Tutorial 4 _ Solution

1 Dr Chong _ Solution Tutorial 4

Solution Tutorial 4 – Chapter 5

BEKC 4873 Advanced Manufacturing Systems

1. Name three production situations in which FMS technology can be applied. (1) Presently, the plant either produces parts in batches or uses manned GT cells and management wants to automate. (2) It must be possible to group a portion of the parts made in the plant into part families, whose similarities permit them to be processed on the machines in the flexible manufacturing system. (3) The parts or products made by the facility are in the mid-volume, mid-variety production range. The appropriate production volume range is 5000 to 75,000 parts per year.

2. What is flexible manufacturing system? A flexible manufacturing system (FMS) is a highly automated GT machine cell, consisting of a group of processing workstations (usually CNC machine tools), interconnected by an automated material handling and storage system, and controlled by a distributed computer system. The reason the FMS is called flexible is that it is capable of processing a variety of different part styles simultaneously at the various workstations, and the mix of part styles and quantities of production can be adjusted in response to changing demand patterns.

3. What are the three capabilities that a manufacturing system must process in order to be flexible? (1) the ability to identify and distinguish among the different incoming part or product styles processed by the system, (2) quick changeover of operating instructions, and (3) quick changeover of physical setup.

4. Name the four tests of flexibility that a manufacturing system must satisfy in order to be classified as flexible? (1) Part variety test. Can the system process different part styles in a non-batch mode? (2) Schedule change test. Can the system readily accept changes in production schedule: changes in either part mix or production quantities? (3) Error recovery test. Can the system recover gracefully from equipment malfunctions and breakdowns, so that production is not completely disrupted? (4) New part test. Can new part designs be introduced into the existing product mix with relative ease?

5. What are the four basic components of a flexible manufacturing system? (1) workstations, (2) material handling and storage system, (3) computer control system, and (4) people who are required to manage and operate the system.

6. What are four benefits that can be expected from a successful FMS installation? (1) increased machine utilization, (2) fewer machines required to accomplish the same amount of work, (3) reduction in factory floor space required, (4) greater responsiveness to change, (5) reduced inventory requirements, (6) lower manufacturing lead times, (7) reduced labor requirements and higher labor productivity, and (8) opportunity for unattended operation.

7. A flexible manufacturing cell consists of two machining workstations plus a load/unload station. The load/unload station is station 1. Station 2 performs milling operations and consists of one server (one CNC milling machine). Station 3 has one server that performs drilling (one CNC drill press). The three stations are connected by a part handling system that has one work carrier. The mean transport time is 2.5 min. The FMC produces three parts, A, B, and C. The part mix fractions and process routings for the three parts are


presented in the table below. The operation frequency fijk = 1.0 for all operations. Determine

(a) maximum production rate of the FMC, (b) corresponding production rates of each product, (c) utilization of each machine in the system (d) number of busy servers at each station.

Part j Part mix pj Operation k Description Station i Process time tijk

A 0.2 1 Load 1 3 min 2 Mill 2 20 min 3 Drill 3 12 min 4 Unload 1 2 min

B 0.3 1 Load 1 3 min 2 Mill 2 15 min 3 Drill 3 30 min 4 Unload 1 2 min

C 0.5 1 Load 1 3 min 2 Drill 3 14 min 3 Mill 2 22 min 4 Unload 1 2 min

(a) WL1 = (3+2)(0.2)(1.0) + (3+2)(0.3)(1.0) + (3+2)(0.5)(1.0) = 5.0 min

WL2 = 20(0.2)(1.0) + 15(0.3)(1.0) + 22(0.5)(1.0) = 19.5 min

WL3 = 12(0.2)(1.0) + 30(0.3)(1.0) + 14(0.5)(1.0) = 18.4 min

nt = 3(0.2)(1.0) + 3(0.3)(1.0) + 3(0.5)(1.0) = 3,

WL4 = 3(2.5) = 7.5 min

Bottleneck station has largest WLi/si ratio:

