Answers to Problems at the End of Chapter 10 10-40
Note: Multiple optimal layouts exist
10.36a
10.36b Product Ranking: B > A > C
2 2 2 2 5 5 5 5
4 4 4 4 4 2 5 5
6 6 3 1 1 2 5 5
6 6 3 1 1 4 2 5
3 3 3 1 1 4 2 5
0.5
Product # of Bays Load Rate
Tj/Sj
(j)
(Sj)(Tj)
A15 10.0667
B 5 1 0.2
C16 10.0625
Note: Each block is 15’ x 15’
54.375 61.875 69.375 76.875 84.375 91.875
54.375 61.875 69.375 76.875 84.375 91.875
54.375 61.875 69.375 76.875 84.375 91.875
58.125 65.625 73.125 80.625 88.125 95.625
65.625 73.125 80.625 88.125 95.625 103.125
73.125 80.625 88.125 95.625 103.125 110.625
0.5
0.25
B B A A C C
B A A A C C
B A A A C C
B A A C C C
A A A C C C
A A C C C C
0.5
0.25
10.37b Product Ranking: Z > X > Y
Distances to/from storage bays are determined by summing the distance traveled from the
receiving door (A) to the center of the storage bay and from the center of the storage bay
to the shipping door (B). Consider the shaded storage bay, shown below. Travel from the
receiving door to the midpoint of the cross-aisle is 92′; travel up the cross-aisle to the
center of the two-way aisle is 52′; travel along the two-way aisle to the center point of the
of the center aisle to the mid-point of a storage bay is 18‘. Hence, the total distance equals
244′ + 18′ + 18′, or 280′.
Product Area # of Bays Load Rate
Tj/Sj
(j)
(ft2)(Sj)(Tj)
X6,400 16 400 25
Y8,800 22 400 18.18
Z2,400 6600 100
528 488 448 408 376 376 376 376 376 376 376
280 280 280 280 280 280 280 280 280 280 280
280 280 280 280 280 280 280 280 280 280 280
528 488 448 408 376 376 376 376 376 376 376
0.5
0.5
Answers to Problems at the End of Chapter 10 10-43
SECTION 10.4
10.38 Initial Layout:
Improvement Procedure: (Note: not all pairwise interchanges are illustrated for each
iteration)
Y Y Y Y Y Y Y Y Y Y Y
Z Z Z X X X X X X X X
Z Z Z X X X X X X X X
Y Y Y Y Y Y Y Y Y Y Y
0.5
0.5
A B C D
P/D Location 30 110 220 305
fij A B C D
A 2,100 1,100 1,000
B400 1,300
C 2,700
dij A B C D
A80 190 275
B110 195
C85
Answers to Problems at the End of Chapter 10 10-44
ITERATION 2:
Exchange A and B: TC = 870,000, Savings = 108,000
No additional savings. Solution is the layout: B – A – D – C.
10.39 Initial Layout:
Improvement Procedure:
ITERATION 1:
Exchange A and B: TC = 1,023,000, Savings = 132,000
C B A D
P/D Location 60 170 250 305
B A D C
P/D Location 50 130 185 270
A B C D
P/D Location 30 110 220 305
fij A B C D
A 1,800 1,200 1,200
B600 1,200
C 1,800
dij A B C D
A80 190 275
B110 195
C85
B A C D
P/D Location 50 130 220 305
P/D Location 30 85 160 270
B D A C
P/D Location 50 125 180 270
Answers to Problems at the End of Chapter 10 10-45
Exchange B and C: Same TC as interchanging A and D
Best interchange is C and D.
ITERATION 2:
Exchange A and B: TC = 825,000, Savings = 132,000
P/D Location 60 170 245 300
P/D Location 30 120 205 280
P/D Location 30 85 160 270
ITERATION 3: The only exchanging B and C will result in a layout that hasn’t been
obtained during the first two iterations.
