The solution of homework #4(ISyE 6202)
The Solution of Homework #6
# Problem 1
Sol)
a) Since the motions across the two axes are independent from each other, the most
appropriate “distance” metric for this context is the Chebychev norm, which
 xi  x j y i  y j 
 required for
essentially measures the maximum time t ij  max 
,
 vx
vy 


traveling between two locations with coordinates xi , yi  and x j , y j .
Let’s label each picking location with a letter from A to G as follows:
y
F
A
G
x
x
C
x
E
x
B
x
D
x
3ft
x
x
3ft
The travel time per one bay and one level are respectively
3
3
tx 
min/ bay, t y 
min/ level . If we assume that the travel time is computed to
200
100
the center of each location, then we can compute the travel time between any two
locations as follows:
-1-
The solution of homework #4(ISyE 6202)
O
-
O
A
B
C
D
E
F
G
A
15/200
-
B
C
D
E
13.5/200 22.5/200 28.5/200 34.5/200
9/200
18/200
24/200
30/200
9/200
15/200
21/200
12/200
12/200
12/200
-
F
33/200
18/200
24/200
18/200
30/200
24/200
-
G
33/200
21/200
24/200
18/200
30/200
18/200
15/200
-
We can apply the closest insertion algorithm as follows:
Sp0 = <O>, Sa0 = {A, B, C, D, E, F, G}, C0 = [A(O), B(O), C(O), D(O), E(O), F(O), G(O)]
1) j* = arg min{CAO, CBO, CCO, CDO, CEO, CFO, CGO}
= arg min{15/200, 13.5/200, 22.5/200, 28.5/200, 34.5/200, 33/200, 33/200} = B
i*=1
Sa1={A, C, D, E, F, G}, Sp1=<O, B>, C1 = [A(B), C(B), D(B), E(B), F(B), G(B)]
2) j* = arg min{CAB, CCB, CDB, CEB, CFB, CGB}
= arg min{9/200, 9/200, 15/200, 21/200, 24/200, 24/200} = A (or C)
i*= arg min{COA+CAB-COB=(15+9-13.5)/200=10.5/200,
CBA+CAO-CBO= (9+15-13.5)/200=10.5/200} = 1 (or 2)
Sa2={C, D, E, F, G}, Sp2=<O, A, B>, C2 = [C(B), D(B), E(B), F(A), G(A)]
3) j* = arg min{CCB, CDB, CEB, CFA, CGA}
= arg min{9/200, 15/200, 21/200, 18/200, 21/200} = C
i*= arg min{ COC+CCA-COA= (22.5+18-15)/200=25.5/200,
CAC+CCB-CAB= (18+9-9)/200=18/200,
CBC+CCO-CBO= (9+22.5-13.5)/200=18/200} = 2 (or 3)
3
Sa ={D, E, F, G}, Sp3=<O, A, C, B>, C3 = [D(C), E(C), F(A, C), G(C)]
4) j* = arg min{CDC, CEC, CFC, CGC}
= arg min{12/200, 12/200, 18/200, 18/200} = D (or E)
*
i = arg min{ COD+CDA-COA= (28.5+24-15)/200=37.5/200,
CAD+CDC-CAC= (24+12-18)/200=18/200,
CCD+CDB-CCB= (12+15-9)/200=18/200,
CBD+CDO-CBO= (15+28.5-13.5)/200=30/200} = 3 (or 2)
Sa4={E, F, G}, Sp4=<O, A, C, D, B>, C4 = [E(C, D), F(A, C), G(C)]
-2-
The solution of homework #4(ISyE 6202)
5) j* = arg min{CEC(D), CFA(C), CGC}
= arg min{12/200, 18/200, 18/200} = E
i*= arg min{ COE+CEA-COA= (34.5+30-15)/200=49.5/200,
CAE+CEC-CAC= (30+12-18)/200=24/200,
CCE+CED-CCD= (12+12-12)/200=12/200,
CDE+CEB-CDB= (12+21-15)/200=18/200,
CBE+CEO-CBO= (21+34.5-13.5)/200=42/200} = 3
Sa5={F, G}, Sp5=<O, A, C, E, D, B>, C5 = [F(A, C), G(C, E)]
6) j* = arg min{CFA(C), CGC(E)}
= arg min{18/200, 18/200} = F (or G)
i*= arg min{ COF+CFA-COA= (33+18-15)/200=36/200,
CAF+CFC-CAC= (18+18-18)/200=18/200,
CCF+CFE-CCE= (18+24-12)/200=30/200,
CEF+CFD-CED= (24+30-12)/200=42/200,
CDF+CFB-CDB= (30+24-15)/200=39/200,
CBF+CFO-CBO= (24+33-13.5)/200=43.5/200} = 2
Sa6={G}, Sp6=<O, A, F, C, E, D, B>, C6 = [G(F)]
7) i*= arg min{ COG+CGA-COA= (33+21-15)/200=39/200,
CAG+CGF-CAF= (21+15-18)/200=18/200,
CFG+CGC-CFC= (15+18-18)/200=15/200,
CCG+CGE-CCE= (18+18-12)/200=24/200,
CEG+CGD-CED= (18+30-12)/200=36/200,
CDG+CGB-CDB= (30+24-15)/200=39/200,
CBG+CGO-CBO= (24+33-13.5)/200=43.5/200} = 2
Sp7=<O, A, F, G, C, E, D, B>
Then the solution is OAFGCEDBO and the total travel time is
(15+18+15+18+12+12+15+13.5)/200 = 118.5/200 min.
