Assignment

“IE 152: Manufacturing, Planning and Design”

Assignment on Systematic Layout Planning

Contributors:Eryka Gaye AdiqueAlexander AlfabeteJasper Jay Andrey

Ken Lester BermudezIsauro Bona IIILiezle Caroche

Jomel Mary DimasakaCherith FuraggananZilin Willian Japay

Dexter LeeJelica Legaspi

Myrian MendozaVictor Martin Mendoza

Aerold MontalboMarianne Nicolas

Timothy John Orleans

September 23, 2010

1

2.10. Eight parts (A to H) are to be manufactured and assembled to produce two products (X and Y). The processing sequences for the parts and assembly sequences for the products given below. The daily production rate for product X is 400 and for product Y is 600.

Required:(a) Determine the daily production rates for each part.(b) Design a product layout for manufacturing and assembling each product.(c) Design a process layout for manufacturing both products.(d) Design a group layout for manufacturing the two products.(e) Construct a from-to chart for the daily flow of parts between processes.(f) Construct an activity relationship chart for each product and combined REL chart for two products.(g) Using the traditional SLP approach, design a REL diagram for each REL chart constructed in part (f)

Part Processing sequence (grouped by machine)A (cast), (tumble), (mill), (turn, thread, drill, tap)B (cast), (tumble), (mill), (turn, drill, tap)C (turn, thread)D (punch), (bend), (drill)E (punch), (bend), (drill)F (punch), (bend), (drill)G (turn)H (turn, thread)

DAILY MACHINE REQUIREMENTS

MachineParts

A B C D E F G HBench mold 3 4.2Tumble mill 1.1 0.7Milling machine 2.4 3.4Turret lathe 3.5 1.7Automatic screw machine 1.3 0.6 0.9Drill press 1.5 2.2 1.1Punch press 0.3 0.5 0.4Press brake 0.8 1.2 1.1

II. Solution:

(a) Required: Average Daily Requirements

Assumption: All eight parts will be used to produce both products X and Y.

Solution:

Parts A to H: 400 (Product X) + 600 (Product Y) = 1,000 units

(b) Required: Product Layout

Reference: 2.6. Flow Analysis and Activity Analysis; 2.6.2 Types of Layout, pages 54-60

2


Diagram:

(c) Required: Process Layout

Reference: 2.6. Flow Analysis and Activity Analysis; 2.6.2 Types of Layout, pages 54-60;

Diagram:

3

(d) Required: Group Layout

Reference: 2.6. Flow Analysis and Activity Analysis; 2.6.2 Types of Layout, pages 54-60;


Diagram

(e) Required: From-To Chart

Assumptions: All eight parts will be used to produce both products X and Y. After Assembly, all products will be delivered to the consumer.

Diagram:

4

To

From StoreBench Mold

Tumble Mill

Milling Machine

Turret Lathe

Automatic Screw Machine

Drill Press

Punch Press

Press Brake Assembly

WarehouseTotal

Store 2000 3000 3000 8000Bench Mold 2000 2000Tumble Mill 2000 2000Milling Machine 2000 2000Turret Lathe 2000 2000Automatic Screw Machine

3000 3000Drill Press 3000 3000Punch Press 3000 3000Press Brake 3000 3000Assembly 8000 8000WarehouseTotal 2000 2000 2000 2000 3000 3000 3000 3000 8000 36000The from-to chart is used in designing both groups and process layouts. Consequently, it is used to analyze the material flow for low-

volume products and product families. The numerical values are based from the daily production rates from each part (ex: 1000 units x 2 [see given] = 2000 for bench mold and so on). The diagram above is designed according to the specific arrangement of the processing of parts in order to assemble the final product. The process started from the stores up to the assembly, and finally to the warehouse.

Reference: 2.6. Flow Analysis and Activity Analysis; 2.6.3 Flow Analysis Approaches, pages 61-66; Figure 2.23

5

(f) Required: Activity Relationship chart; Combined REL Chart


Diagram:

Reference: 2.6.4 Activity Relationship Analysis, pages 67-68; Figure 2.24

(g) Required: REL Diagram


Reference: 2.7 Relationship Diagram pages 69-87; Figures 2.24, 2.25, 2,27

Diagram:

A relationships are considered first, represented by four lines, followed by E which has three lines, and I, with two lines.

2.17 Construct the graph for the block layout shown in Figure 2.17; construct the dual graph and show that it is the same as the REL diagram for the layout.

Given: block diagram shown in Figure 2.17Required: primal graph , dual graphReference: 2.7, Relationship Diagram, page 78

Solution:First, we will assume a dummy activity H. This dummy activity will function as an external department. By doing so, we can have the corner points in the block layout and proceed with the whole process.

The numbers are the corner points in the layout. These corner points are points were at least 3 departments meet, including the exterior department. (Sec. 2.7, Relationship Diagram, page 78). Now the corner points will be joined by lines, called “edges.”

A B

C D

E F G

1

3 4

8 965 7

2

10 11

H

A B

C D

E F G

Figure 2.17

This is the primal graph of the layout constructed by connecting the corner points of the block layout.

This the dual graph or the REL diagram since we consider any shared point to satisfy the requirements of adjacency. (Sec. 2.7, Relationship Diagram, page 78)

1

3 4

8 965 7

2

10 11

1

3 4

8 965 7

2

100

11

A B

C D

EF

G

H

2.20. A job shop has received an order for high precision formed parts. The cost of producing each part is estimated to be $20,000. The customer requires that either four or five good parts be supplied. Each good part sold yield a revenue of $30,000. However, if fewer than four good parts are produced, none will be purchased; if more than five good parts are produced, the excess will not be purchased. The probability of an individual part being acceptable to the customer is 0.80. Determine the optimum batch production quantity given that job shop has agreed to supply the formed parts. Is the situation a profitable one for the job shop? If not, how must the selling price be changed for expected profit to be positive?

Given: Cost of each part = $20,000Each good part sold (revenue) = $30,000Probability of an individual part being acceptable to the customer = 0.80

Required:

Optimum batch production quantitySelling price to be profitable

Reference:

2.8. Space Determination and Availability; 2.8.1. Production Rate Determination; Reject Allowance Problem Approach.

Solution:

The following are the notations that will be used based on the process stated on the handouts.

x=random variable denoting the number of good parts producedp(x)=probability of producing exactly good partsQ=production lot sizeR (Q,x)=revenue resulting from producing Q parts of which exactly x are goodC (Q,x)=cost of producing a lot size Q, of which exactly x are goodP(Q,x)= R (Q,x)- C (Q,x)=profit resulting from producing Q parts, of which exactly x are goodE[R(Q)]=expected revenue resulting from producing Q partsE[C(Q)]= expected cost resulting from producing Q partsE[P(Q)]= expected profit resulting from producing Q parts

The expected profit model is formulated based on the conditions in the problem.

