Answers to Problems at the End of Chapter 10 10-80
As before, {F1(n)} = (-1, -3, 4).
H*1(1) = 0 ∴ H*1(2) = H*1(1) + F1(2) = 0 – 3 = -3 and
H*1(3) = H*1(2) + F1(3) = -3 + 4 = 1. Hence, {H*1(n)} = (0, -3, 1).
10.86c 1) {f1(n)} = (0, 0, -4), {f2(n)} = (-4, 0, 0), {f3(n)} = (0, 0, 4), and
{f4(n)} = (4, 0, 0). {F1(n)} = (0, 0, 0) ∴ {H*1(n)} = (0, 0, 0).
2) {f1(n)} = (-4, 0, 0), {f2(n)} = (0, -4, 0), {f3(n)} = (0, 0, 4), and
{f4(n)} = (4, 0, 0). {F1(n)} = (0, -4, 4) ∴
10.87a k = 33, p = 4, r = 33mod4 = 1. Since r ≠ 0, the answer is yes.
10.87c {f1(n)} = (-2, 0, -2, 0), {f2(n)} = (0, –2, 0, -2), {f3(n)} = (0, 0, 4, 0), and
H3(2) = 2
10.87d B = 4 (Although not asked for, the inventory totals 4 + 8 + 12 + 8, or 32.)
10.87e k = 34, r = 34mod4 = 2. Since r/p is not a proper fraction, it is not clear if the general
k = 35. Hence, the inventory on the conveyor and the carrier capacity will be the
same for any feasible value of k.
10.87f k = 33, p = 4, r = 1
1) {f1(n)} = (-2, 0, -2, 0), {f2(n)} = (0, –2, 0, -2), {f3(n)} = (0, 0, 2, 2),
2) {f1(n)} = (-2, 0, -2, 0), {f2(n)} = (0, –2, 0, -2), {f3(n)} = (2, 0, 2, 0),
sequences.
10.88a k = 62, p = 3, r = 62mod3 = 2, {f1(n)} = (0, -4, 0), {f2(n)} = (3, 3, 3), and
{f3(n)} = (0, -5, 0) ∴ {F1(n)} = (3, –6, 3). H*1(1) = 0, H*1(3) = 3, H*1(2) = -3.
10.88b 1) {f1(n)} = (0, -4, 0), {f2(n)} = (3, 3, 3), {f3(n)} = (-5, 0, 0) ∴ {F1(n)} = (-2, -1, 3).
H*1(1) = 0, H*1(3) = 3, H*1(2) = 2.
2) {f1(n)} = (0, -4, 0), {f2(n)} = (3, 3, 3), {f3(n)} = (0, 0, -5) ∴ {F1(n)} = (3, -1, -2).
H*1(1) = 0, H*1(3) = -2, H*1(2) = -3.
10.88c To minimize inventory on the conveyor, use {f3(n)} = (0, -5, 0) for a value of 26.
10.89a k = 83, p = 3, r = 83mod3 = 2, {f1(n)} = (1, 0, 0), {f2(n)} = (0, 2, 0), {f3(n)} = (0, 0, 1),
{f4(n)} = (0, -4, 0). ∴ {F1(n)} = (1, -2, 1).
10.89c {f1(n)} = (1, 0, 0), {f2(n)} = (0, 2, 0), {f3(n)} = (0, 0, 1), and {f4(n)} = (0, -a, -b).
Therefore, {F1(n)} = (1, 2-a, 1-b).
10.89d {f1(n)} = (1, 0, 1), {f2(n)} = (0, 2, 0), {f3(n)} = (0, -4, 0).
10.90 Using the information and procedure given in Example 10.52, we have the following:
S1 = (27, 26, …, 11), S2 = (34, 33, …, 28), S3 = (51, 50, …, 35), S4 = (10, 9, …, 52), Δ2 = 5,
10.91 a. flat belt driven roller conveyor, TL = 300′, WBR = 33″, RC = 9″, and S = 200 fpm
Based on the WBR and RC combination, LF = 0.9.
b. roller supported belt conveyor, TL = 100′, WBR = 33″, RC = 9″, and S = 300 fpm
Based on the WBR and RC combination, LF = 0.62.
10.92a chain driven roller conveyor, TL = 150′, WBR = 39″, RC = 9″, S = 90 fpm, length of
pallet = 48″, clearance between pallets = 24″, and weight of loaded pallet = 1,200 lbs.
10.92b For this calculation, ipf denotes inches per foot and ipls denotes inches per load segment.
10.93a V-belt driven roller conveyor, TL = 250′, WBR = 30″, and RC = 7.5″ Using linear
interpolation with the data in the table,
10.93b From the weight and length combinations for the cartons shown on page 10-16, notice
0.5464 lbs/in. Further, notice that the weight per inch is not the same as the average
Answers to Problems at the End of Chapter 10 10-85
10.93c The “worst case” condition occurs when the conveyor is loaded with the shortest and
Length Weight
Comb. (in.) (lbs.) lbs/in*
112 5 0.27778
212 10 0.55556
Carton Combinations
10.94a Flat belt driven roller conveyor, WBR = 27″, RC = 6″, tote box length = 24″, tote box
10.94b # full load segments = ⌈(200 x 12)/(24″+6″)⌉ = 80
Exactly 80 load segments will be on the conveyor; the average weight of a load segment
is 30 lbs. Therefore, the average load on the conveyor is 2,400 lbs.