-3-
The solution of homework #4(ISyE 6202)
b) Consider the following figure.
L
Slope=0.5
y
B
P
3ft
x
A
3ft
First of all, notice that if S/R vehicle travels from the aisle picking station at full
v y 100
speed to both directions x and y, it travels along a line L with slope

 0.5 . If
v x 200
our destination is below line L, which means that the slope of travel direction is less than
the one of L, horizontal travel takes longer than vertical travel and the travel time is
dominated by v x . But if the destination is above line L, which means that the slope of
travel direction is greater than the one of L, the travel time is dominated by v y .
Let’s consider the single-line order concerning an item located at bay 7 and level 2 in
the rack. Assuming that we pick this item by visiting the center of the corresponding cell,
the considered picking location is represented by the point P(19.5, 4.5) annotated in the
figure above. Since P is below line L, the travel time to it is determined by the required
19.5
 0.0975 min. Motion on the y-axis can be
motion on the x-axis, and it is equal to
200
carried out with any profile for speed v y which guarantees that the AS/R machine will be
at coordinate Py by the time that it is also at coordinate Px. The following two scenarios
constitute the most extreme ways to achieve this condition, and therefore, they delineate
the region in which any other motion satisfying the above condition can take place:
1) Initially, move only along the x-axis with v x  200 , v y  0 ; at some point A,
start moving also full speed along the y-axis, so that you reach coordinate Py
exactly at the same time that you reach coordinate Px on the x-axis. Notice that
the second part of the motion outlined by this scenario, corresponds to a line
segment AP, which is parallel to line L.
2) Move along line L with speeds v x  200, v y  100 , overshooting the target
coordinate Py on the y-axis; however, when you reach some point B on line L,
reverse directions for the motion on the y-axis, so that the new motion is
-4-
The solution of homework #4(ISyE 6202)
performed with speeds v x  200, v y  100 ; pick point B so that the line
representing the motion during the second phase of this scenario will pass through
the target point P.
The travel region with the same minimum travel time to reach P from the I/O point O, is
limited to the trapezoid OAPB. Next, we characterize analytically this region, by
determining the coordinates for points A and B.
1) Determining the coordinates of A
As it was mentioned above, line AP is parallel to L and passes through P(19.5,
4.5).
Therefore,
the
equation
of
the
line
AP
is
y  4.5  0.5( x  19.5)  y  0.5 x  5.25 . So if we substitute y  0 , we get
A(10.5, 0). Notice that if we choose A(11,0), which is over A(10.5, 0) on the x11 4.5
20


 0.1  0.0975 .
axis, the travel time is time(OA) + time(AP) =
200 100 200
2) Determining the coordinates of B
In this case, line BP passes through point P(19.5, 4.5), but its slope is the equal to
negative slope of L, Therefore, the equation for line BP is
y  4.5  0.5( x  19.5)  y  0.5 x  14.25 and this line intersects with
y  0.5 x . Since B is the intersection point, we get B(14.25, 7.125). Notice that if
we choose B(15, 7.5), which is over B(14.25, 7.125) on the line L, then the travel
15
3
21


 0.105  0.0975 .
time is time(OB) + time(BP) =
200 100 200
So the trapezoid OAPB with O(0,0), A(10.5, 0), P(19.5, 4.5), and B(15, 7.5) is the rack
region that could be visited at zero additional travel cost(time), while filling the item
located at bay 7 and level 2.
