ISyE 3104: Introduction to Supply Chain Modeling:
Manufacturing and Warehousing
Instructor: Spyros Reveliotis
Summer 2006
Solutions for Homework #7
ISYE 3104 Summer 2006
Homework 7 Solution
Chapter 7
35. In a push system, inventory is pushed through the various levels of the supply chain
based on some planning activity, which, in turn, is based to a large extent on forecasting. The
MRP planning framework discussed in class constitutes the most typical example of a pushtype system. A pull system, on the other hand, operates in a more reactive manner, since it
authorizes new production lots as a response to demand occurring to downstream stages of
the supply chain. In general, pull systems are more responsive to disruptions taking place on
the shop-floor, preventing thus the cluttering of the production environment with
unnecessary WIP. On the other hand, they might be unable to respond effectively to frequent
and large fluctuations of the experienced demand, since these response might involve
substantial lead times.
Just-In-Time (JIT) is a paradigm in the manufacturing world that advocates the systematic
and persistent identification of any sources of inefficiency in the production environment and
their mitigation and eventual eradication through the adoption of pertinent operational
policies. JIT attacks sources of inefficiency within the production system itself, but also,
those developing on both the material supply and the market side / boundaries of the system.
A successful implementation of JIT practices typically enables the system to be effectively
meeting its demand while maintaining substantially lower amounts of inventory. In that
sense, the JIT practices are closer to pull-type of systems. However, when coupled with a JIT
philosophy, an MRP-II planning framework can be quite efficient as well.
36. A basic definition of a just-in-time system was provided in the second paragraph of the
response to the previous question. As we saw, JIT advocates the timely production of small
lots in response to the actual needs of the system operation. KANBAN is a manual
information system / shop-floor control mechanism that enables the implementation of a JIToriented production pattern. However, JIT practices could be implemented through other
shop-floor control mechanisms as well; for instance, the CONWIP scheme presented in class,
or even a well-tuned MRP-II framework.
Chapter 8
4.
FCFS
Sequence
Completion Time Flow Time Tardiness
1:20
20
0
1:34
34
0
2:09
69
19
2:19
79
49
Total
202
68
Average
50.5
17
Number of tardy trucks = 2
1
2
3
4
2
ISYE 3104 Summer 2006
Homework 7 Solution
SPT
Sequence
Completion Time Flow Time Tardiness
4
1:10
10
2
1:24
24
1
1:44
44
3
2:19
79
Total
157
Average
39.25
Number of tardy trucks = 2
0
0
19
29
48
12
EDD
Sequence
Completion Time Flow Time Tardiness
1:20
20
1:30
30
1:44
44
2:19
79
Total
173
Average
43.25
Number of tardy trucks = 1
1
4
2
3
0
0
0
29
29
7.25
Remark: As you might have noticed, there is a small discrepancy regarding the concept of
the critical ratio in your textbook, in that the example of pg. 411 uses the inverse of the
quantity defined in the relevant definition of pg. 409. The end result is not affected since in
the first case jobs are prioritized from lowest to highest critical ratio while in the second case
they are prioritized from highest to lowest. In the following solution, we use the notion of
critical ratio in the same way that it is used in the example of pg. 411.
CR
Time: 1:00 p.m.:
Truck
Unloading Time Due Time
1
20
25
2
14
45
3
35
50
4
10
30
CR
25/20 = 1.25*
45/14 = 3.21
50/35 = 1.43
30/10 = 3
Time: 1:20 p.m.:
Truck
Unloading Time Due Time
2
14
25
3
35
30
4
10
10
CR
25/14 = 1.79
30/35 = 0.86*
10/10 = 1
Time: 1:55 p.m.:
Truck
Unloading Time Due Time CR
2
14
-10 -10/14 <0
4
10
-25 -25/10 <0
Sequence: 1-3-4-2
Summary for CR
3
ISYE 3104 Summer 2006
Homework 7 Solution
Sequence
Completion Time Flow Time Tardiness
1:20
20
1:55
55
2:05
65
2:19
79
Total
219
Average
54.75
Number of tardy trucks = 3
1
3
4
2
6.
0
5
35
34
74
18.5
a) SPT minimizes mean flow time. The SPT sequence is 4-2-1-3.
b) EDD minimizes maximum lateness. The EDD sequence is 1-4-2-3.
c) Start with EDD solution:
Truck
Time
Tardiness
1
20
0
4
10
0
2
14
0
3
35
29
The method calls for rejecting the job with the longest processing time and placing it
at the end of the sequence. Since job 3 already has the longest time, the EDD
sequence is the final solution.
26.
a) Since there are two tasks requiring 12 minutes, the minimum cycle time is 12
minutes. The theoretical minimum number of stations is obtained from 70/12 =
5.833. Hence, the theoretical minimum is 6.
b)
Station
Tasks
Idle Time
1
1
0
2
2,3
0
3
6
0
4
4,5,7,9
0
5
8,10
3
6
11
6
7
12
5
c)
1. As discussed in the textbook example, five stations result in 28 units per
line per day so that 4 lines are required.
2. Again, per the textbook example, six stations result in 32.3 units per line
per day so that 3 lines almost gives 100 units per day.
3. The 7-station balance above has C=12 which gives 35 units a day. Three
lines would result in more than 100 units per day.
27.
a)
Task
Positional
Weight
1
100
2
94
3
46
4
43
4
ISYE 3104 Summer 2006
5
6
7
8
9
10
11
12
13
14
15
Homework 7 Solution
37
47
23
20
29
16
16
20
12
8
5
Ranking: 1-2-6-3-4-5-9-7-8-12-10-11-13-14-15
b) 100/30 = 3.33 implying a minimum of 4 stations would be required
Station
Tasks
Idle Time
1
1,2,6,3
2
2
4,5,7,9
1
3
4
8,12,10,11 13,14,15
2
15
c) Start with C=25. A perfect balance would require four stations. Unfortunately this
is not possible.
For C=26 we do find the following 4-station balance.
Station
Tasks
Idle Time
1
1,2,3,5
0
2
4,6,7,8
0
3
9,11,12
1
4
10,13,14,15
3
5