Solutions for Chapter 6
Solutions To Problems From Chapter 6
6.1
a) The Greedy Heuristic Solution is x11=60, x21=40, x22=84, x23=21, x33=56, and x34=69,
at a total cost of $438,940.
b) The optimal solution one obtains from Solver is : x14=60, x21=61, x22=84, x31=39,
x33=77, and x34=9 at a total cost of $381,880. The spreadsheet appears below.
Solution to Problem 6.1
Variables
x11
Values
x12
60
Objective Coeff
x13
x14
x21
x22
0
0
60
40
250 420
380
280
1280
1
1
x23
84
x24
0
990 1440
x31
0
x32
x33
x34
39
0
56
69
1520 1550
1420
1660
1730
Operator
Value
Min
438940
RHS
st
Constraint 1
1
1
Constraint 2
1
1
1
1
Constraint 3
1
Constraint 4
1
Constraint 5
1
1
Constraint 6
1
1
1
1
1
1
Constraint 7
1
1
1
1
1
1
=
120
60
=
124
145
=
164
125
=
139
100
=
84
84
=
56
77
=
129
69
c) The percentage error from using the greedy heuristic is approximately 15%.
6.2
By using very large costs for the routes that are eliminated and resolving the problem,
one finds the new optimal solution is: x11=39, x14=21, x21=61, x22=84, x33=77, x34=78.
The total cost of the solution is 387,730, and the percentage error difference is 30%.
The spreadsheet solution appears below.
Solution to Problem 6.2
Variables
Values
Objective Coeff
x11
x12
39
x13
0
250 1.00E+30
0
x14
21
x21
61
x22
x23
84
x24
0
0
x31
x32
0
0
x33
77
x34 Operator
Value
RHS
48
380 280 1280 990 1.00E+30 1520 1.00E+30 1420 1660 1730
Min
387730
st
Constraint 1
1
1
1
1
Constraint 2
1
1
1
1
Constraint 3
Constraint 4
Constraint 5
Constraint 6
Constraint 7
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
127
1
=
60
60
=
145
145
=
125
125
=
100
100
=
84
84
=
77
77
=
69
69
Production and Operations Analysis, Fourth Edition
6.3
The network for this case looks essentially the same as Figure 6.2, except that we add a
transshipment node between the plants and warehouses, and all flow goes through the
transshipment node in Oklahoma City. If we identify the nodes as follows: 1: Sunnyvale,
2: Dublin, 3: Bangkok, 4: Oklahoma City, 5:Amarillo, 6:Teaneck, 7: Chicago, and 8:
Sioux Falls, we obtain the following balance of flow equations:
 x14  x15  x16 x17 – x18 = – 45
node 1:
node 2:
 x24  x25  x26  x27  x28   120
node 3:
 x34  x35 x36  x37  x38   95
node 4:
x14 + x24 +x34  x43 x46  x47  x48 = 0.
node 5:
x15 + x25 + x35 + x45 = 80
node 6:
x16 + x26 + x36 + x46 = 78
node 7:
x17 + x27 + x37 + x47 = 47
node 8:
x18 + x28 + x38 + x48 = 55.
The spreadsheet with the solution is:
Solution to Problem 6.3
Variables
x14
x15 x16
x17 x25
x26
x35
x36
x47
x49 x57 x58 x59 x67 x68 x69
Operator
RHS
Value
Values
45
0
Obj Funct
76
##
-1
-1
0
0
0
120
82
13
20
170 250 1440 1200 1660 1600
110
0
35
0
95 180 195
0
0
78
55
35 245 145
0
Min
3E+05
st
Node 1
-1
-1
Node 2
-1
-1
Node 3
Node 4
Node 5
Node 6
Node 7
-1
-1
1
-1
1
1
1
1
1
1
-1
-1
-1
-1
1
-1
1
1
Node 8
1
-1
1
1
Node 9
-1
1
1
1
<=
-45
-45
<=
-120
-120
<=
-95
-95
<=
25
25
<=
47
47
<=
-0
0
<=
55
80
<=
78
78
<=
55
55
Note that the objective function value is 294,410. The addition of the transshipment node
provides only small savings over the system without the transshipment point. The savings
are $3390, or about 1.1%.
