name : matr - wzl.rwth-aachen.de · page 1 matr.-no. : pm a task 1 page 1 pm a task 1, 20 points...

PM A Task 1 Name :

Matr.-No. :

PM A Task 1 Page 1

PM A Task 1, 20 Points

You are working as a product manager for a fridge manufacturer. To get new market

share in niche markets you are planning the market launch of a new fridge variant for

your product line as a reaction concerning sinking sales volumes in your existing

market. For this reason you are making a variant analysis to compare two product

concepts for the projected new fridge variant concerning their suitability.

a) A member of the development department with expertise in variant management

gives you the advice, not to trap into the vicious circle of complexity manage-

ment. What does he mean by giving you the advice? Explain the vicious circle of

complexity management and the resulting danger in short sentences. (3 points)

A rising variety in the product line mostly goes along with the sales activities in

niche markets (0.5 points). The costs that depend on the rising variety are get-

ting intransparent (0.5 points). In this case it is no longer traceable, which variant

causes which costs (0.5 points). The higher costs of exotic variants become sub-

sidised by the standard product line (0.5 points). This subsidisation results in a

competitive disadvantage towards competitors that offer a less variant-rich prod-

uct line (0.5 points). As a result of the sinking competitiveness the sales volumes

of the standard product variants decrease and so the vicious circle begins again

(0.5 points).

PM A Task 1 Name :

Matr.-No. :

PM A Task 1 Page 2

b) In the existing product line you offer fridges that are 85 cm, 140 cm and 180 cm

high. The fridges also differ in their energy efficiency, divided in the categories

A+, A and B. The customer can order fridges with glass or metal shelves. The

fridges can optionally have freezing compartments. In this case the customer can

decide between middle or premium class freezing compartments. Between how

many fridge variants could the customer choose by a free combination of all the

named product attributes and specifications? (1 point)

The theoretical number of variants is the product of the number of specifications

for each attribute.

Height: 80 cm, 140 cm or 180 cm > 3 specifications

Energy efficiency category: A+, A or B > 3 specifications

Fridges shelves: glass or metal > 2 specifications

Freezing compartment: without, middle class or premium class > 3 specifications

Theoretical number of variants: 3 * 3 * 2 * 3 = 54

(1 point for the right number)

c) In real the fridge manufacturer offers fewer variants on market. As a restriction of

the marketing, fridges with energy efficiency category A must always have a glass

shelf and a premium class freezer compartment (restriction 1). Fridges with en-

ergy efficiency category B only can have a middle class freezer compartment,

when they also have a metal shelf (restriction 2). And at least fridges that are 80

cm high have no freezer compartment (restriction 3). Please name the number

of variants that can be cancelled in case of each mentioned restriction. Name the

totally number of cancelled variants considering all restrictions together.

(6 points)

PM A Task 1 Name :

Matr.-No. :

PM A Task 1 Page 3

In case of the mentioned restrictions following variants can be cancelled:

Cancelled variants in case of restriction 1:

variantenergy efficiency

categoryshelf

freezer

compartmentheight double named

1 A+ metal shelf without 80 cm



4 A+ metal shelf middle class 80 cm 23

5 A+ metal shelf middle class 140 cm

6 A+ metal shelf middle class 180 cm

7 A+ metal shelf premium class 80 cm 24

8 A+ metal shelf premium class 140 cm

9 A+ metal shelf premium class 180 cm

10 A+ glass shelf without 80 cm



13 A+ glass shelf middle class 80 cm 25

14 A+ glass shelf middle class 140 cm

15 A+ glass shelf middle class 180 cm



categoryshelf

freezer


16 B glass shelf middle class 80 cm 21

17 B glass shelf middle class 140 cm

18 B glass shelf middle class 180 cm



categoryshelf

freezer


19 B metal shelf middle class 80 cm

20 B metal shelf premium class 80 cm

21 B glass shelf middle class 80 cm 16

22 B glass shelf premium class 80 cm

23 A+ metal shelf middle class 80 cm 4

24 A+ metal shelf premium class 80 cm 7

25 A+ glass shelf middle class 80 cm 13

26 A+ glass shelf premium class 80 cm

27 A metal shelf middle class 80 cm

28 A metal shelf premium class 80 cm

29 A glass shelf middle class 80 cm

30 A glass shelf premium class 80 cm

PM A Task 1 Name :

Matr.-No. :

PM A Task 1 Page 4

In case of restriction 1 in sum 15 variants can be cancelled. In case of restriction

2 in sum 3 variants can be cancelled. In case of restriction 3 in sum 13 variants

can be cancelled (1.5 points for the right number for each restriction). By

combination of all restrictions 26 variants can be cancelled (1 point).

d) The projected new fridge variant is an exclusive version of your mostly sold fridge

model „Economy 180-4*“. This model is 180 cm high, has a premium class

freezer compartment, metal shelves and the energy efficiency category A. For this

model you calculated a technical rating XT of 60% and an economical rating

XW of 75%. Please name the right formula to calculate the technical and econom-

ical rating. (1 point)

0.5 points for each right formula

n

i i

n

i i i

G

XT G

1

1

) (

n

i i

n

i i i

G

XW G

1

1

) (

technical rating XT =

economical rating XW =

with n = number of sub functions

with n = number of sub functions

PM A Task 1 Name :

Matr.-No. :

PM A Task 1 Page 5

e) The projected new fridge variant „Economy 180-4* AntiBac“ differs in compari-

son to the old model „Economy 180-4*“ in the fact, that it has a new antibacterial

protection layer. That is the reason why the technical rating of the new variant is a

quarter higher than the old model. By using the following formula you can calcu-

late the maximum allowed manufacturing costs KH of a complete assembled

fridge as a function of the economical rating XW.

KH [€] = 225 - 1,2 * Xw [%]

Please calculate the maximum allowed manufacturing costs of the new variant

„Economy 180-4* AntiBac“ in the case, that by using Kesselring‘s S-graph

(figure 1) the new variant is as advantageous as the old model. (2 points)

0 20 40 60 80 100

020

40

60

80

100

Economical rating

Te

ch

nic

alra

tin

g

Economy 180-4* AntiBac

Economy 180-4*

Figure 1: Kesselring’s S-graph

PM A Task 1 Name :

Matr.-No. :

PM A Task 1 Page 6

The technical rating of the new model is 60% * 1.25 = 75% (0.5 points for the

right result). The economical rating of the new model is 60% as a result of the

circular arc method under the assumption of the indifference of both solutions

(1 point for the right result, a mathematical result is also possible). The max-

imum value for the production cost is 153 € (0.5 points the right result).

f) In table 2 you will find a detailed overview of the used component parts of the

fridge model series with 180cm height, their part numbers and unit costs as well

as the assembly time of the component parts for one person (labour costs 40€/h).