E[P(Q)]= C (Q,x)}p(x)

C (Q,x) = $20,000Q

R (Q,x) = $ 0 x<4$30,000 x=4,5$30,000(5) x>5

E[R (Q)]= 30,000 + 150,000

E[P (Q)]= 30,000 + 150,000 - 20,000Q

Q30,000 150,000

Cost Profit4 49,152.00 - 80,000.00 (30,848.00)5 98,304.00 - 100,000.00 (1,696.00)6 - 39,321.60 120,000.00 (80,678.40)7 - 70,778.88 140,000.00 (69,221.12)8 - 95,944.70 160,000.00 (64,055.30)9 - 116,077.36 180,000.00 (63,922.64)

10 - 132,183.49 200,000.00 (67,816.51)11 - 145,068.39 220,000.00 (74,931.61)12 - 150,000.00 240,000.00 (90,000.00)13 - 150,000.00 260,000.00 (110,000.00)

Based on the computations done, the project will only incur losses, without any profit whatsoever. Therefore, in order to minimize the expected loss, the company should produce 5 units of the high precision parts. By doing so, there would be an expected loss of $ 1,696.00

Solving for selling price to be profitable:

The project will be profitable at the evaluated optimum quantity in the first part of the problem. For Q=5, the project will be first profitable. Hence, the evaluation for the selling price of the high precision parts should be made when Q=5. The following are the computations for the selling price.

Answer:

The optimum batch quantity is 5 with Expected Profit = ($1,696.00). Moreover, the selling price to breakeven is $30,517.57813. In able to have profit the selling price should greater than the breakeven value.

2.22. A firm has received a production order of 10 injection-molded parts. The parts will sell for $4000 each; it will cost $2500 to produce an individual part. The probability of a part meeting the final inspection standards is .80. Any good parts in excess of 10 can be salvaged for $2000; any bad parts produced can be salvaged for $1000. If an insufficient number of good parts are produced, each shortage must be overcome at a cost of $3000 in excess of the cost of production. Determine the number of parts to be scheduled for production to maximize expected profit.

Required:The number of parts to be scheduled for production that would maximize the expected profit.

Given:Production Order: 10 injection-molded partsSelling Price: $4000/partCost: $2500/partProbability of meeting standard: 0.80Salvage value of excess (good) $2000Salvage value of bad part: $1000

Reference:

2.8. Space Requirements and Availability; 2.8.1. Production Rate Determination; Reject Allowance Problem Approach p. 88-91.

Solution:

The first thing that should be done is express the revenue and cost functions based on the given above.

To express the revenue function we divide the conditions set by the problem into cases, and . If the number of good parts produced is from 0 to 10,

the selling price would be $4000 for the 10 good parts made and the salvage value of each bad part would be $1000. On the other hand if the number of good parts produced is from 11 to , there would be an additional of $2000 for every good part produced in excess. Thus,

the revenue function will be as follows:

Where:Q =production lot sizex=random variable denoting the number of good parts produced

To express the cost function, we do the same, as we did to obtain the revenue function we divide the conditions set by the problem into cases, and

. If the number of good parts produced is from 0 to 10, it will cost $2500 to produce each part

of the order quantity and if an insufficient number of good parts are produced, each shortage must be overcome at a cost of $3000 in excess of the cost of production. On the other hand

if the number of good parts produced is from 11 to , only the cost of $2500 to produce each

part of the order quantity is considered. Thus the cost function will be as follows:

Where:

Q =production lot sizex=random variable denoting the number of good parts produced

After simplifying the two cases of the revenue function, and two cases of cost functions above (by combining like elements), we can express the expected revenue and cost functions to be as follows:

Combining the expected revenue function ( and expected cost function

, by subtracting the values of the cost function from that of the revenue function,

we arrive with the expected profit equation:

Based on the expected profit equation above we create a table that will us help find the number of parts to be scheduled for production that would maximize the expected profit. The table below summarizes the expected profit from the lot size of 0 to 17:

Lot Size

ExpectedProfit

0 0 0 0 0 01 -1200 1600 -24000 32000 84002 -3000 4160 -30000 40000 111603 -4500 7232 -30000 40000 127324 -6000 10508.8 -30000 40000 14508.85 -7500 13785.6 -30000 40000 16285.66 -9000 16931.33 -30000 40000 17931.3287 -10500 19867.34 -30000 40000 19367.34088 -12000 22551.7 -30000 40000 20551.69536

9 -13500 24967.61 -30000 40000 21467.6144610 -15000 27115.1 -30000 40000 22115.0981111 -16500 944.8928 -18897856.1 40000 -18873411.2112 -18000 1769.527 -35390530.52 40000 -35366760.9913 -19500 2484.209 -49684181.68 40000 -49661197.4714 -21000 3099.936 -61998711.91 40000 -61976611.9815 -22500 3627.701 -72554023.54 40000 -72532895.8416 -24000 4078.061 -81561222.79 40000 -81541144.7317 -25500 4460.867 -89217342.16 40000 -89198381.29

To determine the optimum production lot size the following table is filled out:

From the table it can be deduced that the optimum lot size to produce that would maximize the expected profit would be 10. This lot size will incur an expected profit of $22,115.09811.

2.23. A customer places an order for 50 units of a product, which your company can manufacture on a single machine. The product is custom-made and you doubt that the order will be repeated. The customer will pay you $1000 per unit of product produced within stated quality specifications, subjected to the following conditions: no more than 55 units would be purchased and the entire shipment will be refused if it contains less than 45 good units.

The cost of manufacturing and delivering the order is estimated to be $800 per unit. The manufacturing process is such that the probability of any unit of product being “out of specification” is 0.15. Because of a lack of instrumentation, you cannot inspect the product; inspection will be done by the customer after the shipment is delivered. It is desired to determine the number of units of product to manufacture to maximize expected profits. Develop a formulation of expected profit as a function of Q, the production lot size.