6.4
a) We solve the problem in units of 100 cars. Total production=100, total requirements =
84. It follows that this is an unbalanced problem supply exceeding demand. To find the
greedy heuristic solution, add a dummy column with high cost. Letting Flint, Fresno, and
Monterrey correspond to sources 1, 2, and 3 and Phoenix, Davenport, Columbia, and the
Dummy column corresponding to sinks 1, 2, 3, and 4, the greedy heuristic yields x 12=28,
128
Solutions for Chapter 6
x13=15, x21=26, and x32=15 at a total cost of 1126 (thousands). This is exactly the same
cost as was obtained in the chapter for the optimal solution. This shows that this problem
has multiple optimal solutions. See spreadsheet below.
Solution to Problem 6.4 b)
Variables
Values
Obj Funct
x11 x12 x13 x21 x22 x23 x31 x32 x33 Operator
0
13
30
26
0
0
0
15
12
8
17
7
14
21
18
22
1
1
1
1
1
1
Value RHS
0
31 min
1126
st
Cons1
Cons2
Cons3
Cons4
1
1
1
Cons5
1
Cons6
1
<
43
43
<
26
26
1<
15
31
=
26
26
=
28
28
1=
30
30
1
1
1
1
1
c) 0%.
d) Replacing the Monterrey/Columbia cost with a high number and resolving the problem
gives exactly the same cost as part b). This is to be expected since x 33 = 0 in the original
optimal solution.
6.5
The spreadsheet solution for this problem appears below. Interpret the node designations
as follows: 1=Flint, 2=Fresno, 3=Monterrey, 4=Sante Fe, 5=Jefferson City, 6=Phoenix,
7=Davenport, 8=Columbia. The value of the objective function at the optimal solution is
1052 (thousands).
Solution to Problem 6. 5
Variables
x14 x15 x24 x25 x34 x35 x46 x47 x48 x56 x57 x58
Operator
Values
0
43
26
0
15
0
26
0
15
0
28
15
Obj Funct
8
6
6
9
9
14
3
8
10
5
5
9
-1
-1
-1
-1
Value
RHS
Min
1052
>=
-43
-43
>=
-26
-26
>=
-15
-31
>=
-7.6E-10
0
>=
-4.3E-11
0
>=
26
26
>=
28
28
>=
30
30
st
Node 1
Node 2
Node 3
Node 4
Node 5
Node 6
Node 7
Node 8
-1
1
1
1
-1
1
1
-1
-1
-1
1
-1
1
-1
-1
1
1
1
1
129
1
Production and Operations Analysis, Fourth Edition
6.6
The solution appears in the spreadsheet below.
Solution to Problem 6. 6
Variables
x13
Values
x14 x15
x16
0
0 5000
0
Obj Funct
182 375
x23
285
460
-1
-1
x24
x25
3500 3500
77
290
-1
-1
x26 x34 x35
0
0
0
x36 x45 x46
3500
245 575 275
Operator
0
0
3500
125 380
90
110
Min
Value
RHS
4407000
st
Node 1
-1
-1
Node 2
Node 3
1
-1
1
Node 4
-1
1
1
Node 5
-1
-1
1
1
-1
1
Node 6
6.7
-1
1
1
1
-1
1
1
1
=
-5000
-5000
=
-7000.0001
-7000
=
0.0001415
0
=
-7.074E-05
0
=
3499.9999
3500
=
8500.0001
8500
The solution appears in the spreadsheet below.
