Datos Unidades Valor Tensión Volt 12
Etapa 5 Recursos Humanos
10-1
Answers to Questions and Problems at the End of Chapter 10 Answers to Questions and Problems at the End of Chapter 10
10.1
10.1 Using the half-sum method: half of the sum of the weights equals 35 x
x--ccoooorrdd wwii 3wwii yy--ccoooorrdd wwii 3wwii
0 15 15 5 30 30
5 25 40>35 10 35 65>35
10 30 70 15 5 70
(x*x*,y*y*) = (5,10), which is the location of existing facility 2
f(10,10) = 15(|10-0|+|10-10|) + 20(|10-5|+|10-10|) + 5(|10-5|+|10-15|) + 30(|10-10|+|10-5|)
f(10,10) = 450
10-2 10.2
10.2 Using the half-sum method: half of the sum of the weights equals 15 x
Any point on the defined line segment is an optimum location.
10.3
10.3 Using the half-sum method: half of the sum of the weights equals 3 x
Any point in the defined square is an optimum location.
10.4
10.4 It is desired to determine the unweighted minimax location, i.e., it is important that the farthest house be as close as possible, regardless of the monetary value of the house.
a
The minimax locations are all the points on the line segment joining the points:
(x1*, y1*) =!(c1-c3, c1+c3+c5) =!(36,44) = (18,22) (x2*, y2*) =!(c2-c4, c2+c4-c5) =!(31,39) = (15.5,19.5)
10-3 10.5
10.5 a. Since the weights are uniformly distributed over the indicated regions, the optimum x-coordinate location will be the point at which no more than half the weight is to the left of the point and no more than half the weight is to the right of the point; likewise, the optimum y-coordinate is the point at which no more than half the weight is below the point and no more than half is above the point. Show below are plots of the cumulative weights along each axis.
10-4 10.5
10.5 b. Since the minimax location is to be determined with all points equally weighted, the rectangular regions can be represented by their corners.
Hence,
The minimax locations are all the points on the line segment joining the points:
(x1*, y1*) =!(c1-c3, c1+c3+c5) =!(16,16) = (8,8) (x2*, y2*) =!(c2-c4, c2+c4-c5) =!(20,12) = (10,6) Hence, the optimum location is inside region A4.
10.6
10.6 Since there are only 3 sites to consider, enumeration is the quickest means of solving the problem.
f(50,50) = 600(|50-20|+|50-25|) + 400(|50-36|+|50-18|) + 500(|50-62|+|50-37|) + 300(|50-50|+|50-56|) + 200(|50-25|+|50-0|) = 80,700 f(30,45) = 600(|30-20|+|45-25|) + 400(|30-36|+|45-18|) +
500(|30-62|+|45-37|) + 300(|30-50|+|45-56|) + 200(|30-25|+|40-0|) =70,50070,500 f(65,28) = 600(|65-20|+|28-25|) + 400(|65-36|+|28-18|) +
500(|65-62|+|28-37|) + 300(|65-50|+|28-56|) + 200(|65-25|+|28-0|) = 76,900 (x*x*,y*y*) = (30,45)
10.7
10.7 a. Using the half-sum method: half of the sum of the weights equals 1200 x
x--ccoooorrdd wwii 3wwii yy--ccoooorrdd wwii 3wwii
5 200 200 5 300 300
15 400 600 10 200 500
25 500 1100 15 400 900
30 600 1700>1200 20 400 1300>1200
35 300 2000 25 500 1800
50 400 2400 30 600 2400
(x*x*,y*y*) = (30,20)
10-5
b. Shown below is a partial contour line which passes through the location for Building 2. Since Building 1 is located inside the contour line and Building 3 is located outside the contour line, the least cost location is Building 1 at (20,20).
