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IQ-IE plot

Name:

Student ID:

300577607

Answer the questions below by filling in/editing the boxes/chart provided.

1. This question will involve using the equation for the isocost line to calculate a set of plot points for an isocost line. Use the prices (pL and pK) and expenditure levels (E1 and E2) from the Parameter table, calculate two sets of points for two isocost lines (in the Table of plot points) for the range of qL shown in the table. Each set of plot points is for an isocost curve (IE1 and IE2). Note, if you calculate a negative number, leave it blank.

2. Add the isocost lines IE1 and IE2 to the chart. Note: qL is on the horizontal axis. (See the article in Tool #1 on how to add data to an existing chart.)

3. This question refers to the production function below the Parameter table - in the form Yn = A(qL^c qK^d). Rearrange this function, using the relevant numbers (A, c, d) from the Parameter table, to get an isoquant curve equation, Qk = F(Ql,c,d,Yn,A).

4. This question will involve using the equation for the isoquant curve and a given level of output to calculate a set of plot points. Using the given level of output indicated (Y1), calculate a set of points (in the Table of plot points) for the range of qL shown in the table. Note, the formula for the first point where qL > 0 for the isoquant curve has been entered in the table.

5. Plot the isoquant curve using the plot points calculated in the question above (as a scatter chart with a smooth line). (I.e. update/add the IQ data series in your chart.)

6. Based on your table and chart, state the cost-minimising production plan, qL*; qK* (i.e. a pair of qL; qK values). Note: the equation for the isoquant can be used to calculate qK* as a function of qL*. In the spreadsheet, if you enter a value for qL in cell B55, the corresponding value for qK will be calculated automatically in cell B56.

7. Explain why the production plan you have identified above is cost-minimising. State the firm's level of expenditure/cost to the given level of output indicated (Y1). (Hint: refer to slide 34, week 4 notes.)

1. Parameter table

2. Table of plot points

Qn 1

Qn 4

0.5

IE1

IE2

IQ1

0.5

239.4

239.71

120

359.78

150

339.00

180

210

Production function:

Y=4(L^0.5 K^0.5)

240

270

Qn 3

300

Isoquant curve:

Chart for Qn 2 and 5

Qn 6

qL*

Slope of isocost line:

-0.50

qK*

Slope of IQ at identified plan:

Qn 7

	1. Parameter table								Production function:
Student ID	c	d	A	pL	pK	E1	E2	Y1
300624284	0.5	0.5	4	1	2	111.72	168.0758387	158	Y=4(L^0.5 K^0.5)
300601840	0.5	0.5	4	1	2	335.88	503.9670143	475	Y=4(L^0.5 K^0.5)
300621040	0.5	0.5	4	1	2	159.81	240.0429635	226	Y=4(L^0.5 K^0.5)
300627980	0.5	0.5	4	1	2	391.74	587.7367153	554	Y=4(L^0.5 K^0.5)
300541140	0.5	0.5	4	1	2	55.86	84.88239419	79	Y=4(L^0.5 K^0.5)
300574441	0.5	0.5	4	1	2	176.07	264.4054614	249	Y=4(L^0.5 K^0.5)
300592188	0.5	0.5	4	1	2	72.12	108.9921679	102	Y=4(L^0.5 K^0.5)
300593522	0.5	0.5	4	1	2	207.89	312.0867569	294	Y=4(L^0.5 K^0.5)
300449209	0.5	0.5	4	1	2	79.9	120.5698756	113	Y=4(L^0.5 K^0.5)
300344572	0.5	0.5	4	1	2	227.69	341.7622614	322	Y=4(L^0.5 K^0.5)
300611656	0.5	0.5	4	1	2	260.22	390.5227711	368	Y=4(L^0.5 K^0.5)
300151761	0.5	0.5	4	1	2	304.06	456.2539096	430	Y=4(L^0.5 K^0.5)
300619051	0.5	0.5	4	1	2	203.65	305.7283316	288	Y=4(L^0.5 K^0.5)
300364831	0.5	0.5	4	1	2	332.34	498.6653964	470	Y=4(L^0.5 K^0.5)
300622453	0.5	0.5	4	1	2	180.31	270.7619223	255	Y=4(L^0.5 K^0.5)
300548873	0.5	0.5	4	1	2	347.9	521.9927788	492	Y=4(L^0.5 K^0.5)
300623999	0.5	0.5	4	1	2	395.98	594.0992367	560	Y=4(L^0.5 K^0.5)
300577607	0.5	0.5	4	1	2	239.71	359.7814647	339	Y=4(L^0.5 K^0.5)
300331564	0.5	0.5	4	1	2	224.15	336.4627526	317	Y=4(L^0.5 K^0.5)
300562778	0.5	0.5	4	1	2	308.3	462.6154522	436	Y=4(L^0.5 K^0.5)
300639883	0.5	0.5	4	1	2	352.14	528.3549049	498	Y=4(L^0.5 K^0.5)
300553762	0.5	0.5	4	1	2	304.06	456.2539096	430	Y=4(L^0.5 K^0.5)
300606900	0.5	0.5	4	1	2	359.92	540.0189183	509	Y=4(L^0.5 K^0.5)
300567140	0.5	0.5	4	1	2	55.86	84.88239419	79	Y=4(L^0.5 K^0.5)