Solution of Problem 6. 7
Variables
Values
Obj Funct
x13
x14 x15
0
x16
x23
0
0
3000
182 375
285
460
-1
-1
x24
x25
3000 4000
77
290
x26 x34
0
x35
x36
x45 x46
Operator
0
0
3000
0
0
4000
245 575
275
125
380
90
110
Min
Value
RHS
3586000.1
st
Node 1
-1
-1
Node 2
Node 3
Node 4
Node 5
Node 6
-1
1
-1
-1
-1
1
1
-1
1
1
-1
1
1
1
-1
-1
1
1
130
-1
1
1
1
>=
-3000.0003 -5000
>=
-6999.9999 -7000
>=
1.116E-08
>=
0
0
0
>=
3000
3000
>=
7000.0002
7000
Solutions for Chapter 6
6.8. The solution appears in the spreadsheet below.
Solution of Problem 6. 8
Variables
x13
x14 x15
0
Values
Obj Funct
st
Node 1
Node 2
Node 3
Node 4
Node 5
Node 6
6.9
x23
x24
x25 x26 x34 x35
0 1500 3500 2000 5000
182 375
-1
x16
-1
285
460
-1
-1
77
0
0 2000
290 245 575 275
-1
1
0
-1
-1
-1
1
1
0
125 380
Operator
-1
90
110
-1
1
1
-1
1
1
-1
1
1
Value
RHS
0 5000
-1
1
1
x36 x45 x46
1
1
Min
4441500
=
-5000 -5000
=
-7000 -7000
=
0
0
=
0
0
=
3500
3500
=
8500
8500
Although the problem asks to compute average sales based on forecasts, it makes more
sense to determine the net demand after taking out on order and in transit stock. Doing so
yields the following Net Demand Forecasts:
Week
0
Net Demand 0
1
0
2
0
3
25
4
12
5
0
6
19
7
85
8
33
The average of these values based on six periods of data is 29. (If one uses the original
data one obtains 33.25 here.) Using this as the value of  in the EOQ formula gives the
___________
EOQ as  2*40*29/.25 = 96. (with the larger value of  onc obtains 103). Ordering in
lots of 96 gives the following DRP Profile:
Weeks:
Sales Forecasts:
In Transit:
On Order:
EOQ placed:
EOQ arrives:
Projected Balance:
0
1
2
3
4
5
6
7
8
22
40
35
60
12
0
19
85
33
40
96
51
18
26
96
26
44
96
35
131
96
71
59
59
Production and Operations Analysis, Fourth Edition
6.10 Again, we will use the net demand data as shown in the solution to problem 6.9. Using the
Silver Meal Algorithm with K=40, h=.25 gives the following solution (starting in period
3): y3 = 56, y7=118, and all other values zero. The resulting DRP Profile is:
Weeks:
0
Sales Forecasts:
In Transit:
On Order:
Order placed:
Order arrives:
Projected Balance:
1
2
3
4
5
6
7
8
22
40
35
60
12
0
19
85
33
0
118
33
0
26
56
26
44
118
35
56
31
19
19
6.11
In order to incorporate a safety stock of 30 units for each period, we just increase the net
demand by 30 units in the first period. Taking into account receipts and on hand, this
gives a requirements vector of (0, 0, 21, 60, 12, 0, 19, 85, 33). The Silver Meal solution
for this set of requirements and K=40 and h=0.25 is y = (0, 0, 93, 0, 0, 0, 137, 0, 0).
6.12
The ROP method assumes a stationary demand pattern, but allows randomness, while
DRP allows for demand variation. The DRP method is conceptually the same as MRP
applied to distribution rather than production. The discussion in Chapter 7 comparing
MRP and traditional ROP methods applies here as well.
6.13 Consider Example 6.4 assuming that the truck capacity is 250 loaves instead of 300
loaves. Since (1,2) is the first pair in the ranking consider linking (1,2) without violating
the constraint. Since the total demand at these two locations is 247, include this link.
Clearly it would be impossible to include any other locations on this route. The next pair
in the ranking which doesn't include either location 1 or 2 is (3,5). These two locations
can be linked since the total demand in these locations is 26 + 110 = 136. Location 4
cannot be included on this route since that brings the demand on this route to 136 + 140
= 276 > 25. Hence location 4 would require a separate route. This solution thus requires
three distinct routes: (1,2), (3,5) and 4.
6.14
The distance matrix for this case is
0
1
2
3
4
1
2
3
4
5
45
35
10
30
15
15
10
35
25
20
30
25
35
20
20
The resulting values of sij are:
132
Solutions for Chapter 6
s12
s13
s14
s15
s23
=
=
=
=
=
70
60
20
50
50
s24
s25
s34
s35
s45
=
=
=
=
=
20
30
20
40
20
It turns out that the ranking of the pairs for this case is exactly the same as the ranking
when using the Euclidean metric, if one chooses to break ties to be consistent. Hence the
resulting solution is the same as the solution of Example 7.11.
6.15
Although this is not marked as a computer problem, the calculations involved with
adding only two additional locations turns out to be tedious. You may wish to
recommend that a computer be used to assist with these calculations.
The first step required in solving this problem is to complete the cost matrix. This means
computing the Euclidean distances from all locations to location 6 (ci6 for i = 0 to 5)
and location 7 (ci7 for i = 0 to 6). The two new columns that result are:
6
0
1
2
3
4
5
6
7
17.0
18.2
19.3
8.2
10.0
8.2
23.2
28.2
32.5
21.4
18.1
7.6
14.2
The next step is to compute the new sij values (11 in all). They are
s16
s26
s36
s46
s56
=
=
=
=
=
32.3
28.1
31.2
14.1
31.2
s17
s27
s37
s47
s57
S67
=
=
=
=
=
=
28.5
21.1
24.2
12.2
38.0
26.0
The next setp is to rank the 21 pairs of locations in decreasing order of the sij values.