One assembly operator is needed for each fridge. The manufacturing costs of the

old fridge model „Economy 180-4*“ are 130 €. You can choose between two so-

lutions to realise the antibacterial protection layer:

Solution 1: You can assemble a new fridge box as a result of an innova-

tive technology (Component part 1.2). The wall is covered with an anti-

bacterial protection layer in comparison to the old fridge box (Component

part 1.1). The layer will be coated during the manufacturing process of the

fridge box.

Solution 2: The interior of the fridge will be lined with an antibacterial pro-

tection layer in the last assembly step.

Which solution fulfils the requirements of exercise part e)? Hint: If you cannot

solve exercise part e) the maximum production costs KH are 150€! Please calcu-

late the manufacturing costs for both alternatives and explain your decision!

(2.5 points)

PM A Task 1 Name :

Matr.-No. :

PM A Task 1 Page 7

No. Costs [€] No. Costs [€] No. Costs [€]

Box 1.1 20 1.2 35 ... ... 3

Compressor 2.1 25 2.2 29 … … 6

Reverse side cooling pipe 3.1 10 3.2 12 3.3 14 10

Reverse side isolation 4.1 6 4.2 3 … … 3

Heat exchanger 5.1 5 5.2 8 ... ... 3

Door seal 6.1 9 6.2 6 ... ... 5

Door 7.1 12 ... ... ... ... 3

Rack 8.1 8 8.2 15 ... ... 1

Antibacterial protection layer 9.1 15 … … ... ... 15

DescriptionComponent Parts Assembly

Time [min.]

Table 2: component parts, costs and assembly time of the fridge model range

The manufacturing costs for the new fridge are 145 € if you choose solution 1 be-

cause you have to add only the additional charge of the new fridge box of 15 € to

the old costs (1 point for the right solution). The manufacturing costs for solu-

tion 2 are 155 €. You have to add the costs for the protection layer of 15 € and

the assembly costs of 10 € (1 point for the right solution). Only solution 1 fulfils

the requirements of exercise part e) (0.5 Points for the right decision).

PM A Task 1 Name :

Matr.-No. :

PM A Task 1 Page 8

g) Which solution can be described by an early variant origination point in terms of

the variant tree and which solution can be described by a late variant origination

point? Please explain your answer! (1.5 points)

Solution 1 has an early variant origination point; solution 2 has a late variant orig-

ination point. (0,5 points). Solution 1 differs between assembly variants with an

antibacterial protection layer (component part 1.1) and assembly variants without

an antibacterial protection layer (component part 1.2) in the first assembly step

(0.5 points). The variance of solution 2 will be created in the last assemby step

through the laying-up with an antibacterial protection layer (0.5 points).

h) Please explain, what kind of product structure design fits the new fridge box

(component part 1.2) most: Differential design or integral design? (1 point)

The new fridge box is related to the integral design (0.5 points), because the

fridge box integrates different product functions (isolation, antibacterial protection)

in one component part (0.5 points).

PM A Task 1 Name :

Matr.-No. :

PM A Task 1 Page 9

i) For the development of a new fridge box the manufacturer has used the TRIZ-

method for the first time. Please name four important characteristics of technical

problem solving, which are used by the TRIZ-method! (2 points)

– A systematic problem analysis leads to problem solving in many cases

(0.5 points)

– The contradiction is a central element of technical problems which lead to

innovations (0.5 points)

– Many problems have existing solutions in different branches and disciplines

with different names (0.5 points)

– The development of technical systems to an ideal product follows specific

rules (0.5 points)

PM A Task 2 Name :

Page 1 Matr.-No. :

PM A Task 2 page 1

PM A Task 2, 20 Points

Due to your excellent education at RWTH Aachen you start a career in a middle-

sized mechanical engineering company. After a short time you are the assistant of

the executive production manager and responsible for all the tasks of material man-

agement and job control.

a) After a while in your new position, you realize that your boss lacks basic

knowledge of material management. Because of that, he asks you to prepare a

short presentation for him in order to clarify the following aspects: (10 Points)

1) Please name three basic planning tasks of materials management. (0,75

Points: 0,25 Points each correct term, PM A L 8 P. 6)

1. Materials Procurement Planning

2. Materials Requirement Planning

3. Materials Stock Planning

2) Please address now possible target conflicts of materials management.

Therefore name at first four requirements of materials management. Indicate

for every requirement the effects on „lot size“ and „stock“ by using only the

„▲“ (upward arrow) and „▼“ (downward arrow). (2 Points: 0,5 points for

each correct requirement AND BOTH effects. If only one require-

ment/effect is missing: 0,25 Points; PM A L 8 P. 7)

Requirement: low capital commitment

Effect on lot size: „▼“

Effect on stock:„▼“

PM A Task 2 Name :

Page 2 Matr.-No. :

PM A Task 2 page 2

Requirement: low requirement for space


Effect on stock:„▼“

Requirement: Rational manufacture

Effect on lot size: „▲“

Effect on stock: „▲“

Requirement: High availibility


Effect on stock: „▲“

3) Please name the five steps of the procedure for stochastic requirements plan-

ning. (1,25 Points: 0,25 Points for each correct term, PM A L 8 P. 20)

1. Record of time series

2. Determine the consumption model

3. Select the method

4. Draw up the demand forecast

5. Evaluate forecast quality

PM A Task 2 Name :

Page 3 Matr.-No. :

PM A Task 2 page 3

4) The Production Manager thanks you for your presentation, but he did not un-

derstand the order point method completely. So please show the effective

stock progression by the time using the table below. Complete the table by en-

tering the progression of the actual stock and the available stock! The starting

stock is at 400 pieces. (1,5 Points: 0,75 points for each complete and cor-

rect column, 0,25 points together for three correct values in a column)

WD Task Actual Stock

(pieces)

Available Stock

(pieces)

82 Starting Stock 400 400

83 Outbound movement

50 pieces

375 375


150 pieces

225 225

86 Reservation of 75

pieces for WD 90

225 150


pieces for WD 91

225 100


75 pieces

150 100


50 pieces

100 100


pieces for WD 97

100 75


25 pieces

75 75

5) Mark the acquired values in the diagram! (1,5 Points: 0,75 points for each

complete and correct graph)

PM A Task 2 Name :

Page 4 Matr.-No. :

PM A Task 2 page 4

Diagram for task 2 a) 5):

8283

8485

8687

8889

9091

9293

9495

9697

98

300

250

200

150

100

50

400

350

Stock

[pieces]

WD

PM A Task 2 Name :

Page 5 Matr.-No. :

PM A Task 2 page 5

6) After your boss has realized, that you are able to make him familiar with the

theoretical basics of material management, he has got another task for you.