300987654	0.5	0.5	4	1	2	260.22	390.5227711	368	Y=4(L^0.5 K^0.5)
300123456	0.5	0.5	4	1	2	180.31	270.7619223	255	Y=4(L^0.5 K^0.5)

	1. Parameter table								Production function:
Student ID	c	d	A	pL	pK	E1	E2	Y1
300624284	0.5	0.5	4	1	2	111.72	168.08	158.00	Y=4(L^0.5 K^0.5)
300601840	0.5	0.5	4	1	2	335.88	503.97	475.00	Y=4(L^0.5 K^0.5)
300621040	0.5	0.5	4	1	2	159.81	240.04	226.00	Y=4(L^0.5 K^0.5)
300627980	0.5	0.5	4	1	2	391.74	587.74	554.00	Y=4(L^0.5 K^0.5)
300541140	0.5	0.5	4	1	2	55.86	84.88	79.00	Y=4(L^0.5 K^0.5)
300574441	0.5	0.5	4	1	2	176.07	264.41	249.00	Y=4(L^0.5 K^0.5)
300592188	0.5	0.5	4	1	2	72.12	108.99	102.00	Y=4(L^0.5 K^0.5)
300593522	0.5	0.5	4	1	2	207.89	312.09	294.00	Y=4(L^0.5 K^0.5)
300449209	0.5	0.5	4	1	2	79.9	120.57	113.00	Y=4(L^0.5 K^0.5)
300344572	0.5	0.5	4	1	2	227.69	341.76	322.00	Y=4(L^0.5 K^0.5)
300611656	0.5	0.5	4	1	2	260.22	390.52	368.00	Y=4(L^0.5 K^0.5)
300151761	0.5	0.5	4	1	2	304.06	456.25	430.00	Y=4(L^0.5 K^0.5)
300619051	0.5	0.5	4	1	2	203.65	305.73	288.00	Y=4(L^0.5 K^0.5)
300364831	0.5	0.5	4	1	2	332.34	498.67	470.00	Y=4(L^0.5 K^0.5)
300622453	0.5	0.5	4	1	2	180.31	270.76	255.00	Y=4(L^0.5 K^0.5)
300548873	0.5	0.5	4	1	2	347.9	521.99	492.00	Y=4(L^0.5 K^0.5)
300623999	0.5	0.5	4	1	2	395.98	594.10	560.00	Y=4(L^0.5 K^0.5)
300577607	0.5	0.5	4	1	2	239.71	359.78	339.00	Y=4(L^0.5 K^0.5)
300331564	0.5	0.5	4	1	2	224.15	336.46	317.00	Y=4(L^0.5 K^0.5)
300562778	0.5	0.5	4	1	2	308.3	462.62	436.00	Y=4(L^0.5 K^0.5)
300639883	0.5	0.5	4	1	2	352.14	528.35	498.00	Y=4(L^0.5 K^0.5)
300553762	0.5	0.5	4	1	2	304.06	456.25	430.00	Y=4(L^0.5 K^0.5)
300606900	0.5	0.5	4	1	2	359.92	540.02	509.00	Y=4(L^0.5 K^0.5)
300567140	0.5	0.5	4	1	2	55.86	84.88	79.00	Y=4(L^0.5 K^0.5)