The ranking one obtains is:
(1,2), (1,3), (2,3), (5,7), (1,5)
(1,6), (3,6), (5,6), (3,5), (1,7)
(2,6), (2,5), (6,7), (3,7), (2,7)
133
Production and Operations Analysis, Fourth Edition
(4,6), (1,4), (3,4), (4,5), (2,4), (4,7).
The final step is to obtain the routes by considering the pairs in sequence and including
new locations until the capacity constraint is exceeded. We assume the same capacity
constraint as in Example 6.4 which is 300 loaves. As before 1, 2, and 3 are on the first
route. This results in a requirement of 273 for this route. This route is then closed. The
next route starts with 5 and 7 with a req't. of 236. This route is then closed. The final
route consists of locations 4 and 6.
While tedious, this example does illustrate one point: that by adding only two additional
locations, the complexity of the calculations goes up considerably, in fact by a factor of
about 2.
6.16
The first step here is to compute the sij values. There will be a total of ten values to
compute. One obtains:
s12
s13
s14
s15
s23
=
=
=
=
=
60
48
10
35
90
s24
s25
s34
s35
s45
=
=
=
=
=
27
63
3
43
15
Next, these pairs are ranked in decreasing order giving:
(2,3), (2,5), (1,2), (1,3), (3,5), (1,5), (2,4), (4,5), (1,4), (3,4)
Now we construct the routes considering each pair in sequence and checking at each
assignment that we do not violate the 50 mile constraint. The first route consists of
locations 2 and 3 at a total distance of 18 miles. Next we include 5 which requires an
additional 20 miles, bringing this route up to 38 miles total. The pairings would
recommend including location 1 next, but the distance separating 5 and 1 is 15 miles
thus violating the constraint. In fact, this route must be closed at this point. The next route
would start with the next pair of unassigned routes, which are 1 and 4, which would
constitute the second route.
6.17
The matrix one obtains of the straight-line distances separating locations is:
0
1
2
3
4
1
14.9
2
26.9
12.1
3
30.2
16.1
8.6
The savings terms are:
134
4
37.0
30.1
28.2
36.7
5
12.2
8.6
18.0
24.0
25.5
Solutions for Chapter 6
s12
s13
s14
s15
s23
=
=
=
=
=
29.7
29.0
21.8
18.5
41.0
s24
s25
s34
s35
s45
=
=
=
=
=
33.8
30.5
37.1
33.8
40.6
This results in the following ranking of the pairs:
(2,3), (4,5), (3,4), (2,4), (3,5), (2,5), (1,2), (1,3), (1,4), (1,5)
First locations 2 and 3 are combined to be on the same route. The req't. of this route is
1,050. Assuming a 1200 gallon capacity this means this route is closed. Locations 1, 4
and 5 can now be included in the second route. The total requirement of this second
route is 1100 gallons.
6.18
Postponement refers to delaying the final configuration of a product as long as possible.
The main advantage is that by delaying, new information becomes available, thus
allowing a better match between products and demand. As a simple example, consider
packing for trip. By delaying the decision of exactly what clothes to pack as long as
possible, one can check the weather at the destination, and only pack what's suitable for
the expected weather conditions.
6.19
In this example, Lincoln and Jaguar would want to delay putting in the engines as long as
possible to see which engine type was preferred by consumers before committing to a
particular mix. While it would probably not be cost effective here, the two companies
could build local configuration facilities that would allow putting in the engines on a
custom order basis.
6.20
Too many suppliers can cause logistics complications. When problems arise, identifying
which supplier is responsible could be difficult. Also, more personnel are required to
manage relationships with suppliers. There are also risks associated with too few
suppliers. For example, if there is only one supplier for a critical part or resource, and
something happens to that supplier, disaster could result. Many years ago, a single
Japanese company supplied all the resin used in semiconductor production worldwide.
When this company experienced a fire, semiconductor production was seriously affected.
6.21
In some cases, it is easier and more efficient for large firms to outsource specific
functions, even when some of those functions are duplicated in-house. One notable
example is the development of the original IBM PC. The project was initiated by IBM in
the early 80’s somewhat secretly. To keep this project separate from the company’s more
traditional business, much of the PC was outsourced (Intel produced the microprocessor
and the operating system was licensed from Microsoft.) In retrospect, one would imagine
that IBM was very sorry it did not develop these key components in-house. Both
Microsoft and Intel grew to be giants, largely as a result of IBM’s decision.
