Hi, I need help with essay on Costing, Budgeting for Projects Accounting. Paper must be at least 1000 words. Please, no plagiarized work!Download file to see previous pages... 9018088 0.000842049 12 8

Hi, I need help with essay on Costing, Budgeting for Projects Accounting. Paper must be at least 1000 words. Please, no plagiarized work!

9018088 0.000842049 12 8 13 9 7.703656443 1.680506618 13 9 14 7 7.414185094 0.171549292 14 7 15 4.791208791 44.00883114 -1.008791209 19.92307692 0.247845443 2.110327908 0.57993311 3.738278384 16.56687898 12 231.5175824 167.6967033 Estimating Learning-Curve Functions 2 2) I used the formula approach and forecast approach to estimate non-linear functions such as the learning-curve function. The predicted value for 15 units is 4.791209. Note: When I used the forecast approach, either there was an error in the formula or the predicted value is wrong. The formula on p.209 is =FORECAST(15,B1:B14,A1:A14), but I came up with a predicted value of 4.32967. But since I must arrive at the value 4.791209, I figured the formula should’ve been =FORECAST(15,B2:B15,A2:A15). I finally succeeded at having the same value 4.791209. I think my formula is more accurate because for 14 units of outputs, the DLH value is already given with a value of 7. The formula for regression analysis is Y = a + bX where a is the constant term and b is the slope. When estimating LCR, I used Solver routine in Excel. I started with an estimated LCR of .80 and entered that figure in cell D2. For getting the predicted DLHs, I used Excel learning-curve formula which is =$B$2*A2^(LN($D$2)/LN(2)). For unit 1 (X), the predicted value is an exact match. It is the same as the actual DLH, 29. The rest of the predicted DLHs for outputs 2-14 were a close match except for 9 and 11 units (in thousands) of outputs. Next, I calculated the “squared-error.” The obvious answer is 0, since there is no error. The predicted DLH is exactly the same as the actual DLH. The formula for “squared-error” in Excel is =(C2-B2)^2 for 1 unit (in thousands). The rest can be calculated in Solver routine by dragging the cells when copying the formula for 1 unit of output. Estimating Learning-Curve Functions 3 3) Shown on the graph with the downward slope, as the total outputs increased, the cumulative units produced decreased. In other words, the more experience there is in producing outputs, the amount of time to perform the task decreases. The plot is consistent with the data given for units produced and DLHs. 4) I estimated the LCR with the “incremental unit-time learning-curve model” using Solver routine and I arrived at 0.69 or .70. In the Solver routine window, I entered the “target cell” as $E$2 which is the “squared error” contained in E2 and the “changing cell” as $D$2 which is the Estimating Learning-Curve Functions 4 LCR contained in D2. The exponent in learning-curve model, b, is -0.51681. So for DLH 29, the incremental unit-time learning-curve model is Y = 29.0*X^-0.51681. 5) The formula for cumulative average time per unit to produce x units is Y = ax^b. where Y = cumulative average time per unit to produce x units, a = the time taken for the first unit of output, x = the cumulative number of units, and b = the index of learning (log LR/log 2).