Executive Summary

Hightec Inc is the manufacturer of transducers that converts gas or liquid pressure into an

electrical signal. It rents a 12000 square foot building which comprises of 4 sections viz office,

engineering, assembly and a machine shop. The employee strength of the firm is 80.

The firm is facing many problems related to the inefficiency of the plant layout. Some of them are:

Due to the space constraint they are not able to purchase the machines which are expected to give a

higher productivity.

Machine shop is over occupied.

A lot of machines are being used in second and third shift which is beyond justification.

There is a lot of handling time due to inefficiency.

Apart from that there are several other problems like the privacy of the employees is being

hampered

Alternatives

Hightec has two alternatives

Build a new plant : one of the alternatives is to build a new plant in an area 19000 square foot. The

costs involved are $100,000 for purchasing the land and $40/ square foot for construction.

Rent an additional building: it contains renting a building of 7000 square foot. The costs involved are

a rent of $2800/month for each building and a one time investment of $15000 for building the

corridor.

Question

Which expansion option would you recommend to Glenn Moore? Justify your position.

Solution

Buying a new plant Renting additional area(yrly)

Land $100,000 Rent $67200

Construction $760,000 Corridor(1 time) $15000

$860,000

Given the cost of capital is 15%

If we opt for building new plant then it will cost $860,000*.15 i.e. $129,000 while renting the yearly

cost is $67,200 so financially renting seems to be the better alternative.

From the above calculations it is clear that building new plant will involve a huge initial cash outflow.

Thus it would be better not to block the money in the form of bulding. Instead we should pay it in

instalment in the form of monthly rent.

Question 2

Design an effective block plan and evaluate it. Cite any qualitative consideration that you believe

make your design attractive.

department1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Area

1 administrative office

I A E U A E O O O O I E O U 3

2 conference room I U U U U U U U U U U U U U 1

3 eng & matl mngmnt

A U I U U O A E E I E E U O 2

4 productn mangr E U I U A A A A A I I E O A 1

5 lunch room U U U U U U U U U U U U U U 2

6 computer A U U A U A X U U U O I U U 1

7 Inventory strage E U O A U A A O O O O U U U 2

8 machine shop O U A A U X A A X I O U U I 6

9 assembly area O U E A U U O A A A I U I A 7

10 cleaning O U E A U U O X A O O U U U 1

11 welding O U I I U U O I A O O U U U 1

12 electronic I U E I U O O O I O O E U U 1

13 sales & accntng E U E E U I U U U U U E O U 2

14 shipping O U U O U U U U I U U U O U 1

15 load test U U O A U U U I A U U U U U 1

32

From the above table Total Closeness Rating (TCR) of each department was calculated .The basis of

calculating TCR was

A E I O U X

125 25 5 1 0 -125

Following were the scores of each department

TCR rating for each department

AE I O U X TCR

1 2 3 2 5 2 0 340

2 0 0 1 0 13 0 5

3 2 4 2 2 4 0 362

4 6 2 3 1 2 0 816

5 0 0 0 0 14 0 0

6 3 0 1 1 8 1 256

7 3 1 0 5 5 0 405

8 4 0 2 2 4 2 262

9 5 1 2 2 4 0 662

10 2 1 0 4 6 1 154

11 1 0 3 4 6 0 144

12 0 2 3 5 4 0 70

13 0 4 1 1 8 0 106

14 0 0 1 3 10 0 8

15 2 0 1 1 10 0 256

30 18 22 36 100 4

Following algorithm is used to decide the placement preference of the department:

1. The first department placed in the layout is the one with the greatest TCR value. If a tie exists, choose the one with more A’s.

2. If a department has an X relationship with he first one, it is placed last in the layout. If atie exists, choose the one with the smallest TCR value.

3. The second department is the one with an A relationship with the first one. If a tie exists,choose the one with the greatest TCR value.

4. If a department has an X relationship with he second one, it is placed next-to-the-last of last in the layout. If a tie exists, choose the one with the smallest TCR value.

5. The third department is the one with an A relationship with one of the placed departments.If a tie exists, choose the one with the greatest TCR value.

