ACCA-F5

Performance Management 55

Limiting FactorsQ1


objective andfeasible region on agraph:

Mc A:3X+2Y = 18Mc B: X+2Y = 10Z=5X+2Y

Total Contribution =20+12=32

Mc A: 3X+2Y = 18

Mc B: X+2Y = 10

X=4, Y=3



ACCA-F5



Equation method:

McA: 3X+2Y=18

McB: X+2Y=10

2X=8X=4, 2Y=6, Y=3



ACCA-F5



X Y MaxS Price ($/u) 220 206DML ($6/Lit) 30

5Lit/u36

6Lit/u24,000 Lit (1st 3 mth, &

2nd 3 mth)

DMM ($7.5/Lit) 456Lit/u

304Lit/u

24,000 Lit (2nd

3 mthonly)

V OH 55 44Total 130 110Contribution ($/u) 90 96Contribution/LtdFactor ($/u)

18 16

Rank 1 2



ACCA-F5



(a) Optimal production plan: Allocate all DML to X with the highestContribution/limiting factor

24000Lit / 5Lit per unit = 4,800 of XTotal contribution for producing 4,800 of X

= $432,000



ACCA-F5



(b) Optimal production plan:Formulate equations for constraints:

DML: 5X+6Y=24000 (x2)DMM: 6X+4Y=24000 (x3)

10X+12Y=48000

18X+12Y=72000

X=3000: Y=1500



ACCA-F5



AR2 GL3 HT4 Max

S Price ($/u) 21.00 28.50 27.30

DM R2 (kg/u) 2 ($5) 3 ($7.5) 3 (7.5) 5500kg

DM R3 (kg/u) 2 ($4) 2.2 ($4.4) 1.6 ($3.2)DL (H/u) 0.6 ($2.4) 1.2 ($4.8) 1.5 ($6)

V OH ($/u) $1.10 $1.30 $1.10

FOH ($/u) 1.50 1.60 1.70

Demand 950 1000 900

Contribution / u $8.5 $10.5 $9.5





ACCA-F5


Limiting FactorsQ4 (b)

4(b) Shadow price of DM R2:

Extra unit of R2 will go to produce HT4:1 extra kg R2 can produce extra 0.33 unit of HT4

= (1kg/3kg per unit HT4)Extra contribution for 0.33 units HT4 = $3.14Shadow price for R2 = $3.14Max price to offer for extra R2 = $5.64 /kg

($2+$3.14)



ACCA-F5


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Chapter 7 PPS