He asks you to recommend a forecasting method for a product with a given

consumption. Please extrapolate the requirement for the coming periods using

the Floating mean value (n = 3) and Exponential smoothing of the 1st order (α

= 0.3) rounded on one position after decimal point. Please determine as well

the difference of your calculated extrapolations from the actual consumptions.

Sum up the single deviations per method. Recommend on basis of the devia-

tion sum one forecasting method. (3 points: 0,25 points for each value and

each sum of deviation, 0,5 for the answer)

Formula Floating mean value: Fehler! Es ist nicht möglich, durch die Be-

arbeitung von Feldfunktionen Objekte zu erstellen.

Formula Exponential smoothing of the 1st order: Fehler! Es ist nicht mög-

lich, durch die Bearbeitung von Feldfunktionen Objekte zu erstellen.

Period t Consumption Vt Forecast Pt+1 Period t Consumption Vt Forecast Pt+1

0 89 90 0 89 90

1 96 97 1 96 97

2 104 96,3 2 104 99,1

3 97 99,0 3 97 98,5

4 105 102,0 4 105 100,5

5 109 103,7 5 109 103,1

Floating mean value (n = 3) Exponential smoothing (α = 0,3)

Difference:

Fl. MV: 0,7+6,0+7,0 = 13,7 ExpSm.: 2,1+6,5+8,5 = 17,1

Recommendation:

The forecasting method of the Floating mean value extrapolates the given con-

sumption better, than the exponential smoothing, since the difference sum is

smaller.

http://dict.leo.org/ende?lp=ende&p=eL4jU.&search=position

http://dict.leo.org/ende?lp=ende&p=eL4jU.&search=after

http://dict.leo.org/ende?lp=ende&p=eL4jU.&search=decimal

http://dict.leo.org/ende?lp=ende&p=eL4jU.&search=point

PM A Task 2 Name :

Page 6 Matr.-No. :

PM A Task 2 page 6

b) Effective dates for capacities are the result of the flow scheduling as a part of

the production requirement planning. These effective dates for capacities are

the basis for the following task of capacity planning. (2 points)

1) Which is the main objective of capacity planning? (0,5 Points) L07 – page

22

Within capacity planning, the distribution of activities among the individual

units of capacity is optimised, under consideration of the load limitations.

2) Which two basic measures exist for realizing the main objective of capacity

planning? (0,5 Points) L07 – page 24

1: capacity harmonisation

2: capacity adjustment

3) During the flow scheduling it is also possible to consider capacity limits or

constraints that occurred due to competing orders. Name and illustrate now

(by using key words) the characteristics of the for this purpose appropriate

scheduling method. How is this scheduling method conducted in combina-

tion with other scheduling methods? (1 Point; 0,25 Points for the right

nomination of the method; 0,5 Points for the right illustration; 0,25

Points for the right appropriation of the method) L07 – page 22

capacity-oriented scheduling

Within capacity-oriented scheduling, the mutual dependency between or-

ders and therefore between capacity limits is considered.

As a rule, at first order-oriented then capacity-oriented scheduling is con-

ducted.

PM A Task 2 Name :

Page 7 Matr.-No. :

PM A Task 2 page 7

c) Your company receives a profitable but also urgent order. The production

manager already calculated the delivery date by forward scheduling. Further-

more he is interested in the latest possible start dates for all parts and the as-

sembly. Calculate these dates by applying the backward scheduling without

taking capacities into account. Please mark for every working step in your dia-

gram on which machine it is performed. Also name all parts and components

that determine the critical path! (2 points: Backward-Scheduling 1,5 points –

0,5 points devaluation per mistake and if machines are not marked, 0,5

points critical path)

In production the following basic assumptions apply:

• The company possesses 3 different types of machines M1, M2 and M3

• The transition time between 2 machines in manufacturing is 2 hours

• The transition time of the individual parts or the modules to final or subas-

sembly is 3 hours

• You have one operator available per machine. In subassembly two workers

can assemble up to two modules simultaneously.

The product has the following structure:

E

BG2BG1

T2T1 T3 T4

E

BG2BG1

T2T1 T3 T4

PM A Task 2 Name :

Page 8 Matr.-No. :

PM A Task 2 page 8

Process sheets for all products and components are shown in following tables:

3 hVM2Subassembly BG210

DurationLocationDescriptionAG

Demand: 1 workerWorkplan BG2

3 hVM2Subassembly BG210

DurationLocationDescriptionAG

Demand: 1 workerWorkplan BG2

7 h H2 Final assembly E 10

Duration Location Description AG

Demand: 1 worker Workplan E

H2 Final assembly E 10


Demand: 1 worker Workplan E

4 h VM1 Subassembly BG1 10


Demand: 1 worker Workplan BG1

VM1 Subassembly BG1 10


Demand: 1 worker Workplan BG1

2 h M3 Drilling 30

1 h M2 Milling 20

2 h M1 10

Duration Location

Workplan T1

Turning

Description AG

M3 Drilling 30

M2 Milling 20

M1 10

Duration Location

Workplan T1

Turning

Description AG

2 h M2 Milling 20

3 h M1 Sawing 10


Workplan T2

M2 Milling 20

M1 Sawing 10


Workplan T2

3 h M2 Milling 20

2 h M1 Sawing 10


Workplan T3

M2 Milling 20

M1 Sawing 10


Workplan T3

5 h M3 Drilling 20

4 h M2 Milling 10


Workplan T4

M3 Drilling 20

M2 Milling 10


Workplan T4

PM A Task 2 Name :

Page 9 Matr.-No. :

PM A Task 2 page 9

0 5 10 15 20 25 30t [h]

T1

T2

T4

BG1

E

Backw

ard

Sc

he

du

lin

gF

orw

ard

Sc

he

du

lin

g

BG2

T3

T1

T2

T4

BG1

E

BG2

T3

M1 M2 M3

M2M1

VM BG1

M2M1

M2 M3

VM BG2

Final Assembly

M2 M3

VM BG2

Final Assembly

M2M1

M1 M2 M3

VM BG1

M2M1

0 5 10 15 20 25 30t [h]