B A D C
P/D Location 50 130 185 270
P/D Location 30 85 170 280
P/D Location 30 110 185 270
10.40 Initial Layout:
Improvement Procedure:
ITERATION 1:
Exchange A and B: TC = 972,000, Savings = 104,000
C B A D
P/D Location 60 170 250 305
D B C A
P/D Location 25 100 210 300
Exchange B and C: Same TC as interchanging A and D
Exchange B and D: TC = 1,080,000, Savings = -4,000
A B D C
P/D Location 30 110 185 270
A B C D
P/D Location 30 110 220 305
B A C D
P/D Location 50 130 220 305
A D C B
P/D Location 30 85 170 280
Answers to Problems at the End of Chapter 10 10-47
ITERATION 2:
Exchange A and B: TC = 835,000, Savings = 104,000
ITERATION 3: The only exchanging B and C will result in a layout that hasn’t been
10.41
B A D C
P/D Location 50 130 185 270
fij A B C D E F
A75 150 075 150
B75 75 075
C150 75 0
D75 150
E75
P/D Location 60 170 245 300
P/D Location 25 100 180 270
P/D Location 30 120 205 280
P/D Location 30 85 160 270
Answers to Problems at the End of Chapter 10 10-48
Improvement Procedure: (Due to the structure of the problem, it is not necessary to
consider all possible pairs. For example, interchanges directly “across the aisle” need not
be considered, since exchanging A&D in the original layout will not change any
distances between any pairs.)
ITERATION 1:
ITERATION 2:
10.42a We assume that cij = 1 i,j. Since the from-to matrix is symmetric there is no need to
convert to a flow-between matrix to start the construction algorithm.
ITERATION 1:
Answers to Problems at the End of Chapter 10 10-49
ITERATION 2:
10.42b We assume that cij = 1 i,j. Since the from-to matrix is symmetric there is no need to
convert to a flow-between matrix to start the construction algorithm.
ITERATION 2:
10.43 The design of the flow chart will depend on the student. However, the flow chart should
10.44 The design of the flow chart will depend on the student. However, the flow chart should
10.45 Research question. The report will depend on the research articles used for the report.
SECTION 10.5
10.46a Shown below is an Excel spreadsheet solution of the problem. To compute the number of
Answers to Problems at the End of Chapter 10 10-51
Q = 60 L = 48 W = 52 c = 8 A = 144
Day Inv 1 2 3 4 5 6 7 8 9 10
160 15 8 5 4 3 3 3 2 2 2
259 15 8 5 4 3 3 3 2 2 2
358 15 8 5 4 3 3 3 2 2 2
457 15 8 5 4 3 3 3 2 2 2
556 14 7 5 4 3 3 2 2 2 2
655 14 7 5 4 3 3 2 2 2 2
754 14 7 5 4 3 3 2 2 2 2
853 14 7 5 4 3 3 2 2 2 2
952 13 7 5 4 3 3 2 2 2 2
10 51 13 7 5 4 3 3 2 2 2 2
11 50 13 7 5 4 3 3 2 2 2 2
12 49 13 7 5 4 3 3 2 2 2 2
13 48 12 6 4 3 3 2 2 2 2 2
14 47 12 6 4 3 3 2 2 2 2 2
15 46 12 6 4 3 3 2 2 2 2 2
16 45 12 6 4 3 3 2 2 2 2 2
17 44 11 6 4 3 3 2 2 2 2 2
18 43 11 6 4 3 3 2 2 2 2 2
Row Depth (x)
Number of Storage Rows Required to Accommodate the Inventory Level
Answers to Problems at the End of Chapter 10 10-52
Day Inv 1 2 3 4 5 6 7 8 9 10
