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ENGG*3070 – Assignment No. 4 (Due Nov. 24, 2011) Flexible Manufacturing systems (FMS) 1. A FMS consists of three stations plus a load/unload station. Station 1 loads and unloads parts using two...

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ENGG*3070 – Assignment No. 4 (Due Nov.
24, 2011)

Flexible Manufacturing systems (FMS)
1. 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 work ca
iers 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, and (c) average utilization of the system sU .


Part j Part mix pj Operation
k
Descriptio
n
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 XXXXXXXXXXLoad 1 4 min
2 H. Mill 2 10 min
3 Drill 4 9 min
4 Unload 1 3 min
D XXXXXXXXXXLoad 1 4 min
2 V. Mill 3 18 min
3 Drill 4 8 min
4 Unload 1 3 min



2. Use the extended bottleneck model to solve Problem No. 1 with the following number of
parts in the system: (a) N = 5 parts, (b) N = 8 parts, and (c) N = 12 parts. Also determine the
manufacturing lead time for the three cases of N in (a), (b), and (c).
3. A flexible manufacturing system is used to produce four parts. The FMS consists of one
load/unload station and two automated processing stations (processes X and Y). The
number of servers for each station type is to be determined. The FMS also includes an
automated conveyor system with individual carts to transport parts between servers. The
carts move the parts from one server to the next, drop them off, and proceed to the next
delivery task. Average time required per transfer is 3.5 minutes. The following table
summarizes the FMS:
Station 1 Load and unload Number of human servers (workers) to be
determined
Station 2 Process X Number of automated servers to be determined
Station 3 Process Y Number of automated servers to be determined
Station 4 Transport system Number of carts to be determined
All parts follow the same routing, which is 1 → 2 → 3 → 1. The product mix and
processing times at each station are presented in the table below:
Product j Product mix
pj
Station
1
Station
2
Station
3
Station
1
A 0.1 3 min 15 min 25 min 2 min
B 0.3 3 min 40 min 20 min 2 min
C 0.4 3 min 20 min 10 min 2 min
D 0.2 3 min 30 min 5 min 2 min
Required production is 10 parts per hour, distributed according to the product mix
indicated. Use the bottleneck model to determine (a) the minimum number of servers
at each station and the minimum number of carts in the transport system that are
equired to satisfy production demand and (b) the utilization of each station for the
answers above.
Inventory Management
4. A manufacturing firm located in Guelph produces an item in a three-month time supply.
An analyst, attempting to introduce a more logical approach to selecting run quantities,
has obtained the following estimates of characteristics of the item
?? = 400 ??????????/????
?? = $5
?? = $4 ?????? 100 ??????????
?? = 0.25$/$/????
Note: Assume that the production rate is much larger than D.
a) What is the economic order quantity of the item?
) What is the time between consecutive replenishments of the item when the EOQ is
used.
c) The production manager insists that the ?? = $5 figure is only a guess. Therefore, he
insists on using his simple three-month supply rule. Indicate home much you would find
the range of ?? values for which the EOQ (based on ?? = $5) would be preferabel (in
terms of a lower total replenishement and ca
ying costs) to the three-month supply.
5. The supplier of a product wants to discourage large quantity purchases. Suppose that all
of the assumption of the basic EQQ apply except that a reverse quantity discount is
applicable, that is, the unit variable cost is given by
?? = �
???? 0 < ?? < ????
????(1 + ??) ???? ≤ ??
XXXXXXXXXXwhere ?? > 0
a) Write an expression (or expressions) for the total relevant cost per year as a function
of the order quantity ??. Introduce (and define) whatever other symbols you feel are
necessary.
) Using graphical sketches, indicate the possible positions of the best order quantity
(as is done on Slide 18 - pages 11-12 for the regular discount case).
c) What is the best order quantity for an item with the following characteristics.
Demand rate = 50,000 units/yr
Fixed setup cost per replenishment = $10
?? = $1.0/unit, ?? = 0.005
Selling price = $1.44/unit
Ca
ying charge = 0.20$/$/yr
???? = 1500 units
Note: The variable cost ?? applies to all-units (not incrementally)
    Flexible Manufacturing systems (FMS)
    Inventory Management
Answered Same Day Nov 21, 2021

Solution

Sonam answered on Nov 22 2021
127 Votes
2.
MLT1 = 7.0 + 8.7 + 10.5 + 11.25 + 12.6 = 50.05 minutes
Rp* = 0.1778 pc/min from Problem1
N* = 0.1778(50.05) = 8.9
(a) For N = 5 < N* = 8.9,
MLT1 = 50.05 minutes, and Tw = 0 Rp = 5/50.05 = 0.0999 pc/min = 5.99 pc/hour
U1 = (7.0/2)*(0.0999) = 0.350 = 35.0%
U2 = (8.7/2)*(0.0999) = 0.435 = 43.5%
U3 = (10.5/3)*(0.0999) = 0.350 = 35.0%
U4 = (11.25/2)*(0.0999) = 0.562 = 56.2%
U5 = (12.6/3)*(0.0999) = 0.420 = 42.0%
Us = [(2* 0.35) + (2* 0.435) + (3*0.35) + (2*0.562)]/9 = 0.416 = 41.6%
(b) N = 8 < N* = 8.9,
Rp = 8/50.05 = 0.1598 pc/min = 9.59 pc/hour
MLT1 =50.05 minutes, and Tw = 0.
U1 = (7.0/2)(0.1598) = 0.559 = 55.9%
U2 = (8.7/2)(0.1598) = 0.695 = 69.5%
U3 = (10.5/3)(0.1598) = 0.559 = 55.9%
U4 = (11.25/2)(0.1598) = 0.899 = 89.9%
U5 = (12.6/3)(0.1598) = 0.671 = 67.1%
Us = [(2* 0.559) + (2* 0.695) + (3* 0.559) +...
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