300987654	0.5	0.5	4	1	2	260.22	390.52	368.00	Y=4(L^0.5 K^0.5)
300123456	0.5	0.5	4	1	2	180.31	270.76	255.00	Y=4(L^0.5 K^0.5)

Name:	300624284						Student ID:	300577607

Answer the questions below by filling in/editing the boxes/chart provided.
1. This question will involve using the equation for the isocost line to calculate a set of plot points for an isocost line. Use the prices (pL and pK) and expenditure levels (E1 and E2) from the Parameter table, calculate two sets of points for two isocost lines (in the Table of plot points) for the range of qL shown in the table. Each set of plot points is for an isocost curve (IE1 and IE2). Note, if you calculate a negative number, leave it blank.

																	2. Table of plot points	1	2	3	4
																300454859	X	IC1	IC2	BC
2. Add the isocost lines IE1 and IE2 to the chart. Note: qL is on the horizontal axis. (See the article in Tool #1 on how to add data to an existing chart.)																	0	#N/A	#N/A	#DIV/0!
																	1	#N/A	#N/A	#DIV/0!
3. This question refers to the production function below the Parameter table - in the form Yn = A(qL^c qK^d). Rearrange this function, using the relevant numbers (A, c, d) from the Parameter table, to get an isoquant curve equation, Qk = F(Ql,c,d,Yn,A).																	2	#N/A	#N/A	#DIV/0!
																	3	#N/A	#N/A	#DIV/0!
																	4	#N/A	#N/A	#DIV/0!
4. This question will involve using the equation for the isoquant curve and a given level of output to calculate a set of plot points. Using the given level of output indicated (Y1), calculate a set of points (in the Table of plot points) for the range of qL shown in the table. Note, the formula for the first point where qL > 0 for the isoquant curve has been entered in the table.																	5	#N/A	#N/A	#DIV/0!
																	6	#N/A	#N/A	#DIV/0!
																	7	#N/A	#N/A	#DIV/0!
											Mark	Out of:					#REF!	#N/A	#N/A	#REF!
5. Plot the isoquant curve using the plot points calculated in the question above (as a scatter chart with a smooth line). (I.e. update/add the IQ data series in your chart.)										1	-	3					#REF!	#N/A	#N/A	#REF!
										2	1.00	1
6. Based on your table and chart, state the cost-minimising production plan, qL; qK (i.e. a pair of qL; qK values). Note: the equation for the isoquant can be used to calculate qK* as a function of qL*. In the spreadsheet, if you enter a value for qL in cell B55, the corresponding value for qK will be calculated automatically in cell B56.										3	-	2
										4	1.00	1
										5	1.00	1
7. Explain why the production plan you have identified above is cost-minimising. State the firm's level of expenditure/cost to the given level of output indicated (Y1). (Hint: refer to slide 34, week 4 notes.)										6		1
										7	1.00	1
										Total	4.00	/10
					179.8907324	-0.5	*$E28
1. Parameter table			2. Table of plot points	Qn 1		Qn 4
c	0.5		X	IE1	IE2	IQ1		MRTS
d	0.5		0	119.9	179.9	#N/A	#N/A	#DIV/0!				y	106	3	4
A	4		30.00	104.9	164.9	239.4	FALSE	-3.50					106	1	2
pL	1		60	89.9	149.9	119.7	FALSE	-1.50
pK	2		90	74.9	134.9	79.8	FALSE	-0.83
E1	239.71		120	59.9	119.9	59.9	FALSE	-0.50
E2	359.78		150	44.9	104.9	47.9	FALSE	-0.30
Y1	339.00		180	29.9	89.9	39.9	FALSE	-0.17
			210	14.9	74.9	34.2	FALSE	-0.07
Production function:	Y=4(L^0.5 K^0.5)		240	#N/A	59.9	29.9	#N/A	#N/A
			270	#N/A	44.9	26.6	#N/A	#N/A
Qn 3			300	#N/A	29.9	23.9	#N/A	#N/A
Isoquant curve:	K=[Y/(4 L^0.5)]^(1/0.5)								1.00

Chart for Qn 2 and 5									1.00
									1.00

Qn 6
qL*	119.9	Slope of isocost line:			-0.50
qK*	59.9	Slope of IQ at identified plan:			-0.50