10.95a RC = 6″, WBR = 33″, carton weight = 35 lbs., carton length = 30″, spacing between
10.95b HP = [BV + LF(TL) + FF(L)](S)/14,000
10.95c HP = [BV + LF(TL) + FF(L)](S)/14,000
Answers to Problems at the End of Chapter 10 10-87
10.96a Roller supported belt conveyors, TL1 = 150′,TL2 = 100′, WBR = 27″, RC = 10″, S1 = 300
10.96b Number of full load segments on 1st conveyor = ⌈(150)(12)/(24+12)⌉ = 50
There is no partial segment on 1st conveyor
10.97a TL1 = TL2 = 100′., S1 = 150 fpm, S2 = 250 fpm, WBR = 33″, and RC = 6″
10.97b For the totes to be processed at the same rates by both conveyors, the clearance between
tote boxes on the 2nd conveyor, c2, must be such that S1/(20+c1) = S2/(20+c2) = 150/26 =
10.97c Maximum RC for load stability. Recall, at least 3 rollers should be under a tote box at all
10.98a Roller-supported belt conveyor: TL = 200′, WBR = 27″, RC = ,
c = 6″
A sequence of 11 consecutive cartons will be defined as a “train segment.” Based on the
load distribution, a train segment will consist of the following sequence of cartons: A, A,
B, B, C, C, C, D, D, D, E. Including the clearance between cartons, each train segment is
10.98b To determine the heaviest loading condition on the conveyor, it is necessary to determine
Answers to Problems at the End of Chapter 10 10-89
factor of 1.2).
10.99b f(BW,HP) = $300BW + $5,000HP
Here, we assume the HP value is for the total horsepower requirement, including safety
10.100a Dry sand; conveyor length = 75′; elevation = 24′; BW = 30″; S = 600 fpm.
10.101a Lumpy, abrasive steel trimmings; conveyor length = 50′; elevation = 20′; BW = 30″;
10.101b Smax = 100 + 8.333BW = 100 + 8.333(30) = 350 fpm < 400 fpm. Therefore, the speed
10.102a Unsized, washed gravel; horizontal distance = 232.55′; elevation = 59.33′; BW = 24″;
S = 400 fpm; density = 75 lb/ft3
10.103a ; rate = 100 tph; density = 60 lb/ft3
10.104a C ; elevation = 30′; density = 30 lb/ft3
10.105b Delivered capacity = 48.5(400/100) = 194 tph
@ 150 tph, ΔHP to convey material horiz = 1.4 + (43.185/50)(1.6 – 1.4) = 1.573 hp
@ 200 tph, ΔHP to convey material horiz = 1.8 + (43.185/50)(2.1 – 1.8) = 2.059 hp
Thus, @ 194 tph, ΔHP to convey material horiz = 1.573 +(44/50)(2.059 – 1.573) =
2.00 hp
10.105d
10.106 The From-To matrix can be determined from the problem statement. Note that stations 2,
3, and 9 are I/O stations. The From-To Matrix is as follows:
Empty travel time can be determined from figure 10.59b where σij is equal to the
rectilinear distance in grids from station i to station j multiplied by 5 seconds/grid.
Each loaded trip required 2(0.15) = 0.30 min for pick-up and deposit, so
the total loaded travel time from station i to station j is τij = σij + 0.30 min/trip. By
Delivered
Capacity
Answers to Problems at the End of Chapter 10 10-93
The AGV can meet throughput since 0.6065 < 1. The fraction of time the AGV travels
empty while busy is αe/ρ = 0.2199/0.6065 = .3626 (roughly one-third of the time).
SECTION 10.9
10.107 (M|M|10):(GD|∞|∞)
λ = 10/hr, μ = 2/hr, c = 10, ρ = 10/20 = 0.5
10.108 (M|M|1):(GD|N|∞)
λ = 30/hr, μ = 50/hr, c = 1 ρ = 30/50 = 0.6
10.109 (M|M|1):(GD|N|∞)
λ = 2/min, μ = 3.333/min, c = 1 ρ = 0.6
10.111 (M|G|c):(GD|c|∞)
below, verifying that 4 is the answer.
10.112 (M|G|1):(GD|∞|∞)
10.113 (M|M|1):(GD|∞|∞)
10.114a (M|M|2):(GD|∞|∞)
10.114b Lq = ρ(cρ)cP0/[c!(1 – ρ)2]
c
Pc
10.115 (M(b)|M|1):(GD|∞|∞)
10.116 (M(b)|M|1):(GD|∞|∞)
E(b) = 0.25(1) + 0.5(2) + 0.25(3) = 2.0, λ = 20/hr, μ = 60/hr, ρ = 20(2)/60 = 2/3
10.117a c = 1, μ = 12/hr
10.117b (D|M|1):(GD|∞|∞)
10.117c The arrival and service rates behave like an (M|M|1):(GD|3|3) queue with K = 3, λ =
5/hr, μ = 12/hr, and ρ = 5/12 = 0.41667. As shown below, L = 1.244.
n
λnμnPnnPn
10.118a (M|M|1):(GD|5|∞) variation
10.118b (M|M|1):(GD|5|∞) variation. If P0 = 1, what is the value of P1? Since P1 = 2P0, then P1
10.119a (M|M|2):(GD|8|8) λ = 0.1/hr, μ = 0.5/hr, c = 2, K = 8
10.119b (M|M|2):(GD|8|8) λ = 0.1/hr, μ = 0.5/hr, c = 2, K = 8. Lq = 0.39776
n
λnμnPnPn
nλnμnPnPnPn
# Waiting
(m)
mPn