6. The procedure continues until all departments have been placed.

(Source: www.personal.psu.edu)

Following were the placement orders obtained

Department TCR Plcmnt order

1 administrative office340

6

2 conference room 5 12

3 eng & matl mngmnt 362 5

4 productn mangr 816 1

5 lunch room 0 13

6 Computer 256 14

7 inventory storage 405 3

8 machine shop 262 4

9 assembly area 662 2

10 Cleaning 154 15

11 Welding 144 8

12 Electronic 70 10

13 sales & accntng 106 9

14 Shipping 8 11

15 load test 256 7

Following algorithm is considered for placing the departments

1. The department with placement order 1 is placed first somewhere in the middle of the space available.

2. If a block is fully adjacent to a block where any department is placed then it is given a score of 125 and if it is partial adjacent(diagonal blocks)then it is given a score of ½ of 125 that is 62.5

3. For each position, Weighted Placement (WP) is the sum of the numerical values for all pairs of adjacent departments.

4. The first department selected is placed in the middle.

5. The new department is located based on the greatest WP value.

6. In case of tie the department at the west in the counterclockwise direction is select(Source: www.personal.psu.edu)

Following are the various iterations

62.5125 62.5

125 4 12562.5 125 62.5

62.5187.5 187.5 62.5

187.5 9 4 125187.5 7 312.5 62.5

62.5 125 62.5 0

62.5187.5 250 187.5 62.5

187.5 3 9 4 187.562.5 312.5 7 8 187.5

0 62.5 187.5 187.5 62.5

62.5187.5 250 187.5 62.5

187.5 3 9 4 187.562.5 1 7 8 187.562.5 312.5 15 312.5 62.5

0 62.5 125 62.5 0

62.5187.5 187.5 62.5

125 9 4 12562.5 187.5 187.5 62.5

62.5187.5 187.5 62.5

187.5 9 4 187.5187.5 7 8 187.5

62.5 187.5 187.5 62.5

62.5187.5 250 187.5 62.5

187.5 3 9 4 187.562.5 1 7 8 187.562.5 187.5 250 187.5 62.5

62.5187.5 250 187.5 62.5

187.5 3 9 4 187.562.5 1 7 8 187.562.5 11 15 13 187.562.5 187.5 250 187.5 62.5

62.5187.5 187.5 62.5

187.5 14 12 312.5 62.5250 3 9 4 187.5250 1 7 8 250

187.5 11 15 13 187.562.5 187.5 250 187.5 62.5

062.5 187.5 250 187.5 62.5

62.5 312.5 14 12 2 187.5125 5 3 9 4 250

62.5 312.5 1 7 8 250 0 187.5 11 15 13 187.5 0 62.5 187.5 250 187.5 62.5

62.5187.5 250 250 187.5 62.5

187.5 6 14 12 2 187.5125 5 3 9 4 250

62.5 312.5 1 7 8 250 0 187.5 11 15 13 187.5 0 62.5 187.5 250 187.5 62.5

62.5187.5 250 250 187.5 62.5

062.5 125 62.5 0

62.5 312.5 12 312.5 62.5187.5 3 9 4 187.5

250 1 7 8 250187.5 11 15 13 187.5

62.5 187.5 250 187.5 62.5

62.5187.5 250 62.5 62.5

187.5 14 12 2 187.5250 3 9 4 250250 1 7 8 250

187.5 11 15 13 187.562.5 187.5 250 187.5 62.5

187.5 6 14 12 2 187.5250 5 3 9 4 250

187.5 10 1 7 8 25062.5 312.5 11 15 13 187.5

0 62.5 187.5 250 187.5 62.5

Thus final block plan comes out to be

614 12 2

5 3 9 410 1 7 8 11 15 13

Based on the space available for each department and the block plan developed following is the

recommended facility layout

613 13 9 9 12 2

5 1 15 9 9 9 4

5 1 14 3 9 9 8 8 8

10 1 11 3 8 8 8 7 7

Following are the advantages of this plant

Cleaning area is kept away from the machine shop

The departments of production viz machine shop, assembly room as well as the conference room is

kept adjacent to production managers cabin

Computer is kept near to sales and accounting and administrative area

Inventory storage area is very close to the machine shop and to the assembly area

Sales and accounting department is close to the administrative area

Lunch room is kept separated away at one of the corners.