Answers to Problems at the End of Chapter 10 10-60
10.52b Q = 35, s = 0. Using Excel to solve by enumerating over x gives the following result.
10.53b Q = 30, L = 48″, W = 40″, c = 4″, r = 4″, f = 9″, A = 96″, z = 7.
10.53c Q = 30, L = 48″, W = 40″, c = 4″, r = 4″, f = 9″, A = 96″, z = 7.
10.54a
10.54b Q = 30, L = 60″, W = 38″, c = 3″, f = 12″, r = 4″, A = 72″, z = 8
Q = 35 L = 42 W = 48 c = 4
A = 96 f = 12 r = 6 z = 6
x = 1 2 3 4 5 6 7 8 9 10
ν = 35 18 12 9 7 6 5 5 4 4
SBS = 17,856 13,201 11,798 11,208 10,912 10,841 10,788 10,939 10,969 11,195
xDLSS = 7
Q = 30 L = 60 W = 38 c = 10 A = 156 z = 4
x = 1 2 3 4 5 6 7 8 9 10 11 12
xz = 4 8 12 16 20 24 28 32 36 40 44 48
y = 8 4 3 2 2 2 2 1 1 1 1 1
SBS = 196.3 158.4 154.8 155.5 168.0 175.2 177.1 186.0 206.0 226.0 246.0 266.0
x*BS = 3
x = 1 2 3 4 5 6 7 8 9 10
ν = 30 15 10 8 6 5 5 4 4 3
SDL = 65.9 54.0 50.9 50.1 49.9 50.3 51.3 52.2 53.4 53.5
xDL = 5
Answers to Problems at the End of Chapter 10 10-61
10.54d SDD = (W + 1.5c + 0.5r)[2L + 0.5(A + f)](Q + 2)/4z
10.54e STD = ν(W + 1.5c + 0.5r)[3L + 0.5(A + f)](2Q – 3ν + 3)/2Qz
10.55a Q = 250, L = 36″, W = 48″, c = 4″, r = 4″, f = 12″, A = 60″, z = 15. Computing the value
of SDL yields the following results. The lane depth of 5 is the worst of the three.
10.55b As shown below, 15 is the best lane depth of the three choices.
10.55d Recall, the deep lane storage equation for floor space required can be obtained directly
Answers to Problems at the End of Chapter 10 10-62
Answers to Problems at the End of Chapter 10 10-63
10.56a
Answers to Problems at the End of Chapter 10 10-64
10.56b Q1 = 50, L = 42″, W = 48″, c = 3″, f = 10″, r = 4″, A = 72″, z = 5
SDD = (Q/2)(W+1.5c+0.5r)[2L+0.5(A+f)][2(Q+s)-Q+2]/2(Q+s)z
SDD = 359.55 sq. ft.
10.56c Refer to the solution to 10.55(d). For Q = 50, x1 = 10, and x2 = 5, using SOLVER we
Answers to Problems at the End of Chapter 10 10-65
For Q = 150, x1 = 10, and x2 = 5, using SOLVER we obtain y1 = 12, y2 = 6, and SDL =
1481.94/5 = 296.39 sq. ft.
Answers to Problems at the End of Chapter 10 10-66
For Q = 150, x1 = 15, and x2 = 5, using SOLVER we obtain y1 = 7, y2 = 9, and SDL =
1475.94/5 = 295.19 sq. ft.
Answers to Problems at the End of Chapter 10 10-67
For Q = 150, x1 = 15, and x2 = 10, using SOLVER we obtain y1 = 4, y2 = 9, and SDL =
1475.94/5 = 295.19 sq. ft.
Answers to Problems at the End of Chapter 10 10-68
= 10, y1 = 4, x2 = 5, and y2 = 2, 110.05 sq. ft. will be required, on average, to store 50 unit
10.57 Research Question. Very different results can be obtained from the Web search. A good
SECTION 10.6
10.58 We have L = 250 ft, H = 55 ft, th = 0.781 min, tv = 0.733, T = 0.781 min, and Q = 0.939.
10.59 We have th = 0.75 min, tv = 0.625 min, T = 0.75 min, Q = 0.833. Therefore, TSC = 1.523
Answers to Problems at the End of Chapter 10 10-69
Let x be the maximum percentage of single-command operations performed such that the
utilization of the S/R machine does not exceed 85%. To meet the required performance of
10.61a We have th = 0.73 min, tv = 0.875 min, T = 0.875 min, Q = 0.834. Therefore, TSC = 1.88
min per single-command cycle and TDC = 3.05 min per dual-command cycle. Since each
10.61b Assuming that 100% of the storages are performed with dual-command cycles, each S/R
10.62a By (10.59) and (10.60) L = 233.33 ft and H = 58 ft. Therefore, th = 0.78 min, tv = 0.97
Answers to Problems at the End of Chapter 10 10-70
10.62b Compute the cost of racks for v = 53.33 ft3, w= 3,000 lb, n = 12. Assuming α = $30, by
(10.65) we obtain CRO = $167.70. Thus, the total rack cost is equal to
167.70×50×12×2×8 = $1,609,954.30.
10.63 By (10.59) and (10.60) we can determine the length and height of the racks (in feet) for
the three choices:
Next, the single-command and dual-command cycle times (in minutes) for the three
choices can be computed as follows:
Forty percent of the operations will be single-command based. Thus, the system is
Choice L H
(a) 300 36
(b) 336 36
(c) 376 36
Choice
thtvT Q TSC TDC
(a) 0.667 0.45 0.667 0.675 1.368 2.234
(b) 0.747 0.45 0.747 0.602 1.437 2.326
(c) 0.836 0.45 0.836 0.538 1.517 2.43
Answers to Problems at the End of Chapter 10 10-71
10.64 By (10.59) and (10.60) we have L = 310 ft and H = 70 ft, for which th = 0.729 min,
satisfy the utilization constraint.