Given:

Order= 50 unitsSelling price= $1000 per unitFor customer to purchase the product where x is the random variable denoting the number of good products

producedManufacturing and delivering cost= $800 per unitProbability of out of specification= 0.15

Required:

Formulation of expected profit as a function of the production lot size,

Reference:

2.8. Space Requirements and Availability; 2.8.1. Production Rate Determination: Reject Allowance Problem Approach p. 89-91

Solution:

Definition of notation:X = random variable denoting the number of good products producedP(x) = probability of producing exactly x good castingsQ = production lot sizeR(Q,X) = revenue resulting from producing Q products of which exactly X are goodC(Q,X) = cost of producing a lot size of Q of which exactly X are goodP(Q,X) = profit resulting from producing Q products of which exactly X are good

=

E[R(Q)]= expected revenue resulting from producing Q castingsE[C(Q)]= expected cost resulting from producing Q castingsE[P(Q)]= expected profit resulting from producing Q castings

From our given we get the probability

Since the probability of out of specification is given, to get the probability of producing good units (p(x)) we would subtract the given probability from one

We then formulate our revenue and cost functions based on the conditions given in the problem.

We then make the expected profit function since all needed data are solved. The formula of expected profit is given as:

Equation 2.1 page 90

For the problem since is not a function of good units produced, the expected profit

can be computed as follow:

The function of expected revenue would be

The function of expected cost would be

Therefore, the expected profit could be computed as follows:

Conclusion:

To solve or the production rate we must first identify and distinguish all given values and data to us. We then analyze each data and properly denote them in their proper notation. We must also check and formulate the given constraints.

2.27. Consider a toaster that toasts one side of each two pieces of bread at the same time. It takes two hands to insert or remove each slice. To turn the slice over, it is necessary to push the toaster door all the way down and allow the spring to bring it back. Thus both slices can be turned at the same time, but only one slice can be inserted or removed at one time. The time required to toast one side of a slice of bread is 0.50 minute. The time required to turn a slice over is 0.02 minute. It takes 0.05 minute to remove a toasted slice and place it on a plate. The time required to secure a piece of bread and place it in the toaster is 0.05 minute. Determine the minimum amount of time required to toast three slices of bread on both sides. Begin with three untoasted slices of bread on a plate and end with all three slices of bread toasted and placed on a plate. Illustrate your solution with a multiple activity chart.

Given:Operation Time Toast one side of a slice of bread 0.05 min.Turn a slice over 0.02 min.Remove a toasted slice and place it on a plate 0.05 min.Secure a piece of bread and place it in the toaster 0.05 min.

Required:Minimum amount of time required to toast three slice of bread on both sides

Reference:2.6. Flow Analysis and Activity Analysis; 2.6.3. Flow Analysis Approaches;

Multiproduct Process Charts.

Solution:

Multiple Activity Chart (Present)Product: Present Proposed Saving

Bread Cycle Time

(min)

Worker 2.34 Process: Machine 2.34

Toasting Working Worker 0.34 Machine 2.04

Machine: Idle Toaster Worker 2.00

Machine 0.30 Utilization

Worker 14.53% Machine 87.18%

TimeWorker Machine

Time(min) (min)

0.05Secure 2 pieces of bread and place it in the toaster

Idle Time

0.05

0.10 0.10

0.15

Idle Time

Toast one side of 2 slices of bread

0.15

0.20 0.20

0.25 0.25

0.30 0.30

0.35 0.35

0.40 0.40

0.45 0.45

0.50 0.50

0.55 0.55

0.60 0.60

0.62 Turn 2 slices of bread over Turn 2 slices of bread over 0.62

0.67

Idle Time


0.67

0.72 0.72

0.77 0.77

0.82 0.82

0.87 0.87

0.92 0.92

0.97 0.97

1.02 1.02

1.07 1.07

1.12 1.12

1.17Remove 2 toasted slices of bread and place it on a plate

Idle Time

1.17

1.22 1.22

1.27Secure a piece of bread and place it in the toaster

1.27

1.32

Idle Time

Toast one side of a slice of bread

1.32

1.37 1.37

1.42 1.42

1.47 1.47

1.52 1.52

1.57 1.57

1.62 1.62

1.67 1.67

1.72 1.72

1.77 1.77

1.79 Turn a slice of bread over Turn a slice of bread over 1.79

1.84

Idle Time

Toast one side of a slice of bread

1.84

1.89 1.89

1.94 1.94

1.99 1.99

2.04 2.04

2.09 2.09

2.14 2.14

2.19 2.19

2.24 2.24

2.29 2.29

2.34Remove toasted slice of bread and place it on a plate

Idle Time 2.34

Multiple Activity Chart (Proposed)Product: Present Proposed Saving

Bread Cycle Time

(min)

Worker 2.34 1.94 0.40Process: Machine 2.34 1.94 0.40

Toasting Working Worker 0.34 0.44 -0.10Machine 2.04 1.54 0.50

Machine: Idle Toaster Worker 2.00 1.50 0.50

Machine 0.30 0.40 -0.10Utilization

Worker 14.53% 22.68% -8.15% Machine 87.18% 79.38% 7.80%

TimeWorker Machine

Time(min) (min)

0.05Secure 2 pieces of bread and place it in the toaster

Idle Time

0.05

0.10 0.10

0.15

Idle Time


0.15

0.20 0.20

0.25 0.25

0.30 0.30

0.35 0.35

0.40 0.40

0.45 0.45

0.50 0.50

0.55 0.55

0.60 0.60

0.62 Turn a slice of bread over Turn a slice of bread over

0.67Remove other slice of bread and temporarily place it on a plate

0.72Secure another piece of bread and place it in the toaster

0.77

Idle Time


0.82

0.87

0.92

0.97

1.02

1.07

1.12

1.17

1.22

1.27Remove toasted slice of bread and place it on a plate

Idle Time

1.29 Turn other slice of bread over Turn other slice of bread over

1.34Put back half toasted slice of bread in the toaster

Idle Time

1.39

Idle Time


1.39

1.44 1.44

1.49 1.49

1.54 1.54

1.59 1.59

1.64 1.64

1.69 1.69

1.74 1.74

1.79 1.79

1.84 1.84

1.89Remove 2 toasted slices of bread and place it on a plate

Idle Time

1.89

1.94 1.94

1.99 1.99

2.04 2.04

2.09 2.09

2.14 2.14

2.19 2.19

2.24 2.24

2.29 2.29

2.34 2.34

Conclusion:The minimum amount of time required to toast three slices of bread is 1.94 minutes.

2.28 Presently, on operator is tending five identical machines. Each machine is used to produce similar products. With five machines it has been observed that during a cycle a total of 20 machine-minutes is consumed waiting for a service to begin. In the past, it was observed that the operator was idle for 10 minutes each cycle when assigned to

tend three machines. If machining time is 25 minutes and independent operator time is 1 minute per machine cycle, what is the value for concurrent activity? If C1 equals $8 per hour and C2 equals $25 per hour, what is the economic assignment? What is the value of the minimum cost per unit produced?