Generalization
For any other item located in the rack below the line L, we can apply the same reasoning
presented above for point P. For items located above the line L, the gist of the reasoning
presented above remains the same, but the roles of the x and y axes will be interchanged.
Finally, notice for a point P located on line L, the aforementioned trapezoid OAPB
collapses to the line segment OP.
-5-
The solution of homework #4(ISyE 6202)
# Problem 2
a) A numbering scheme based on Sierpinski’s space-filling curve and a picking tour
based on the bay-numbering scheme
If we super-impose Sierpinski’s space-filling curve on the considered picking area, we
can get a numbering scheme and a picking tour based on the bay-numbering scheme,
where the picking tour is constructed based on the minimum length tour between two
consecutive locations, as follows:
2m
12m
35
34
33
30
29
28
15
14
13
12
7
6
5
7
37
32
31
26
27
36
38
25
24
22
6
16
11
x
9
8
4
39
42
43
x
23
21
40
41
44
x
45
46
20 100
19 101
103 102
104 105
1 106 1
17
18
10
3
2
7
1
54
53
49
48
x
47
x
56
52 2
x
50
51
73
6
99 74
98 76
97 96
94 95
x
93 92
55
3 57
64
x
71
6 70
58
62
63
x
72
69
75
77
x
78
91
89
80
x
79
84
85
90
9
7
7
Docking station
b) (i) Why a bay-numbering scheme based on Sierpinski’s space filling curve
might not be the most appropriate for the considered picking area ?
As we mentioned in class, a numbering scheme based on Sierpinski’s space filling
curve essentially assumes/characterizes proximity according to Euclidean metric.
However, in an aisle-based picking environment, the Euclidean distance is not the
most appropriate metric for characterizing proximity, since locations that are
physically close to each other might still require significant traveling through the
aisles for consecutive picking among them. As a more concrete example, consider
-6-
59
60
61
65
66
67
68
81
82
83
86
87
88
The solution of homework #4(ISyE 6202)
the locations 39, 40, 41, 42, 43, 44 and imagine that corresponding route for a
picking list that picks from all these locations. In fact, the presented route
demonstrates this type of inefficiency in the picking from the locations numbered
78 and 79 in the figure.
(ii) An alternative efficient bay-numbering scheme
Such a scheme must consider the aisle structure supporting the picking operation.
Hence, in the considered case, the picking area can be divided into two subregions, upper and lower picking area, defined by the middle cross-aisle. Then,
each of these two regions is numbered according to a numbering scheme based on
the S-shaped heuristic.
(iii) Efficiency of the new number scheme
Based on the S-shaped heuristic, we can develop a numbering scheme and a new
picking route with less travel distance as follows:
12m
23
21
19
17
15
13
12
10
8
6
4
2
1
2m
7
22
20
18
16
14
24
26
28
30
32
25 42
27 40
29 38
x x
31 36
33 5 34
7
1
43
41
39
37
x
35
x
44
46
x
48
50
52
2
1
6
105
103
101
99
97
7
88
90
92
94
96
1
87 86
89 84
91 82
93 80
x
95 78
62
60
58
x
56
54
4
7
11 106
9 104
x 102
7
5 100
3 98
45
47
49
x
51
53
6
85
83
x
81
79
77
66
x
68
70
72
74
Docking station
If we let the height of the picking area be 12m and the length between aisle be 2m, we
can show the improvement of solution by computing picking tours with minimum length
-7-
64
63
61
59
57
55
65
67
69
71
73
75
76
The solution of homework #4(ISyE 6202)
for each case in part a) and b). All computations are based on the assumption that we
travel in the middle of the aisle so that the travel length from bottom to top is 11m.


The length of the solution in part a) = 62m
The length of the solution in part b) = 54m
Notice that the efficiency of the new tour came essentially from the elimination of the
“backtracking” taking place in the original tour due to the impertinent numbering of
locations 78 and 79.
-8-
The solution of homework #4(ISyE 6202)
Problem 3
Sol)
a) Seed algorithm
 Seed selection: The largest number of aisles to be visited
 Order addition: To minimize the number of new aisles
 Sort while pick: Orders are picked on a picking cart with three separate
compartments. So the maximum size of batching is 3.
 Picking routes are designed according to the S-shaped heuristic
 Assumption
- Each compartment is big enough to contain any number of one order.
- All computation of travel length is based on the assumption that we travel in
the middle of the aisle so that the travel length from bottom to top is 11m.
