IT_Y4_new

# Reset t 1 if n 2 let m max i 1 i 2 and reset ss n 1 m

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Reset: t 1 = . If n > 2: Let m = max (i 1 , i 2 ) and Reset: SS = (n - 1, m + 1, i 2 ). Generate Y 1 and reset t 1 = t + Y 1 . Case 3: Case 2: SS = (n, i 1 , i 2 ) and t 2 < t A, t 2 < t 1 Reset: t = t 2 . Reset: C 2 = C 2 + 1. Collect output data D(i 2 ) = t. If n = 1: Reset: SS = (0). Reset: t 2 = . If n = 2: Reset: SS = (1, i 1 , 0). Reset: t 2 = . If n > 2: Let m = max (i 1 , i 2 ) and Reset: SS = (n - 1, i 1 , m + 1). Generate Y 2 and reset t 2 = t + Y 2 . 5. Consider a shop that stocks a particular type of product that it sells for a price of r per unit. Customers demand in accordance with a Poisson process with rate λ and the amount demanded by each one is a random variable having distribution G. In order to meet demands, the shopkeeper must keep an amount of the product on hand and if the on-hand becomes low, additional units are ordered from distributor. If present inventory level is x and no order if outstanding, then if x<s, the amount s-x is ordered. The cost of ordering y units of product is a specified function c(y) and it takes l units time until order is delivered. In addition, the shop pays an inventory holding cost of h per unit item per b unit time. Whenever customers demands more of the product than is presently available, than the amount on hand is sold and the remainder of the order is order is lost to the shop. Define variables and events to analyze this model and estimate the shop’s expected profit up to fixed time T. (10 marks) solution (An Inventory Model) Time variables t System State Variables (x, y) x = the amount of inventory on hand. y = the amount of inventory on order. Counter variables C = the total amount of ordering costs by t. H = the total amount of inventory holding costs by t. R = the total amount of revenue earned by t.

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Event List: t 0 , t 1 t 0 = the arrival time of next customer t 1 = the time at which the order being filled will be delivered. 9 if there is no order, t 1 = . Case 2: t 0 < t 1 Reset: H = H + ( t 0 – t)xh. Reset: t = t 0 . Generate D, a random variable having distribution G. (D is the demand of customer that arrived at time t 0 .) Let ω = min (D, x) be the amount of order that can be filled. Reset: R = R + ω r. Reset: x = x – ω . If x < s and y = 0, then reset y = S – x, t 1 = t + L. Generate U and reset t0 = t – 1/ λ log(U). Case 2: t 1 t 0 Reset: H = H + ( t 1 – t)xh. Reset: t = t 1 . Reset: C = C + c(y). Reset: x = x + y. Reset: y = 0, t 1 = . 6. Suppose that the different policyholders of a casually insurance company generate claims according to independent Poisson process with a common rate λ , and that each claim amount has distribution F. Suppose also that new customers sign up according to a Poisson process with rate v, and each existing policyholder pays the insurance firm at a fixed rate C per unit time. This system states with n customer and initial capital a 0, find then. Define the variable and enters for this system and estimate the probability that the firm’s capital is always non negative at all times up to time T. (20 marks) solution (Insurance Risk Model) Time variables t System State Variables (n, a) n = the number of policyholders.
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