T1

T2

T4

BG1

E

Backw

ard

Sc

he

du

lin

gF

orw

ard

Sc

he

du

lin

g

BG2

T3

T1

T2

T4

BG1

E

BG2

T3

M1 M2 M3

M2M1

VM BG1

M2M1

M2 M3

VM BG2

Final Assembly

M2 M3

VM BG2

Final Assembly

M2M1

M1 M2 M3

VM BG1

M2M1

Parts / Compontens on critical path: T4 – BG2 - E

d) The production manager is very satisfied with your work. Nonetheless he ob-

serves that you have not regarded the machine capacities for scheduling the or-

der. Which machine(s) does he see as bottleneck machine(s) within the backward

scheduling? (0,5 points)

Bottleneck machines (Backward scheduling): M1, M2, M3

PM A Task 2 Name :

Page 10 Matr.-No. :

PM A Task 2 page 10

Order

Code

1234

20C20A20A20B

20A20B40B20C

20B20C

XX

08.08.200906.08.200915.08.200910.08.2009

21

43

1 2 3

Targeted

Start DateSequence

Job sequence with load (h)

and work systemsOrder

Code

1234

20C20A20A20B

20A20B40B20C

20B20C

XX

08.08.200906.08.200915.08.200910.08.2009

21

43

1 2 3

Targeted

Start DateSequence

Job sequence with load (h)

and work systems

e) One Strategy in the production control is the Load-Oriented Order Release (BoA).

The following table 1 gives you the devaluation factor your company uses. In ta-

ble 2 the orders after the backward scheduling are given with their due dates and

the necessary ressources. Please use table 2 and 3 to release as many orders

with the BoA principle as possible! (3,5 points: 0,5 points for the correct order,

0,5 points each order in the BoA, 0,25 each correct order release)

1 2 3 4 5

Devaluation

factor1 0,5 0,25 0,125 0,0625

Operation

Table 1 – Devaluation factors

Table 2 – Orders after backward scheduling

Solution:

Table 3 – Result of the Load-Oriented Order Release

Calculation:

Devaluated Work Load (DWL) = Work Load * Devaluation Factor

(Remaining Capacity = Capacity – Devaluated Work Load)

1. Line, Order 2: Machine sequence: ABC

Order

der

-

code

2

1

4

3

20 (9)

10 (

(

- 1)

- ( - )

20 ( - 21)

10 (51)

5 (46)

20 (26)

20 (26)

5 (17)

20 ( - 3)

10 ( - 13)

- ( - )

F

F

X

X

1

2

3

4

A (29) B (61) C (22)

Release

info (F/X) Sequence

Work systems with

remaining capacity (h) Order

der

-

code

2

1

4

3

20 (9)

10 (

(

- 1)

- ( -1)

20 ( - 21)

10 (51)

5 (46)

20 (26)

- (46)

5 (17)

20 ( - 3)

10 ( - 13)

- ( -3)

F

F

X

X

1

2

3

4

A (29) B (61) C (22)

Release

info (F/X) Sequence

Work systems with

remaining capacity (h)

PM A Task 2 Name :

Page 11 Matr.-No. :

PM A Task 2 page 11

DWL A: 20 * 1 =20 Remaining Capacity A: 29 - 20 = 9

DWL B: 20 * 0,5 = 10 Remaining Capacity B: 61 - 10 = 51

DWL C: 20 * 0,25 = 5 Remaining Capacity C: 22 - 5 = 17

2. Line, Order 1: Machine Sequence: CAB

DWL C: 20 * 1 =20 Remaining Capacity C: 17 - 20 = -3 (blocked)

DWL A: 20 * 0,5 = 10 Remaining Capacity A: 9 - 10 = -1 (blocked)


3. Line, Order 4: Machine Sequence: BC

DWL B: 20 * 1 = 20 Remaining Capacity B: 46 - 20 = 26

DWL C: no calculation necessary

4. Line, Order 3: Machine Sequence: AB

DWL A: no calculation necessary

DWL B: no calculation necessary

f) Due to computer problems it is required to revise the already released order se-

quence one day after the first order was released for production. The responsible

shop manager tells you, that (solely) the first job of the first released order is al-

ready finished. Use this information to revise the Load-Oriented Order Release. Is

it now possible to release additional orders? (2 points; 1 point for the right cal-

culation, 1 point for the right order release)

Solution:

Calculation:

1. Line, Order 2: Machine sequence: ABC

DWL A: - =0 Remaining Capacity A: 29


DWL C: 20 * 0,5 = 10 Remaining Capacity C: 22 - 10 = 12

Order -

code

2

1

4

3

- (29)

)

10 (19)

- (19)

)

20 ( - 1)

20 (41)

5 (36)

20 (16)

20 ( - 4)

10 (12)

20 ( - 8)

10 ( - 18)

- ( - 8)

F

F

X

F

1

2

3

4

A (29) B (61) C (22)

Release

info (F/X) Sequence

Work systems with

remaining capacity (h) Order -

code

2

1

4

3

- (29)

)

10 (19)

- (19)

)

20 ( - 1)

20 (41)

5 (36)

20 (16)

20 ( 16)

10 (12)

20 ( - 8)

10 ( - 18)

- ( - 8)

F

F

X

F

1

2

3

4

A (29) B (61) C (22)

Release

info (F/X) Sequence

Work systems with

remaining capacity (h)

PM A Task 2 Name :

Page 12 Matr.-No. :

PM A Task 2 page 12

2. Line, Order 1: Machine Sequence: CAB

DWL C: 20 * 1 =20 Remaining Capacity C: 12 - 20 = -8 (blocked)

DWL A: 20 * 0,5 = 10 Remaining Capacity A: 29 - 10 = 19


3. Line, Order 4: Machine Sequence: BC


DWL C: no calculation necessary

4. Line, Order 3: Machine Sequence: AB

DWL A: 20 * 1= 20 Remaining Capacity A: 19 - 20 = -1 (blocked)


PM I Task 3 Name :

Page 1 Matr.-Nr. :

PM I Task 3 Page 1

PM I Task 3, 10 points

You are an employee in the process optimization department at a supplier for the

automotive industry and working at different production sites. Frequent interruption of

the assembly process, caused by missing parts, at the existing production facility in

Ludwigshafen requires you to adjust the number of kanban containers in the existing

kanban system.

a) You started your work by collecting data on the withdrawal and refill times of the

past few days (table 3-1). Based on this data please calculate the number of kan-

ban containers needed in order to prevent production stops. Please show the

used equations and round the result up to zero digits (3 points).