- After picking the last item and there is no more item to pick, then we return to
the docking station by Just taking the shortest way back.
From Figure 3, we have the following information:
order
# of items
# of aisles
a
7
3
b
4
2
c
4
2
d
4
3
e
6
4
f
2
2
1) 1st seed : e
order to be added
a
b
c
d
f
g
h
# of new aisles
1
0
1
0
1
0
0
selection
*
*
 Batch1 = {b, d, e}
 Travel distance = 56(=4+11+2+11+2+11+2+11+2)
2) 2nd seed : a
order to be added
c
f
g
h
# of new aisles
0
0
1
1
 Batch2 = {a, c, f}
-9-
selection
*
*
g
4
3
h
1
1
The solution of homework #4(ISyE 6202)
 Travel distance = 60(=4+11+6+11+2+11+11+4)
3) 3rd seed : g
 Batch3 = {g, h}
 Travel distance = 44(=4+11+2+11+2+7+7)
So the total travel distance = 56 + 60 + 44 = 160.
b) Savings algorithm
 Assumption
- Travel time is proportional to the travel distance so that we can use the travel
distance to select a batch instead of travel time.
- All computation of travel length is based on the assumption that we travel in
the middle of the aisle so that the travel length from bottom to top is 11m.
- After picking the last item and there is no more item to pick, then we return to
the docking station by Just taking the shortest way back.
 Following is the computation of the travel distance required for picking an order.
order
a
b
c
d
e
f
g
h
Travel distance
4+11+6+11+2+8+8+4=54
4+11+2+11+2=30
2+11+2+11+4=30
4+11+2+11+2+5+5=40
4+11+2+11+2+11+2+11+2=56
2+11+2+11+4=30
4+11+2+11+2+7+7=44
2+10+10+2=24
 Following is the computation of the travel distance required for picking each pair
of orders and the distance savings.
order
(a,b)
(a,c)
(a,d)
(a,e)
(a,f)
(a,g)
(a,h)
(b,c)
(b,d)
(b,e)
(b,f)
Travel distance
4+11+2+11+4+11+2+11+4=60
4+11+6+11+2+11+11+4=60
4+11+2+11+2+11+2+11+2+8+8+4=76
4+11+2+11+2+11+2+11+2+8+8+4=76
4+11+6+11+2+8+8+4=54
4+11+2+11+2+11+2+11+2+8+8+4=76
4+11+2+11+4+11+4+11+4=62
4+11+2+11+4+11+2+11+4=60
4+11+2+11+2+5+5=40
4+11+2+11+2+11+2+11+2=56
4+11+2+11+4+11+2+11+4=60
- 10 -
Distance savings
54+30-60=24
54+30-60=24
54+40-76=18
54+56-76=34
54+30-54=30
54+44-76=22
54+24-62=16
30+30-60=0
30+40-40=30
30+56-56=30
30+30-60=0
Max
The solution of homework #4(ISyE 6202)
(b,g)
(b,h)
(c,d)
(c,e)
(c,f)
(c,g)
(c,h)
(d,e)
(d,f)
(d,g)
(d,h)
(e,f)
(e,g)
(e,h)
(f,g)
(f,h)
(g,h)
4+11+2+11+2+7+7=44
4+11+2+11+2=30
4+11+2+11+2+11+2+11+2+11+11+4=82
4+11+2+11+2+11+2+11+2+11+11+4=82
2+11+2+11+4=30
4+11+2+11+2+11+2+11+2+11+11+4=82
2+11+4+11+2+11+11+4=56
4+11+2+11+2+11+2+11+2=56
4+11+2+11+2+11+2+11+2+5+5+4=70
4+11+2+11+2+7+7=44
4+11+2+11+2+5+5=40
4+11+2+11+2+11+2+11+2+5+5+4=70
4+11+2+11+2+11+2+11+2=56
4+11+2+11+2+11+2+11+2=56
4+11+2+11+2+11+2+11+2+5+5+4=70
2+11+4+11+2+5+5+4=44
4+11+2+11+2+7+7=44
30+44-44=30
30+24-30=24
30+40-82= -12
30+56-82=4
30+30-30=30
30+44-82= -8
30+24-56= -2
40+56-56=40
40+30-70=0
40+44-44=40
40+24-40=24
56+30-70=16
56+44-56=44
56+24-56=24
30+44-70=4
30+24-44=10
44+24-44=24
*
The maximum savings is achieved by merging of order e and g. Let batch1={e,
g}. It is interesting to notice that the merging of order c with d, g, or h results in the
negative savings. So, they should not be merged in a batch.
