ISyE 6201: Manufacturing Systems
Instructor : Spyros Reveliotis
Solutions for Homework #5
ISYE 6201 Fall 2006
Homework 7 Solution
A. Questions Chapter 4
17. Relatively constant volume and mix are essential to kanban because, under this
production authorization mechanism, we seek to support a certain production rate by capping
appropriately the system WIP. However, we can achieve this effect only when the system is
operated in “steady state”. An unstable production volume and/or product mix constitutes a
departure from the steady state regime which compromises the aforementioned relationship
between the applied WIP levels and target throughput. A potential remedy to this problem
would be to maintain very large WIP buffers, but this approach essentially defeats the whole
notion and purpose of JIT production.
19. Pull systems authorize production in response to a change in plant status (e.g, the release
of a KANBAN at a certain station), while push systems schedule production with a process
exogenous to the plant (e.g., an MRP-generated production schedule).
20. In a push system, the best place for the bottleneck is at the front of the line, since this
would (i) simplify the scheduling of this bottleneck (in a bottleneck-based scheduling
framework according to the Theory of Constraints) and (ii) it would tend to prevent “WIP
explosions” because the rest of the line would be able to clear out whatever it received from
the bottleneck. In a pull system, the location of the bottleneck is less important, since
releases will be tied to the amount of WIP in the system (and hence in front of the bottleneck)
regardless of where it is located.
21. In the following, “K” indicates kanban and “M” indicates MRP. Then, based on the
response of Question 17 above, a pertinent selection is as follows:
 Auto plant producing three styles of vehicle – K
 Custom job shop – M
 Circuit board plant with 40,000 active part numbers – M
 Circuit board plant with 12 active part numbers – K
 One assembly line where all parts are purchased – M
B. Problems
Chapter 3, Problem 11.
For item A, the 120 units on hand cover the demands for periods 1 and 2. The first demand
occurs in period 3. Thus, we could plan to receive either
i.
49 units with 0 part-periods,
ii. 49 + 42 = 91 units with 42 part-periods,
iii. 49 + 42 + 84 = 175 units with 42 + 2*84 = 210 part-periods, or
iv.
49 + 42 + 84 + 86 = 261 units with 42 + 2*84 + 3*86 = 468 part-periods
Since the ratio of setup cost to carrying cost is 200 we need go no further and choose a lot
size of 175. This covers demand for periods 3, 4 and 5.
We now decide how much to receive in period 6. The choices are
i.
86 units with 0 part-periods,
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ISYE 6201 Fall 2006
Homework 7 Solution
ii. 86 + 7 = 93 units with 7 part-periods,
iii. 93 +18 = 111 units with 7 + 2*18 = 43 part-periods,
iv.
111 + 49 = 160 units with 43 + 3*49 = 190 part-periods, or
v. 160 + 30 = 190 units with 190 + 4*30 = 310 part-periods.
The choice is to combine periods 6, 7, 8 and 9 and to make 160 units. This covers through
period 9.
The quantity to receive in period 10 is simply 30 units.
The MRP tableau for Item A:
Item A
Week
Demand
On Hand
120
Net Demand
Plan Order Receipts
Plan Order Releases
1
41
79
0
0
175
2
44
35
0
0
0
3
84
-49
49
175
0
4
42
42
0
160
5
84
84
0
0
6
86
86
160
0
7
7
7
0
0
8
18
18
0
30
9
49
49
0
0
10
30
30
30
0
We are now ready to process all parts with a low level code of 1. The only item is item 400.
400
Week
Demand
Sch Receipts
On Hand
200
Net Demand
Plan Order Receipts
Plan Order Releases
1
175
2
25
25
3
4
160
25 -135
135
135
135
5
-
6
-
7
-
8
30
30
30
9
-
10
-
30
Part 400 is composed of one unit of 200 and one of 300, both raw materials. Thus, these
parts are of level code 2.