Also use the following data:

- a working day is 10 hours

- each kanban container contains 15 parts

- formula for calculating the standard deviation:

N

i

i xxN

S1

2

1

1

withdrawal

day / hour

refill

day / hour

difference [h]

6/ 4.82 6/ 9.13 4.31

7/ 2.65 7/ 8.21 5.56

8/ 7.45 9/ 2.16 4.71

7/ 9.71 8/ 1.12 1.41

7/ 0.26 7/ 3.76 3.5

8/ 0.05 8/ 8.83 8.78

Table 3-1: Observed withdrawal and refill times

1. Calculation of the average time between withdrawal and refill of the kanban

inventory at the assembly station (0,5 points)

hhhhhhh

t

N

tt

t

N

i

EiAi

71.46

78.85.341.171.456.531.4

1

PM I Task 3 Name :

Page 2 Matr.-Nr. :

PM I Task 3 Page 2

2. Calculation of the standard deviation (0,5 points)

withdrawal

day / hour

refill

day / hour

difference [h] tti

6/ 4.82 6/ 9.13 4.31 -0.4

7/ 2.65 7/ 8.21 5.56 0.85

8/ 7.45 9/ 2.16 4.71 0

7/ 9.71 8/ 1.12 1.41 -3.3

7/ 0.26 7/ 3.76 3.5 -1.21

8/ 0.05 8/ 8.83 8.78 4.07

44.207.421.13.385.04.016

1

1

1

22222

1

2

S

xxN

SN

i

i

3. Calculation of part consumption (0,5 points)

5 kanban containers with 15 parts each were used between 6/4.82 and 8/7.45

in 22.63h there were x(22.63h) = 75 parts used

in 10h (= 1 day) there were x(10h) = 33.14 parts used

4. Calculation of the average refill quantity for each refill (0,5 points)

partsh

partshxtx h 61.15314.371.41

5. Calculation of the average deviation of the needed refill quantity (0,5 points)

partsh

partshxss htx 09.8314.344.21

6. Calculation of the needed quantity of kanban containers at the workstation

(0,5 points)

container

container

parts

parts

containerperpartsno

Xcontainersno

partspartspartssxX x

58.1

15

7.23

..

7.2309.861.15

2 kanban containers are needed at the workstation

(Exercise 9, task 1.c)

PM I Task 3 Name :

Page 3 Matr.-Nr. :

PM I Task 3 Page 3

After completing your work in Ludwigshafen you are assigned to the site in Braun-

schweig. Here you have to draw the value stream map of a process and design new

concepts for it. In the first step you have drawn the current value stream map shown

in figure 3-1.

Fig. 3-1: Current value stream map

Le

ad t

ime

da

ys

Pro

cessin

g tim

e

sec.

Assem

bly

CT

=90 s

ec.

ST

=5

min

AT

=2700

0 s

ec.

Sh

ippin

g

CT

=50 s

ec.

ST

=0 m

in

AT

=2700

0 s

ec.

21

Mill

ing

CT

=95 s

ec.

ST

=8 m

in

AT

=2700

0 s

ec.

1

Tu

rnin

g

CT

=80 s

ec.

ST

=5 m

in

AT

=2700

0 s

ec.

1

Sa

win

g

CT

=40 s

ec.

ST

=5 m

in

AT

=2700

0 s

ec.

1

Trim

min

g

CT

=30 s

ec.

ST

=90 m

in

AT

=2700

0 s

ec.

1

5 d

ays

2 d

ays

2 d

ays

2 d

ays

3 d

ays

Su

pp

lier

Cu

sto

me

r

125.0

00 ite

ms/y

ear

1x w

eekly

Weekly

ord

er

Weekly

ord

er

Pro

duction p

lannin

g

2-w

eek p

lan

Oth

er

part

s

10 d

ays

Roun

d s

teell

10 d

ays

Ste

el pro

file

s

10 d

ays

Every

3 d

ays

Grin

din

g

CT

=60 s

ec.

ST

=10 m

in

AT

=2700

0 s

ec.

15 d

ays

Be

nd

ing

CT

=75 s

ec.

ST

=5 m

in

AT

=2700

0 s

ec.

1

CT

=35 s

ec.

ST

=5 m

in

AT

=2700

0 s

ec.

12 d

ays

Pu

nch

ing

2 d

ays

5 d

ays

Weekly

pla

nnin

g

Le

ad t

ime

da

ys

Pro

cessin

g tim

e

sec.

Assem

bly

CT

=90 s

ec.

ST

=5

min

AT

=2700

0 s

ec.

Sh

ippin

g

CT

=50 s

ec.

ST

=0 m

in

AT

=2700

0 s

ec.

21

Mill

ing

CT

=95 s

ec.

ST

=8 m

in

AT

=2700

0 s

ec.

1

Tu

rnin

g

CT

=80 s

ec.

ST

=5 m

in

AT

=2700

0 s

ec.

1

Sa

win

g

CT

=40 s

ec.

ST

=5 m

in

AT

=2700

0 s

ec.

1

Trim

min

g

CT

=30 s

ec.

ST

=90 m

in

AT

=2700

0 s

ec.

1

5 d

ays

2 d

ays

2 d

ays

2 d

ays

3 d

ays

Su

pp

lier

Cu

sto

me

r

125.0

00 ite

ms/y

ear

1x w

eekly

Weekly

ord

er

Weekly

ord

er

Pro

duction p

lannin

g

2-w

eek p

lan

Oth

er

part

s

10 d

ays

Roun

d s

teell

10 d

ays

Ste

el pro

file

s

10 d

ays

Every

3 d

ays

Grin

din

g

CT

=60 s

ec.

ST

=10 m

in

AT

=2700

0 s

ec.

15 d

ays

Be

nd

ing

CT

=75 s

ec.

ST

=5 m

in

AT

=2700

0 s

ec.

1

CT

=35 s

ec.

ST

=5 m

in

AT

=2700

0 s

ec.

12 d

ays

Pu

nch

ing

2 d

ays

5 d

ays

Weekly

pla

nnin

g

Assem

bly

CT

=90 s

ec.

ST

=5

min

AT

=2700

0 s

ec.

Sh

ippin

g

CT

=50 s

ec.

ST

=0 m

in

AT

=2700

0 s

ec.

21

Mill

ing

CT

=95 s

ec.

ST

=8 m

in

AT

=2700

0 s

ec.

1

Tu

rnin

g

CT

=80 s

ec.

ST

=5 m

in

AT

=2700

0 s

ec.

1

Sa

win

g

CT

=40 s

ec.

ST

=5 m

in

AT

=2700

0 s

ec.

1

Trim

min

g

CT

=30 s

ec.

ST

=90 m

in

AT

=2700

0 s

ec.

