ct4 october 2010 Q12

[neha3]

A pet shop has four glass tanks in which snakes for sale are held. The shop can stock
at most four snakes at any one time because:
• if more than one snake were held in the same tank, the snakes would attempt to
eat each other and
• having snakes loose in the shop would not be popular with the neighbours
The number of snakes sold by the shop each day is a random variable with the
following distribution:
Number of Snakes Potentially Sold Probability
in Day (if stock is sufficient)
None 0.4
One 0.4
Two 0.2
If the shop has no snakes in stock at the end of a day, the owner contacts his snake
supplier to order four more snakes. The snakes are delivered the following morning
before the shop opens. The snake supplier makes a charge of C for the delivery.
(i) Write down the transition matrix for the number of snakes in stock when the
shop opens in a morning, given the number in stock when the shop opened the
previous day. [2]
(ii) Calculate the stationary distribution for the number of snakes in stock when
the shop opens, using your transition matrix in part (i). [4]
(iii) Calculate the expected long term average number of restocking orders placed
by the shop owner per trading day. [2]
CT4 S2010—7
If a customer arrives intending to purchase a snake, and there is none in stock, the sale
is lost to a rival pet shop.

Please can anyone explain me this question... I get the transaction matrix, however couldnt understand the rest of the answers...?

 

[neha3]

Can any one pls urgently help on this query

 

[Krithika]

Hi,

I tried solving the question and got stuck at iv).

I guess you were able to solve i) and ii) and iii) has asked for the expected number of restocking in a long term... which is straight forward...

from the matrix - 1-> 4 and 2 -> 4 to be considered that is 0.6 and 0.2 (others need not be considered as there won't be more than 2 snakes sold per day, so no need of restocking).

iv) now, when we need to calculate lost of sales, we need the probability of running out of stock, ie 1-(0.6+0.2) = 0.2. The point I am not getting is that why the solution considers only the long run impact from pi1 alone and not pi2????

i think others parts are straight forward from this.

Hope this helps to some extent... shall wait for some better explanation.

Regards,
Krithika

 

[didster]

iv) You lose a sale if you otherwise would have sold more than you have in stock.
So only case is if you have one in stock and would have otherwise sold 2.
(if you start with 2 or more and you sell 2 or less you always have enough stock, ditto if you sell none in any event)
Answer is long term probability of having 1 snake on the day times the probability of selling 2.


v) Same thing as earlier but with different transition matrix
vi) general logic in report is good enough I think so you need to ask more specif questions if it doesn't make sense.
vii) just compare costs of restocking plus lost sales under two scenarios (if you restock with less than 2 at end of day you never lose sales no just higher restocking fee) and identify what conditions minimize costs

 

[neha3]

Hi Didstar,

Thanks for the explaination. Please can u elaborate ur first sentence, that will be be helpfull for me to understand better.

 

[didster]

You sell either 0, 1 or 2.
If you have 2 or more, you must have enough stock, so no lost sales
You never have zero since you would have restocked the night before.
So you only consider where you start with 1. Again you're ok if you dont sell or sell 1.
So need to consider starting with 1 and needing to sell 2.

ps you may get a better response to your other question (and others in general) if you specify what's the problem you're having. Don't just quote the examiner's question.