Station WLi/si ratio

1 (load/unload) 5.0/1 = 5.0 min

2 (mill) 19.5/1 = 19.5 min ← Bottleneck

3 (drill) 18.4/1 = 18.4 min

4 (material handling) 7.5/1 = 7.5 min

Bottleneck is station 2: Rp* = 1/19.5 = 0.05128 pc/min = 3.077 pc/hr

(b) RpA = 0.05128(0.2) = 0.01026 pc/min = 0.6154 pc/hr

RpB = 0.05128(0.3) = 0.01538 pc/min = 0.9231 pc/hr

RpC = 0.05128(0.5) = 0.02564 pc/min = 1.5385 pc/hr

(c) U1 = (5.0/1)(0.05128) = 0.256 = 25.6%

U2 = (19.5/1)(0.05128) = 1.0 = 100%

U3 = (18.4/1)(0.05128) = 0.944 = 94.4%

U4 = (7.5/1)(0.05128) = 0.385 = 38.5%

(d) BS1 = (5.0)(0.05128) = 0.256 servers

BS2 = (19.5)(0.05128) = 1.0 servers

BS3 = (18.4)(0.05128) = 0.944 servers

BS4 = (7.5)(0.05128) = 0.385 servers


8. Solve the P7 except the number of servers at station 2 (CNC milling machines) = 3 and the number of servers at station 3 (CNC drill process) = 2. Note that with the increase in the number of machines, the FMC is now a FMS.

(a) WL1 = (3+2)(0.2)(1.0) + (3+2)(0.3)(1.0) + (3+2)(0.5)(1.0) = 5.0 min

WL2 = 20(0.2)(1.0) + 15(0.3)(1.0) + 22(0.5)(1.0) = 19.5 min

WL3 = 12(0.2)(1.0) + 30(0.3)(1.0) + 14(0.5)(1.0) = 18.4 min

nt = 3(0.2)(1.0) + 3(0.3)(1.0) + 3(0.5)(1.0) = 3,

WL4 = 3(2.5) = 7.5 min

Bottleneck station has largest WLi/si ratio:



2 (mill) 19.5/3 = 6.5 min

3 (drill) 18.4/2 = 9.2 min ← Bottleneck



(b) RpA = 0.1087(0.2) = 0.02174 pc/min = 1.3043 pc/hr

RpB = 0.1087(0.3) = 0.03261 pc/min = 1.9565 pc/hr

RpC = 0.1087(0.5) = 0.05435 pc/min = 3.261 pc/hr

(c) U1 = (5.0/1)(0.1087) = 0.544 = 54.4%

U2 = (19.5/3)(0.1087) = 0.706 = 70.6%

U3 = (18.4/2)(0.1087) = 1.00 = 100%

U4 = (7.5/1)(0.1087) = 0.815 = 81.5%

(d) BS1 = (5.0)(0.1087) = 0.544 servers

BS2 = (19.5)(0.1087) = 2.12 servers

BS3 = (18.4)(0.1087) = 2.0 servers

BS4 = (7.5)(0.1087) = 0.815 servers

9. A FMS consists of three stations plus a load/unload station. Station 1 loads and unloads parts using two servers (material handling workers). Station 2 performs horizontal milling operations with two servers (identical CNC horizontal milling machines). Station 3 performs vertical milling operations with three servers (identical CNC vertical milling machines). Station 4 performs drilling operations with two servers (identical drill presses). The machines are connected by a part handling system that has three (3) work carriers and a mean transport time = 3.5 min. The FMS produces four parts, A, B, C, and D, whose part mix fractions and process routings are presented in the table below. The operation frequency fijk = 1.0 for all operations. Determine

(a) maximum production rate of the FMS

(b) utilization of each machine in ths system

(c) average utilization of the system sU .

Part j Part mix pj Operation k Description Station i Process time tijk

A 0.2 1 Load 1 4 min


2 H. Mill 2 15 min

3 V Mill 3 14 min

4 Drill 4 13 min

5 Unload 1 3 min

B 0.2 1 Load 1 4 min

2 Drill 4 12 min

3 H. Mill 2 16 min

4 V. Mill 3 11 min

5 Drill 4 17 min

6 Unload 1 3 min

C 0.25 1 Load 1 4 min

2 H. Mill 2 10 min

3 Drill 4 9 min

4 Unload 1 3 min

D 0.35 1 Load 1 4 min

2 V. Mill 3 18 min

3 Drill 4 8 min

4 Unload 1 3 min

(a) WL1 = (4+3)(0.2)(1.0) + (4+3)(0.2)(1.0) + (4+3)(0.25)(1.0) + (4+3)(0.35)(1.0) = 7.0 min