1

5 d

ays

2 d

ays

2 d

ays

2 d

ays

3 d

ays

Su

pp

lier

Cu

sto

me

r

125.0

00 ite

ms/y

ear

1x w

eekly

Weekly

ord

er

Weekly

ord

er

Pro

duction p

lannin

g

2-w

eek p

lan

Oth

er

part

s

10 d

ays

Roun

d s

teell

10 d

ays

Ste

el pro

file

s

10 d

ays

Oth

er

part

s

10 d

ays

Roun

d s

teell

10 d

ays

Ste

el pro

file

s

10 d

ays

Every

3 d

ays

Grin

din

g

CT

=60 s

ec.

ST

=10 m

in

AT

=2700

0 s

ec.

15 d

ays

Be

nd

ing

CT

=75 s

ec.

ST

=5 m

in

AT

=2700

0 s

ec.

1

CT

=35 s

ec.

ST

=5 m

in

AT

=2700

0 s

ec.

12 d

ays

Pu

nch

ing

2 d

ays

5 d

ays

Weekly

pla

nnin

g

PM I Task 3 Name :

Page 4 Matr.-Nr. :

PM I Task 3 Page 4

b) Where is the bottleneck (pacemaker process) in the value stream? (1 point)

Bottleneck:

Milling_(0,5 points)___________________________ _____

Reason:

The milling workstation has the longest cycle time. All other workstations

have lower work contents and thus shorter cycle times. (0,5 points)

(Exercise 9, task 2.2.a)

c) Please calculate the output of the workstation „Turning“ during 1 shift by taking

into account the setup time. Use an average lot size of 6 parts. Please indicate

the used equations! Take note of the following (1,5 points for the right result and

indicating the calculation formulae):

Shift length Ls: 8 h/shift

Break time Tb: 1 h/shift (30min lunch break, 2x15min break)

Lot size L: 6 parts after each setup

1. Calculation of the number of lots per shift

shift

lots

s

lotlot

parts

part

s

h

s

shift

h

shift

h

M

STLCT

TbLsM

31,32

min60

min5680

360018

2. Calculation of the number of parts per shift

shift

partsN

shift

parts

lot

parts

shift

lotsN

LMN

193

86,193631,32

(0,5 points for the time per lot or per part; 0,5 points for the division of the time

per shift by the time per lot or per part; 0,5 points for the correct result) (Exercise

9, task 2.2.a)

PM I Task 3 Name :

Page 5 Matr.-Nr. :

PM I Task 3 Page 5

d) Please calculate the production lead time and the processing time of the entire

value stream. Write this in the time axis in figure 3-1. (1 point for right production

lead time and 1 point for right processing time).

Le

ad

tim

e

29

da

ys

Pro

cessin

gtim

e

50

5 [sec.]

Assem

bly

CT

= 9

0 s

ec.

ST

=5 m

in

AT

= 2

700

0 s

ec.

Ship

pin

g

CT

= 5

0 s

ec.

ST

= 0

min

AT

= 2

700

0 s

ec.

21

Mill

ing

CT

= 9

5 s

ec.

ST

= 8

min

AT

= 2

700

0 s

ec.

1

Turn

ing

CT

= 8

0 s

ec.

ST

= 5

min

AT

= 2

700

0 s

ec.

1

Saw

ing

CT

= 4

0 s

ec.

ST

= 5

min

AT

= 2

700

0 s

ec.

1

Trim

min

g

CT

= 3

0 s

ec.

ST

= 9

0 m

in

AT

= 2

700

0 s

ec.

1

5 d

ays

2 d

ays

2 d

ays

2 d

ays

3 d

ays

Supplie

rC

usto

mer

125.0

00 ite

ms/y

ear

1x w

eekly

Weekly

ord

er

Weekly

ord

er

Pro

duction

pla

nnin

g

2-w

eek p

lan

Oth

er

part

s

10 d

ays

Roun

d s

teell

10 d

ays

Ste

el pro

file

s

10 d

ays

Every

3 d

ays

Grindin

g

CT

= 6

0 s

ec.

ST

= 1

0 m

in

AT

= 2

700

0 s

ec.

15 d

ays

Bendin

g

CT

= 7

5 s

ec.

ST

= 5

min

AT

= 2

700

0 s

ec.

1

CT

= 3

5 s

ec.

ST

= 5

min

AT

= 2

700

0 s

ec.

12 d

ays

Punchin

g

2 d

ays

5 d

ays

Weekly

pla

nnin

g

10

da

ys

5 d

ays

5 d

ays

2 d

ays

2 d

ays

2 d

ays

3 d

ays

40

se

c.

80

se

c.

95

se

c.

60

se

c.

18

0 s

ec.

50

se

c.

Le

ad

tim

e

29

da

ys

Pro

cessin

gtim

e

50

5 [sec.]

Assem

bly

CT

= 9

0 s

ec.

ST

=5 m

in

AT

= 2

700

0 s

ec.

Ship

pin

g

CT

= 5

0 s

ec.

ST

= 0

min

AT

= 2

700

0 s

ec.

21

Mill

ing

CT

= 9

5 s

ec.

ST

= 8

min

AT

= 2

700

0 s

ec.

1

Turn

ing

CT

= 8

0 s

ec.

ST

= 5

min

AT

= 2

700

0 s

ec.

1

Saw

ing

CT

= 4

0 s

ec.

ST

= 5

min

AT

= 2

700

0 s

ec.

1

Trim

min

g

CT

= 3

0 s

ec.

ST

= 9

0 m

in

AT

= 2

700

0 s

ec.

1

5 d

ays

2 d

ays

2 d

ays

2 d

ays

3 d

ays

Supplie

rC

usto

mer

125.0

00 ite

ms/y

ear

1x w

eekly

Weekly

ord

er

Weekly

ord

er

Pro

duction

pla

nnin

g

2-w

eek p

lan

Oth

er

part

s

10 d

ays

Roun

d s

teell

10 d

ays

Ste

el pro

file

s

10 d

ays

Every

3 d

ays

Grindin

g

CT

= 6

0 s

ec.

ST

= 1

0 m

in

AT

= 2

700

0 s

ec.

15 d

ays

Bendin

g

CT

= 7

5 s

ec.

ST

= 5

min

AT

= 2

700

0 s

ec.

1

CT

= 3

5 s

ec.

ST

= 5

min

AT

= 2

700

0 s

ec.

12 d

ays

Punchin

g

2 d

ays

5 d

ays

Weekly

pla

nnin

g

10

da

ys

5 d

ays

5 d

ays

2 d

ays

2 d

ays

2 d

ays

3 d

ays

40

se

c.

80

se

c.

95

se

c.

60

se

c.

18

0 s

ec.

50

se

c.

Assem

bly

CT

= 9

0 s

ec.

ST

=5 m

in

AT

= 2

700

0 s

ec.

Ship

pin

g

CT

= 5

0 s

ec.