 Now we can compute the travel distance and the time savings considering batch1
and the remaining orders as follows.
order
(a,b)
(a,c)
(a,d)
(a,e,g)
(a,f)
(a,h)
(b,c)
(b,d)
(b,e,g)
(b,f)
(b,h)
(c,d)
(c,e,g)
(c,f)
(c,h)
(d,e,g)
(d,f)
(d,h)
(e,g,f)
Travel distance
4+11+2+11+4+11+2+11+4=60
4+11+6+11+2+11+11+4=60
4+11+2+11+2+11+2+11+2+8+8+4=76
4+11+2+11+2+11+2+11+2+8+8+4=76
4+11+6+11+2+8+8+4=54
4+11+2+11+4+11+4+11+4=62
4+11+2+11+4+11+2+11+4=60
4+11+2+11+2+5+5=40
4+11+2+11+2+11+2+11+2=56
4+11+2+11+4+11+2+11+4=60
4+11+2+11+2=30
4+11+2+11+2+11+2+11+2+11+11+4=82
4+11+2+11+2+11+2+11+2+11+11+4=82
2+11+2+11+4=30
2+11+4+11+2+11+11+4=56
4+11+2+11+2+11+2+11+2=56
4+11+2+11+2+11+2+11+2+5+5+4=70
4+11+2+11+2+5+5=40
4+11+2+11+2+11+2+11+2+5+5+4=70
- 11 -
Distance savings
54+30-60=24
54+30-60=24
54+40-76=18
54+56-76=34
54+30-54=30
54+24-62=16
30+30-60=0
30+40-40=30
30+56-56=30
30+30-60=0
30+24-30=24
30+40-82= -12
30+56-82=4
30+30-30=30
30+24-56= -2
40+56-56=40
40+30-70=0
40+24-40=24
56+30-70=16
Max
*
The solution of homework #4(ISyE 6202)
(e,g,h)
(f,h)
4+11+2+11+2+11+2+11+2=56
2+11+4+11+2+5+5+4=44
56+24-56=24
30+24-44=10
The maximum savings is obtained by merging of order d,e, and g. Since the
number of orders is 3, this constitutes one batch as batch1 = {d, e, g} with travel
distance 56.
 Now we have orders {a, b, c, f, h} and can compute the travel distance and the
time as follows.
order
(a,b)
(a,c)
(a,f)
(a,h)
(b,c)
(b,f)
(b,h)
(c,f)
(c,h)
(f,h)
Travel distance
4+11+2+11+4+11+2+11+4=60
4+11+6+11+2+11+11+4=60
4+11+6+11+2+8+8+4=54
4+11+2+11+4+11+4+11+4=62
4+11+2+11+4+11+2+11+4=60
4+11+2+11+4+11+2+11+4=60
4+11+2+11+2=30
2+11+2+11+4=30
2+11+4+11+2+11+11+4=56
2+11+4+11+2+5+5+4=44
Distance savings
54+30-60=24
54+30-60=24
54+30-54=30
54+24-62=16
30+30-60=0
30+30-60=0
30+24-30=24
30+30-30=30
30+24-56= -2
30+24-44=10
Max
*
*
The maximum savings is taken by merging of order a and f or c and f. We can choose one
of them. Let’s choose a and f as a batch2 = {a, f}.
 Then we have the following computation for possible pairs of orders.
order
(a,f,b)
(a,f,c)
(a,f,h)
(b,c)
(b,h)
(c,h)
Travel distance
4+11+2+11+4+11+2+11+4=60
4+11+6+11+2+11+11+4=60
4+11+2+11+4+11+2+11+4=60
4+11+2+11+4+11+2+11+4=60
4+11+2+11+2=30
2+11+4+11+2+11+11+4=56
Distance savings
54+30-60=24
54+30-60=24
54+24-60=18
30+30-60=0
30+24-30=24
30+24-56= -2
Max
*
*
*
The maximum savings is obtained by merging of order (a, f) and b or (a, f) and c. But if
we select batch2 = {a, b, f}, then the savings from batch3 = {c, h} is -2. Otherwise, by
selecting batch2 = {a, c, f}, the savings from batch3 = {b, h} is 24. So let’s choose batch2
= {a, c, f} and batch3 = {b, h} with travel distances 60 and 30 respectively.
So we have Batch1 = {d, e, g}, Batch2 = {a, c, f}, and Batch3 = {b, h} with the total
travel distance is 160.