10-6 10.8
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10-18 10.20
10.20 a. Average distance item i travels between dock k and its storage region, Ri =3 dk j / Ai
j0 Ri
b. Average cost of trans porting item i between dock k and its storage region, Ri = wi k3 dk j / Ai
j0 Ri
c. Integer programming formulation
m p
Minimize 3 i = 1 k = 1 j3 wi k (30 d Rik j / Ai) subject to:
m
3 Ai = n i = 1 n
3 x i j = Ai ú i j = 1
m
3 x i j = 1 ú j i = 1
x i j = 0, 1
10-19 1
100..2211 PrroPodduucctt AArreeaa ##BBaayyss LLooaad d RRaattee TT j j / S / S j j
(j) (ft2) (S j) (T j )
A 4,375 7 500 500/7 = 71.43
B 7,500 12 600 600/12 = 50.00
C 1,500 3 700 700/3 = 233.33
D 6,250 10 200 200/10 = 20.00
Product Ranking: C > A > B > D
(multiple optimum layouts exist)
10-20 1
100..2222 PrroPodduucctt AArreeaa ##BBaayyss LLooaad d RRaattee TT j j / S / S j j
(j) (ft2) (S j) (T j )
1 2,400 6 600 100
2 3,200 8 400 50
3 2,000 5 800 160
4 2,800 7 400 57.14
5 4,000 10 400 40
6 1,600 4 800 200
Product Ranking: 6 > 3 > 1 > 4 > 2 > 5
(multiple optimum layouts exist)
10-21 1
100..2233 PrroPodduucctt ##BBaayyss LoLoaad d RRaattee TT j j / S / S j j
(j) (S j) (T j )
A 15 1 0.0667
B 5 1 0.20
C 16 1 0.0625
Product Ranking: B > A > C
10-22
(multiple optimum layouts exist) 10.24
10.24 Distances to/from storage bays are determined by summing the distance traveled from the receiving door (A) to t he 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 mid-point 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 storage bay is 54'; travel from the mid-point of the aisle to the mid-point of the storage bay is 14'. Hence, a total of 212' is required to storage a load in the storage bay. Travel from the storage bay to the shipping door requires 14' from bay to aisle, 202' along the aisle to the mid-point of the farthest one-way aisle, 52' down the aisle to the mid-point of the center aisle, and 4' to the shipping door for a total distance of 272'. The total distance traveled to/from the storage bay is 488', as shown. We do not compute the distance traveled by the lift truck after storage and before retrieval since we do not know its destination and srcin points, respectively. Once the distance is known for one storage bay, the other distances can be easily calculated.
10-23
The distances to/from storage bays located along the center aisle are determined as follows: the distance from from door A to door B is 244'; the distance from the mid-point of the center aisle to the mid-point of a storage bay is 18'. Hence, the total distance equals 244' + 18' + 18', or 280'.
P
Prroodduucctt AArreeaa ##BBaayyss LoLoaad d RRaattee TT j j / S / S j j (j) (ft2) (S j) (T j )
X 6,400 16 300 300/16 = 18.75
Y 8,800 22 400 400/22 = 18.18
Z 2,400 6 500 500/6 = 83.33
Product Ranking: Z > X > Y
10-24 10.25
10.25 Create Distance Matrix:
10-25 10.26
10.26 a. Shown below is an Excel spreadsheet solution of the problem. To compute the number of storage rows required to accommodate the designated inventory level the ROUNDUP Function is used, e.g., if the inventory level is shown in cell D6 and the row depth is given in cell F5, then if the product is stacked 4 levels high, the following entry would be entered: =ROUNDUP(D6/(4*F5),0).
10-26 10.26
10.26 (continued)
To determine the average number of storage rows required over the life of a storage lot (0), the value obtained using the SUM function over the range of storage rows required is divided by the number of entries, e.g.,
=SUM(F6:F65)/60.
Likewise, to determine the average amount of floor space required during the life of the storage lot (S), the following was used:
=F69*(K1+M1)*(F5*I1+0.5*O1), where F69 is the value of0, K1 is the value of W, M1 is the value of c, F5 is the value of x, I1 is the value of L, and O1 is the value of A.
Alternately, the following equation can be used to compute the value of SBS: SBS = y(W + c)(xL + 0.5A)[2Q - xyz + xz]/2Q, where y is obtained
10-27
using the ROUNDUP function. Shown below is an Excel solution to the problem. In enumerating over x, SBS is minimized with x equal to 5.
b. Using a continuous approximation,
xcBS = [AQ/2Lz]!= [144(60)/2(48)(4)]! = 4.74
Therefore, rounding off the value to the nearest integer gives xcBS = 5.
10.27
10.27 L = 42", W = 48", c = 4", r = 6", f = 12", A = 96", z = 7 a. Q = 300
xcDLSS = [(A + f)(Q)/2L]!= [(96 + 12)(300)/2(42)]!= 19.64 xcDLSS = 20
Using Excel to solve by enumerating over x gives the same result, as shown below.
b. Q = 35, s = 0
Using Excel to solve by enumerating over x gives the following result.
10-28 10.28
10.28 L = 40", W = 48", c = 8", A = 156", z = 5 a. Q = 300
xcBS = [AQ/2Lz]!= [156(300)/2(40)(5)]! = 10.82
Therefore, rounding off the value to the nearest integer gives xcBS = 11.
Using Excel to solve by enumeration gives x*BS = 10, as shown below.
Excel’s SUMPRODUCT function was used to calculate the value of0; the values in the percentage column were multiplied by the number of storage rows required to accommodate the inventory level; the sum of the products is displayed in the0 row. Knowing the value of0, S is easily determined by multiplying by the size of the foot-print of a row, (W + c)(xL + 0.5A).
Developing a spreadsheet allows analyses of changes in inventory profiles over lifetimes of production lots to be performed easily.