ST

= 0

min

AT

= 2

700

0 s

ec.

21

Mill

ing

CT

= 9

5 s

ec.

ST

= 8

min

AT

= 2

700

0 s

ec.

1

Turn

ing

CT

= 8

0 s

ec.

ST

= 5

min

AT

= 2

700

0 s

ec.

1

Saw

ing

CT

= 4

0 s

ec.

ST

= 5

min

AT

= 2

700

0 s

ec.

1

Trim

min

g

CT

= 3

0 s

ec.

ST

= 9

0 m

in

AT

= 2

700

0 s

ec.

1

5 d

ays

2 d

ays

2 d

ays

2 d

ays

3 d

ays

Supplie

rC

usto

mer

125.0

00 ite

ms/y

ear

1x w

eekly

Weekly

ord

er

Weekly

ord

er

Pro

duction

pla

nnin

g

2-w

eek p

lan

Oth

er

part

s

10 d

ays

Roun

d s

teell

10 d

ays

Ste

el pro

file

s

10 d

ays

Every

3 d

ays

Grindin

g

CT

= 6

0 s

ec.

ST

= 1

0 m

in

AT

= 2

700

0 s

ec.

15 d

ays

Bendin

g

CT

= 7

5 s

ec.

ST

= 5

min

AT

= 2

700

0 s

ec.

1

CT

= 3

5 s

ec.

ST

= 5

min

AT

= 2

700

0 s

ec.

12 d

ays

Punchin

g

2 d

ays

5 d

ays

Weekly

pla

nnin

g

Assem

bly

CT

= 9

0 s

ec.

ST

=5 m

in

AT

= 2

700

0 s

ec.

Ship

pin

g

CT

= 5

0 s

ec.

ST

= 0

min

AT

= 2

700

0 s

ec.

21

Mill

ing

CT

= 9

5 s

ec.

ST

= 8

min

AT

= 2

700

0 s

ec.

1

Turn

ing

CT

= 8

0 s

ec.

ST

= 5

min

AT

= 2

700

0 s

ec.

1

Saw

ing

CT

= 4

0 s

ec.

ST

= 5

min

AT

= 2

700

0 s

ec.

1

Trim

min

g

CT

= 3

0 s

ec.

ST

= 9

0 m

in

AT

= 2

700

0 s

ec.

1

5 d

ays

2 d

ays

2 d

ays

2 d

ays

3 d

ays

Supplie

rC

usto

mer

125.0

00 ite

ms/y

ear

1x w

eekly

Weekly

ord

er

Weekly

ord

er

Pro

duction

pla

nnin

g

2-w

eek p

lan

Oth

er

part

s

10 d

ays

Roun

d s

teell

10 d

ays

Ste

el pro

file

s

10 d

ays

Oth

er

part

s

10 d

ays

Roun

d s

teell

10 d

ays

Ste

el pro

file

s

10 d

ays

Every

3 d

ays

Grindin

g

CT

= 6

0 s

ec.

ST

= 1

0 m

in

AT

= 2

700

0 s

ec.

15 d

ays

Bendin

g

CT

= 7

5 s

ec.

ST

= 5

min

AT

= 2

700

0 s

ec.

1

CT

= 3

5 s

ec.

ST

= 5

min

AT

= 2

700

0 s

ec.

12 d

ays

Punchin

g

2 d

ays

5 d

ays

Weekly

pla

nnin

g

10

da

ys

5 d

ays

5 d

ays

2 d

ays

2 d

ays

2 d

ays

3 d

ays

40

se

c.

80

se

c.

95

se

c.

60

se

c.

18

0 s

ec.

50

se

c.

10

da

ys

5 d

ays

5 d

ays

2 d

ays

2 d

ays

2 d

ays

3 d

ays

40

se

c.

80

se

c.

95

se

c.

60

se

c.

18

0 s

ec.

50

se

c.

Current value stream map (incl. lead time)

(Exercise 9, task 2.1.d)

PM I Task 3 Name :

Page 6 Matr.-Nr. :

PM I Task 3 Page 6

e) After analysing the weak point you start optimizing each process and design a

whole new future value stream map (figures 3-2 and 3-3). Which alternative do

you present to your manager? (2,5 points)

Alternative 1

Alternative 2

Please justify your answer (0,5 points for each valid justification):

1. Production planning is done at the bottleneck (assembly) and only in one place in

the process

2. Upstream processes are controlled by PULL, downstream processes are con-

trolled by FIFO

3. Higher transparency of inventory due to correct kanban control

4. Shorter delivery cycles allow more flexibility in purchasing

Shorter lead time

0,5 points for the right alternative; 0,5 points for each right justification (Exercise

9, task 2.2.c-g)

PM I Task 3 Name :

Page 7 Matr.-Nr. :

PM I Task 3 Page 7

Bild 3-2: Future value stream: alternative 1

Bild 3-3: Future value stream: Alternative 2

Manufacturing

Takt= 85 sec.

CT= 80 sec.

ST= durschnitl.

AT = 27000 sec.

Manufacturing

Takt= 85 sec.

CT= 80 sec.

ST= durschnitl.

AT = 27000 sec.

1 day

1 day

Sawing

CT= 40 sec.

ST= 5 min

AT = 27000 sec.

1

CT= 30 sec.

ST= 90 min

AT = 27000 sec.

Trimming

11 day

1 day

Assembly

CT= 90 sec.

ST=5 min

AT = 27000 sec.

2

Shipping

CT= 50 sec.

ST= 0 min

AT = 27000 sec.

1

Manufac-

turing and

Assembly

Packaging

SupplierCustomer

125.000 items/ year

3x wöchentlich

Weekly order Weekly order

Daily

Other parts

10 days

Round steel

10 days

Steel profiles

10 days

Production planning

Manufacturing

Takt= 85 sec.

CT= 80 sec.

ST= durschnitl.

AT = 27000 sec.

Manufacturing

Takt= 85 sec.

CT= 80 sec.

ST= durschnitl.

AT = 27000 sec.

Manufacturing

Takt= 85 sec.

CT= 80 sec.

ST= durschnitl.

AT = 27000 sec.

Manufacturing

Takt= 85 sec.

CT= 80 sec.

ST= durschnitl.

AT = 27000 sec.

1 day1 day

1 day1 day

Sawing

CT= 40 sec.

ST= 5 min

AT = 27000 sec.

1

Sawing

CT= 40 sec.

ST= 5 min

AT = 27000 sec.

1

CT= 30 sec.

ST= 90 min

AT = 27000 sec.

Trimming

1

CT= 30 sec.

ST= 90 min

AT = 27000 sec.

Trimming

11 day

1 day

Assembly

CT= 90 sec.