36 25 7 4 3 2 2 2 1 1 1 1
37 24 6 3 2 2 2 1 1 1 1 1
38 23 6 3 2 2 2 1 1 1 1 1
39 22 6 3 2 2 2 1 1 1 1 1
40 21 6 3 2 2 2 1 1 1 1 1
41 20 5 3 2 2 1 1 1 1 1 1
42 19 5 3 2 2 1 1 1 1 1 1
43 18 5 3 2 2 1 1 1 1 1 1
44 17 5 3 2 2 1 1 1 1 1 1
45 16 4 2 2 1 1 1 1 1 1 1
46 15 4 2 2 1 1 1 1 1 1 1
47 14 4 2 2 1 1 1 1 1 1 1
48 13 4 2 2 1 1 1 1 1 1 1
49 12 3 2 1 1 1 1 1 1 1 1
50 11 3 2 1 1 1 1 1 1 1 1
51 10 3 2 1 1 1 1 1 1 1 1
52 9 3 2 1 1 1 1 1 1 1 1
53 8 2 1 1 1 1 1 1 1 1 1
54 7 2 1 1 1 1 1 1 1 1 1
55 6 2 1 1 1 1 1 1 1 1 1
56 5 2 1 1 1 1 1 1 1 1 1
57 4 1 1 1 1 1 1 1 1 1 1
58 3 1 1 1 1 1 1 1 1 1 1
59 2 1 1 1 1 1 1 1 1 1 1
60 1 1 1 1 1 1 1 1 1 1 1
η = 8.00 4.27 3.00 2.40 2.00 1.80 1.60 1.47 1.40 1.33
S = 57,600 43,008 38,880 38,016 37,440 38,880 39,168 40,128 42,336 44,160
x* = 5
Number of Storage Rows Required to Accommodate the Inventory Level
Row Depth (x)
Answers to Problems at the End of Chapter 10 10-53
10.46c
10.47a L = 48″, W = 40″, c = 8″, A = 156″, z = 5
Q = 300
Q = 60 L = 48 W = 52 c = 8 A = 144
z = 4 4 4 4 4 4 4 4 4 4
x = 1 2 3 4 5 6 7 8 9 10
xz = 4 8 12 16 20 24 28 32 36 40
y = 15 8 5 4 3 3 3 2 2 2
SBS = 57,600 43,008 38,880 38,016 37,440 38,880 39,168 40,128 42,336 44,160
x* = 5
Answers to Problems at the End of Chapter 10 10-54
in inventory profiles over lifetimes of production lots to be performed easily.
10.47b Q = 50. Using enumeration yields the results shown below.
10.47c Q = 100, x1 = 6, x2 = 3. As shown, y1 = 2, y2 = 3, and SBS = 262.90 sq. ft.
x = 1 2 3 4 5 6 7 8 9 10 11 12
xz = 5 10 15 20 25 30 35 40 45 50 55 60
y = 10 5 4 3 2 2 2 2 2 1 1 1
SBS = 36,344 26,544 24,394 23,990 23,352 24,931 26,062 26,746 26,981 26,768 29,008 31,248
x*BS = 5
10.47d Q = 50 and x = 3.
10.47e S = 28,512 sq. ft.
10.48b Using a continuous approximation, with Q = 300,
Q = 50 L = 40 W = 48 c = 8 A = 156 z = 5
Inv Percent 1 2 3 4 5 6 7 8 9 10
50 20% 10 5 4 3 2 2 2 2 2 1
45 20% 9 5 3 3 2 2 2 2 1 1
40 15% 8 4 3 2 2 2 2 1 1 1
35 10% 7 4 3 2 2 2 1 1 1 1
30 10% 6 3 2 2 2 1 1 1 1 1
25 5% 5 3 2 2 1 1 1 1 1 1
20 5% 4 2 2 1 1 1 1 1 1 1
15 5% 3 2 1 1 1 1 1 1 1 1
10 5% 2 1 1 1 1 1 1 1 1 1
55% 1 1 1 1 1 1 1 1 1 1
η = 7.05 3.75 2.70 2.20 1.75 1.65 1.55 1.40 1.20 1.00
S = 46,586 33,180 29,938 29,322 27,244 29,383 31,074 31,203 29,434 26,768
Row Depth (x)
Number of Storage Rows Required to Accommodate the Inventory Level
x*BS = 5
Q = 300 L = 42 W = 48 c = 8 A = 156 z = 3
x = 10 11 12 13 14 15 16 17 18 19 20 21
xz = 30 33 36 39 42 45 48 51 54 57 60 63
y = 10 10 9 8 8 7 7 6 6 6 5 5
SBS = 153,384 152,712 152,531 152,356 152,168 152,645 152,880 153,014 154,123 154,526 154,224 155,904
x*BS = 14
10.48c Using a continuous approximation, with Q = 300 and s = 30
10.48d Using enumeration for Q =30 and s = 5 gives x*BSSS = 5, as shown below.
10.48e Q = 60, s = 0, x1 = 8, x2 = 4. As shown, y1 = 2, y2 = 1, and SBS = 276.73 sq. ft.