- 12 -
The solution of homework #4(ISyE 6202)
Extra Credit
Problem 1
Since the warehouse layout is the same as the one we dealt with in class, we can have the
same 7 possible characterizations for an Lj(- or +) PTS and use the same notations.
Consider the following graph-based representation of the underlying topology. All
computation of travel length is based on the assumption that we travel in the middle of
the aisle so that the travel length from bottom to top is 11m.
a1
a2
2
a3
a4
a5
1
3
7
11
3
7
11
4
0
4
1
3
2
b1
b2
b3
b4
b5
Then the application of Ratliff & Rosenthal’s algorithm for the computation of the
minimal length picking tour evolves as follows:
1) In the following tables, the leftmost column represents the classes of the starting subtour, used for visiting the nodes involved in the previous stage, and the topmost row
represents the classes for the sub-tour constructed for the considered stage, through
the addition to the starting sub-tour of new nodes, according to one of the possible
schemes enumerated in the paper (e.g., I-i, I-ii, II-I, etc.) Each table cell represents the
length of the resulting sub-tour and the applied extension scheme. For example, in the
table for L2-, cell (1,1) states that a sub-tour for L2-, belonging to class (U,U,1C), can
be constructed from a sub-tour for L1+, belonging to class (U,U,1C), by applying
extension scheme II-i. The length of the new sub-tour is 15. The last line identifies
the minimum-length sub-tour constructed for each class, the class of the starting subtour, and the applied extension scheme.
2) L1- : None
3) L1+
To
From
-
1.(U,U,1C)
2.(E,0,1C)
3.(0,E,1C)
4.(E,E,1C)
5.(E,E,2C)
6.(0,0,0C)
7.(0,0,1C)
11, I-i
22, I-ii
0, I-iii
22, I-v
-
-
-
- 13 -
The solution of homework #4(ISyE 6202)
4) L2To
From
1.(U,U,1C)
2.(E,0,1C)
3.(0,E,1C)
4.(E,E,1C)
5.(E,E,2C)
6.(0,0,0C)
7.(0,0,1C)
min
1.(U,U,1C)
2.(E,0,1C)
3.(0,E,1C)
4.(E,E,1C)
5.(E,E,2C)
6.(0,0,0C)
7.(0,0,1C)
11+4=15, II-i
22+8=30,II-iv
0+8=8,II-iv
22,II-v
0,II-v
22,I-v
30, 4, II-iv
8, 3, II-iv
0, 3, II-v
4.(E,E,1C)
5.(E,E,2C)
22+4=26, II-ii
0+4=4, II-iii
22+4=26,II-iii
2+8=30,II-iv
26, 2, II-ii
26, 4, II-ii
4, 3, II-iii
2.(E,0,1C)
3.(0,E,1C)
22+4=26, II-ii
15, 1, II-i
5) L2+
To
From
1.(U,U,1C)
1.(U,U,1C)
15+20=35,I-ii
15+16=31,I-iii
15+16=31,I-iv
15+14=29,I-iv
15+22=37,I-v
2.(E,0,1C)
26+11=37,I-i
3.(0,E,1C)
4+11=15,I-i
4.(E,E,1C)
5.(E,E,2C)
6.(0,0,0C)
7.(0,0,1C)
6.(0,0,0C)
7.(0,0,1C)
15+11=26,I-i
26+22=48,I-v
26+20=46,I-ii
4+16=20,I-iii
4+22=26,I-v
26+16=42,I-iii
26+16=42,I-iv
26+14=40,I-iv
4+20=24,I-ii
4+16=20,I-iv
4+14=18,I-iv
30+20=50,I-ii
30+16=46,I-iii
30+16=46,I-iv
30+14=44,I-iv
30+11=41,I-i
8+22=30,I-v
8+20=28,I-ii
8+16=24,I-iii
8+16=24,I-iv
8+14=22,I-iv
20, 3, I-iii
26, 3, I-v
26, 1, I-i
18, 3, I-iv
3.(0,E,1C)
4.(E,E,1C)
5.(E,E,2C)
8+11=19,I-i
6.(0,0,0C)
7.(0,0,1C)
min
15, 3, I-i
