[Guest]

A trader X was involved in the buying and selling of MCS shares overfive trading days. At the beginning of the first day, the MCS share waspriced at Rs 100, while at the end of the fifth day it was priced at Rsa. At the end of each day, the MCS share price either went up by Rs 10,or else, it came down by Rs 10. Trader X took buying and sellingdecisions at the end of each trading day. The beginning price of MCSshare on a given day was the same as the ending price of the previousday. X started with enough number of shares and amount of cash toexhaust 5 days of trading.

Also known is that -

Each day ifthe price went up, X sold 10 shares of MCS at the closing price. On theother hand, each day if the price went down, he bought 10 shares at theclosing price.

How much money he ended up (more or less wrt to the amount he started on Day 1) after 5th day of trading?

if, a =100

and

a = 90

[Guest]

How much money he ended up (more or less wrt to the amount he started on Day 1) after 5th day of trading?

if, a =100

Well, a can not be 100! Its 5 days of trading so at the end of 5 days, the final price can not be the same as the starting price as we are looking at odd number of changes.

Now, i can start thinking about a = 90 though

[Guest]

Lets say the trader started with X number of shares

Since in order for the end price to be 90, the stock must have risen twice and fallen thrice, the trader must have bought in all 30 shares and sold 20 shares. So he must end up with X + 10 shares

For the price variations, buying and selling of stocks, the trader sold the stocks twice at prices higher by 10 and bought the stocks thrice at prices lower by 10.

The net difference should then be Rs 700. (The two sales brings in additional 200, and three buys cost 900 extra; so 900 - 200 = 700)

So the trader should end up with 10 stocks more and Rs 700 less. Since the price of share at the end of 5th day is 90, his portfolio value is higher than what he started with because 10 shares @ 90 per share (900) is higher than the Rs 700 that he has less.

Net gain in the portfolio should then be Rs 200.

[Guest]

DeGee's solution is sure correct.

But I suppose OP seeks a shorter way for more trade days???

Then 

if number of trade days is n -->

If a=100 --> number of sells=number of buys (n is even)

and each sell+buy pair causes a 100 Bs gain.

so n/2 * 100 = net gain

If a=90 --> number of sells=number of buys -1 (n is odd)

and each sell+buy pair causes a 100 Bs gain.

so (n-1)/2 * 100 = net gain

[Guest]

A trader X was involved in the buying and selling of MCS shares overfive trading days. At the beginning of the first day, the MCS share waspriced at Rs 100, while at the end of the fifth day it was priced at Rsa. At the end of each day, the MCS share price either went up by Rs 10,or else, it came down by Rs 10. Trader X took buying and sellingdecisions at the end of each trading day. The beginning price of MCSshare on a given day was the same as the ending price of the previousday. X started with enough number of shares and amount of cash toexhaust 5 days of trading.

Also known is that -

Each day if the price went up, X sold 10 shares of MCS at the closing price. On theother hand, each day if the price went down, he bought 10 shares at theclosing price.

How much money he ended up (more or less wrt to the amount he started on Day 1) after 5th day of trading?

if, a =100

and

a = 90

correction :

A trader X was involved in the buying and selling of MCS shares over five trading days. At the beginning of the first day, the MCS share was priced at Rs 100, while at the end of the fifth day it was priced at Rs a. At the end of each day, the MCS share price either went up by Rs10,or else, it came down by Rs 10. Trader X took buying and selling decisions at the end of each trading day. The beginning price of MCS share on a given day was the same as the ending price of the previous day. X started with enough number of shares and amount of cash to exhaust 5 days of trading.

Also known is that -

Each day if the price went up, X sold 10 shares of MCS at the closing price. On the other hand, each day if the price went down, he bought 10 shares att he closing price.

How much money he ended up (more or less wrt to the amount he started on Day 1) after 5th day of trading?

if, a =110

and

a = 90

Edited August 25, 2009 by guru_bhai

[Guest]

1 confusion.........

For a = 110

there are 3 U (price ups) and 2 D (Price downs) . Net there are 5!/3!2! = 10 cases possible.

Accordingly he will have 10 shares less after 5 days of trading.

Now in each of these ten cases the net difference is Rs 1300 more than original amount.

Now sth i cant understand is why in all these 10 cases the ans is 1300.

Plz help...

[Guest]

1 confusion.........

For a = 110

there are 3 U (price ups) and 2 D (Price downs) . Net there are  5!/3!2! = 10 cases possible.

Accordingly he will have 10 shares less after 5 days of trading.

Now in each of these ten cases the net difference is  Rs 1300 more than original amount.

Now sth i cant understand is why in all these 10 cases the  ans is 1300.

Plz help...

There must be something that I missed.

How can you gain 'Rs 1300' after  5 days of trading when a=110?

I suggest, as in my first post, you can gain (n-1)/2*100 =200 Rs.

(though I don't know what Rs  means, maybe a money unit??)

[Guest]

There must be something that I missed.

How can you gain 'Rs 1300' after 5 days of trading when a=110?

I suggest, as in my first post, you can gain (n-1)/2*100 =200 Rs.

(though I don't know what Rs means, maybe a money unit??)

In the first place we cant find gain, since we dont know what was the amount with which the trader started trading, we also dont know the initial number of shares he had , so we dont know the value of the shares he had initially (Only thing we know is that he had enough amount and shares to survive 5 days of trading)

The question only asks the difference between the amount he started and the amount he ended.

the answers for which are

for a =110 : DIFFERENCE : Rs 1300

for a =90 : DIFFERENCE : Rs 700

Rs is just a unit for money used in INDIA ,so thats not important at all.