6.22
The bullwhip refers to the increase in variability as one moves up the supply chain. It has
135
Production and Operations Analysis, Fourth Edition
been observed in practice in several contexts, and can result in highly variable demand
patterns on the factory even for items in which customer demand is fairly stable. The idea
behind the term "bullwhip" is that a small, but well timed, flick of the wrist results in a
very effecive and highly charged snap at the end of the whip.
6.23
Eliminating the middleman should reduce the bullwhip effect, since there are fewer
intervening layers in the system. However, there are some circumstances where
middlemen are not the cause of the problem. For example, one cause of high variability is
“phantom” orders. This is a result of shortage gaming, discussed in the section. Another
situation in which there might be high variability of orders is when demand is very hard
to predict, such as is the case for fashion items.
6.24
The causes listed for the bullwhip effect include:
1. Demand forecast updating
2. Order batching
3. Price fluctuations
4. Shortage gaming
The recommended remedies are:
1. Information sharing
2. Channel alignment
3. Price stabilization
4. Disincentives for shortage gaming.
It is unclear if any of these remedies could be implemented and what effect they might
have within existing supply chain distribution structures.
6.25
The popularity of EDI can be attributed primarily to: 1) sunk cost of investment in EDI
systems, both in terms of dollars and time, and 2) improved security afforded by EDI
over web-based transactions systems. For new entrants to the market, web based systems
are likely to be less expensive and easier to implement, by avoiding the complex EDI
protocols.
6.26
E-tailing has grown into a multibillion dollar business in only a few years. E-tailing web
sites require a much lower level of investment, so entry into the marketplace is much
easier. However, the other side of the coin is that unless these web sites can distinguish
themselves in some way, there is a high risk of failure. For example, Amazon.com was
one of the first e-tailers, and continues to be very successful. But at this point, it would be
difficult for new web-based book sellers to break into the market.
6.27
Risk pooling refers the decrease of relative variation in multilevel inventory systems in
which distribution centers are used to pool the risk at demand points. The risk pooling
effect is a consequence of the fact that adding independent random variables, variances
rather than standard deviations, are additive.
136
Solutions for Chapter 6
6.28
If a system gets big enough, (say over 1,000 stores), the number of DC’s required to
serve these stores can get large as well. By inserting an additional layer between the DC’s
and the factory, one can do risk pooling from the DC’s. Usually, only very large scale
systems would benefit from such a structure.
6.29
The two factors that determine the risk pooling benefit include the variance of sales and
the number of stores served by the DC. That is, risk pooling has a greater effect when the
variance of sales is high, and when there is a larger number of stores (or more generally,
demand points).
6.30
Certainly China should emerge as one of the most important new markets in coming
years. Not only is China the most populous country in the world, but investments of
billions of dollars by Japan and the United States, among others, are being poured into
China. As the standard of living improves, and (hopefully) the government allows the
economic system to migrate towards free enterprise, the need and desire among Chinese
for Western goods will increase as well. The second most populous country in the world,
India, also has the potential for being a significant market. Another emerging market is
Latin America. However, progress in both Latin America, and to a greater extent, Africa,
is hampered by political instability.
6.31
In the opinion of this writer, this trend is probably not going to decline. Even so, factors
that could slow or reverse the trend include: 1) world-wide economic depression leading
to the demise of these firms. 2) war 3) political problems in the emerging new markets
discussed in the answer to problem 6.30.
6.32
Manufacturing has always migrated to places where wage rates are lower. This occurred
in the United States in the 1940’s and 1950’s when textile and clothing manufacturing
moved from the northeast to the southeast. Today, much of the movement is offshore to
Asia and Latin America. Some writers suggest that this is a poor choice. That is, they
claim that there is too much emphasis placed on direct labor costs, and not enough on
other factors, such as stability and communication between manufacturing and design.
This migration would probably halt if wage rates equalized worldwide. (This is unlikely
to occur any time soon, however.)
6.33
Supply chains cannot react as fast when long distances separate manufacturing facilities
from other parts of the corporation. Supply chains are also hampered by less developed
infrastructures. Poor roads, airports, and port facilities can significantly impair the flow
of goods. Finally, implementation of such initiatives as JIT purchasing can be impossible
when suppliers and customers are separated by long distances.
137