46, 2, I-ii
6) L3To
From
1.(U,U,1C)
2.(E,0,1C)
3.(0,E,1C)
4.(E,E,1C)
5.(E,E,2C)
6.(0,0,0C)
7.(0,0,1C)
1.(U,U,1C)
2.(E,0,1C)
15+4=19,II-i
46+4=50,II-ii
26+4=30,II-ii
20+4=24,II-iii
26+4=30,II-iii
46+8=54,II-iv
20+8=28,II-iv
26+8=34,II-iv
18+8=26,II-iv
- 14 -
46,II-v
20,II-v
26,II-v
The solution of homework #4(ISyE 6202)
min
19, 1, II-i
30, 4, II-ii
24, 3, II-iii
34, 4, II-iv
26, 5, II-iv
3.(0,E,1C)
4.(E,E,1C)
5.(E,E,2C)
20, 3, II-v
7) L3+
To
From
1.(U,U,1C)
2.(E,0,1C)
3.(0,E,1C)
19+14=33,I-ii
19+8=27,I-iii
19+22=41,I-v
30+11=41,I-i
24+11=35,I-i
4.(E,E,1C)
34+11=45,I-i
5.(E,E,2C)
26+11=37,I-i
6.(0,0,0C)
7.(0,0,1C)
min
27, 1, I-iii
1.(U,U,1C)
2.(E,0,1C)
6.(0,0,0C)
7.(0,0,1C)
6.(0,0,0C)
7.(0,0,1C)
19+11=30,I-i
30+14=44,I-ii
24+8=32,I-iii
44, 2, I-ii
30+22=52,I-v
24+22=46,I-v
34+14=48,I-ii
34+8=42,I-iii
34+22=56,I-v
30+8=38,I-iii
24+14=38,I-ii
26+22=48,I-v
26,14=40,I-ii
26+8=34,I-iii
32, 3, I-iii
30, 1, I-i
34, 5, I-iii
3.(0,E,1C)
4.(E,E,1C)
5.(E,E,2C)
8) L4To
From
1.(U,U,1C)
2.(E,0,1C)
3.(0,E,1C)
4.(E,E,1C)
5.(E,E,2C)
6.(0,0,0C)
7.(0,0,1C)
min
1.(U,U,1C)
2.(E,0,1C)
27+4=31,II-i
44+4=48,II-ii
30+4=34,II-ii
32+4=36,II-iii
30+4=34,II-iii
44+8=52,II-iv
32+8=40,II-iv
44,II-v
32,II-v
30,II-v
30+8=38,II-iv
34+8=42,II-iv
31, 1, II-i
34, 4, II-ii
34, 4, II-iii
38, 4, II-iv
40, 3, II-iv
30, 4, II-v
9) L4+
To
From
1.(U,U,1C)
1.(U,U,1C)
31+16=47,I-ii
31+20=51,I-iii
31+8=39,I-iv
31+22=53,I-v
2.(E,0,1C)
34+11=45,I-i
3.(0,E,1C)
34+11=45,I-i
4.(E,E,1C)
38+11=49,I-i
5.(E,E,2C)
40+11=51,I-i
2.(E,0,1C)
3.(0,E,1C)
4.(E,E,1C)
5.(E,E,2C)
31+11=42,I-i
34+22=56,I-v
34+16=50,I-ii
34+20=54,I-iii
34+22=56,I-v
38+16=54,I-ii
38+20=58,I-iii
38+8=46,I-iv
38+22=60,I-v
40+22=62,I-v
- 15 -
34+20=54,I-iii
34+8=42,I-iv
34+16=50,I-ii
34+8=42,I-iv
40+16=56,I-ii
6.(0,0,0C)
7.(0,0,1C)
The solution of homework #4(ISyE 6202)
40+20=60,I-iii
40+8=48,I-iv
6.(0,0,0C)
7.(0,0,1C)
min
39, 1, I-iv
50, 2, I-ii
54, 3, I-iii
42, 1, I-i
42, 2, I-iv
42, 3, I-iv
10) L5To
From
1.(U,U,1C)
2.(E,0,1C)
3.(0,E,1C)
4.(E,E,1C)
5.(E,E,2C)
6.(0,0,0C)
7.(0,0,1C)
min
1.(U,U,1C)
2.(E,0,1C)
3.(0,E,1C)
4.(E,E,1C)
5.(E,E,2C)
6.(0,0,0C)
7.(0,0,1C)
39+4=43,II-i
50+4=54,II-ii
42+4=46,II-ii
50+8=58,II-iv
54+8=62,II-iv
54+4=58,II-iii
42+4=46,II-iii
50,II-v
54,II-v
42,II-v
42+8=50,II-iv
42+8=50,II-iv
43, 1, II-i
46, 4, II-ii
46, 4, II-iii
50, 4, II-iv
50, 5, II-iv
42, 4, II-v
11) L5+
To
From
1.(U,U,1C)
2.(E,0,1C)
3.(0,E,1C)
43+22=65,I-iv
43+0=43,I-vi
46+11=57,I-i
46+11=57,I-i
4.(E,E,1C)
50+11=61,I-i
5.(E,E,2C)
6.(0,0,0C)
7.(0,0,1C)
min
50+11=61,I-i
1.(U,U,1C)
43, 1, I-vi
2.(E,0,1C)
3.(0,E,1C)
4.(E,E,1C)
5.(E,E,2C)
6.(0,0,0C)
43+11=54,I-i
46+0=46,I-vi
46+0=46,I-vi
46, 2, I-vi
46, 3, I-vi
46+22=68,I-v
46+22=68,I-v
50+22=72,I-v
50+0=50,I-vi
50+22=72,I-v
50+0=50,I-vi
50, 4, I-vi
50, 5, I-vi
42+0=42,I-vi
42, 7, I-vi
The following tables summarize the above computation.