10.49a Q = 200, L = 5′, W = 4′, A = 10′, z = 5, c = not stated. (Note: it is not necessary to know
Answers to Problems at the End of Chapter 10 10-57
10.49b
1 2 3 4 5 6 7 8 9 10
200 0.10 40 20 14 10 8 7 6 5 5 4
190 0.02 38 19 13 10 8 7 6 5 5 4
180 0.02 36 18 12 9 8 6 6 5 4 4
170 0.02 34 17 12 9 7 6 5 5 4 4
160 0.02 32 16 11 8 7 6 5 4 4 4
150 0.02 30 15 10 8 6 5 5 4 4 3
140 0.02 28 14 10 7 6 5 4 4 4 3
130 0.04 26 13 9 7 6 5 4 4 3 3
120 0.04 24 12 8 6 5 4 4 3 3 3
110 0.04 22 11 8 6 5 4 4 3 3 3
100 0.04 20 10 7 5 4 4 3 3 3 2
90 0.04 18 9 6 5 4 3 3 3 2 2
80 0.06 16 8 6 4 4 3 3 2 2 2
70 0.06 14 7 5 4 3 3 2 2 2 2
60 0.06 12 6 4 3 3 2 2 2 2 2
50 0.08 10 5 4 3 2 2 2 2 2 1
40 0.08 8 4 3 2 2 2 2 1 1 1
30 0.08 6 3 2 2 2 1 1 1 1 1
20 0.08 4 2 2 1 1 1 1 1 1 1
10 0.08 2 1 1 1 1 1 1 1 1 1
Number of Storage Rows Required to Accommodate the Inventory Level
Row Depth (x)
Inventory
Level
Prob.
Answers to Problems at the End of Chapter 10 10-58
10.50d Q = 30, L = 60″, W = 36″, c = 24″, A = 156″, z = 3, s = 0.
10.51a Q = 30, L = 3′, W = 4′, c = 1′, A = 8′, z = 3
10.51c The optimum storage depth is 5, with an average floor space requirement of 158.3 sq. ft.
x = 10 11 12 13 14 15 16 17 18 19 20 21
xz = 30 33 36 39 42 45 48 51 54 57 60 63
y = 10 10 9 8 8 7 7 6 6 6 5 5
SBS = 1,204.5 1,196.9 1,193.4 1,190.3 1,187.3 1,189.7 1,190.3 1,190.3 1,197.9 1,200.2 1,197.0 1,209.3
x*BS = 14
x = 12 13 14 15 16 17 18 19 20 21 22 23 24
xz = 36 39 42 45 48 51 54 57 60 63 66 69 72
y = 9 8 8 7 7 6 6 6 5 5 5 5 5
SBS = 1,275.0 1,269.5 1,266.0 1,264.1 1,263.4 1,263.8 1,265.0 1,267.0 1,269.5 1,272.7 1,276.2 1,280.2 1,284.5
x*BSSS = 16
x = 1 2 3 4 5 6 7 8 9 10 11 12 13
xz = 3 6 9 12 15 18 21 24 27 30 33 36 39
y = 10 5 4 3 2 2 2 2 2 1 1 1 1
SBS = 316.3 247.5 232.9 231.9 236.3 243.3 252.0 261.6 271.8 282.5 293.5 304.8 316.3
x*BSSS = 4
Inv % 1 2 3 4 5 6 7 8 9 10
30 44% 10 5 4 3 2 2 2 2 2 1
25 11% 9 5 3 3 2 2 2 2 1 1
20 11% 7 4 3 2 2 2 1 1 1 1
15 11% 5 3 2 2 1 1 1 1 1 1
10 11% 4 2 2 1 1 1 1 1 1 1
511% 2 1 1 1 1 1 1 1 1 1
η = 7.44 3.89 3.00 2.33 1.67 1.67 1.56 1.56 1.44 1.00
S = 260.6 194.4 195.0 186.7 158.3 183.3 194.4 217.8 223.9 170.0
x* = 5
Number of Storage Rows Required to Accommodate the Inventory Level
Row Depth (x)
10.51d With Q = 50, we assume the inventory pattern is 51 – k, for days 1, … 50. In other words,
10.52a Q = 300
xcDLSS = [(A + f)(Q)/2L]½ = [(96 + 12)(300)/2(42)]½ = 19.64