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ZOO505 Assignment 2 Solution and Discussion

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Fall 2019 – ZOO505- Cell and Molecular Biology

Assignment No. 2 Marks: 10
Fall 2019
Due Date: 14-01-2020
Instructions
A. You are suggested to read the recommended review article and write a brief summary (One Page only) in your own words.
o Read the entire article thoroughly and highlight or underline the major points.
o Write down any significant facts or interesting details in your own words.
o Do not simply “rephrase” exact statements by swapping a few words around. Instead, completely rewrite the information.
o Cheating/copying from any source will result in “F” grade in the course.
o The title of the summary should be in bold Times New Roman with font size 14.
o The text of the summary should be in Time New Roman with font size 12 and 1.5 line spacing.
o No assignment will be accepted after the due date.
o If you have any question please contact: [email protected]
Review Article:
“Eukaryotic Chromosome DNA Replication: Where, When, and How?”
Review article is available in Download section of LMS. Download the pdf file titled with the name of article as mentioned above, for writing summary.
Note: Students should read this article and prepare one-page summary in MS word file and upload on your LMS before deadline 14-01-2020.
Outcomes:
• This assignment will help to enhance reading skills and enable you to comprehend and analyze the scientific texts.
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ECO404 Assignment Idea Sol 2019
Case 1:
Competition is a subject always open to debate in health care, and pharmaceuticals are no exception. Price competition in the pharmaceuticals supplies industry is growing rapidly in the increasingly price-conscious new millennium. Ferozsons is one of the leading companies in Pakistan maintaining exclusive agreements with a number of international partners for distribution, selling and co-manufacturing of surgical products. During the past year, Ferozsons sold 7 million (700, 000, 0) units of surgical products at a price of $13.50, for total revenues of $85 million. During the current year, sale of units of surgical products have fallen from 7 million (700, 000, 0) units to 4.6 million (460, 000, 0) units following a competitor price cut from $12.95 to $10.50 per unit.
Requirement:
a. Calculate the arc cross price elasticity of demand for surgical products of Ferozesons.
b. Marketing Director of surgical system in Ferozsons, projects that unit sale will recover from 4.6 million (460, 000, 0) units to 5.8 million (580, 000, 0) units if Surgical Systems reduces its own price from $13.50 to $12.50 per unit. Calculate Surgical Systems’ implied arc price elasticity of demand. (Marks: 4+4)
Case 2:
PEL prides itself on innovative home appliances geared towards the modern consumer. With the start of winter season PEL has announced 2% discount off the average price of refrigerators sold during the month of October in an effort to reduce excess end-of-themodel-year inventory. Customer response was wildly enthusiastic, with unit sales rising by 20% over the previous month’s level.
Requirement:
a. Calculate the point price elasticity of demand for PEL refrigerators sold during the month of August.
b. Calculate the profit-maximizing price per unit if PEL has Rs.30, 600 marginal selling costs. (Marks: 3+4)

Answer (A)
Arc Price elasticity of demand (responsiveness of demand) formula is
% ∆Q/%∆Price = ((Q2-Q1/average quantity)*100)/(p2-p1/average price)*100
Competitive price cut Surgical products of Ferozson,s quantity demand
$ 12.95 70,00,000
$10.50 46,00,000

Arc elasticity of demand = (4600000-7000000/5800000)*100 = -2400000/5800000 = -41.37
(10.50-12.95/11.72)*100 = -2.45/11.72 = -20.90
Arc elasticity of demand = -41.37/ -20.90
= 1.97
Note: If price of surgical products of competitive company decreases by 1% then quantity demand of surgical products of Ferozsons will be decrease by 1.97%

(B)
Price changes in ferozsons quantity demand changes
P1 13.50 Q1 4600,000
P2 12.50 Q2 5800,000

Arc price elasticity= ( 5800,000-4600,000/5200000)*100 = 23.07%
= (12.50-13.50/13) * 100 = -7.69%
Arc price elasticity= 23.07/-7.69
= -3 absolute value is 3
Note: If ferozsons decreases price by 1 % , quantity demand increases by 3 % . If ferozsons decreases price by 10% ,quantity demand will increase by 30%.

