免費幫你chegg解鎖
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In the given Scenario there are two main problems
1) Thomas cannot enter into a contract after certain a limit for a single transaction amount.
2) Facilitation Fee is illegal cannot be shown as facilitation fee
Here Thomas has to procure more materials from suppliers as the demand is increasing and also he has to procure them for below market prices.One thing he has searched for options and found out few alternatives and came to a conclusion facilitation fee is lower compared to alternate suppliers .So he can go with alternate supplier by paying consulting fee or he can treat facilitation fee as miscellaneous expenses which may be found out eventually through audit but for time it will do the job .
Alternatively he can convey it to the Management of all the possible option and do accordingly.
1) Thomas cannot enter into a contract after certain a limit for a single transaction amount.
2) Facilitation Fee is illegal cannot be shown as facilitation fee
Here Thomas has to procure more materials from suppliers as the demand is increasing and also he has to procure them for below market prices.One thing he has searched for options and found out few alternatives and came to a conclusion facilitation fee is lower compared to alternate suppliers .So he can go with alternate supplier by paying consulting fee or he can treat facilitation fee as miscellaneous expenses which may be found out eventually through audit but for time it will do the job .
Alternatively he can convey it to the Management of all the possible option and do accordingly.
Schema:
Course (c-name, room, instructor)
Enrollment (student-name, c-name, grade)
Ans
(a) Database file store in the disk in one of the following forms
1.Sequential File Organization or Ordered Index File ( Index SequentialAccess method )
2.B+tree file,
3. Hash file,
4. Flat File structure
The most common methods are the B+ tree file structure and Index SequentialAccess method (ISAM). Hash file structure is most efficient among them. Hash file generally takes O(1) to find an item. I suggest you use Sequential File Organization or Ordered Index File for the given schema. Because it will search efficiently between two given relations. Course name and student names attributes belongs to different tables(relation). To search run this type of query you have to link between two tables using the primary key and foreign key. Sequential File Organization or Ordered Index File structure use two types of indexing
1. Dense Index: One index for one file.
2.Sparse Index: One index for more than one file.
(b) There are three main three types of indexing
1.Cluster Indexing
2. Secondary Indexing (Non -Cluster )
3.Multilevel.
Most efficient (faster) is cluster indexing. It will decrease retrieval cost. because of mone than one record related to the same key. One thing you have to remember that it must be an ordered datafile. Here Instructors and students are in a different relation so before making a query it is necessary to join both relations
Secondary indexing is not faster as cluster indexing. Multilevel indexing used on a large amount of data.
Course (c-name, room, instructor)
Enrollment (student-name, c-name, grade)
Ans
(a) Database file store in the disk in one of the following forms
1.Sequential File Organization or Ordered Index File ( Index SequentialAccess method )
2.B+tree file,
3. Hash file,
4. Flat File structure
The most common methods are the B+ tree file structure and Index SequentialAccess method (ISAM). Hash file structure is most efficient among them. Hash file generally takes O(1) to find an item. I suggest you use Sequential File Organization or Ordered Index File for the given schema. Because it will search efficiently between two given relations. Course name and student names attributes belongs to different tables(relation). To search run this type of query you have to link between two tables using the primary key and foreign key. Sequential File Organization or Ordered Index File structure use two types of indexing
1. Dense Index: One index for one file.
2.Sparse Index: One index for more than one file.
(b) There are three main three types of indexing
1.Cluster Indexing
2. Secondary Indexing (Non -Cluster )
3.Multilevel.
Most efficient (faster) is cluster indexing. It will decrease retrieval cost. because of mone than one record related to the same key. One thing you have to remember that it must be an ordered datafile. Here Instructors and students are in a different relation so before making a query it is necessary to join both relations
Secondary indexing is not faster as cluster indexing. Multilevel indexing used on a large amount of data.
Rn = ? (bottom: i=1, top: n) f(xi)delta x]
= ? (bottom: i=1, top: n) [ (-(3i/n)^2 / 2 - 7) 3/n ]
= ? (bottom: i=1, top: n) [ (-(27i^2/(2n^3) - 21/n ]
= -27/(2n^3)? (bottom: i=1, top: n) [ i^2 ] - 21/n ? (bottom: i=1, top: n) [ 1 ]
=-27/(2n^3)*n(n+1)(2n+1)/6 - 21/n*n
= -9/(4n^2)*(2n^2+3n+1) - 21
The inf of this will be when n->infinity so
= -9/4 * 2 - 21
= -51/2
= -25.5
Which is what the integral should be if you double check.
View comments (1)
= ? (bottom: i=1, top: n) [ (-(3i/n)^2 / 2 - 7) 3/n ]
= ? (bottom: i=1, top: n) [ (-(27i^2/(2n^3) - 21/n ]
= -27/(2n^3)? (bottom: i=1, top: n) [ i^2 ] - 21/n ? (bottom: i=1, top: n) [ 1 ]
=-27/(2n^3)*n(n+1)(2n+1)/6 - 21/n*n
= -9/(4n^2)*(2n^2+3n+1) - 21
The inf of this will be when n->infinity so
= -9/4 * 2 - 21
= -51/2
= -25.5
Which is what the integral should be if you double check.
View comments (1)
秒回 仲有冇人
Answer
12 year treasury bond
IP12 = 2*2 + 3*4 + 4*6 / 2+4+6 = 3.333%
r12 = r + IP12 + MRP12
4+3.33+0.02*(12-1)
7.55
7 Year corporate A rated bond
IP7 = 2*2 + 3*4 + 4*1 / 2+4+1 = 2.857%
r7 = r + IP7 + MRP7+ DRP7+LP7
4+2.857+0.02*(7-1)+(0.98 – 0.30) + 0.30
7.96%
Graph :
Rates
Year Treasury A Corp Spread LP DRP
1 5.37 7.2 1.83 0.7 1.13
2 5.47 7.3 1.83 0.7 1.13
3 5.67 7.48 1.83 0.7 1.13
4 5.71 7.5 1.83 0.7 1.13
5 5.6 7.47 1.83 0.7 1.13
10 5.75 7.58 1.83 0.7 1.13
20 6.33 8.16 1.83 0.7 1.13
30 5.94 7.7 1.83 0.7 1.13
12 year treasury bond
IP12 = 2*2 + 3*4 + 4*6 / 2+4+6 = 3.333%
r12 = r + IP12 + MRP12
4+3.33+0.02*(12-1)
7.55
7 Year corporate A rated bond
IP7 = 2*2 + 3*4 + 4*1 / 2+4+1 = 2.857%
r7 = r + IP7 + MRP7+ DRP7+LP7
4+2.857+0.02*(7-1)+(0.98 – 0.30) + 0.30
7.96%
Graph :
Rates
Year Treasury A Corp Spread LP DRP
1 5.37 7.2 1.83 0.7 1.13
2 5.47 7.3 1.83 0.7 1.13
3 5.67 7.48 1.83 0.7 1.13
4 5.71 7.5 1.83 0.7 1.13
5 5.6 7.47 1.83 0.7 1.13
10 5.75 7.58 1.83 0.7 1.13
20 6.33 8.16 1.83 0.7 1.13
30 5.94 7.7 1.83 0.7 1.13