WL2 = 15(0.2)(1.0) + 16(0.2)(1.0) + 10(0.25)(1.0) = 8.7 min

WL3 = 14(0.2)(1.0) + 11(0.2)(1.0) + 18(0.35)(1.0) = 11.3 min

WL4 = 13(0.2)(1.0) + (12+17)(0.2)(1.0) + 9(0.25)(1.0) + 8(0.35)(1.0) = 13.45 min

nt = 4(0.2) + 5(0.2) + 3(0.25) + 3(0.35) = 3.6, WL5 = 3.6(3.5) = 12.6 min



2 (drill) 8.7/2 = 4.35 min

3 (H. mill) 11.3/3 = 3.77 min

4 (. mill) 13.45/2 = 6.725 min ← Bottleneck



(b) U1 = (7.0/2)(0.1487) = 0.520 = 52.0%

U2 = (8.7/2)(0.1487) = 0.647 = 64.7%

U3 = (11.3/3)(0.1487) = 0.560 = 56.0%

U4 = (13.45/2)(0.1487) = 1.00 = 100%

U5 = (12.6/2)(0.1487) = 0.624 = 62.4%

(c) sU = ( ) ( ) ( ) ( )2 0.520 2 0.647 3 0.560 2 1.0

9

+ + + = 0.668 = 66.8%

10. A semi-automated flexible manufacturing cell is used to produce three products. The products are made by two automated processing stations followed by an assembly station. There is also a load/unload station. Material handling between stations in the FMC is accomplished by mechanized carts that move tote bins containing the particular components to be processed and then assembled into a given product. The carts transfer tote bins between stations. In this way the carts are kept busy while the tote bins are queued in front of the workstations. Each tote bin remains with the product throughout processing and assembly. The details of the FMC can be summarized as follows:


Station Description Number of servers

1 Load and unload 2 human workers 2 Process X 1 automated server 3 Process Y 1 automated server 4 Assembly 2 human workers 5 Transport Number of carriers to be determined.

The product mix fractions and station processing times for the parts are presented in the

table below. The same station sequence is followed by all products: 1 → 2 → 3 → 4 → 1.

Product j Product mix pj Station 1 Station 2 Station 3 Station 4 Station 1

A 0.35 3 min 9 min 7 min 5 min 2 min B 0.25 3 min 5 min 8 min 5 min 2 min C 0.40 3 min 4 min 6 min 8 min 2 min

The average cart transfer time between stations is 4 minutes.

a. What is the bottleneck station in the FMC, assuming that the material handling

system is not the bottleneck?

b. At full capacity, what is the overall production rate of the system and the rate for

each product?

c. What is the minimum number of carts in the material handling system required to

keep up with the production workstations?

d. Compute the overall utilization of the FMC.

e. What recommendations would you make to improve the efficiency and/or reduce the

cost of operating the FMC?

(a) WL1 = (3+2)(0.35 + 0.25 + 0.4) = 5.0 min

WL2 = 9(0.35)(1.0) + 5(0.25)(1.0) + 4(0.4)(1.0) = 6.0 min

WL3 = 7(0.35)(1.0) + 8(0.25)(1.0) + 6(0.4)(1.0) = 6.85 min

WL4 = 5(0.35)(1.0) + 5(0.25)(1.0) + 8(0.4)(1.0) = 6.2 min

nt = 4 for all parts, WL5 = 4(4) = 16.0 min



2 (process X) 6.0/1 = 6.0 min

3 (process Y) 6.85/1 = 6.85 min ← Bottleneck

4 (assembly) 6.2/2 = 3.1 min

5 (transport) 16.0/nc

(b) Bottleneck is station 3: Rp* = 1/6.85 = 0.146 pc/min = 8.76 pc/hr

RpA = 0.35(8.76) = 3.07 pc/hr

RpB = 0.25(8.76) = 2.19 pc/hr

RpC = 0.4(8.76) = 3.50 pc/hr

(c) Minimum number of carts required in system: 16.0/nc = 6.85

Rearranging, nc = 16/6.85 = 2.34 → Use 3 carts

(d) U1 = (5.0/2)(0.146) = 0.365 = 36.5%

U2 = (6.0/1)(0.146) = 0.876 = 87.6%

U3 = (6.85/1)(0.146) = 1.0 = 100%

U4 = (6.2/2)(0.146) = 0.453 = 45.3%


U5 = (16.0/3)(0.146) = 0.779 = 77.9%

sU = 2 0 365 1 0 876 1 10 2 0 453

6

( . ) ( . ) ( . ) ( . )+ + + = 0.585 = 58.5%

(e) Recommendations: (1) reduce number of servers at station 1 to 1 server, and (2) reduce

number of servers at station 4 to 1 server.

Tutorial 4 _ Solution

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