Given:Current: 5 machines

One operator <---> 5 identical machines Total idle time of machines: 20 minutes per cycle = Im

Machining time: 25 minutes = t Individual operator time: 1 minute per machine cycle = b C1 = $8 per hour C2 = $25 per hour

Previous: 3 machines Idle time of operator: 10 minutes per cycle = Io

Required:(a) Value of concurrent activity, a(b) Economic assignment, m(c) Value of the minimum cost per unit produced, TC(m)

Solution:

(Topic: Space Requirements and Availability; Employee Requirements page 93-97)

Assumption:

The data for three-machine assignment were past data and they were not considered in the computations.

Im = 20/5 = 4 mins The total idle time of machines per cycle is 20 minutes so

we divide it first by the total number of machines.

Im = m(a + b) – (a + t) Equation 2.6 (p.96)4 = 5(a + 1) – (a + 25) We use this equation to get the value of the concurrent 4 = 5a + 5 – a – 25 activity.4 = 4a – 2024 = 4aa = 6 minutes

n’ = a + t Equation 2.3 (p.94) a + b We use this equation to determine the number of machines= 6 + 25 = 4.4286 to assign an operator for neither machine nor operator 6 + 1 idle time.

n’ = 4.4286If m = 4; m ≤ n’TC(m) = (C1 + mC2)(a + t) Equation 2.8 (p.96)

m We use this equation to calculate the cost per unit TC(4) = (8 + 4[25])(6 + 25) produced, based on an assignment of m machines

4 per operator; in this case, m=4 TC(4) = $837

If m = 5; m>n’TC(m) = (C1 + mC2)(a + b) Equation 2.8 (p.96)TC(5) = (8 + 2[25])(6 + 1) We use this equation to calculate the cost per unit TC(5) = $931 produced, based on an assignment of m machines

per operator; in this case, m=5

Φ = TC(n) Equation 2.9 (p.96) TC(n + 1) We use this equation to determine which is more

economicΦ = $837 between an n or n+1 machine assignment

$931Φ = $0.8990

Since Φ < 1, we assign n = 4 machines per operator which will give the minimum cost per unit produced of TC(4)=$837.

2.30. With a multiple activity chart show how one operator can handle two machines of type A and one machine of type B during a repeating cycle using the following data.

Given:

Machine A Machine BActivity Time(min) Activity Time(min)Load 1 Load 1 ½Inspect ½ Inspect 0Travel ½ Travel ½Machining 7 Machining 6 ½Unload 1 Unload 1

Req’d:

A multiple activity chart showing a continuous cycle of one worker operating 2 Machine A and 1 Machine B

Reference:

2.6. Flow Analysis and Activity Analysis; 2.6.3. Flow Analysis Approaches; Multiproduct Process Charts.

Solution:

First step is to analyze the whole process first before developing the multiple process activity chart.

Multiple Activity Chart

Product: Process: Machine(s):

N/A Machining 2 Machine A and 1 Machine B

TimeOperator Machine A1

TimeMachine A2

TimeMachine B

Time(min) (min) (min) (min)

1.0 Load (Machine A1)

Idle Time 1.0

Machining

1.0

Machining

1.0

2.0Inspect (Machine A1)

Machining

2.0

2.0

2.0Travel (Machine A2)

3.0 Unoad (Machine A2)

3.0

Idle Time3.0

3.0


4.0

4.0

4.0


5.0

Machining

5.0

5.0Travel (Machine B)

6.0 Unload (Machine B)

6.0

6.0

Idle Time

6.0

7.0Load (Machine B)

7.0

7.0

7.0

8.0

8.0

8.0


Machining

9.0Unload (Machine A1)

Idle Time9.0

9.0

9.0


10.0

10.0

10.0


Machining

11.0

11.0


12.0Unload (Machine A2)

12.0

Idle Time12.0

12.0


13.0

13.0

13.0

14.0Inspect (Machine A2) 14.0

Machining

14.0

14.0Travel (Machine B)

15.0 Unload (Machine B)

15.0

15.0

Idle Time

15.0

16.0Load (Machine B)

16.0

16.0

16.0

17.0 17.0

17.0

17.0Travel (Machine A1) Machining

2.34. The Acme Chemical Company has a drug-line conveyor system that operates between two manufacturing buildings. Large hoppers are attached to the conveyor to transport

chemical mixes from building A to building B for further batch processing. Empty hoppers are returned to building A on the same loop conveyor. Empty hoppers are cleaned automatically on the return trip to A. Activities and times for this process are:

Activities Average Time (min)

Unload empty hoppers from conveyor at A, fill 40 and place on conveyorTravel from A to B 10Unload full hopper at B, empty, and place 50 on conveyorTravel from B to A (including cleaning) 15

To ensure that chemical mixes are always available at building B, what is the minimum number of hoppers required? If the chemical mix cannot stay in the hopper longer than 20 minutes from the time loading is completed and unloading begins, what is the maximum number of hoppers that could be used? Demonstrate graphically that your

solution is feasible.

Given:Activities Average Time (min)

Unload empty hoppers from conveyor at A, fill 40 and place on conveyorTravel from A to B 10Unload full hopper at B, empty, and place 50 on conveyorTravel from B to A (including cleaning) 15

Required: a.) To ensure that chemical mixes are always available at building B, what is the

minimum number of hoppers required?

b.) If the chemical mix cannot stay in the hopper longer than 20 minutes from the time loading is completed and unloading begins, what is the maximum number of hoppers that could be used?

Reference:2.8. Space Requirements and Availability; 2.8.3. Employee Requirements; Identical

activities p. 93-97.

Concepts Used:Employee Requirement

Solution:a.) To be able to ensure that chemical mixes are always available at building B we must equate the idle time of the operator in building B to zero, thus implying that chemical mixes are always available.

, thus

Where:

=idle operator time during a repeating cycle

=production lot size

=random variable denoting the number of good parts produced

Computing for we use the equation 2.3.:

Where:=concurrent activity time

=independent machine activity

=independent operator activity time

Using the given: =90 (40+50: unload and fill hopper at A + unload and empty

hopper at B), =25 (10+15: travel from A to B + travel from B to A.), = 0 (there is no

independent operator activity time given) we come up with the following equation:

Since and a fraction of a hopper cannot be assigned we round up to

the next integer, which is 2, thus . Therefore the repeating cycle time can be

computed as:

equation 2.4.

Where:=number of machines assigned to an operator.