Qn 7
									1.00

Name:	#N/A				Student ID:	#REF!	300624284	Rounding		Filename	SID	Last Name	First Name	Comment	Mark	Q1	Q2	Q3	Q4	Q5	Q6	Q7	Q1Tot	Q2Tot	Q3Tot	Q4Tot	Q5Tot	Q6Tot	Q7Tot	Total	Pct
Comment:	#N/A								0	RE	300624284	Bae	Yukyung		0.00								2	1	1	1	1	1	1	8	0%
										RE	300601840	Boylan Caffery	Naomi		0.00								2	1	1	1	1	1	1	8	0%
										RE	300621040	Compton	Chippy		0.00								2	1	1	1	1	1	1	8	0%
										RE	300627980	Cooper	Toni		0.00								2	1	1	1	1	1	1	8	0%
										RE	300541140	Gaur	Pratika		0.00								2	1	1	1	1	1	1	8	0%
										RE	300574441	Inayat	Khadijah		0.00								2	1	1	1	1	1	1	8	0%
	Mark	Out of:								RE	300592188	Jiang	Shan		0.00								2	1	1	1	1	1	1	8	0%
Q1	#DIV/0!	2	#DIV/0!							RE	300593522	Kasai	Kyoko		0.00								2	1	1	1	1	1	1	8	0%
Q2	1.0	1	100%							RE	300449209	Kaur	Sahil Preet		0.00								2	1	1	1	1	1	1	8	0%
Q3	1.0	1	100%							RE	300344572	Keegan	Chris		0.00								2	1	1	1	1	1	1	8	0%
Q4	#DIV/0!	1	#DIV/0!							RE	300611656	Le	Hoa		0.00								2	1	1	1	1	1	1	8	0%
Q5	1.0	1	100%							RE	300151761	Majer	Joel		0.00								2	1	1	1	1	1	1	8	0%
Q6		1	#VALUE!	qL*	119.9					RE	300619051	Matsui	Yuko		0.00								2	1	1	1	1	1	1	8	0%
Q7	1.0	1	100%	qK*	59.9					RE	300364831	O'Connor	Sarah		0.00								2	1	1	1	1	1	1	8	0%
Total	#DIV/0!	/8								RE	300622453	Park	Cheol Joo		0.00								2	1	1	1	1	1	1	8	0%
										RE	300548873	Pujara	Megha		0.00								2	1	1	1	1	1	1	8	0%
Breakdown by question (1=correct, 0=incorrect)				129.02-0.5*$E28	-0.5	*$E28				RE	300623999	Qiu	Lily		0.00								2	1	1	1	1	1	1	8	0%
	Q1.		Q4.							RE	300577607	Shrestha	Shraddha		0.00								2	1	1	1	1	1	1	8	0%
	IE1	IE2	IQ1							RE	300331564	Singh	Sushil		0.00								2	1	1	1	1	1	1	8	0%
1	#N/A	#N/A		IQ1() incorrect,						RE	300562778	Wang	Bing Zheng		0.00								2	1	1	1	1	1	1	8	0%
2	#N/A	#N/A		IQ1() incorrect,						RE	300639883	Wright	Jason		0.00								2	1	1	1	1	1	1	8	0%
3	#N/A	#N/A	#N/A							RE	300553762	Yang	Aenon		0.00								2	1	1	1	1	1	1	8	0%
4	#N/A	#N/A	#N/A							RE	300606900	Zhao	Crystal		0.00								2	1	1	1	1	1	1	8	0%
5	#N/A	#N/A	#N/A							RE	300567140	Zhou	Yiran		0.00								2	1	1	1	1	1	1	8	0%
6	#N/A	#N/A	#N/A
7	#N/A	#N/A	#N/A								300987654	USER	Default	Well done. Q6. You have not quite correctly identified the cost-minimising production plan. That plan occurs at the tangency point of an IC and the IQ, i.e. the slopes of these two curves should be the same at that point, which you have not quite achieved with the plan noted. It should be: (119.9,59.9).	#DIV/0!	#DIV/0!	1	1	#DIV/0!	1		1	2	1	1	1	1	1	1	8	#DIV/0!
8	#N/A	#N/A	#N/A								300123456	USER	Default	Well done. Q6. You have not quite correctly identified the cost-minimising production plan. That plan occurs at the tangency point of an IC and the IQ, i.e. the slopes of these two curves should be the same at that point, which you have not quite achieved with the plan noted. It should be: (119.9,59.9).	8.00	2	1	1	1	1	1	1	2	1	1	1	1	1	1	8	100%
9	#N/A	#N/A	#N/A
10	#N/A	#N/A	#N/A											Well done. Q6. You have not quite correctly identified the cost-minimising production plan. That plan occurs at the tangency point of an IC and the IQ, i.e. the slopes of these two curves should be the same at that point, which you have not quite achieved with the plan noted. It should be: (119.9,59.9).	#DIV/0!	#DIV/0!	1	1	#DIV/0!	1		1	2	1	1	1	1	1	1	8	#DIV/0!
11	#N/A	#N/A	#N/A
														Well done.
Q2.
	1.0
														Table 1 - it is not clear why you have changed the range for Ql (in cells E28:E38) in Table 1.
Q3.														It looks like you may have struggled with this one.
	1.0													Once again your assignment was unmarkable. The version you submitted appears to have been converted from another programme (Numbers?). It did not have your Student ID number entered. And bar two or three numbers, there did not appear to be any content submitted.