PTS-class
7.(0,0,1C)
1.(U,U,1C)
L1-
L1 +
11, -, I-i
2.(E,0,1C)
22, -, I-ii
3.(0,E,1C)
0, -, I-iii
L2 15, 1, II-i
26, 2, II-ii
26, 4, II-ii
4, 3, II-iii
4.(E,E,1C)
22, -, I-v
30, 4, II-iv
8, 3, II-iv
5.(E,E,2C)
6.(0,0,0C)
7.(0,0,1C)
0, 3, II-v
- 16 -
L2 +
15, 3, I-i
L3 19, 1, II-i
46, 2, I-ii
30, 4, II-ii
20, 3, I-iii
26, 1, I-i
26, 3, I-v
18, 3, I-iv
24, 3, II-iii
34, 4, II-iv
26, 5, II-iv
20, 3, II-v
The solution of homework #4(ISyE 6202)
PTS-class
1.(U,U,1C)
2.(E,0,1C)
3.(0,E,1C)
4.(E,E,1C)
L3 +
27, 1, I-iii
44, 2, I-ii
32, 3, I-iii
30, 1, I-i
L4 31, 1, II-i
34, 4, II-ii
34, 4, II-iii
38, 4, II-iv
5.(E,E,2C)
34, 5, I-iii
40, 3, II-iv
6.(0,0,0C)
7.(0,0,1C)
L4 +
39, 1, I-iv
50, 2, I-ii
54, 3, I-iii
42, 1, I-i
42, 2, I-iv
42, 3, I-iv
30, II-v
L5 43, 1, II-i
46, 4, II-ii
46, 4, II-iii
50, 4, II-iv
L5 +
43, 1, I-vi
46, 2, I-vi
46, 3, I-vi
50, 4, I-vi
50, 5, II-iv
50, 5, I-vi
42, 4, II-v
42, 7, I-vi
From the L5+ column in the above table, we can see that the minimum-length picking
tour has a length equal to 42, and belongs to class (0,0,1C). Furthermore, this tour was
obtained from a sub-tour belonging to the class (0,0,1C) by applying the extension
scheme I-vi. This sub-tour was, in turn, obtained from a sub-tour of class (E,E,1C) by
applying the extension scheme II-v. Working backward in a similar fashion all the
columns of the above table, we eventually get the minimum-length tour depicted in the
following figure:
a1
a2
2
a3
a4
a5
1
3
7
11
3
7
11
4
0
4
1
3
2
b1
b2
b3
b4
b5
Remarks: We notice that, in the general case, one must check the feasibility of the
resulting tour, before it is accepted as the minimum-length tour for considered set of
picking locations. To exemplify this need in a more concrete manner, suppose there is no
aisle 5 in the considered problem. Then the final stage in the above computations is L4+,
and the obtained tables indicate that the minimum-length tour belongs to the class
(U,U,1C) with a total length of 39 units. Furthermore, this tour is obtained from a subtour belonging to the class (U,U,1C) by applying the extension scheme I-iv. The
complete construction of this tour will reveal that it is infeasible, if one is to start his/her
trip from the node corresponding to the annotated I/O point. Hence, in this case, we need
to consider the tour with the next smallest length. A feasible minimum-length tour is that
belonging to the class (E,E,1C) with length 42, and, in fact, it is the one depicted in the
above figure.
- 17 -