Thus, substituting =50 (only considering the activity time at B) and m=2 we

get:

To show graphically that the above solution is feasible a multiple activity chart is created as shown:

Legend:O-B: Operator at building BM: Machine (followed by number)U: UnloadingL: Loading (done at building A)R: automatic run (conveyor)

Thus the minimum number of hoppers required is 2.Fig.1 multiple activity chart for operator at building B

b.)To compute for the maximum number of hoppers to be used we equate the (idle time per machine) to 10 minutes. This is because the maximum allowable time for the chemical to be transported from building A to B is said to be 20 minutes, but since the average time for transporting the chemical from building A to B is 10 minutes, we subtract 10 minutes from the maximum 20, giving us 10 minutes of idle time.

, thus

And since:equation2.6.

We first equate m from the above equation substituting , and then

equate for m:

We then substitute this m to the appropriate equation when .

equation 2.4.

Substituting the given ( =90, =20, = 0):

To get , we utilize the following equation:

To show graphically that the above solution is feasible a multiple activity chart is created as shown:

Legend:L-A: Loading done at building AU-B: Unloading done at building B

M: Machine (followed by number)R: automatic run (conveyor) : Idle time

Thus the maximum number of hoppers is 3

Fig.2 multiple activity chart with idle time of 5 minutes

2.37. Automatic palletizers are used to palletize cases of finished goods at production plant. The palletizer places a a given number of layers of cartons on each pallet, which each layer having a specified number of cartons. The time required for the palletizer to perform a palletizing cycle is 2 minutes per pallet.

An operator is responsible for programming and setting up the palletizer for each “palletizing run” based on the dimensions of the case. A palletizing run consists of a batch of 100 pallets. Hence, the palletizer can operate without interruption for 200 minutes before programming and setup are required. It takes the operator 5 minutes to program and set up the palletizer.

Additionally, the operator must supply empty pallets to the palletizer. The empty pallets are placed in a magazine with a capacity to hold 25 pallets. It requires 6 minutes for operator to travel to the empty pallet storage area, return with 25 pallets, and places the 25 empty pallets in the magazine.

Determine the number of palletizers to be assigned to an operator without creating idle time for the palletizer. Assume negligible travel time between palletizers.

Given:

To accomplish one palletizing run,

b0 = 6 minutes; where b0 = time to unload/load 25 pallets a = 5 minutes setup time t = 200 minutes for 1 batch of 100 pallets

Required:

Number of palletizers to be assigned to operator

Reference:

2.8. Space Requirements and Availability; 2.8.3. Employee Requirements; Indentitical Activity p. 93-97.

Solution:

(Topic: Space Requirements and Availability; Employee Requirements P 93-97)

Given that the palletizer can operate without interruption, thus, t = 200 minutes having 100 pallets in a batch. Such the case, the total concurrent activity time can be computed through:

Using the equation:

equation 2.3.

Where:

n’ = number of machines to assign to an operator without idle timea = concurrent activity timeb = independent operator activityt = independent machine activity time(a + b) = time units for an operator(a + t) = time to complete production cycle

Substituting the given values. Therefore,

Since the computed n’ is not a whole number thus, Im = 0 means m ≤ n’.Finally, we can say that 7 machines should be assigned to

2.39. An operator is currently operating three identical machines. The operator utilized 78% of the repeating cycle. Concurrent activity equals 8 minutes and independent operator activity equals 5 minutes. Each operator costs $10 per hour and each machine costs $12 per hour.

(a) What is the cost per unit produced based on the minimum-cost assignment?(b) For what range of values for concurrent activity will the economic assignment equal three machines?

Given:m (number of machines assigned an operator)= 3 machinesT (repeating cycle- operator)= 78%a (concurrent activity time)= 8 minutesb (independent operator activity time)= 5 minutesC1 (cost per operator-hour)= $10C2 (cost per machine-hour)= $12

Required:

a. Cost per unit produced based on minimum cost assignment (TC)b. Range of values for concurrent activity where economic assignment will equal to

threemachines.

Reference:

2.8. Space Requirements and Availability; 2.8.3. Employee Requirements; Identical Activity p. 93-97.

Solution:Solve for (a)Required: TC

The formula for TC is:

and

C1, C2, m, a, b and percentage repeating cycle of operator are given. To solve for n’ and TC, we need to find t.

m≤n’m>n’

Equation 2.7.

Solve for t Io= T(repeating cycle-machine) (for m ≤ n’)Io= (1-0.78)Tc

Io= (0.22) Tc

Since, Tc= a + t (for m ≤ n’)

Io= (a + t) – m (a + t) (for m ≤ n’) Then,

(a + t) – m (a + b) = 0.22 (a + t)(8 + t) – 3 (8 + 5) = 0.22 (8 + t)t = 42 minutes

To know the formula we should use in getting the TC, we should solve for n.

Solve for n

= 3.846 machines

Since, )

and= 3.846

then,

Solve for so that we could know whether to use n or n + 1 Since

= C1/C2

Then,

= 10/12

Solving for ,

Since

Solve for TC which is required for (a)Use the formula

Equation 2.3.

Equation 2.10.

Equation 2.9.

Solve for (b)

For the system to use 3 machines, should be less than 1. Since

And

minutes < a

Conclusion:

The values for the concurrent activity time should start at 8.29686 minutes so that the economic assignment equal three machines.

Equation 2.10.

Equation 2.3.

2.40. The KLM Job Shop has requested that a new layout be designed for their operation in Alpharetta, Georgia. There are 12 departments involved. The department areas (in square feet) and activity relationships for the job shop are summarized in Figure 2. 40. Design a block layout using the SLP approach.

Given: Values assigned to each relationship are as follows:

A = 10000, E = 1000, I = 100, O = 10, U = 0, AND X = -10000

The REL chart is summarized in this table:Table 2.40.1. Summary of relationships of KLM Job Shop.DEPT.