The real thing that is puzzling me is that in all the 10 cases(mentioned in my prev post) the difference is coming out to be 1300 and i dont have a mathematical reason for this.

So if some one can prove why this happening via some equation i will be grateful to him.

It will also help me understand on which kind of equations similar phenomena of constant difference occurs.

Thank You

Anurag Kesarwani

Edited August 26, 2009 by guru_bhai

[Guest]

The real thing that is puzzling me is that in all the 10 cases(mentioned in my prev post) the difference is coming out to be 1300 and i dont have a mathematical reason for this.

So if some one can prove why this happening via some equation i will be grateful to him.

It will also help me understand on which kind of equations similar phenomena of constant difference occurs.

Thank You

Anurag Kesarwani

The reason why it is the same in each case is clearer if you look at relative gains. No matter what number of shares and money you start with, under the given conditions, the total gain will always be the same.

See my previous post to understand why...

Here's some more explanation:

When you sell shares, your gain from the previous position is 10 per share and when you buy, you buy at a cost of -10 from the previous position.

In all, there are 3 ups and 2 downs.

Each cycle of up and down gives you additional 100. Since there are 3 ups, the last up in addition to the 100 additional amount, gives you 1000. So your net gain is 1300 when a = 110.

In case of 2 ups and 3 downs, the 2 cycles of ups and downs gave you 200. And the final down made you spend 100 less. But the value of those last shares is -100 because a = 90, which cancels the saving from last purchase. So your net gain was 200 when a = 90

Edited August 26, 2009 by DeeGee

[superprismatic]

In the first place we cant find gain, since we dont know what was the amount with which the trader started trading, we also dont know the initial number of shares he had , so we dont know the value of the shares he had initially (Only thing we know is that he had enough amount and shares to survive 5 days of trading)

The question only asks the difference between the amount he started and the amount he ended.

the answers for which are

for a =110 : DIFFERENCE : Rs 1300

for a =90 : DIFFERENCE : Rs 700

Rs is just a unit for money used in INDIA ,so thats not important at all.

The real thing that is puzzling me is that in all the 10 cases(mentioned in my prev post) the difference is coming out to be 1300 and i dont have a mathematical reason for this.

So if some one can prove why this happening via some equation i will be grateful to him.

It will also help me understand on which kind of equations similar phenomena of constant difference occurs.

Thank You

Anurag Kesarwani

Consider the pair (P,A) where P is the stock price and A is the total monetary accumulation

trader X has so far in his trading. This pair starts off to be (100,0) at the beginning

of the first day. Let D be the function which takes (P,A) to its new value at the end of

the day if the stock falls 10 points. We have that D[(P,A)]=(P-10,A+100-10P). Similarly,

if we let U be the function which takes (P,A) to its new value at the end of the day if the

stock rises 10 points. We have that U[(P,A)]=(P+10,A+100+10P). So,

     U[D[(P,A)]]=U[(P-10,A+100-10P)]=((P-10)+10,(A+100-10P)+100+10(P-10))=(P,A+100)

[/code]


and,

[code]

     D[U[(P,A)]]=D[(P+10,A+100+10P)]=((P+10)-10,(A+100+10P)+100-10(p+10))=(P,A+100)

which shows that

     U[D[(P,A)]]=D[U[(P,A)]]

[/code]

which tells us that D and U commute, as functions, with each other.  So, in your example

of 3 Ups and 2 Downs, the order doesn't matter.  With a=110,  you will always end with

the same accumulation (Rs 1300).  If you require more of an explanation, feel free to

ask and I will try my best to explain further.

[Guest]

At my previous posts I had calculated net gain. But I see that you're involved with only Rs (money in your pocket).

So I modify my formula:

assume p0=initial cost of a share and p1=last cost of a share

(n-1)/2 * 100 = net gain

money in pocket= net gain + p1*(p1-p0)

if p0=100 and p1=90 then money=(5-1)/2 * 100 + 90 * (90-100) = 200 - 900 = -700 Rs

if p1=110 then money=(5-1)/2 * 100 + 110*(110-100)=200 + 1100 = 1300 Rs

Bye

[Guest]

Consider the pair (P,A) where P is the stock price and A is the total monetary accumulation

trader X has so far in his trading. This pair starts off to be (100,0) at the beginning

of the first day. Let D be the function which takes (P,A) to its new value at the end of

the day if the stock falls 10 points. We have that D[(P,A)]=(P-10,A+100-10P). Similarly,

if we let U be the function which takes (P,A) to its new value at the end of the day if the

stock rises 10 points. We have that U[(P,A)]=(P+10,A+100+10P). So,

     U[D[(P,A)]]=U[(P-10,A+100-10P)]=((P-10)+10,(A+100-10P)+100+10(P-10))=(P,A+100)

and,

     D[U[(P,A)]]=D[(P+10,A+100+10P)]=((P+10)-10,(A+100+10P)+100-10(p+10))=(P,A+100)

which shows that

     U[D[(P,A)]]=D[U[(P,A)]]

which tells us that D and U commute, as functions, with each other. So, in your example

of 3 Ups and 2 Downs, the order doesn't matter. With a=110, you will always end with

the same accumulation (Rs 1300). If you require more of an explanation, feel free to

ask and I will try my best to explain further.

Thank You superprismatic for such a nice explanation

you are a genius

Edited August 27, 2009 by guru_bhai