10.29
10.29 Q = 30, L = 48", W = 40", c = 4", r = 4", f = 12", A = 96", z = 7 a. SSD = (W + 1.5c + 0.5r)[L + 0.5(A + f)](Q + 1)/2z
SSD = (40 + 6 + 2)[48 + 0.5(96 + 12)](31)/2(7) SSD = 10,841.14 sq. in.
b. SDD = (W + 1.5c + 0.5r)[2L + 0.5(A + f)](Q + 2)/4z SDD = (40 + 6 + 2)[96 + 0.5(96 + 12)](32)/4(7) SDD = 8,228.57 sq. in.
c. STD =<(W + 1.5c + 0.5r)[3L + 0.5(A + f)](2Q - 3< + 3)/2Qz where< =j Q/3 k =j 30/3 k = 10
(note:j xx k denotes the largest integer# xx).
STD = 10(40 + 6 + 2)[144 + 0.5(96 + 12)](60 - 30 + 3)/60(7) STD = 7,467.43 sq. in.
10-29 10.30
10.30 a. Q = 30, L = 48", W = 50", c = 10" A = 156", z = 3
b. Using a continuous approximation, with Q = 300, x
c
BS = [AQ/2Lz]
!
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!
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Therefore, rounding off the value to the nearest integer gives xcBS = 13.
Enumerating over values of x, as shown below the optimum value is actually 14.
c. Using a continuous approximation, with Q = 300 and s = 30 xcBSSS = [A(Q + 2s)/2Lz]! = [156(300 + 60)/2(48)(3)]! = 13.96 Therefore, rounding off the value to the nearest integer gives xcBSSS = 14.
Enumerating over values of x, as shown below the optimum value is actually 14.
d. Using enumeration for Q =30 and s = 5 gives x*BSSS = 5, as shown below.
10-30 10.31
10.31 a. Q = 30, L = 36", W = 60", c = 12", A = 156", z = 3
b. Q = 300, L = 36", W = 60", c = 12", A = 156", z = 3
c. Q = 300, L = 36", W = 60", c = 12", A = 156", z = 3 , s = 30
d. Q = 30, L = 60", W = 36", c = 24", A = 156", z = 3, s = 0
The best configuration is 36" x 60" x 48".
10-31 10.32
10.32 Q = 30, L = 3', W = 4', c = 1', A = 8', z = 3 The problem solution is shown below.
a. For x = 2, the average amount of floor space required in the freezer is 194.4 sq. ft.
b. For x = 3, the average amount of floor space required in the freezer is 195.0 sq. ft.
(Notice that the optimum storage depth is 5, with an average floor space requirement of 158.3 sq. ft.)
10.33
10.33 a. Q = 30, L = 60", W = 38", c = 10", A = 156", z = 4
b. Q = 30, L = 60", W = 38", c = 3", f = 12", r = 4 ", A = 72", z = 8 xcDL = [(A + f)(Q)/2L]!= [(72 + 12)(30)/2(60)]!= 4.58 xcDL = 5 (Enumeration yields the same result, as shown below.)
10-32 c. Q = 30, L = 60", W = 38", c = 5", f = 12", r = 5 ", A = 96", z = 6
SSD = (W + 1.5c + 0.5r)[L + 0.5(A + f)](Q + 1)/2z SSD = (38 + 7.5 + 2.5)[60 + 0.5(96 + 12)](31)/2(6)(144) SSD = 98.17 sq. ft.
d. SDD = (W + 1.5c + 0.5r)[2L + 0.5(A + f)](Q + 2)/4z SDD = (38 + 7.5 + 2.5)[120 + 0.5(96 + 12)](32)/4(6)(144) SDD = 77.33 sq. ft.
e. STD =<(W + 1.5c + 0.5r)[3L + 0.5(A + f)](2Q - 3< + 3)/2Qz where< =j Q/3 k =j 30/3 k = 10
STD = 10(38 + 7.5 + 2.5)[180 + 0.5(96 + 12)](60 - 30 + 3)/60(6)(144) STD = 71.50 sq. ft.
10.34
10.34 a. 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.
b. As shown below, 15 is the best lane depth of the three choices.
c. Q = 250, s = 50, L = 36", W = 48", c = 4", f = 12", r = 4 ", A = 96", z = 6 SDD = (Q/2)(W+1.5c+0.5r)[2L+0.5(A+f)][2(Q+s)-Q+2]/2(Q+s)z SDD = 125(48 + 6 + 2)[72 + 0.5(96 + 12)](352)/2(300)(6)(144) SDD = 598.89 sq. ft.
10-33 10.35
10.35 a. Q = 200, L = 5', W = 4', A = 10', z = 5, c = not stated
(Note: it is not necessary to know the values of W and c in order to determine the optimum row depth.)
Using a continuous approximation, with Q = 200, xcBS = [AQ/2Lz]!= [10(200)/2(5)(5)]! = 6.32
Therefore, rounding off the value to the nearest integer gives xcBS = 6.
Enumerating over values of x, demonstrates this is not a convex function, as shown below. There are two local optima at x = 6 and x = 8. If only values of x = 5, 6, and 7 were considered, the alternate optimum at x = 8 would be missed.
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