Department Summary TCR

1 2 3 4 5 6 7 8 9 10 11 12 A E I O U X

1 - I U X U U U U U U U U 0 0 1 0 8 2 -19900

2 I - U U U U I U U U U U 0 0 2 0 9 0 200

3 U U - A I E E U E U I I 1 3 3 0 4 0 13, 300

4 X Y A - U I X U E I I I 1 1 4 0 3 2 -8600

5 X U I U - I X U U I E I 0 1 4 0 4 2 -18600

6 U U E I I - E I E I I I 0 3 6 0 2 0 3600

7 U I E X X E - E U A U U 1 3 1 0 4 2 -6900

8 U U U U U I E - U E I U 0 2 2 0 7 0 2200

9 U U E E U E U U - U U U 0 3 0 0 8 0 3000

10 U U U I I I A E I - I I 1 1 3 0 6 0 11300

11 U U I I E I U I U U - U 0 0 4 0 7 0 1400

12 U U I I I I U U U U U - 0 0 4 0 7 0 400

Required: Block Layout

Solution: Based on the procedure of assigning the order of placement given discussed in

Section 2 .7. 4, the order of departments in their corresponding numbers are as follows in:1st. 3 because it has the highest TCR2nd. 4 because of its A relationship with 33rd. 6 because it has E relationship with 3 and a TCR value of 36004th. 9 because of its E relationship with 3, 4,and 6 and a TCR value of 30005th. 7 because of its E relationship with 3 and 66th. 10 because of its A relationship with 7

7th .8 because of E relationship with 7 and 10 and TCR value of 2,2008th. 11 because of its I relationship with 6 and 4 and TCR value of 1400. 9th. 12 because its TCR value is 40010th. 2 because of TCR value of 20011th. 5 because of X relationship with 712th. 1 because of its X relationship with 4rd

Next is the computation for the relative locations for each department where positions 1,7, 3 are directly adjacent and will have a value equal to the TCR of the next department and 2,4,6,8 are partially adjacent and will be computed as TCR/2 of the next department.

WPV1. 10,000 -- position one is chosen because of the western edge rule2. (10,000/2) = 5,0003. 10,0004. (10,000/2) = 5,0005. 10,0006. (10,000/2) = 5,0007. 10,0008. (10,000/2) = 5,000

On the other positions, same steps will apply.

1. 1002. (100/2) = 503. 100 + (1,000/2) = 6004. 1,000 + (100/2) = 1,0505. (1,000/2) = 500 6. 1,0007. (1,000/2) = 500 8. 1,000 + (100/2) = 1,0509. 100 + (1000/2) = 60010. (100/2) = 50

1. 1,0002. (1,000/2) = 5003. 1,000 + 1,000 + (1,000/2) = 2,5004. (1,000/2) = 5005. 1,0006. (1,000/2) = 5007. 1,000 + (1,000/2) = 1,5008. 1,000 + (1,000/2) = 1,5009. (1,000/2) = 50010. (1,000 + (1,000/2) = 1,50011. 1,000 + (1,000/2) = 1,50012. (1,100/2) = 500

3

8 7 6

1 5

2 3 4

1. -10,000 + 0 = -10,0002. (-10,000/2) + 0 = -5,0003. 04. 0 + (1,000/2) = 5005. 0 + 1,000 = 1,0006. (1,000/2) = 5007. 1,000 + (1,000/2) = 1,5008. (1,000/2) + 1,000= 1,5009. (1,000/2) = 50010. (-10,000/2) + 1,000 = -4,00011. -10,000 + (1,000/2) = -9,50012. (-10,000/2) = -5,000

1. 100 + 0 = 1002. (100/2) = 503. 04. 0 + (100/2) =505. 0 + 100 = 1006. (100/2) + 10,000 = 10,0507. (10,000/2) = 5,0008. 10,0009. (10,000/2) = 5,00010. (100/2) + 10,000 + 0 = 10,05011. 012. 0 + (100/2) = 5013. 100 + 0 = 10014. (100/2) = 50

1. 02. 03. 04. 0 + (100/2) = 505. 100 + (1,000/2) + 1,000 = 1,6006. (100/2) = 507. 1,0008. (1,000/2) = 5009. 1,000 + 1,000 = 2,00010. ,000 + 1,000 = 2,00011. (1,000/2) = 50012. 1,000 + (100/2) = 1,050

13. 014. 015. 016. 0

1. 100 + 0 =1002. (100/2) + 0 = 503. 04. 0 + (100/2) = 505. 100 + 0 + 0 = 1006. 07. 0 + (100/2) = 508. 100 + 0 = 1009. (100/2) = 5010. 10011. (100/2) = 5012. 100 + 0 + 0 = 10013. 014. 0 + (100/2) + 100 = 15015. (100/2) = 5016. 100 + (100/2) 15017. 10018. (100/2) = 50

1. 100 + 0 = 1002. (100/2) + 0 = 503. 04. 0 + (100/2) = 505. 100 + 0 + 0 =1006. 07. 08. 09. 010. 011. 012. 013. 014. 015. 0 + (100/2) = 5016. 100 + (100/2) = 15017. 100 + (100/2) = 150

18. (100/2) = 50

1. 02. 03. 04. 05. 0 + (100/2) + 0 = 506. 07. 08. 09. 010. 011. 012. 0 + 0 + 100 + 0 + 10013. 0 + (100/2)14. 015. 016. 017. 018. 019. 020. 0

1. 02. 03. 04. (100/2) + 0 = 505. 100 + (-10,000/2) + 100 = 4,8006. (100/2) = 507. 1008. (100/2) = 509. 010. 011. 012. 013. 0 + (-10,000/2) + 1,000 = -4,00014. (1,000/2) = 50015. 1,000 + (100/2) + 100 = 1,15016. (100/2) =5017. 100

2

12

11

810

6 79

34

19

18

15

1

2 11

17

6

4 5

9

16

7

3 10

1213

8

20 14

18. (100/2) =5019. 100 +0 + (100/2) =15020. 0

1. 10,000 + 0 = -10,0002. (-10,000/2) = -5,0003. 04. 05. 06. 07. 1008. 09. 010. (100/2) = 5011. 10012. (100/2) = 5013. 100 + (-10,000/2) = -4,90014. -10,00015. (-10,000/2) = -5,00016. -10,00017. (-10,000/2) = -5,00018. 019. 020. 0

Final Block Layout from Relationship Diagramming Process

Proposed Layout with corresponding space requirement

2.44. An auto-parts warehouse in Flyspeck, Texas, has required that a new layout be designed for their main warehouse located in metropolitan Flyspeck. The warehouse has 10

major activity “centers”. The current building has the dimensions of 150x225 ft. Other pertinent data are summarized in Figure P2.44. Using SLP, design a layout to be contained in the current building.

Figure P2.44

Solution

Given:

Required: Design new layout using SLP

Solution:

1. Using graph based process from the activity relationship chart given, form an Activity Relationship Graph/Diagram. (2.7.3, pp. 71).