10.66 Sixty percent of the operations are performed on a dual-command basis. Therefore, the
system is required to perform a total of 60 dual-command cycles and 80 single-command
10.67 We need to minimize the expected single-command travel time, which is given by the
Choice Rack Cost Total Cost
(a) $1,706,016 $3,706,016
(b) $1,719,664 $3,519,664
$1,800,000
$2,000,000
S/R Machine Cost
10.68
Consider the rack formed by region AB: T = 0.405, Q = 0.926, and E(SCAB) = 0.5208
min. Consider next the rack formed by region A: T = 0.375, Q = 0.08, and
10.69a Let the random point be given by (x,y) where 0 ≤ x ≤ 1 and 0 ≤ y ≤ Q. The travel time to
10.69b Let the two random points be given by (x1,y1) and (x2,y2) where 0 ≤ xi ≤ 1, and
10.70a
Answers to Problems at the End of Chapter 10 10-73
10.70b Considering the entire rack and following the same procedure as in part (a) to compute
10.70c
For the fast movers, consider the “upper” part of the rack (i.e., the 125′×20‘ region above
the dashed line): th = 0.25, tv = 0.25, T = 0.25, Q = 1. Thus, E(SCUpper) =0.3333. Since
the “lower” part of the rack (the 125’x20′ region below the dashed line) has the same
Answers to Problems at the End of Chapter 10 10-74
10.71a
Consider first the rack formed by region DC (a 20 ft by 300 ft region): T = 0.666,
Q = 0.333, E(SCDC) = 0.691358 min.
10.71b For the entire rack we have: T = 0.888, Q = 1, E(SC) = 1.1851852 min. Solving the
10.72 Let the shaded area be region C. Consider first the entire rack: T = 1.25, Q = 0.30,
Answers to Problems at the End of Chapter 10 10-75
Considering the rack formed by the shaded area, we have 166.666/400 = 0.4167 min for
10.73 To show that the system will meet the required throughput we need to first compare λr
and λs. Since λr > λs we simply need to satisfy (10.98). Using (10.98) we have
non-negativity requirement of qs and qr. The utilization is 0.912, or 91.2%.
10.74a We have th = 0.8889, tv = 0.8889, T = 0.8889, Q = 1. By (10.75)
10.75b Similar to part (a), average time per trip is 6.2627 min, and the throughput capacity is
76.64 picks/hour.
10.77a S = 7000(18) = 126,000 sq. ft.; R = 215 picks/hr; n = 9 stops/trip; p = 1.5 min/stop; K =
1.20 min; hv = 400 fpm; vv = 100 fpm; t = 1.63; q = 1.255; and u = 0.31614.
10.77c T = 0.4743, Q = 1. By (10.75) we have E(P,X1, …,X8,D) = 1.3031 min. Hence, the
10.78a-c Assuming a square-in-time rack (Q = 1), the results are summarized as follows:
Rack surface area for one side of an aisle is equal to 300,000 divided by (2 times the
10.78d Although seeking a “trend” with three data points is highly undesirable, we might say
10.79 Using a computer to generate results for M = 1 to 50, E(D) appears to be convex with
Part
Surface area
on one side
of an aisle
(sq. ft.)
T (Min)
Rack
Length
(ft)
Rack
Height
(ft)
E(…) by
(10.75)
(min)
Travel Time
per trip
(min)
Throughput
capacity
(picks/hr)
a 50,000 1.118 447.2 118.8 3.28 13.68 131.58
b 37,500 0.9682 387.28 96.82 2.84 13.24 181.26
c 30,000 0.866 346.4 86.6 2.54 12.94 231.82
Answers to Problems at the End of Chapter 10 10-78
10.80 Let E(D’) be the expected distance traveled per trip under the traversal policy with the
units away. Since E(D1‘) equals twice the distance to the farthest pick, we obtain:
We use ALGORITHM 2 (pg. 640). In STEP 4 of the first iteration we have
10.81c Expected system utilization (= 215/235.488) = 91.3%. Thus, the effective picker
M
Warehouse
Width
Aisle
Length
E(D1) E(D2) E(D)
24 240 60.00 406.15 605.90 1,012.06
25 250 57.60 423.08 586.50 1,009.57
26 260 55.38 440.00 568.27 1,008.27
27 270 53.33 456.92 551.13 1,008.05
28 280 51.43 473.85 534.97 1,008.81
29 290 49.66 490.77 519.72 1,010.49
30 300 48.00 507.69 505.30 1,012.99
Answers to Problems at the End of Chapter 10 10-79
10.83 In order to obtain a 100% picker utilization, i.e. E(PU) = 1.0, we need p ≥ t2 = k2 + C =
r2 + C = (q/ )2 + C.
10.84 First, we establish that we need to improve E(PU). Next, we need to identify the current
bottleneck; there are two possible bottlenecks: 1. The picker, or 2. The S/R machine. In
SECTION 10.8
Note: even though r/p is not a proper fraction, the solution procedure presented in the
10.86a k = 32; p = 3, r = 32mod3 = 2
{f1(n)} = (-5, 0, 0), {f2(n)} = (0, -3, 0), {f3(n)} = (0, 0, 4), and