200
Week
1
Demand
485
Sch Receipts
On Hand
300 -185
Net Demand
185
Plan Order Receipts
Plan Order Releases
2
3
200
-
-
4
320
160
5
30
100
30
6
-
7
-
8
60
60
9
-
10
-
For part 200, the first uncovered demand occurs in period 1 and is equal to 185. Since there
is a scheduled receipt to arrive in period 3, it must be expedited to period 1. This may or
may not be possible but it is better than releasing a new order in period 1 with a short leadtime. Similarly, the scheduled receipt in period 5 must also be expedited. For the rest of the
demand that are not covered by these two scheduled receipts, we can plan productions as
usual and allow the regular lead-time. The final schedule is shown below:
200
Week
Demand
Sch Receipts
Adj Sch Receipts
On Hand
300
Net Demand
Plan Order Receipts
Plan Order Releases
1
485
2
3
4
320
200
200
15
15
205
100
15 -205
205
205
30
5
30
100
30
30
6
-
7
-
8
60
60
60
9
-
10
-
60
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ISYE 6201 Fall 2006
Homework 7 Solution
For Part 300, the first scheduled receipt combined with the on-hand inventory covers all
demand. Consequently, the second scheduled receipt is canceled and does not appear in the
“Adj Sch Receipts” row (although notice that such cancellations of scheduled receipts
might not be always possible; e.g., depending on the applied contracts, nature of the
required processing, etc.)
300
Week
Demand
Sch Receipts
Adj Sch Receipts
On Hand
140
Net Demand
Plan Order Receipts
Plan Order Releases
1
135
2
0
3
0
4
0
100
5
0
0
0
5
0
0
0
5
0
0
0
5
0
0
0
5
30
6
0
7
0
100
8
0
9
0
10
0
100
75
0
0
0
75
0
0
0
75
0
0
0
75
0
0
0
75
0
0
0
75
0
0
0
Chapter 16, Problem 1.
t
(a)
max
m
{r
t 1 i 1
i
S it  hi I it  k itVit }
s.t.
d it  S it  d it i, t
m
a
i 1
ij
X it  c jt , j , t
I it  I it 1  X it  Vit  S it , i, t
v it  Vit  vit , i, t
X it , S it , I it ,Vit  0, i, t
(b) Replace the constraints on Vit with
t
V
t 1
it
 v i , i
(c) Replace the constraints on Vit with
Vi  Vit  1.2Vi , i, t
where Vi is a variable to be chosen by the LP (and is constrained to be nonnegative).
(d) Models like these can help illuminate the impacts of different types of supplier contracts.
Chapter 16, Problem 4.
(a)
max
s.t.
12( X H 1  X H 2  X H 3 )  10( X S1  X S 2  X S 3 )  7( X E1  X E 2  X E 3 )
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ISYE 6201 Fall 2006
Homework 7 Solution
X H 1  X H 2  X H 3  700
X S 1  X S 2  X S 3  900
X E1  X E 2  X E 3  450
21X H 1  17 X S 1  14 X E1  10000
21X H 2  17 X S 2  14 X E 2  7000
21X H 3  17 X S 3  14 X E 3  4200
X H 1  X S 1  X E1  550
X H 2  X S 2  X E 2  750
X H 3  X S 3  X E 3  225
X ij  0, i  H , S , E ; j  1,2,3
(b) Replace the capacity constraints (the last three constraints not counting nonnegativity)
with the following (note that F represents the fraction of capacity at each plant):
X H 1  X S 1  X E1  550 F
X H 2  X S 2  X E 2  750 F
X H 3  X S 3  X E 3  225 F
0  F 1
(c) To ensure at least 50% heavy duty batteries, we want
X H1  X H 2  X H 3
X H1  X H 2  X H 3
 0.5
 X S 1  X S 2  X S 3  X E1  X E 2  X E 3
With a bit of algebra, we can convert this to the following linear constraint, which can be
added to the formulation of part (a).
X H 1  X H 2  X H 3  X S 1  X S 2  X S 3  X E1  X E 2  X E 3  0
Chapter 16, Problem 6.