Case : 2
(A)
Pel refrigerators:
%∆Q = 20%
%∆P= -2%
Point Price elasticity of demand = ((Q2-Q1/Q1)*100)/(p2-p1/P1)*100
= .20/-.02
Ped = -10 absolute value is 10

(B)
έ= elasticity
Marginal cost = 30600
έ = -10

Profit maximizing price = Mc (έ/ έ+1) = Mc/ (1+1/ έ)
= 30600 (-10/-10+1
=306001.111111
P = 34000
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Solution (A)
The formula to calculate the execution time : Execution Time = IC  CPI  T
IC for Load / Store Instructions
IC for ALU instructions
IC for Control instructions
= Total Instructions  Ratio of Load / Store Instructions = 200  0.35
= 70 instructions
= Total Instructions  Ratio of ALU Instructions = 200  0.55
= 110 instructions
= Total Instructions  Ratio of Control Instructions = 200  0.10
= 20 instructions
Now, we will calculate the total clock cycles required to execute each type of instructions
Total Clock Cycles for Load / Store
Total Clock Cycles for ALU
Total Clock Cycles for Control
= IC for Load / Store  CPI for Load / Store = 70  2.5
= 175 clock cycles
= IC for ALU  CPI for ALU = 110  1.25
= 137.5 clock cycles
= IC for control  CPI for control = 20  3
= 60 clock cycles
Time required (in seconds) for each clock cycle (T)=1/CPU frequency 1 / 500  106 = 0.002  10−6 seconds
= 210−9 seconds
Now finally, we will calculate the execution time
Execution Time (ET ) = Total Clock Cycles  1/ CPU Frequency
= (175 + 137.5 + 60)  (1/ 500  106 ) seconds
= 372.5210−9seconds 1/500106 =210−9seconds = 745  10−9 seconds
= 745 nanoseconds

Solution (B)
If decrease the average CPI for ALU by 20%, the new average CPI
New CPI for ALU = 1.25  (100−20)/100 = 1.25  0.8
= 1 CPI
If average CPI for Load / Store instruction is increased by 10%, new average CPI New CPI for Load / Store = 2.5  (100 +10)/100
= 2.5  1.1 = 2.75 CPI
Hence, new execution time will be
ExecutionTime(E.T) = (702.75+1101+203)x(1/500106)seconds
= (192.5+110+60)/(5108)seconds
= 362.5  2  10−9 seconds = 725  10−9 seconds
= 725 nanoseconds

Q. 2 Solution:

Solution A (0-Address Code)
PUSH B
PUSH C
ADD ; gives B+C PUSH A
MUL ; gives A(B+C) PUSH 2
PUSH A
MUL ; gives 2A PUSH C
MUL ; gives 2AC
PUSH B
DIV ; gives 2AC/B
SUB ; gives A(B+C) - 2AC/B
PUSH C
PUSH C
MUL ; gives C2
ADD ; gives POP D
A(B+C) - 2AC/B + C2

Solution A (1-Address Code)
LDA C MULA C STA X
LDA A MULA C MULA 2
; loads the value stored at memory location C in Accumulator ; gives C2
; stores C2 at memory location X
; loads the value stored at memory location A in Accumulator ; gives AC
; gives 2AC
DIVA B ADDA X STA Y
LDA B ADDA C MULA A SUB Y STA D
; gives 2AC/B
; adding 2AC/B with C2 stored in X gives 2AC/B + C2 ; stores 2AC/B + C2 at memory location Y
; loads the value stored at memory location B in Accumulator ; gives (B+C)
; gives A(B+C)
; subtracts 2AC/B + C2 from A(B+C)
; stores the result at memory location D

Fall 2019_CS501_1_SOL.pdf
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Assignment No: 01
Question #1: Find the root of the equation, Perform three iteration of the equation,
ln (x −1) + sinx =0 by using Newton Raphson method.

Ans: Let f(x) = ln(x+1) + sinx = 0 and f(x) = 1/(x-1) + cosx

F (1.5) = ln(0.5) + (1.5) = - 0.0667

F(2) = ln(1) + sin(2) = 0.035

Since f (1.5) f (2) < 0 so roots lies in interval [1.5, 2]

Let x0 = 1.75 . x0 can be taken in the interval any real number [ 1.5 , 2 ], we let mid point

of this interval .
As we know Newton Raphson method is

Xn+1 = xn – f ( xn ) / f(xn)
First iteration
X1 = x0 –f(x0) / f(x0) = 1.75 - f(1.75) / f(1.75)
= 1.75 – (-0.2571 / 2.3329) = 1.8602
Second iteration:
X2 = x1 - f(x) / f(x) = 1.8602 –[ f(1.8602) / f(1.8602)]
= 1.8602 - ( -0.1181 / 2.1620 ) = 1.9148
Third iteration:
X3 = x2- f(x2) / f(x2) = 1.9148 –f(1.9148) / f(1.9148)
= 1.9148 – [-0.0556/2.0926]
= 1.9414
Question #2: Solve the system of linear equations with the help of Gaussian elimination method.
x + y + z = 6;2x − y + z = 3;x + z = 4

ANS: In Gaussian elimination method we convert the augmented matrix into reduce

Echelon form therefore,

Augmented matrix is

R2- 2R1 , R3 – R1

-1R2 , -1R3
R23
R3-3R2

X + Y+ Z = 6 ;………………….(1)
Y = 2,
Z = 3
Put into eq (1),
we get X = 1 ,
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ZOO505 Assignment 2 Solution and Discussion

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