														Q2. Part way there. Calcuated in table but IE lines not added to chart. Q3, Q6, Q7. Not answered.
Q5.														Q2. You have plotted IE2 but not IE1 (and IQ1 twice). Q6. You have not correctly identified the cost-minimising production plan; this may, in part, reflect the plotting error in Q2. That plan occurs at the tangency point of an IC and the IQ, i.e. the slopes of these two curves should be the same at that point, which you have not quite achieved with the plan noted. It should be: (52,26).
	1.0													Q3 onwards not answered/calculated.
														Q3. Qk=[Y/(4 Ql^0.5)]^2
Q6.														Q3. The IQ eauation is: QK=[Y/(4 QL^0.5)]^2. Q6. You have not quite correctly identified the cost-minimising production plan. That plan occurs at the tangency point of an IC and the IQ, i.e. the slopes of these two curves should be the same at that point, which you have not quite achieved with the plan noted. It should be: (52,26).
														Q6. You have not quite correctly identified the cost-minimising production plan. That plan occurs at the tangency point of an IC and the IQ, i.e. the slopes of these two curves should be the same at that point, which you have not quite achieved with the plan noted. It should be: (40,20).
														Q6. You have not quite correctly identified the cost-minimising production plan. That plan occurs at the tangency point of an IC and the IQ, i.e. the slopes of these two curves should be the same at that point, which you have not quite achieved with the plan noted. It should be: (94,47). Also, it looks like you have reversed the quantity of labour and capital, i.e. you have noted (49.1,90); (90,49.1) would be closer.
Q7.														Q7. No answer.
	1.0													Q7. Your explanation is not correct - "the cost is minimum where the two curves intersects"; as noted above, the cost-minimising plan is a tangency point.
														Q7. Your argument/explanatoin is not clear or sufficiently detailed. The criterion "MR=MC" relates to ouptput choice, not input choice, and is irrelevant to this question.
														Q7. Your argument is not clear. Your statement is not linked to/does not reference the production plan you have identified or explained why it is cost-minimising.
														Q7. The final explanation was not sufficiently clear to merit full marks. How did you identify the cost minimising plan? Is "the marginal product of each dollar… equalised across variable inputs" for any point on the IE or the IQ? What particular point does this occur at?
														Q7. The final explanation was not sufficiently clear to merit full marks. How did you identify the cost minimising plan? What particular point does this occur at? What is the relationship between the relative cost/marginal productivity of the inputs at the tangency point?
														Q7. The final explanation was not of sufficient depth to merit full marks. How did you identify the cost minimising plan? What is the relationship between the relative cost/marginal productivity of the inputs at this tangency point?
														Q7. The final explanation was not of sufficient depth to merit full marks. Why does the "production plan on the point of tangency" minimise the cost of production for the given production target? What is the relationship between the relative cost/marginal productivity of the inputs at this tangency point?
														Q7. The final explanation is not quite accurate, "At this point, the marginal productivity of labour will also be equal to the marginal productivity of capital." Rather, the MPL per dollar spent equals the MPK per dollar spent at this point.
														Q7. The final explanation would have benefitted from further discussion of the relationship between the relative cost/marginal productivity of the inputs at the identified plan.