Activity Relationship Graph

The given graph shows a graphical activity relationship diagram. The graph is non planar since it has many overlapping lines. Also, according Sec. 2.7.3, no more than 3N-6 pairs of activities can be adjacent. Thus, applying the equation:

Combination of activities that has an:

A relationship – (2,3), (7,8)E relationship – (1,2), (3,4), (4,5)I relationship – (1,9), (2,4), (2,5), (2,9), (3,5), (3,9), (4,6), (5,6)O relationship – (1,10), (2,6), (2,10), (3,6), (3,7), (3,10), (4,7), (4,9), (4,10), (5,7), (5,9),

(5,10), (6,7), (6,9), (6,10), (7,9), (7,10), (8,9), (8,10), (9,10)U relationship – (1,3), (1,4), (1,5), (1,6), (1,7), (1,8), (3,8), (4,8), (5,8), (6,8)X relationship – (2,7), (2,8)

Since: N=10 A = 2 ; E = 3; I = 8 ; O=20; TOTAL = 32 adjacent activities

3N-6 = 3(10) – 6 = 24Since 32>24, hence, the graph is non-planar.

Since this is a non-planar graph, we can use the relationship diagramming process in order to develop a block diagram, for the given figure P2.44. (2.7.4, pp. 83). Total closeness rating can be derived using the activity chart given above.

Dept Department Summary *TCR1 2 3 4 5 6 7 8 9 10 A E I O U X

# Activity1 Office2 Counter3 Part Bins4 Muffler Bins5 Tailpipe racks6 Paint room7 Storage8 Receiving9 Lounge10 Restroom

A E I O U X/ / / / / --

1 - E U U U U U U I O 0 1 1 1 6 0 11102 E - A I I O X X I O 1 1 3 2 0 2 -86803 U A - E I O O U I O 1 1 2 3 2 0 112304 U I E - E I O U O O 0 2 2 3 2 0 22305 U I I E - I O U O O 0 1 3 3 2 0 13306 U O O I I - O U O O 0 0 2 5 2 0 2507 U X O O O O - A O O 1 0 0 6 1 1 608 U X U U U U A - O O 1 0 0 2 5 1 209 I I I O O O O O - O 0 0 3 6 0 0 360

10 O O O O O O O O O - 0 0 0 9 0 0 90

TCR or total closeness ratio computations are shown below. Given that the assigned closeness ratings are: A=10000, E=1000, I=100, O=10, U=0, and X=-10000, the following computations were solved.

Dept 1 = 1*1000 + 1*100 + 1*10 + 6*0 = 1110Dept 2 = 1*10000 + 1*1000 + 3*100 + 2*10 + 2*-10000 = -8680Dept 3 = 1*10000 + 1*1000 + 2*100 + 3*10 + 2*0 = 11230Dept 4 = 2*1000 + 2*100 + 3*10 + 2*0 = 2230Dept 5 = 1*1000 + 3*100 + 3*10 + 2*0 = 1330Dept 6 = 2*100 + 5*10 + 2*0 = 250Dept 7 = 1*10000 + 6*10 + 1*0+ 1*-10000 = 60Dept 8 = 1*10000 + 2*10 + 5*0+ 1*-10000 = 20Dept 9 = 3*100 + 6*10 = 360Dept 10 = 9*10 = 90

Following the steps given in 2.7.4 pp 83-87, we can determine the placement sequence. The first department is the one having the greatest TCR. Dept 3 = 11230. The second department placed is Dept 2, since Dept 2 is the only department that has an A relationship with Dept 3. An X relationship exist in Dept 2 between Dept 8 and Dept 7. since Dept 8 has a lower TCR compared to Dept 7, Dept 8 will be taken as the last department to be placed. For the third department, since there is no A relationship exist with the arranged department, then E relationship will now be considered. Taking the highest TCR among the E relationship that exist we take Dept 4. Following the procedure given, we would arrive in this arrangement of departments:

3, 2, 4, 5, 1, 9, 6, 10, 7, 8

Having generated the placement sequence, we now determine the relative locations of the departments. Its placement will be determined by the weighted placement value. The steps in page 85 sec 2.7.4 show the procedure in obtaining the relative locations of each department. Applying these steps will arrive in these computations:

1.

1,3,7,5 = 10,0002,4,6,8 = 5,000

Dept 2 = loc1

2.

1 = 1002, 10 = 100/2 = 503, 9 = 100 + 1000/2 = 6004, 8 = 100/2 + 1000 = 10505, 7 = 1000/2 = 5006 = 1000

Dept 4 = loc 4

3.

1 = 1002, 9, 12 = 100/2 = 503 = 100 + 100/2 + 1000 = 11504,6 = 1000/2 = 5005 = 10007 = 1000 + 100/2 = 10508 = 1000/2 + 100 = 60010,11 = 100 + 100/2 = 150

Dept 5 = loc 3

4.

1, 11 = 1000 + 0/2 = 10002, 10 = 1000/2 + 0 = 5003, 6, 9 = 0 = 04,5,7,8 = 0 + 0/2 = 012 = 1000/2 = 500

Dept 1 = loc 1

5.

1 = 1002 = 100/2 = 503 = 100 + 100/2 + 10 = 1604,7 = 10/2 = 55,6 = 10 + 10/2 = 158 = 100/2 + 10 = 609 = 100 + 10/2 = 10510, 14 = 100/2 = 5011,13 = 100 + 100/2 = 15012 = 100 + 100/2 + 100/2 = 200

Dept 9 = loc 12

6.

1 = 02,16 = 0/23 = 0 + 10/5 + 100 = 1004 = 100/2 = 505,6 = 100 + 100/2 = 1507 = 100/2 = 508 = 100 + 10/2 = 1059 = 10 + 100/2 = 6010,12,14 = 10/2 = 511 = 10 + 10/2 + 10 = 2513 = 1015 = 10 + 0 + 10/2 = 15

Dept 6 = loc 5

7.

1,6,15 = 102,5,7,9,12,14,16,18 = 10/2 = 53,8,13,17 = 10 +10/2 +10 = 254,10,11 = 10 + 10/2 = 15

Dept 10 = loc 3

8.

1 = 0 + 10/2 = 52 = 10 + 0/2 = 103 = 10/2 = 54,8 = 10 + 10/2 + 10 = 255,7,9,12,14,16 = 10/2 = 56,15 = 1010,11 = 10 + 10/2 = 1513 = 10 +10 + (-10000/2) = -498017 = 0 + 10 + (-10000/2) = -499018 = 0/2

Dept 7 = loc 4

9.

1 = 0 +10/2 = 52 = 10 + 0/2 +10000/2 = 50103 = 10000 + 10/2 = 100054 = 10000/2 = 50005 = 10000 + 0/2 = 100006 = 0 + 10000/2 = 50007,9,12,18 = 0/28 = 0 + 0 + 0/2 = 010,11 = 0 + 0/2 = 013,17 = 0 + 10 + (-10000/2) = -499014,16 = 5

Dept 8 = loc

Relative Location of Departments

Now, that the 10 departments are placed relatively with each other, (as we consider the different relationships that exist among them) we could now consider the space requirements to formulate the block diagram of the required layout.