ST=5 min

AT = 27000 sec.

2

Assembly

CT= 90 sec.

ST=5 min

AT = 27000 sec.

2

Shipping

CT= 50 sec.

ST= 0 min

AT = 27000 sec.

1

Shipping

CT= 50 sec.

ST= 0 min

AT = 27000 sec.

1

Manufac-

turing and

Assembly

Packaging

Manufac-

turing and

Assembly

Packaging

SupplierCustomer

125.000 items/ year

3x wöchentlich


Daily

Other parts

10 days

Round steel

10 days

Steel profiles

10 days

Other parts

10 days

Round steel

10 days

Steel profiles

10 days

Production planning

Manufacturing

Takt= 85 sec.

CT= 80 sec.

ST= avg.

AT = 27000 sec.

Manufacturing

Takt= 85 sec.

CT= 80 sec.

ST= avg.

AT = 27000 sec.

Sawing

CT= 40 sec.

ST= 5 min

AT = 27000 sec.

1

CT= 30 sec.

ST= 90 min

AT = 27000 sec.

Trimming

11 day

1 day

Assembly

CT= 90 sec.

ST=5 min

AT = 27000 sec.

2

Shipping

CT= 50 sec.

ST= 0 min

AT = 27000 sec.

1

Manufac-

turing and

Assembly

Packaging

SupplierCustomer

125.000 items/ year

3x wöchentlich


Every 3 days

Other parts

10 days

Round steel

10 days

Steel profiles

10 days

Production planning

1 day

2 days

Manufacturing

Takt= 85 sec.

CT= 80 sec.

ST= avg.

AT = 27000 sec.

Manufacturing

Takt= 85 sec.

CT= 80 sec.

ST= avg.

AT = 27000 sec.

Manufacturing

Takt= 85 sec.

CT= 80 sec.

ST= avg.

AT = 27000 sec.

Manufacturing

Takt= 85 sec.

CT= 80 sec.

ST= avg.

AT = 27000 sec.

Sawing

CT= 40 sec.

ST= 5 min

AT = 27000 sec.

1

Sawing

CT= 40 sec.

ST= 5 min

AT = 27000 sec.

1

CT= 30 sec.

ST= 90 min

AT = 27000 sec.

Trimming

1

CT= 30 sec.

ST= 90 min

AT = 27000 sec.

Trimming

11 day

1 day

Assembly

CT= 90 sec.

ST=5 min

AT = 27000 sec.

2

Assembly

CT= 90 sec.

ST=5 min

AT = 27000 sec.

2

Shipping

CT= 50 sec.

ST= 0 min

AT = 27000 sec.

1

Shipping

CT= 50 sec.

ST= 0 min

AT = 27000 sec.

1

Manufac-

turing and

Assembly

Packaging

Manufac-

turing and

Assembly

Packaging

SupplierCustomer

125.000 items/ year

3x wöchentlich


Every 3 days

Other parts

10 days

Round steel

10 days

Steel profiles

10 days

Production planning

1 day

2 days

PM I Task 4 Name :

Page 1 Matr.-No. :

PM I Task 4 Page 1

PM I Task 4, 10 Points

You are assistant to the managing board in a medium sized company. Your boss is

responsible for the strategy development. In the next meeting of the board one of the

main topics is the business strategy. You support him in his preparation.

a) First he wants to present a reference framework that structures the strategy man-

agement according to a specific logic. He has already made a draft of the most

important picture of that framework. What is the name of that framework? (You

are not asked to fill in the five blanks.) (0,5 Points)

Initiation Positioning

Creation of

valueChange

Performance

Measurement

Initiation Positioning

Creation of

valueChange

Performance

Measurement

b) In addition he has heard of six different pathways through the framework. You

draw the six pathways and explain that the pathways can be ordered according to

their intention. What are the names of the different intentions? (1,5 Points)

General Management Navigator 0,5 Punkte für richtige Bezeichnung

PM I Task 4 Name :

Page 2 Matr.-No. :

PM I Task 4 Page 2

A

E F

CB

D

A

E F

CB

D

1. intention: strategic orientation

2. intention: invention of the business

3. intention: mobilisation

c) Attach two pathways to each intention. (3 Points)

1. intention: A E

2. intention: B C

3. intention: F D

d) In the meeting of the board your boss wants to talk about the advantages and

disadvantages of the different production sites. He asks you to prepare a tech-

nical and quantity based view as well as an economic and value based view of

the site in Portugal in comparison to the site in Aachen (Productivity and Efficien-

cy). He gives you the following information: The wages in Portugal are only one

quarter, but twice as many employees are needed in comparison to Germany to

produce the same amount (200.000 pcs.). The costs of material in Germany are

50% of the total costs. In absolute figures the costs of materials are the same in

both sites. The overhead costs are insignificant. The sales revenue is 15 Mio. € in

Germany, in Portugal only 60% of that. There is no change of inventory. Calculate

both key figures for the site in Aachen and in Portugal depending on the wages in

Germany (WG), the number of employees in Germany (XG) and the total costs in

Germany (CG) and compare them. Choose the number of employees as a refer-

ence for the technical and quantity based view, for the value based view choose

the costs. Calculate everything as far as possible. (5 Points)

0,5 Punkte pro richtiger Bezeichnung

0,5 Punkte pro richtiger Zuordnung eines Pfades

(Buchstaben) zum richtigen Zweck

PM I Task 4 Name :

Page 3 Matr.-No. :

PM I Task 4 Page 3

0,5 Punkte pro richtiger Bezeichnung

0,5 Punkte pro richtiger Zuordnung eines Pfades

(Buchstaben) zum richtigen Zweck

Productivity: G: P: G/P: = 2 Efficiency: G: P: G/P: 1,67*

200.000 Stk. 200.000 Stk.

15 Mio. € 9 Mio. € 0,5 KD + 1/2 XD * LD

0,5 Punkte richtige Produktivität G

0,5 Punkte richtige Produktivität P

0,5 Punkte richtiges Ergebnis G/P

0,5 Punkte richtige Wirtschaftlichkeit G

0,5 Punkte richtige Wirtschaftlichkeit P

Jeweils ,0,5 Punkte Nenner in G und P richtig zusammengefasst

1 Punkt richtige Formel D/P

0,5 Punkte richtiges Zahlenergebnis G/P

0,5 KD + XD * LD 0,5 KD + XD * LD

3/4 KD KD 1,25

XD 2 XD

0,5 KD + 1/2 XD * LD

name : matr - wzl.rwth-aachen.de · page 1 matr.-no. : pm a task 1 page 1 pm a task 1, 20 points...

Documents