S, (3) C in U is a foreign key to T, (4) these relations do not contain size (i.e., the number of rows) of each relation is as follows: TR = 4000, rs = 3000, rT = 2000, rU = 1000. ssume that all attributes of the relations are of the same length and we joining X e {R, S, T,U} and Y E {R, S,T,U} can be expressed as: k(ry ·cx +rY CY)
S, (3) C in U is a foreign key to T, (4) these relations do not contain size (i.e., the number of rows) of each relation is as follows: TR = 4000, rs = 3000, rT = 2000, rU = 1000. ssume that all attributes of the relations are of the same length and we joining X e {R, S, T,U} and Y E {R, S,T,U} can be expressed as: k(ry ·cx +rY CY)
Computer Networking: A Top-Down Approach (7th Edition)
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![Problem 6. (10 points) Consider four relations R(A, Z), S(B, A), T(C, B) and U(D, C) (the
primary keys are underlined). Assume that (1) A in S is a foreign key to R, (2) B in T is a foreign
key to S, (3) C in U is a foreign key to T, (4) these relations do not contain any null value, and
(5) the size (i.e., the number of rows) of each relation is as follows:
rR = 4000, rs = 3000, rT = 2000, ru = 1000.
Also, assume that all attributes of the relations are of the same length and we use hash join, so the
cost of joining X € {R, S, T,U} and Y E {R, S,T,U} can be expressed as:
k(rx cx +ry ·cy)
where k is a constant, rx and cx denote the number of rows and the number of columns of X,
respectively, and ry and cy denote the number of rows and the number of columns of Y (we ignore
the cost of producing the output relation).
Under the above assumptions, find the lowest cost plan for computing R É SAT MU using
dynamic programming and left-deep join trees. You need to complete the following table while
finding the best plans (e.g., in the form of ((O X D) M 0) M O in the last line) and associated costs.
Subquery
Size
Cost
BestPlan
RMS
3000
14000k = (4000 2+ 3000· 2)k]| RMS
RMT
RMU
SAT
SMU
TXU
RASMT
RMSMU
RMTWU
SATWU
RASATMU](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd6615adb-b827-41b9-af7e-ce5cda050521%2Fe5a4a24e-1ec3-45ae-af0f-97c76f6011fd%2F7ckm5j_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Problem 6. (10 points) Consider four relations R(A, Z), S(B, A), T(C, B) and U(D, C) (the
primary keys are underlined). Assume that (1) A in S is a foreign key to R, (2) B in T is a foreign
key to S, (3) C in U is a foreign key to T, (4) these relations do not contain any null value, and
(5) the size (i.e., the number of rows) of each relation is as follows:
rR = 4000, rs = 3000, rT = 2000, ru = 1000.
Also, assume that all attributes of the relations are of the same length and we use hash join, so the
cost of joining X € {R, S, T,U} and Y E {R, S,T,U} can be expressed as:
k(rx cx +ry ·cy)
where k is a constant, rx and cx denote the number of rows and the number of columns of X,
respectively, and ry and cy denote the number of rows and the number of columns of Y (we ignore
the cost of producing the output relation).
Under the above assumptions, find the lowest cost plan for computing R É SAT MU using
dynamic programming and left-deep join trees. You need to complete the following table while
finding the best plans (e.g., in the form of ((O X D) M 0) M O in the last line) and associated costs.
Subquery
Size
Cost
BestPlan
RMS
3000
14000k = (4000 2+ 3000· 2)k]| RMS
RMT
RMU
SAT
SMU
TXU
RASMT
RMSMU
RMTWU
SATWU
RASATMU
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You haven't given me the solution for the last row. Attached is image where the last row is circled that needs to have a solution.
![Assume that we have four relations R(A, B), S(A,C), T(A,D) and
U(A, E) and the size (i.e., the number of rows) of each relation is as follows:
3000, TT= 4000, ru 2000.
TR = 1000, rs =
Also, assume that all attributes of the relations are of the same length and we use hash join, so the
cost of joining X € {R, S, T, U} and Y € {R, S, T, U} is approximately:
k(rx cx+ry CY)
where k is a constant, rx and cx denote the number of rows and the number of columns of
X, respectively, and ry and cy denote the number of rows and the number of columns of Y.
Finally, assume that the size of a join is always 0.2% of the size of the cross product (i.e., rx=y=
0.002.TX.TY).
Under the above assumptions, find the lowest cost plan for computing R▷ S▷ T▷ U using
dynamic programming and left-deep join trees. You need to complete the following table while
finding the best plans (e.g., in the form of ((O)) in the last line) and associated costs.
Subquery
RAS
RAT
RAU
SAT
SAU
TAU
RAS T
RASAU
RATAU
SATMI
RAS TAU
Size
6000 8000k
Cost
Best Plan
[= (1000-2+ 3000-2)k] RS](https://content.bartleby.com/qna-images/question/e7ddc10c-4670-40fd-b02c-6a60c5fcc2f2/4364b6af-273b-4719-b169-9b975b1c4fb3/1etfbel_processed.png)
Transcribed Image Text:Assume that we have four relations R(A, B), S(A,C), T(A,D) and
U(A, E) and the size (i.e., the number of rows) of each relation is as follows:
3000, TT= 4000, ru 2000.
TR = 1000, rs =
Also, assume that all attributes of the relations are of the same length and we use hash join, so the
cost of joining X € {R, S, T, U} and Y € {R, S, T, U} is approximately:
k(rx cx+ry CY)
where k is a constant, rx and cx denote the number of rows and the number of columns of
X, respectively, and ry and cy denote the number of rows and the number of columns of Y.
Finally, assume that the size of a join is always 0.2% of the size of the cross product (i.e., rx=y=
0.002.TX.TY).
Under the above assumptions, find the lowest cost plan for computing R▷ S▷ T▷ U using
dynamic programming and left-deep join trees. You need to complete the following table while
finding the best plans (e.g., in the form of ((O)) in the last line) and associated costs.
Subquery
RAS
RAT
RAU
SAT
SAU
TAU
RAS T
RASAU
RATAU
SATMI
RAS TAU
Size
6000 8000k
Cost
Best Plan
[= (1000-2+ 3000-2)k] RS
Solution
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