Space Requirement

Number Department Name

Area Number of unit squares

Template dimension

1 Office 1,250 50 12.5 x 42 Counter 2,500 100 10 x 103 Parts Bins 10,000 400 20 x 20 4 Muffler bins 2,500 100 10 x 105 Tailpipe racks 6,250 250 25 x 106 Paint Room 3,000 120 12 x 107 Storage 5,000 200 20 x 108 Receiving 2,500 100 10 x 109 Lounge 500 20 4 x 5

10 Rest Room 250 10 2x 5 Scale: 1unit square = 5ft x 5ft = 25sq. ft.

Template dimension was used to estimate the block diagram shown in the figure below. (Sec 2.9 in pg. 101)

Block Diagram of the New Layout

Many alternative layout can be formed using the space requirements and the relative locations of each department. One of the alternative layout that can be deduced using the given in the problem is shown in the figure below. It is estimated that the total land area to be used is 33,750 sq.ft.

Alternative block layout

Following the SLP approach, we can arrive to different alternative block layout. Choosing the alternative layout to be used depends on the objective of designing the new layout.

2.47. An activity relationship chart is shown in Figure P2.47 for the Rickety Furniture Company. Construct a REL diagram for the plant using the graph-based process. Based on the space requirements given, design a block layout.

Activity Area

Total 19, 800 ft2

Figure P2.47

Required: a) REL diagramb) Block layout given the space requirements

Reference:

2.7. Relationship Diagram; 2.7.3. Graph-Based Process

Solution:

Activities with A relationship: (1,10), (7,3), (7,2), (6,3), (6,2), (5,4)Activities with X relationship: (10,7), (10,3), (9,1)Activities with E relationship: (9,10), (2,9), (5,2), (4,2)Activities with I relationship: (10,2), (9,8), (8,2), (8,1), (7,6), (1,2)Activities with O relationship: (9,5), (8,5), (3,2)

In making the REL diagram of the plant of Rickety Furniture Company given the said relationships, the activities with A relationship were the first one to be considered because

IU

O UE

UU

UU

I

XA

EU

A AA

UU

AA

UU

I

I

UU

UU

O

EU

UO

UU

I

IX

UU

U

UE

X

an A relationship means that the pairs of activities are absolutely necessary to be placed side by side with each other. Shown below is the REL diagram of activities with A relationship namely: (1,10), (7,3), (7,2), (6,3), (6,2), (5,4).

The block layout of the pair of activities with A relationship is not applicable since there are independent pairs of activities.

After plotting the pairs of activities with A relationship, the pairs with X relationship will be analyzed next. Since pairs (10,7), (10,3), (9,1) have X relationships, these pairs should not be connected in the REL diagram because X relationship indicates that the pair of activities are undesirable to be placed next with each other.

Plotting of pairs of activities with E relationship will then follow. An E relationship indicates that there is an especially important link between the activities. The REL diagram below plots the activities with A and E relationships.

The block layout for the above REL diagram is shown below.

Incorporating the activities with I relationship to the currently presented REL diagram resulted with the REL diagram shown below.

1 10

77

2

9

6

1

3 4

5

The block layout of the REL diagram which includes the activities that have A, E, and I is shown below.

And lastly, to complete the REL diagram of the given activity and their relationship with one another, the pairs of activities with O relationship are plotted to the precious REL diagram. The REL diagram below shows the final REL diagram which considers the different relationships that exist between the activities, from A relationship to O relationship.

3

7

6

8

1 10

92

5

4

The final block layout that follows the space requirement is shown below.

Assumption: 1 unit square= 10ft x 10ft = 100 ft2

Conclusion:

The ideal plant layout of Rickety Furniture Company is clearly illustrated by the block layout derived from the REL diagram of activities that have A,X, E, I and O relationships.

2.48. An Electronic company has acquired 5000ft-2 of space in a shopping center for its

regional service facility. The space is configured 50 ft x 100 ft, with common carrier access along one of its long walls. Customers enter the facility along one of the short walls. Based on the activity relationship chart shown in Figure P2.48, use the graph-based process to construct a REL diagram for the service facility. Using the space requirements given, design a block layout.

Figure P2.48

Given:

Total Area = 5,000 ft2

The REL chart for the facility.

Required:

REL diagram for the service facility using the graph-based approach.

Reference:

2.7. Relationship Diagram; 2.7.3. Graph-Based Process p. 71-83.

Activity Area

Total 5,000ft2

Solution:

(Topic: Relationship Diagram; Graph – Based Approach p 71-83)

First, we summarize the REL chart, taking into account the relationships of activities.

A: (1,3) (4,5) (4,9) (3,9) (7,8)E: (2,3) (3,6) (3,10) (4,6) (4,10) (5,7) (6,7) (9,10)I: (1,9) (2,5) (2,7) (2,8) (3,8) (6,8)O: (1,4) (3,7) (5,6)U: (1,2) (1,5) (1,6) (1,10) (2,4) (2,6) (2,9) (2,10) (3,4) (3,5) (3,8) (4,7) (5,8) (5,9)

(5,10) (6,9) 6,10) (7,9) (7,10) (8,9) (8,10)

Then, we take into account the A relationships and obtain a primal and dual graph for this and develop the block layout. The red dash lines represent the dual graph for the primal graph. In making the dual graph, the edges of the dual graph should be drawn wherein faces share the same edge of the primal graph. The block layout for this graph could not be made since there are two groups of activities. It cannot be combined into one only because of the considerations on their closeness ratings.

Second, are the A and E relationships. In making the dual graph, assuming that all the activities should be located inside the layout, there should not be any external activity unless specified. Therefore we should adjust the dual graph edges so as to meet this requirement. The graph shown below does not have sufficient number of corner points needed to construct the block layout. The red nodes with letters are the corner points that are drawn for the block layout. By intuition, the number of red nodes wouldn’t be enough to materialize a block layout.

Third, we consider the A, E and I relationships. Same as in the last situation, the graph should be made that there won’t exist an external activity or department.

Next is taking into account the A, E, I and O relationships. This would be the final graph that should be considered for the block layout. This is because the fourth closeness rating (U) will not have an effect on the graph of the dual and primal.

The block layout of the activities described is as follows. Though this layout does not represent the required area for each activity, it clearly shows the use of the dual graph and the extracted corner points.

Lastly would be making the design layout with the considerations of the total area of the facility.

Assignment

Documents

products x

diagram e

product layout reference

product x

activity analysis

relationship diagram

process layout reference

types of layout