(a) Let
1,
Yj  
0,
if we lease machine j
if we do not
The problem formulation:
Max 150( X A1  X A2  X A3 )  225( X B1  X B 2  X B3 )  20000Y1  22000Y2  18000Y3
Subject to:
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ISYE 6201 Fall 2006
Homework 7 Solution
X A1  X A 2  X A3  200
X B1  X B 2  X B 3  100
0.5 X A1  1.2 X B1  80Y1
0.4 X A 2  1.2 X B 2  80Y2
0.6 X A3  0.8 X B 3  80Y3
X Aj , X Bj  0, j  1,2,3
Y j  {0,1}, j  1,2,3
(b) Simply add constraint Y1  Y2  1.
C. Extra credit (30%)
1. Consider a schedule Sthat violates the EDD rule for some pair of consecutive jobs. There
exist consecutive jobs, j and k, that are not in EDD order, i.e. cj < ck and dj > dk. Let Ls be
the maximum job lateness for S. By definition, Ls≥ lateness (k) = ck - dk.
If j and k are swapped in the schedule, we have the new completion times cj’= ck and ck’ =
ck - tj
before
swapping
job 1 job 2
job j
job k
tj
tk
cj
c0
after
swapping
job 1 job 2
job k
job j
tk
tj
c0
c'k
time
ck
time
ck
New lateness of job j = ck - dj < ck - dk≤ Ls
New lateness of job k = ck - tj - dk ≤ ck - dk≤ Ls
Since the lateness of all other jobs remain the same, the maximum lateness of the new
schedule must be less than or equal to Ls. It follows that scheduling jobs according to the
EDD rule gives an optimal schedule to the min-max lateness problem.
By definition the tardiness Tj of a job j is equal to max{0,Lj}, i.e., equal to the job lateness,
if the latter is positive, and zero, otherwise. Hence, to minimize the max tardiness, it is
adequate to minimize the max lateness: If the minimized max lateness happens to be
negative, then, we have a schedule with zero tardiness across all jobs. If the minimized
max lateness is positive, then this defines also the minimum max tardiness (since any other
schedule will have an even greater max lateness).
Hence, the EDD rule also gives an optimal schedule to the min-max tardiness problem.
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ISYE 6201 Fall 2006
Homework 7 Solution
2. Let t[j] and c[j] be the processing time and the completion time, respectively, of the jth job
j
on a given schedule. Then c[ j ]   t[i ] , for j = 1, …, n, and the average cycle time
i 1
n
1
cj
n j 1

1 n
 c[ j ]
n j 1

1 n j
 t[i ]
n j 1 i 1
1
t[1]  (t[1]  t[2] )  ...  (t[1]  t[2]  ...  t[ n] )
n
1
 [nt[1]  (n  1)t[ 2 ]  ...  t[ n ] ]
n
Clearly, the above expression is minimized by matching the largest weights, n, (n-1), etc,
with the smallest possible processing times.

3. By the Little’s Law, WIP = TH * CT. Since the processing time tj’s are given, the
n
throughput n
is fixed. Hence, minimizing the average job cycle time CT is equivalent to
t j
j 1
minimizing the average WIP. It implies that a schedule that follows the SPT rule will also
have a minimum average WIP.
n
Another way to approach this question is to interpret
c
j 1
j
as the cumulative WIP level.
n
From part (2), we know that  c j  [nt[1]  (n  1)t[ 2]  ...  t[ n ] ] . Meanwhile, we know that
j 1
from time 0 to t[1], the WIP level is n jobs; in the next t[2] unit of time, the WIP level is n-1;
and so on.
WIP
n
n-1
n-2
n-3
…
t[1]
t[2]
time
t[3]
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ISYE 6201 Fall 2006
Homework 7 Solution
n
Therefore, the average WIP level is given by
n
c t
j 1
j
j 1
j
and is minimized when the total
(or equivalently the average) cycle time is minimized.
4. The average job lateness
1 n
 (c j  d j )
n j 1
Since dj’s are given, minimizing the average job lateness
1 n
 (c j  d j ) is equivalent to
n j 1
1 n
 c j . It follows from the result 2 above that SPT
n j 1
also minimizes the average job lateness.
minimizing the average job cycle time
8